Iterables#
- class sympy.utilities.iterables.NotIterable[source]#
Use this as mixin when creating a class which is not supposed to return true when iterable() is called on its instances because calling list() on the instance, for example, would result in an infinite loop.
- sympy.utilities.iterables.binary_partitions(n)[source]#
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.
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]
References
[R1032]TAOCP 4, section 7.2.1.5, problem 64
- sympy.utilities.iterables.bracelets(n, k)[source]#
Wrapper to necklaces to return a free (unrestricted) necklace.
- sympy.utilities.iterables.capture(func)[source]#
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'
- sympy.utilities.iterables.common_prefix(*seqs)[source]#
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]
- sympy.utilities.iterables.common_suffix(*seqs)[source]#
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]
- sympy.utilities.iterables.connected_components(G)[source]#
Connected components of an undirected graph or weakly connected components of a directed graph.
- Parameters:
G : tuple[list, list[tuple[T, T]]
A tuple consisting of a list of vertices and a list of edges of a graph whose connected components are to be found.
Examples
Given an undirected graph:
graph { A -- B C -- D }
We can find the connected components using this function if we include each edge in both directions:
>>> from sympy.utilities.iterables import connected_components >>> V = ['A', 'B', 'C', 'D'] >>> E = [('A', 'B'), ('B', 'A'), ('C', 'D'), ('D', 'C')] >>> connected_components((V, E)) [['A', 'B'], ['C', 'D']]
The weakly connected components of a directed graph can found the same way.
Notes
The vertices of the graph must be hashable for the data structures used. If the vertices are unhashable replace them with integer indices.
This function uses Tarjan’s algorithm to compute the connected components in \(O(|V|+|E|)\) (linear) time.
References
- sympy.utilities.iterables.filter_symbols(iterator, exclude)[source]#
Only yield elements from \(iterator\) that do not occur in \(exclude\).
- Parameters:
iterator : iterable
iterator to take elements from
exclude : iterable
elements to exclude
- Returns:
iterator : iterator
filtered iterator
- sympy.utilities.iterables.flatten(iterable, levels=None, cls=None)[source]#
Recursively denest iterable containers.
>>> from sympy 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 import Basic, S >>> class MyOp(Basic): ... pass ... >>> flatten([MyOp(S(1), MyOp(S(2), S(3)))], cls=MyOp) [1, 2, 3]
adapted from https://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks
- sympy.utilities.iterables.generate_bell(n)[source]#
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 thenext_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
andgenerate_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 ]
References
- sympy.utilities.iterables.generate_derangements(s)[source]#
Return unique derangements of the elements of iterable
s
.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, 2])) [[2, 2, 0, 1], [2, 2, 1, 0]] >>> list(generate_derangements([0, 1, 1])) []
- sympy.utilities.iterables.generate_involutions(n)[source]#
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.
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
References
- sympy.utilities.iterables.generate_oriented_forest(n)[source]#
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.
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]]
References
[R1041]T. Beyer and S.M. Hedetniemi: constant time generation of rooted trees, SIAM J. Computing Vol. 9, No. 4, November 1980
- sympy.utilities.iterables.group(seq, multiple=True)[source]#
Splits a sequence into a list of lists of equal, adjacent elements.
Examples
>>> from sympy 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
- sympy.utilities.iterables.has_dups(seq)[source]#
Return True if there are any duplicate elements in
seq
.Examples
>>> from sympy import has_dups, 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
- sympy.utilities.iterables.has_variety(seq)[source]#
Return True if there are any different elements in
seq
.Examples
>>> from sympy import has_variety
>>> has_variety((1, 2, 1)) True >>> has_variety((1, 1, 1)) False
- sympy.utilities.iterables.ibin(n, bits=None, str=False)[source]#
Return a list of length
bits
corresponding to the binary value ofn
with small bits to the right (last). If bits is omitted, the length will be the number required to representn
. 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]
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']
- sympy.utilities.iterables.iproduct(*iterables)[source]#
Cartesian product of iterables.
Generator of the Cartesian product of iterables. This is analogous to itertools.product except that it works with infinite iterables and will yield any item from the infinite product eventually.
Examples
>>> from sympy.utilities.iterables import iproduct >>> sorted(iproduct([1,2], [3,4])) [(1, 3), (1, 4), (2, 3), (2, 4)]
With an infinite iterator:
>>> from sympy import S >>> (3,) in iproduct(S.Integers) True >>> (3, 4) in iproduct(S.Integers, S.Integers) True
See also
- sympy.utilities.iterables.is_palindromic(s, i=0, j=None)[source]#
Return True if the sequence is the same from left to right as it is from right to left in the whole sequence (default) or in the Python slice
s[i: j]
; else False.Examples
>>> from sympy.utilities.iterables import is_palindromic >>> is_palindromic([1, 0, 1]) True >>> is_palindromic('abcbb') False >>> is_palindromic('abcbb', 1) False
Normal Python slicing is performed in place so there is no need to create a slice of the sequence for testing:
>>> is_palindromic('abcbb', 1, -1) True >>> is_palindromic('abcbb', -4, -1) True
See also
sympy.ntheory.digits.is_palindromic
tests integers
- sympy.utilities.iterables.is_sequence(i, include=None)[source]#
Return a boolean indicating whether
i
is a sequence in the SymPy sense. If anything that fails the test below should be included as being a sequence for your application, set ‘include’ to that object’s type; multiple types should be passed as a tuple of types.Note: although generators can generate a sequence, they often need special handling to make sure their elements are captured before the generator is exhausted, so these are not included by default in the definition of a sequence.
