```
from __future__ import print_function, division
from collections import defaultdict
from itertools import combinations, permutations, product, product as cartes
import random
from operator import gt
from sympy.core.decorators import deprecated
from sympy.core import Basic, C
# this is the logical location of these functions
from sympy.core.compatibility import (
as_int, combinations_with_replacement, default_sort_key, is_sequence,
iterable, ordered, xrange
)
from sympy.utilities.enumerative import (
multiset_partitions_taocp, list_visitor, MultisetPartitionTraverser)
[docs]def flatten(iterable, levels=None, cls=None):
"""
Recursively denest iterable containers.
>>> from sympy.utilities.iterables import flatten
>>> flatten([1, 2, 3])
[1, 2, 3]
>>> flatten([1, 2, [3]])
[1, 2, 3]
>>> flatten([1, [2, 3], [4, 5]])
[1, 2, 3, 4, 5]
>>> flatten([1.0, 2, (1, None)])
[1.0, 2, 1, None]
If you want to denest only a specified number of levels of
nested containers, then set ``levels`` flag to the desired
number of levels::
>>> ls = [[(-2, -1), (1, 2)], [(0, 0)]]
>>> flatten(ls, levels=1)
[(-2, -1), (1, 2), (0, 0)]
If cls argument is specified, it will only flatten instances of that
class, for example:
>>> from sympy.core import Basic
>>> class MyOp(Basic):
... pass
...
>>> flatten([MyOp(1, MyOp(2, 3))], cls=MyOp)
[1, 2, 3]
adapted from http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks
"""
if levels is not None:
if not levels:
return iterable
elif levels > 0:
levels -= 1
else:
raise ValueError(
"expected non-negative number of levels, got %s" % levels)
if cls is None:
reducible = lambda x: is_sequence(x, set)
else:
reducible = lambda x: isinstance(x, cls)
result = []
for el in iterable:
if reducible(el):
if hasattr(el, 'args'):
el = el.args
result.extend(flatten(el, levels=levels, cls=cls))
else:
result.append(el)
return result
[docs]def unflatten(iter, n=2):
"""Group ``iter`` into tuples of length ``n``. Raise an error if
the length of ``iter`` is not a multiple of ``n``.
"""
if n < 1 or len(iter) % n:
raise ValueError('iter length is not a multiple of %i' % n)
return list(zip(*(iter[i::n] for i in xrange(n))))
[docs]def reshape(seq, how):
"""Reshape the sequence according to the template in ``how``.
Examples
========
>>> from sympy.utilities import reshape
>>> seq = list(range(1, 9))
>>> reshape(seq, [4]) # lists of 4
[[1, 2, 3, 4], [5, 6, 7, 8]]
>>> reshape(seq, (4,)) # tuples of 4
[(1, 2, 3, 4), (5, 6, 7, 8)]
>>> reshape(seq, (2, 2)) # tuples of 4
[(1, 2, 3, 4), (5, 6, 7, 8)]
>>> reshape(seq, (2, [2])) # (i, i, [i, i])
[(1, 2, [3, 4]), (5, 6, [7, 8])]
>>> reshape(seq, ((2,), [2])) # etc....
[((1, 2), [3, 4]), ((5, 6), [7, 8])]
>>> reshape(seq, (1, [2], 1))
[(1, [2, 3], 4), (5, [6, 7], 8)]
>>> reshape(tuple(seq), ([[1], 1, (2,)],))
(([[1], 2, (3, 4)],), ([[5], 6, (7, 8)],))
>>> reshape(tuple(seq), ([1], 1, (2,)))
(([1], 2, (3, 4)), ([5], 6, (7, 8)))
>>> reshape(list(range(12)), [2, [3], set([2]), (1, (3,), 1)])
[[0, 1, [2, 3, 4], set([5, 6]), (7, (8, 9, 10), 11)]]
"""
m = sum(flatten(how))
n, rem = divmod(len(seq), m)
if m < 0 or rem:
raise ValueError('template must sum to positive number '
'that divides the length of the sequence')
i = 0
container = type(how)
rv = [None]*n
for k in range(len(rv)):
rv[k] = []
for hi in how:
if type(hi) is int:
rv[k].extend(seq[i: i + hi])
i += hi
else:
n = sum(flatten(hi))
hi_type = type(hi)
rv[k].append(hi_type(reshape(seq[i: i + n], hi)[0]))
i += n
rv[k] = container(rv[k])
return type(seq)(rv)
[docs]def group(seq, multiple=True):
"""
Splits a sequence into a list of lists of equal, adjacent elements.
Examples
========
>>> from sympy.utilities.iterables import group
>>> group([1, 1, 1, 2, 2, 3])
[[1, 1, 1], [2, 2], [3]]
>>> group([1, 1, 1, 2, 2, 3], multiple=False)
[(1, 3), (2, 2), (3, 1)]
>>> group([1, 1, 3, 2, 2, 1], multiple=False)
[(1, 2), (3, 1), (2, 2), (1, 1)]
See Also
========
multiset
"""
if not seq:
return []
current, groups = [seq[0]], []
for elem in seq[1:]:
if elem == current[-1]:
current.append(elem)
else:
groups.append(current)
current = [elem]
groups.append(current)
if multiple:
return groups
for i, current in enumerate(groups):
groups[i] = (current[0], len(current))
return groups
[docs]def multiset(seq):
"""Return the hashable sequence in multiset form with values being the
multiplicity of the item in the sequence.
Examples
========
>>> from sympy.utilities.iterables import multiset
>>> multiset('mississippi')
{'i': 4, 'm': 1, 'p': 2, 's': 4}
See Also
========
group
"""
rv = defaultdict(int)
for s in seq:
rv[s] += 1
return dict(rv)
[docs]def postorder_traversal(node, keys=None):
"""
Do a postorder traversal of a tree.
This generator recursively yields nodes that it has visited in a postorder
fashion. That is, it descends through the tree depth-first to yield all of
a node's children's postorder traversal before yielding the node itself.
Parameters
==========
node : sympy expression
The expression to traverse.
keys : (default None) sort key(s)
The key(s) used to sort args of Basic objects. When None, args of Basic
objects are processed in arbitrary order. If key is defined, it will
be passed along to ordered() as the only key(s) to use to sort the
arguments; if ``key`` is simply True then the default keys of
``ordered`` will be used (node count and default_sort_key).
Yields
======
subtree : sympy expression
All of the subtrees in the tree.
Examples
========
>>> from sympy.utilities.iterables import postorder_traversal
>>> from sympy.abc import w, x, y, z
The nodes are returned in the order that they are encountered unless key
is given; simply passing key=True will guarantee that the traversal is
unique.
>>> list(postorder_traversal(w + (x + y)*z)) # doctest: +SKIP
[z, y, x, x + y, z*(x + y), w, w + z*(x + y)]
>>> list(postorder_traversal(w + (x + y)*z, keys=True))
[w, z, x, y, x + y, z*(x + y), w + z*(x + y)]
"""
if isinstance(node, Basic):
args = node.args
if keys:
if keys != True:
args = ordered(args, keys, default=False)
else:
args = ordered(args)
for arg in args:
for subtree in postorder_traversal(arg, keys):
yield subtree
elif iterable(node):
for item in node:
for subtree in postorder_traversal(item, keys):
yield subtree
yield node
[docs]def interactive_traversal(expr):
"""Traverse a tree asking a user which branch to choose. """
from sympy.printing import pprint
RED, BRED = '\033[0;31m', '\033[1;31m'
GREEN, BGREEN = '\033[0;32m', '\033[1;32m'
YELLOW, BYELLOW = '\033[0;33m', '\033[1;33m'
BLUE, BBLUE = '\033[0;34m', '\033[1;34m'
MAGENTA, BMAGENTA = '\033[0;35m', '\033[1;35m'
CYAN, BCYAN = '\033[0;36m', '\033[1;36m'
END = '\033[0m'
def cprint(*args):
print("".join(map(str, args)) + END)
def _interactive_traversal(expr, stage):
if stage > 0:
print()
cprint("Current expression (stage ", BYELLOW, stage, END, "):")
print(BCYAN)
pprint(expr)
print(END)
if isinstance(expr, Basic):
if expr.is_Add:
args = expr.as_ordered_terms()
elif expr.is_Mul:
args = expr.as_ordered_factors()
else:
args = expr.args
elif hasattr(expr, "__iter__"):
args = list(expr)
else:
return expr
n_args = len(args)
if not n_args:
return expr
for i, arg in enumerate(args):
cprint(GREEN, "[", BGREEN, i, GREEN, "] ", BLUE, type(arg), END)
pprint(arg)
print
if n_args == 1:
choices = '0'
else:
choices = '0-%d' % (n_args - 1)
try:
choice = raw_input("Your choice [%s,f,l,r,d,?]: " % choices)
except EOFError:
result = expr
print()
else:
if choice == '?':
cprint(RED, "%s - select subexpression with the given index" %
choices)
cprint(RED, "f - select the first subexpression")
cprint(RED, "l - select the last subexpression")
cprint(RED, "r - select a random subexpression")
cprint(RED, "d - done\n")
result = _interactive_traversal(expr, stage)
elif choice in ['d', '']:
result = expr
elif choice == 'f':
result = _interactive_traversal(args[0], stage + 1)
elif choice == 'l':
result = _interactive_traversal(args[-1], stage + 1)
elif choice == 'r':
result = _interactive_traversal(random.choice(args), stage + 1)
else:
try:
choice = int(choice)
except ValueError:
cprint(BRED,
"Choice must be a number in %s range\n" % choices)
result = _interactive_traversal(expr, stage)
else:
if choice < 0 or choice >= n_args:
cprint(BRED, "Choice must be in %s range\n" % choices)
result = _interactive_traversal(expr, stage)
else:
result = _interactive_traversal(args[choice], stage + 1)
return result
return _interactive_traversal(expr, 0)
[docs]def ibin(n, bits=0, str=False):
"""Return a list of length ``bits`` corresponding to the binary value
of ``n`` with small bits to the right (last). If bits is omitted, the
length will be the number required to represent ``n``. If the bits are
desired in reversed order, use the [::-1] slice of the returned list.
If a sequence of all bits-length lists starting from [0, 0,..., 0]
through [1, 1, ..., 1] are desired, pass a non-integer for bits, e.g.
'all'.
If the bit *string* is desired pass ``str=True``.
Examples
========
>>> from sympy.utilities.iterables import ibin
>>> ibin(2)
[1, 0]
>>> ibin(2, 4)
[0, 0, 1, 0]
>>> ibin(2, 4)[::-1]
[0, 1, 0, 0]
If all lists corresponding to 0 to 2**n - 1, pass a non-integer
for bits:
>>> bits = 2
>>> for i in ibin(2, 'all'):
... print(i)
(0, 0)
(0, 1)
(1, 0)
(1, 1)
If a bit string is desired of a given length, use str=True:
>>> n = 123
>>> bits = 10
>>> ibin(n, bits, str=True)
'0001111011'
>>> ibin(n, bits, str=True)[::-1] # small bits left
'1101111000'
>>> list(ibin(3, 'all', str=True))
['000', '001', '010', '011', '100', '101', '110', '111']
"""
if not str:
try:
bits = as_int(bits)
return [1 if i == "1" else 0 for i in bin(n)[2:].rjust(bits, "0")]
except ValueError:
return variations(list(range(2)), n, repetition=True)
else:
try:
bits = as_int(bits)
return bin(n)[2:].rjust(bits, "0")
except ValueError:
return (bin(i)[2:].rjust(n, "0") for i in range(2**n))
[docs]def variations(seq, n, repetition=False):
"""Returns a generator of the n-sized variations of ``seq`` (size N).
``repetition`` controls whether items in ``seq`` can appear more than once;
Examples
========
variations(seq, n) will return N! / (N - n)! permutations without
repetition of seq's elements:
>>> from sympy.utilities.iterables import variations
>>> list(variations([1, 2], 2))
[(1, 2), (2, 1)]
variations(seq, n, True) will return the N**n permutations obtained
by allowing repetition of elements:
>>> list(variations([1, 2], 2, repetition=True))
[(1, 1), (1, 2), (2, 1), (2, 2)]
If you ask for more items than are in the set you get the empty set unless
you allow repetitions:
>>> list(variations([0, 1], 3, repetition=False))
[]
>>> list(variations([0, 1], 3, repetition=True))[:4]
[(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1)]
See Also
========
sympy.core.compatibility.permutations
sympy.core.compatibility.product
"""
if not repetition:
seq = tuple(seq)
if len(seq) < n:
return
for i in permutations(seq, n):
yield i
else:
if n == 0:
yield ()
else:
for i in product(seq, repeat=n):
yield i
[docs]def subsets(seq, k=None, repetition=False):
"""Generates all k-subsets (combinations) from an n-element set, seq.
A k-subset of an n-element set is any subset of length exactly k. The
number of k-subsets of an n-element set is given by binomial(n, k),
whereas there are 2**n subsets all together. If k is None then all
2**n subsets will be returned from shortest to longest.
Examples
========
>>> from sympy.utilities.iterables import subsets
subsets(seq, k) will return the n!/k!/(n - k)! k-subsets (combinations)
without repetition, i.e. once an item has been removed, it can no
longer be "taken":
>>> list(subsets([1, 2], 2))
[(1, 2)]
>>> list(subsets([1, 2]))
[(), (1,), (2,), (1, 2)]
>>> list(subsets([1, 2, 3], 2))
[(1, 2), (1, 3), (2, 3)]
subsets(seq, k, repetition=True) will return the (n - 1 + k)!/k!/(n - 1)!
combinations *with* repetition:
>>> list(subsets([1, 2], 2, repetition=True))
[(1, 1), (1, 2), (2, 2)]
If you ask for more items than are in the set you get the empty set unless
you allow repetitions:
>>> list(subsets([0, 1], 3, repetition=False))
[]
>>> list(subsets([0, 1], 3, repetition=True))
[(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)]
"""
if k is None:
for k in range(len(seq) + 1):
for i in subsets(seq, k, repetition):
yield i
else:
if not repetition:
for i in combinations(seq, k):
yield i
else:
for i in combinations_with_replacement(seq, k):
yield i
[docs]def numbered_symbols(prefix='x', cls=None, start=0, *args, **assumptions):
"""
Generate an infinite stream of Symbols consisting of a prefix and
increasing subscripts.
Parameters
==========
prefix : str, optional
The prefix to use. By default, this function will generate symbols of
the form "x0", "x1", etc.
cls : class, optional
The class to use. By default, it uses Symbol, but you can also use Wild or Dummy.
start : int, optional
The start number. By default, it is 0.
Returns
=======
sym : Symbol
The subscripted symbols.
"""
if cls is None:
# We can't just make the default cls=C.Symbol because it isn't
# imported yet.
cls = C.Symbol
while True:
name = '%s%s' % (prefix, start)
yield cls(name, *args, **assumptions)
start += 1
[docs]def capture(func):
"""Return the printed output of func().
`func` should be a function without arguments that produces output with
print statements.
>>> from sympy.utilities.iterables import capture
>>> from sympy import pprint
>>> from sympy.abc import x
>>> def foo():
... print('hello world!')
...
>>> 'hello' in capture(foo) # foo, not foo()
True
>>> capture(lambda: pprint(2/x))
'2\\n-\\nx\\n'
"""
from sympy.core.compatibility import StringIO
import sys
stdout = sys.stdout
sys.stdout = file = StringIO()
try:
func()
finally:
sys.stdout = stdout
return file.getvalue()
[docs]def sift(seq, keyfunc):
"""
Sift the sequence, ``seq`` into a dictionary according to keyfunc.
OUTPUT: each element in expr is stored in a list keyed to the value
of keyfunc for the element.
Examples
========
>>> from sympy.utilities import sift
>>> from sympy.abc import x, y
>>> from sympy import sqrt, exp
>>> sift(range(5), lambda x: x % 2)
{0: [0, 2, 4], 1: [1, 3]}
sift() returns a defaultdict() object, so any key that has no matches will
give [].
>>> sift([x], lambda x: x.is_commutative)
{True: [x]}
>>> _[False]
[]
Sometimes you won't know how many keys you will get:
>>> sift([sqrt(x), exp(x), (y**x)**2],
... lambda x: x.as_base_exp()[0])
{E: [exp(x)], x: [sqrt(x)], y: [y**(2*x)]}
If you need to sort the sifted items it might be better to use
``ordered`` which can economically apply multiple sort keys
to a squence while sorting.
See Also
========
ordered
"""
m = defaultdict(list)
for i in seq:
m[keyfunc(i)].append(i)
return m
[docs]def take(iter, n):
"""Return ``n`` items from ``iter`` iterator. """
return [ value for _, value in zip(xrange(n), iter) ]
[docs]def dict_merge(*dicts):
"""Merge dictionaries into a single dictionary. """
merged = {}
for dict in dicts:
merged.update(dict)
return merged
[docs]def common_prefix(*seqs):
"""Return the subsequence that is a common start of sequences in ``seqs``.
>>> from sympy.utilities.iterables import common_prefix
>>> common_prefix(list(range(3)))
[0, 1, 2]
>>> common_prefix(list(range(3)), list(range(4)))
[0, 1, 2]
>>> common_prefix([1, 2, 3], [1, 2, 5])
[1, 2]
>>> common_prefix([1, 2, 3], [1, 3, 5])
[1]
"""
if any(not s for s in seqs):
return []
elif len(seqs) == 1:
return seqs[0]
i = 0
for i in range(min(len(s) for s in seqs)):
if not all(seqs[j][i] == seqs[0][i] for j in xrange(len(seqs))):
break
else:
i += 1
return seqs[0][:i]
[docs]def common_suffix(*seqs):
"""Return the subsequence that is a common ending of sequences in ``seqs``.
>>> from sympy.utilities.iterables import common_suffix
>>> common_suffix(list(range(3)))
[0, 1, 2]
>>> common_suffix(list(range(3)), list(range(4)))
[]
>>> common_suffix([1, 2, 3], [9, 2, 3])
[2, 3]
>>> common_suffix([1, 2, 3], [9, 7, 3])
[3]
"""
if any(not s for s in seqs):
return []
elif len(seqs) == 1:
return seqs[0]
i = 0
for i in range(-1, -min(len(s) for s in seqs) - 1, -1):
if not all(seqs[j][i] == seqs[0][i] for j in xrange(len(seqs))):
break
else:
i -= 1
if i == -1:
return []
else:
return seqs[0][i + 1:]
[docs]def prefixes(seq):
"""
Generate all prefixes of a sequence.
Examples
========
>>> from sympy.utilities.iterables import prefixes
>>> list(prefixes([1,2,3,4]))
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]]
"""
n = len(seq)
for i in xrange(n):
yield seq[:i + 1]
[docs]def postfixes(seq):
"""
Generate all postfixes of a sequence.
Examples
========
>>> from sympy.utilities.iterables import postfixes
>>> list(postfixes([1,2,3,4]))
[[4], [3, 4], [2, 3, 4], [1, 2, 3, 4]]
"""
n = len(seq)
for i in xrange(n):
yield seq[n - i - 1:]
[docs]def topological_sort(graph, key=None):
r"""
Topological sort of graph's vertices.
Parameters
==========
``graph`` : ``tuple[list, list[tuple[T, T]]``
A tuple consisting of a list of vertices and a list of edges of
a graph to be sorted topologically.
``key`` : ``callable[T]`` (optional)
Ordering key for vertices on the same level. By default the natural
(e.g. lexicographic) ordering is used (in this case the base type
must implement ordering relations).
Examples
========
Consider a graph::
+---+ +---+ +---+
| 7 |\ | 5 | | 3 |
+---+ \ +---+ +---+
| _\___/ ____ _/ |
| / \___/ \ / |
V V V V |
+----+ +---+ |
| 11 | | 8 | |
+----+ +---+ |
| | \____ ___/ _ |
| \ \ / / \ |
V \ V V / V V
+---+ \ +---+ | +----+
| 2 | | | 9 | | | 10 |
+---+ | +---+ | +----+
\________/
where vertices are integers. This graph can be encoded using
elementary Python's data structures as follows::
>>> V = [2, 3, 5, 7, 8, 9, 10, 11]
>>> E = [(7, 11), (7, 8), (5, 11), (3, 8), (3, 10),
... (11, 2), (11, 9), (11, 10), (8, 9)]
To compute a topological sort for graph ``(V, E)`` issue::
>>> from sympy.utilities.iterables import topological_sort
>>> topological_sort((V, E))
[3, 5, 7, 8, 11, 2, 9, 10]
If specific tie breaking approach is needed, use ``key`` parameter::
>>> topological_sort((V, E), key=lambda v: -v)
[7, 5, 11, 3, 10, 8, 9, 2]
Only acyclic graphs can be sorted. If the input graph has a cycle,
then :py:exc:`ValueError` will be raised::
>>> topological_sort((V, E + [(10, 7)]))
Traceback (most recent call last):
...
ValueError: cycle detected
.. seealso:: http://en.wikipedia.org/wiki/Topological_sorting
"""
V, E = graph
L = []
S = set(V)
E = list(E)
for v, u in E:
S.discard(u)
if key is None:
key = lambda value: value
S = sorted(S, key=key, reverse=True)
while S:
node = S.pop()
L.append(node)
for u, v in list(E):
if u == node:
E.remove((u, v))
for _u, _v in E:
if v == _v:
break
else:
kv = key(v)
for i, s in enumerate(S):
ks = key(s)
if kv > ks:
S.insert(i, v)
break
else:
S.append(v)
if E:
raise ValueError("cycle detected")
else:
return L
[docs]def rotate_left(x, y):
"""
Left rotates a list x by the number of steps specified
in y.
Examples
========
>>> from sympy.utilities.iterables import rotate_left
>>> a = [0, 1, 2]
>>> rotate_left(a, 1)
[1, 2, 0]
"""
if len(x) == 0:
return []
y = y % len(x)
return x[y:] + x[:y]
[docs]def rotate_right(x, y):
"""
Right rotates a list x by the number of steps specified
in y.
Examples
========
>>> from sympy.utilities.iterables import rotate_right
>>> a = [0, 1, 2]
>>> rotate_right(a, 1)
[2, 0, 1]
"""
if len(x) == 0:
return []
y = len(x) - y % len(x)
return x[y:] + x[:y]
[docs]def multiset_combinations(m, n, g=None):
"""
Return the unique combinations of size ``n`` from multiset ``m``.
Examples
========
>>> from sympy.utilities.iterables import multiset_combinations
>>> from itertools import combinations
>>> [''.join(i) for i in multiset_combinations('baby', 3)]
['abb', 'aby', 'bby']
>>> def count(f, s): return len(list(f(s, 3)))
The number of combinations depends on the number of letters; the
number of unique combinations depends on how the letters are
repeated.
>>> s1 = 'abracadabra'
>>> s2 = 'banana tree'
>>> count(combinations, s1), count(multiset_combinations, s1)
(165, 23)
>>> count(combinations, s2), count(multiset_combinations, s2)
(165, 54)
"""
if g is None:
if type(m) is dict:
if n > sum(m.values()):
return
g = [[k, m[k]] for k in ordered(m)]
else:
m = list(m)
if n > len(m):
return
try:
m = multiset(m)
g = [(k, m[k]) for k in ordered(m)]
except TypeError:
m = list(ordered(m))
g = [list(i) for i in group(m, multiple=False)]
del m
if sum(v for k, v in g) < n or not n:
yield []
else:
for i, (k, v) in enumerate(g):
if v >= n:
yield [k]*n
v = n - 1
for v in range(min(n, v), 0, -1):
for j in multiset_combinations(None, n - v, g[i + 1:]):
rv = [k]*v + j
if len(rv) == n:
yield rv
[docs]def multiset_permutations(m, size=None, g=None):
"""
Return the unique permutations of multiset ``m``.
Examples
========
>>> from sympy.utilities.iterables import multiset_permutations
>>> from sympy import factorial
>>> [''.join(i) for i in multiset_permutations('aab')]
['aab', 'aba', 'baa']
>>> factorial(len('banana'))
720
>>> len(list(multiset_permutations('banana')))
60
"""
if g is None:
if type(m) is dict:
g = [[k, m[k]] for k in ordered(m)]
else:
m = list(ordered(m))
g = [list(i) for i in group(m, multiple=False)]
del m
do = [gi for gi in g if gi[1] > 0]
SUM = sum([gi[1] for gi in do])
if not do or size is not None and (size > SUM or size < 1):
if size < 1:
yield []
return
elif size == 1:
for k, v in do:
yield [k]
elif len(do) == 1:
k, v = do[0]
v = v if size is None else (size if size <= v else 0)
yield [k for i in range(v)]
elif all(v == 1 for k, v in do):
for p in permutations([k for k, v in do], size):
yield list(p)
else:
size = size if size is not None else SUM
for i, (k, v) in enumerate(do):
do[i][1] -= 1
for j in multiset_permutations(None, size - 1, do):
if j:
yield [k] + j
do[i][1] += 1
def _partition(seq, vector, m=None):
"""
Return the partion of seq as specified by the partition vector.
Examples
========
>>> from sympy.utilities.iterables import _partition
>>> _partition('abcde', [1, 0, 1, 2, 0])
[['b', 'e'], ['a', 'c'], ['d']]
Specifying the number of bins in the partition is optional:
>>> _partition('abcde', [1, 0, 1, 2, 0], 3)
[['b', 'e'], ['a', 'c'], ['d']]
The output of _set_partitions can be passed as follows:
>>> output = (3, [1, 0, 1, 2, 0])
>>> _partition('abcde', *output)
[['b', 'e'], ['a', 'c'], ['d']]
See Also
========
combinatorics.partitions.Partition.from_rgs()
"""
if m is None:
m = max(vector) + 1
elif type(vector) is int: # entered as m, vector
vector, m = m, vector
p = [[] for i in range(m)]
for i, v in enumerate(vector):
p[v].append(seq[i])
return p
def _set_partitions(n):
"""Cycle through all partions of n elements, yielding the
current number of partitions, ``m``, and a mutable list, ``q``
such that element[i] is in part q[i] of the partition.
NOTE: ``q`` is modified in place and generally should not be changed
between function calls.
Examples
========
>>> from sympy.utilities.iterables import _set_partitions, _partition
>>> for m, q in _set_partitions(3):
... print('%s %s %s' % (m, q, _partition('abc', q, m)))
1 [0, 0, 0] [['a', 'b', 'c']]
2 [0, 0, 1] [['a', 'b'], ['c']]
2 [0, 1, 0] [['a', 'c'], ['b']]
2 [0, 1, 1] [['a'], ['b', 'c']]
3 [0, 1, 2] [['a'], ['b'], ['c']]
Notes
=====
This algorithm is similar to, and solves the same problem as,
Algorithm 7.2.1.5H, from volume 4A of Knuth's The Art of Computer
Programming. Knuth uses the term "restricted growth string" where
this code refers to a "partition vector". In each case, the meaning is
the same: the value in the ith element of the vector specifies to
which part the ith set element is to be assigned.
At the lowest level, this code implements an n-digit big-endian
counter (stored in the array q) which is incremented (with carries) to
get the next partition in the sequence. A special twist is that a
digit is constrained to be at most one greater than the maximum of all
the digits to the left of it. The array p maintains this maximum, so
that the code can efficiently decide when a digit can be incremented
in place or whether it needs to be reset to 0 and trigger a carry to
the next digit. The enumeration starts with all the digits 0 (which
corresponds to all the set elements being assigned to the same 0th
part), and ends with 0123...n, which corresponds to each set element
being assigned to a different, singleton, part.
This routine was rewritten to use 0-based lists while trying to
preserve the beauty and efficiency of the original algorithm.
Reference
=========
Nijenhuis, Albert and Wilf, Herbert. (1978) Combinatorial Algorithms,
2nd Ed, p 91, algorithm "nexequ". Available online from
http://www.math.upenn.edu/~wilf/website/CombAlgDownld.html (viewed
November 17, 2012).
"""
p = [0]*n
q = [0]*n
nc = 1
yield nc, q
while nc != n:
m = n
while 1:
m -= 1
i = q[m]
if p[i] != 1:
break
q[m] = 0
i += 1
q[m] = i
m += 1
nc += m - n
p[0] += n - m
if i == nc:
p[nc] = 0
nc += 1
p[i - 1] -= 1
p[i] += 1
yield nc, q
[docs]def multiset_partitions(multiset, m=None):
"""
Return unique partitions of the given multiset (in list form).
If ``m`` is None, all multisets will be returned, otherwise only
partitions with ``m`` parts will be returned.
If ``multiset`` is an integer, a range [0, 1, ..., multiset - 1]
will be supplied.
Examples
========
>>> from sympy.utilities.iterables import multiset_partitions
>>> list(multiset_partitions([1, 2, 3, 4], 2))
[[[1, 2, 3], [4]], [[1, 2, 4], [3]], [[1, 2], [3, 4]],
[[1, 3, 4], [2]], [[1, 3], [2, 4]], [[1, 4], [2, 3]],
[[1], [2, 3, 4]]]
>>> list(multiset_partitions([1, 2, 3, 4], 1))
[[[1, 2, 3, 4]]]
Only unique partitions are returned and these will be returned in a
canonical order regardless of the order of the input:
>>> a = [1, 2, 2, 1]
>>> ans = list(multiset_partitions(a, 2))
>>> a.sort()
>>> list(multiset_partitions(a, 2)) == ans
True
>>> a = range(3, 1, -1)
>>> (list(multiset_partitions(a)) ==
... list(multiset_partitions(sorted(a))))
True
If m is omitted then all partitions will be returned:
>>> list(multiset_partitions([1, 1, 2]))
[[[1, 1, 2]], [[1, 1], [2]], [[1, 2], [1]], [[1], [1], [2]]]
>>> list(multiset_partitions([1]*3))
[[[1, 1, 1]], [[1], [1, 1]], [[1], [1], [1]]]
Counting
========
The number of partitions of a set is given by the bell number:
>>> from sympy import bell
>>> len(list(multiset_partitions(5))) == bell(5) == 52
True
The number of partitions of length k from a set of size n is given by the
Stirling Number of the 2nd kind:
>>> def S2(n, k):
... from sympy import Dummy, binomial, factorial, Sum
... if k > n:
... return 0
... j = Dummy()
... arg = (-1)**(k-j)*j**n*binomial(k,j)
... return 1/factorial(k)*Sum(arg,(j,0,k)).doit()
...
>>> S2(5, 2) == len(list(multiset_partitions(5, 2))) == 15
True
These comments on counting apply to *sets*, not multisets.
Notes
=====
When all the elements are the same in the multiset, the order
of the returned partitions is determined by the ``partitions``
routine. If one is counting partitions then it is better to use
the ``nT`` function.
See Also
========
partitions
sympy.combinatorics.partitions.Partition
sympy.combinatorics.partitions.IntegerPartition
sympy.functions.combinatorial.numbers.nT
"""
# This function looks at the supplied input and dispatches to
# several special-case routines as they apply.
if type(multiset) is int:
n = multiset
if m and m > n:
return
multiset = list(range(n))
if m == 1:
yield [multiset[:]]
return
# If m is not None, it can sometimes be faster to use
# MultisetPartitionTraverser.enum_range() even for inputs
# which are sets. Since the _set_partitions code is quite
# fast, this is only advantageous when the overall set
# partitions outnumber those with the desired number of parts
# by a large factor. (At least 60.) Such a switch is not
# currently implemented.
for nc, q in _set_partitions(n):
if m is None or nc == m:
rv = [[] for i in range(nc)]
for i in range(n):
rv[q[i]].append(multiset[i])
yield rv
return
if len(multiset) == 1 and type(multiset) is str:
multiset = [multiset]
if not has_variety(multiset):
# Only one component, repeated n times. The resulting
# partitions correspond to partitions of integer n.
n = len(multiset)
if m and m > n:
return
if m == 1:
yield [multiset[:]]
return
x = multiset[:1]
for size, p in partitions(n, m, size=True):
if m is None or size == m:
rv = []
for k in sorted(p):
rv.extend([x*k]*p[k])
yield rv
else:
multiset = list(ordered(multiset))
n = len(multiset)
if m and m > n:
return
if m == 1:
yield [multiset[:]]
return
# Split the information of the multiset into two lists -
# one of the elements themselves, and one (of the same length)
# giving the number of repeats for the corresponding element.
elements, multiplicities = zip(*group(multiset, False))
if len(elements) < len(multiset):
# General case - multiset with more than one distinct element
# and at least one element repeated more than once.
if m:
mpt = MultisetPartitionTraverser()
for state in mpt.enum_range(multiplicities, m-1, m):
yield list_visitor(state, elements)
else:
for state in multiset_partitions_taocp(multiplicities):
yield list_visitor(state, elements)
else:
# Set partitions case - no repeated elements. Pretty much
# same as int argument case above, with same possible, but
# currently unimplemented optimization for some cases when
# m is not None
for nc, q in _set_partitions(n):
if m is None or nc == m:
rv = [[] for i in range(nc)]
for i in range(n):
rv[q[i]].append(i)
yield [[multiset[j] for j in i] for i in rv]
[docs]def partitions(n, m=None, k=None, size=False):
"""Generate all partitions of integer n (>= 0).
Parameters
==========
``m`` : integer (default gives partitions of all sizes)
limits number of parts in parition (mnemonic: m, maximum parts)
``k`` : integer (default gives partitions number from 1 through n)
limits the numbers that are kept in the partition (mnemonic: k, keys)
``size`` : bool (default False, only partition is returned)
when ``True`` then (M, P) is returned where M is the sum of the
multiplicities and P is the generated partition.
Each partition is represented as a dictionary, mapping an integer
to the number of copies of that integer in the partition. For example,
the first partition of 4 returned is {4: 1}, "4: one of them".
Examples
========
>>> from sympy.utilities.iterables import partitions
The numbers appearing in the partition (the key of the returned dict)
are limited with k:
>>> for p in partitions(6, k=2):
... print(p)
{2: 3}
{1: 2, 2: 2}
{1: 4, 2: 1}
{1: 6}
The maximum number of parts in the partion (the sum of the values in
the returned dict) are limited with m:
>>> for p in partitions(6, m=2):
... print(p)
...
{6: 1}
{1: 1, 5: 1}
{2: 1, 4: 1}
{3: 2}
Note that the _same_ dictionary object is returned each time.
This is for speed: generating each partition goes quickly,
taking constant time, independent of n.
>>> [p for p in partitions(6, k=2)]
[{1: 6}, {1: 6}, {1: 6}, {1: 6}]
If you want to build a list of the returned dictionaries then
make a copy of them:
>>> [p.copy() for p in partitions(6, k=2)]
[{2: 3}, {1: 2, 2: 2}, {1: 4, 2: 1}, {1: 6}]
>>> [(M, p.copy()) for M, p in partitions(6, k=2, size=True)]
[(3, {2: 3}), (4, {1: 2, 2: 2}), (5, {1: 4, 2: 1}), (6, {1: 6})]
Reference:
modified from Tim Peter's version to allow for k and m values:
code.activestate.com/recipes/218332-generator-for-integer-partitions/
See Also
========
sympy.combinatorics.partitions.Partition
sympy.combinatorics.partitions.IntegerPartition
"""
if n < 0:
raise ValueError("n must be >= 0")
if m == 0:
raise ValueError("m must be > 0")
m = min(m or n, n)
if m < 1:
raise ValueError("maximum numbers in partition, m, must be > 0")
k = min(k or n, n)
if k < 1:
raise ValueError("maximum value in partition, k, must be > 0")
if m*k < n:
return
n, m, k = as_int(n), as_int(m), as_int(k)
q, r = divmod(n, k)
ms = {k: q}
keys = [k] # ms.keys(), from largest to smallest
if r:
ms[r] = 1
keys.append(r)
room = m - q - bool(r)
if size:
yield sum(ms.values()), ms
else:
yield ms
while keys != [1]:
# Reuse any 1's.
if keys[-1] == 1:
del keys[-1]
reuse = ms.pop(1)
room += reuse
else:
reuse = 0
while 1:
# Let i be the smallest key larger than 1. Reuse one
# instance of i.
i = keys[-1]
newcount = ms[i] = ms[i] - 1
reuse += i
if newcount == 0:
del keys[-1], ms[i]
room += 1
# Break the remainder into pieces of size i-1.
i -= 1
q, r = divmod(reuse, i)
need = q + bool(r)
if need > room:
if not keys:
return
continue
ms[i] = q
keys.append(i)
if r:
ms[r] = 1
keys.append(r)
break
room -= need
if size:
yield sum(ms.values()), ms
else:
yield ms
[docs]def binary_partitions(n):
"""
Generates the binary partition of n.
A binary partition consists only of numbers that are
powers of two. Each step reduces a 2**(k+1) to 2**k and
2**k. Thus 16 is converted to 8 and 8.
Reference: TAOCP 4, section 7.2.1.5, problem 64
Examples
========
>>> from sympy.utilities.iterables import binary_partitions
>>> for i in binary_partitions(5):
... print(i)
...
[4, 1]
[2, 2, 1]
[2, 1, 1, 1]
[1, 1, 1, 1, 1]
"""
from math import ceil, log
pow = int(2**(ceil(log(n, 2))))
sum = 0
partition = []
while pow:
if sum + pow <= n:
partition.append(pow)
sum += pow
pow >>= 1
last_num = len(partition) - 1 - (n & 1)
while last_num >= 0:
yield partition
if partition[last_num] == 2:
partition[last_num] = 1
partition.append(1)
last_num -= 1
continue
partition.append(1)
partition[last_num] >>= 1
x = partition[last_num + 1] = partition[last_num]
last_num += 1
while x > 1:
if x <= len(partition) - last_num - 1:
del partition[-x + 1:]
last_num += 1
partition[last_num] = x
else:
x >>= 1
yield [1]*n
[docs]def has_dups(seq):
"""Return True if there are any duplicate elements in ``seq``.
Examples
========
>>> from sympy.utilities.iterables import has_dups
>>> from sympy import Dict, Set
>>> has_dups((1, 2, 1))
True
>>> has_dups(range(3))
False
>>> all(has_dups(c) is False for c in (set(), Set(), dict(), Dict()))
True
"""
if isinstance(seq, (dict, set, C.Dict, C.Set)):
return False
uniq = set()
return any(True for s in seq if s in uniq or uniq.add(s))
[docs]def has_variety(seq):
"""Return True if there are any different elements in ``seq``.
Examples
========
>>> from sympy.utilities.iterables import has_variety
>>> has_variety((1, 2, 1))
True
>>> has_variety((1, 1, 1))
False
"""
for i, s in enumerate(seq):
if i == 0:
sentinel = s
else:
if s != sentinel:
return True
return False
[docs]def uniq(seq, result=None):
"""
Yield unique elements from ``seq`` as an iterator. The second
parameter ``result`` is used internally; it is not necessary to pass
anything for this.
Examples
========
>>> from sympy.utilities.iterables import uniq
>>> dat = [1, 4, 1, 5, 4, 2, 1, 2]
>>> type(uniq(dat)) in (list, tuple)
False
>>> list(uniq(dat))
[1, 4, 5, 2]
>>> list(uniq(x for x in dat))
[1, 4, 5, 2]
>>> list(uniq([[1], [2, 1], [1]]))
[[1], [2, 1]]
"""
try:
seen = set()
result = result or []
for i, s in enumerate(seq):
if not (s in seen or seen.add(s)):
yield s
except TypeError:
if s not in result:
yield s
result.append(s)
if hasattr(seq, '__getitem__'):
for s in uniq(seq[i + 1:], result):
yield s
else:
for s in uniq(seq, result):
yield s
[docs]def generate_bell(n):
"""Return permutations of [0, 1, ..., n - 1] such that each permutation
differs from the last by the exchange of a single pair of neighbors.
The ``n!`` permutations are returned as an iterator. In order to obtain
the next permutation from a random starting permutation, use the
``next_trotterjohnson`` method of the Permutation class (which generates
the same sequence in a different manner).
Examples
========
>>> from itertools import permutations
>>> from sympy.utilities.iterables import generate_bell
>>> from sympy import zeros, Matrix
This is the sort of permutation used in the ringing of physical bells,
and does not produce permutations in lexicographical order. Rather, the
permutations differ from each other by exactly one inversion, and the
position at which the swapping occurs varies periodically in a simple
fashion. Consider the first few permutations of 4 elements generated
by ``permutations`` and ``generate_bell``:
>>> list(permutations(range(4)))[:5]
[(0, 1, 2, 3), (0, 1, 3, 2), (0, 2, 1, 3), (0, 2, 3, 1), (0, 3, 1, 2)]
>>> list(generate_bell(4))[:5]
[(0, 1, 2, 3), (0, 1, 3, 2), (0, 3, 1, 2), (3, 0, 1, 2), (3, 0, 2, 1)]
Notice how the 2nd and 3rd lexicographical permutations have 3 elements
out of place whereas each "bell" permutation always has only two
elements out of place relative to the previous permutation (and so the
signature (+/-1) of a permutation is opposite of the signature of the
previous permutation).
How the position of inversion varies across the elements can be seen
by tracing out where the largest number appears in the permutations:
>>> m = zeros(4, 24)
>>> for i, p in enumerate(generate_bell(4)):
... m[:, i] = Matrix([j - 3 for j in list(p)]) # make largest zero
>>> m.print_nonzero('X')
[XXX XXXXXX XXXXXX XXX]
[XX XX XXXX XX XXXX XX XX]
[X XXXX XX XXXX XX XXXX X]
[ XXXXXX XXXXXX XXXXXX ]
See Also
========
sympy.combinatorics.Permutation.next_trotterjohnson
References
==========
* http://en.wikipedia.org/wiki/Method_ringing
* http://stackoverflow.com/questions/4856615/recursive-permutation/4857018
* http://programminggeeks.com/bell-algorithm-for-permutation/
* http://en.wikipedia.org/wiki/
Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm
* Generating involutions, derangements, and relatives by ECO
Vincent Vajnovszki, DMTCS vol 1 issue 12, 2010
"""
n = as_int(n)
if n < 1:
raise ValueError('n must be a positive integer')
if n == 1:
yield (0,)
elif n == 2:
yield (0, 1)
yield (1, 0)
elif n == 3:
for li in [(0, 1, 2), (0, 2, 1), (2, 0, 1), (2, 1, 0), (1, 2, 0), (1, 0, 2)]:
yield li
else:
m = n - 1
op = [0] + [-1]*m
l = list(range(n))
while True:
yield tuple(l)
# find biggest element with op
big = None, -1 # idx, value
for i in range(n):
if op[i] and l[i] > big[1]:
big = i, l[i]
i, _ = big
if i is None:
break # there are no ops left
# swap it with neighbor in the indicated direction
j = i + op[i]
l[i], l[j] = l[j], l[i]
op[i], op[j] = op[j], op[i]
# if it landed at the end or if the neighbor in the same
# direction is bigger then turn off op
if j == 0 or j == m or l[j + op[j]] > l[j]:
op[j] = 0
# any element bigger to the left gets +1 op
for i in range(j):
if l[i] > l[j]:
op[i] = 1
# any element bigger to the right gets -1 op
for i in range(j + 1, n):
if l[i] > l[j]:
op[i] = -1
[docs]def generate_involutions(n):
"""
Generates involutions.
An involution is a permutation that when multiplied
by itself equals the identity permutation. In this
implementation the involutions are generated using
Fixed Points.
Alternatively, an involution can be considered as
a permutation that does not contain any cycles with
a length that is greater than two.
Reference:
http://mathworld.wolfram.com/PermutationInvolution.html
Examples
========
>>> from sympy.utilities.iterables import generate_involutions
>>> list(generate_involutions(3))
[(0, 1, 2), (0, 2, 1), (1, 0, 2), (2, 1, 0)]
>>> len(list(generate_involutions(4)))
10
"""
idx = list(range(n))
for p in permutations(idx):
for i in idx:
if p[p[i]] != i:
break
else:
yield p
[docs]def generate_derangements(perm):
"""
Routine to generate unique derangements.
TODO: This will be rewritten to use the
ECO operator approach once the permutations
branch is in master.
Examples
========
>>> from sympy.utilities.iterables import generate_derangements
>>> list(generate_derangements([0, 1, 2]))
[[1, 2, 0], [2, 0, 1]]
>>> list(generate_derangements([0, 1, 2, 3]))
[[1, 0, 3, 2], [1, 2, 3, 0], [1, 3, 0, 2], [2, 0, 3, 1], \
[2, 3, 0, 1], [2, 3, 1, 0], [3, 0, 1, 2], [3, 2, 0, 1], \
[3, 2, 1, 0]]
>>> list(generate_derangements([0, 1, 1]))
[]
See Also
========
sympy.functions.combinatorial.factorials.subfactorial
"""
p = multiset_permutations(perm)
indices = range(len(perm))
p0 = next(p)
for pi in p:
if all(pi[i] != p0[i] for i in indices):
yield pi
@deprecated(
useinstead="bracelets", deprecated_since_version="0.7.3")
[docs]def unrestricted_necklace(n, k):
"""Wrapper to necklaces to return a free (unrestricted) necklace."""
return necklaces(n, k, free=True)
[docs]def necklaces(n, k, free=False):
"""
A routine to generate necklaces that may (free=True) or may not
(free=False) be turned over to be viewed. The "necklaces" returned
are comprised of ``n`` integers (beads) with ``k`` different
values (colors). Only unique necklaces are returned.
Examples
========
>>> from sympy.utilities.iterables import necklaces, bracelets
>>> def show(s, i):
... return ''.join(s[j] for j in i)
The "unrestricted necklace" is sometimes also referred to as a
"bracelet" (an object that can be turned over, a sequence that can
be reversed) and the term "necklace" is used to imply a sequence
that cannot be reversed. So ACB == ABC for a bracelet (rotate and
reverse) while the two are different for a necklace since rotation
alone cannot make the two sequences the same.
(mnemonic: Bracelets can be viewed Backwards, but Not Necklaces.)
>>> B = [show('ABC', i) for i in bracelets(3, 3)]
>>> N = [show('ABC', i) for i in necklaces(3, 3)]
>>> set(N) - set(B)
set(['ACB'])
>>> list(necklaces(4, 2))
[(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 1),
(0, 1, 0, 1), (0, 1, 1, 1), (1, 1, 1, 1)]
>>> [show('.o', i) for i in bracelets(4, 2)]
['....', '...o', '..oo', '.o.o', '.ooo', 'oooo']
References
==========
http://mathworld.wolfram.com/Necklace.html
"""
return uniq(minlex(i, directed=not free) for i in
variations(list(range(k)), n, repetition=True))
[docs]def bracelets(n, k):
"""Wrapper to necklaces to return a free (unrestricted) necklace."""
return necklaces(n, k, free=True)
[docs]def generate_oriented_forest(n):
"""
This algorithm generates oriented forests.
An oriented graph is a directed graph having no symmetric pair of directed
edges. A forest is an acyclic graph, i.e., it has no cycles. A forest can
also be described as a disjoint union of trees, which are graphs in which
any two vertices are connected by exactly one simple path.
Reference:
[1] T. Beyer and S.M. Hedetniemi: constant time generation of \
rooted trees, SIAM J. Computing Vol. 9, No. 4, November 1980
[2] http://stackoverflow.com/questions/1633833/oriented-forest-taocp-algorithm-in-python
Examples
========
>>> from sympy.utilities.iterables import generate_oriented_forest
>>> list(generate_oriented_forest(4))
[[0, 1, 2, 3], [0, 1, 2, 2], [0, 1, 2, 1], [0, 1, 2, 0], \
[0, 1, 1, 1], [0, 1, 1, 0], [0, 1, 0, 1], [0, 1, 0, 0], [0, 0, 0, 0]]
"""
P = list(range(-1, n))
while True:
yield P[1:]
if P[n] > 0:
P[n] = P[P[n]]
else:
for p in xrange(n - 1, 0, -1):
if P[p] != 0:
target = P[p] - 1
for q in xrange(p - 1, 0, -1):
if P[q] == target:
break
offset = p - q
for i in xrange(p, n + 1):
P[i] = P[i - offset]
break
else:
break
[docs]def minlex(seq, directed=True, is_set=False, small=None):
"""
Return a tuple where the smallest element appears first; if
``directed`` is True (default) then the order is preserved, otherwise
the sequence will be reversed if that gives a smaller ordering.
If every element appears only once then is_set can be set to True
for more efficient processing.
If the smallest element is known at the time of calling, it can be
passed and the calculation of the smallest element will be omitted.
Examples
========
>>> from sympy.combinatorics.polyhedron import minlex
>>> minlex((1, 2, 0))
(0, 1, 2)
>>> minlex((1, 0, 2))
(0, 2, 1)
>>> minlex((1, 0, 2), directed=False)
(0, 1, 2)
>>> minlex('11010011000', directed=True)
'00011010011'
>>> minlex('11010011000', directed=False)
'00011001011'
"""
is_str = isinstance(seq, str)
seq = list(seq)
if small is None:
small = min(seq, key=default_sort_key)
if is_set:
i = seq.index(small)
if not directed:
n = len(seq)
p = (i + 1) % n
m = (i - 1) % n
if default_sort_key(seq[p]) > default_sort_key(seq[m]):
seq = list(reversed(seq))
i = n - i - 1
if i:
seq = rotate_left(seq, i)
best = seq
else:
count = seq.count(small)
if count == 1 and directed:
best = rotate_left(seq, seq.index(small))
else:
# if not directed, and not a set, we can't just
# pass this off to minlex with is_set True since
# peeking at the neighbor may not be sufficient to
# make the decision so we continue...
best = seq
for i in range(count):
seq = rotate_left(seq, seq.index(small, count != 1))
if seq < best:
best = seq
# it's cheaper to rotate now rather than search
# again for these in reversed order so we test
# the reverse now
if not directed:
seq = rotate_left(seq, 1)
seq = list(reversed(seq))
if seq < best:
best = seq
seq = list(reversed(seq))
seq = rotate_right(seq, 1)
# common return
if is_str:
return ''.join(best)
return tuple(best)
[docs]def runs(seq, op=gt):
"""Group the sequence into lists in which successive elements
all compare the same with the comparison operator, ``op``:
op(seq[i + 1], seq[i]) is True from all elements in a run.
Examples
========
>>> from sympy.utilities.iterables import runs
>>> from operator import ge
>>> runs([0, 1, 2, 2, 1, 4, 3, 2, 2])
[[0, 1, 2], [2], [1, 4], [3], [2], [2]]
>>> runs([0, 1, 2, 2, 1, 4, 3, 2, 2], op=ge)
[[0, 1, 2, 2], [1, 4], [3], [2, 2]]
"""
cycles = []
seq = iter(seq)
try:
run = [next(seq)]
except StopIteration:
return []
while True:
try:
ei = next(seq)
except StopIteration:
break
if op(ei, run[-1]):
run.append(ei)
continue
else:
cycles.append(run)
run = [ei]
if run:
cycles.append(run)
return cycles
[docs]def kbins(l, k, ordered=None):
"""
Return sequence ``l`` partitioned into ``k`` bins.
Examples
========
>>> from sympy.utilities.iterables import kbins
The default is to give the items in the same order, but grouped
into k partitions without any reordering:
>>> from __future__ import print_function
>>> for p in kbins(list(range(5)), 2):
... print(p)
...
[[0], [1, 2, 3, 4]]
[[0, 1], [2, 3, 4]]
[[0, 1, 2], [3, 4]]
[[0, 1, 2, 3], [4]]
The ``ordered`` flag which is either None (to give the simple partition
of the the elements) or is a 2 digit integer indicating whether the order of
the bins and the order of the items in the bins matters. Given::
A = [[0], [1, 2]]
B = [[1, 2], [0]]
C = [[2, 1], [0]]
D = [[0], [2, 1]]
the following values for ``ordered`` have the shown meanings::
00 means A == B == C == D
01 means A == B
10 means A == D
11 means A == A
>>> for ordered in [None, 0, 1, 10, 11]:
... print('ordered = %s' % ordered)
... for p in kbins(list(range(3)), 2, ordered=ordered):
... print(' %s' % p)
...
ordered = None
[[0], [1, 2]]
[[0, 1], [2]]
ordered = 0
[[0, 1], [2]]
[[0, 2], [1]]
[[0], [1, 2]]
ordered = 1
[[0], [1, 2]]
[[0], [2, 1]]
[[1], [0, 2]]
[[1], [2, 0]]
[[2], [0, 1]]
[[2], [1, 0]]
ordered = 10
[[0, 1], [2]]
[[2], [0, 1]]
[[0, 2], [1]]
[[1], [0, 2]]
[[0], [1, 2]]
[[1, 2], [0]]
ordered = 11
[[0], [1, 2]]
[[0, 1], [2]]
[[0], [2, 1]]
[[0, 2], [1]]
[[1], [0, 2]]
[[1, 0], [2]]
[[1], [2, 0]]
[[1, 2], [0]]
[[2], [0, 1]]
[[2, 0], [1]]
[[2], [1, 0]]
[[2, 1], [0]]
See Also
========
partitions, multiset_partitions
"""
def partition(lista, bins):
# EnricoGiampieri's partition generator from
# http://stackoverflow.com/questions/13131491/
# partition-n-items-into-k-bins-in-python-lazily
if len(lista) == 1 or bins == 1:
yield [lista]
elif len(lista) > 1 and bins > 1:
for i in range(1, len(lista)):
for part in partition(lista[i:], bins - 1):
if len([lista[:i]] + part) == bins:
yield [lista[:i]] + part
if ordered is None:
for p in partition(l, k):
yield p
elif ordered == 11:
for pl in multiset_permutations(l):
pl = list(pl)
for p in partition(pl, k):
yield p
elif ordered == 00:
for p in multiset_partitions(l, k):
yield p
elif ordered == 10:
for p in multiset_partitions(l, k):
for perm in permutations(p):
yield list(perm)
elif ordered == 1:
for kgot, p in partitions(len(l), k, size=True):
if kgot != k:
continue
for li in multiset_permutations(l):
rv = []
i = j = 0
li = list(li)
for size, multiplicity in sorted(p.items()):
for m in range(multiplicity):
j = i + size
rv.append(li[i: j])
i = j
yield rv
else:
raise ValueError(
'ordered must be one of 00, 01, 10 or 11, not %s' % ordered)
```