See also: iterable
Examples
>>> from sympy.utilities.iterables import is_sequence >>> from types import GeneratorType >>> is_sequence([]) True >>> is_sequence(set()) False >>> is_sequence('abc') False >>> is_sequence('abc', include=str) True >>> generator = (c for c in 'abc') >>> is_sequence(generator) False >>> is_sequence(generator, include=(str, GeneratorType)) True
- sympy.utilities.iterables.iterable(i, exclude=(<class 'str'>, <class 'dict'>, <class 'sympy.utilities.iterables.NotIterable'>))[source]#
Return a boolean indicating whether
i
is SymPy iterable. True also indicates that the iterator is finite, e.g. you can call list(…) on the instance.When SymPy is working with iterables, it is almost always assuming that the iterable is not a string or a mapping, so those are excluded by default. If you want a pure Python definition, make exclude=None. To exclude multiple items, pass them as a tuple.
You can also set the _iterable attribute to True or False on your class, which will override the checks here, including the exclude test.
As a rule of thumb, some SymPy functions use this to check if they should recursively map over an object. If an object is technically iterable in the Python sense but does not desire this behavior (e.g., because its iteration is not finite, or because iteration might induce an unwanted computation), it should disable it by setting the _iterable attribute to False.
See also: is_sequence
Examples
>>> from sympy.utilities.iterables import iterable >>> from sympy import Tuple >>> things = [[1], (1,), set([1]), Tuple(1), (j for j in [1, 2]), {1:2}, '1', 1] >>> for i in things: ... print('%s %s' % (iterable(i), type(i))) True <... 'list'> True <... 'tuple'> True <... 'set'> True <class 'sympy.core.containers.Tuple'> True <... 'generator'> False <... 'dict'> False <... 'str'> False <... 'int'>
>>> iterable({}, exclude=None) True >>> iterable({}, exclude=str) True >>> iterable("no", exclude=str) False
- sympy.utilities.iterables.kbins(l, k, ordered=None)[source]#
Return sequence
l
partitioned intok
bins.Examples
The default is to give the items in the same order, but grouped into k partitions without any reordering:
>>> from sympy.utilities.iterables import kbins >>> 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 is either None (to give the simple partition of 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_flag in [None, 0, 1, 10, 11]: ... print('ordered = %s' % ordered_flag) ... for p in kbins(list(range(3)), 2, ordered=ordered_flag): ... 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
- sympy.utilities.iterables.least_rotation(x, key=None)[source]#
Returns the number of steps of left rotation required to obtain lexicographically minimal string/list/tuple, etc.
Examples
>>> from sympy.utilities.iterables import least_rotation, rotate_left >>> a = [3, 1, 5, 1, 2] >>> least_rotation(a) 3 >>> rotate_left(a, _) [1, 2, 3, 1, 5]
References
- sympy.utilities.iterables.minlex(seq, directed=True, key=None)[source]#
Return the rotation of the sequence in which the lexically smallest elements appear first, e.g. \(cba \rightarrow acb\).
The sequence returned is a tuple, unless the input sequence is a string in which case a string is returned.
If
directed
is False then the smaller of the sequence and the reversed sequence is returned, e.g. \(cba \rightarrow abc\).If
key
is not None then it is used to extract a comparison key from each element in iterable.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'
>>> minlex(('bb', 'aaa', 'c', 'a')) ('a', 'bb', 'aaa', 'c') >>> minlex(('bb', 'aaa', 'c', 'a'), key=len) ('c', 'a', 'bb', 'aaa')
- sympy.utilities.iterables.multiset(seq)[source]#
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
- sympy.utilities.iterables.multiset_combinations(m, n, g=None)[source]#
Return the unique combinations of size
n
from multisetm
.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)
- sympy.utilities.iterables.multiset_derangements(s)[source]#
Generate derangements of the elements of s in place.
Examples
>>> from sympy.utilities.iterables import multiset_derangements, uniq
Because the derangements of multisets (not sets) are generated in place, copies of the return value must be made if a collection of derangements is desired or else all values will be the same:
>>> list(uniq([i for i in multiset_derangements('1233')])) [[None, None, None, None]] >>> [i.copy() for i in multiset_derangements('1233')] [['3', '3', '1', '2'], ['3', '3', '2', '1']] >>> [''.join(i) for i in multiset_derangements('1233')] ['3312', '3321']
- sympy.utilities.iterables.multiset_partitions(multiset, m=None)[source]#
Return unique partitions of the given multiset (in list form). If
m
is None, all multisets will be returned, otherwise only partitions withm
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:
>>> from sympy.functions.combinatorial.numbers import stirling >>> stirling(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 thenT
function.
- sympy.utilities.iterables.multiset_permutations(m, size=None, g=None)[source]#
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
- sympy.utilities.iterables.necklaces(n, k, free=False)[source]#
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) withk
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) {'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
[R1045]Frank Ruskey, Carla Savage, and Terry Min Yih Wang, Generating necklaces, Journal of Algorithms 13 (1992), 414-430; https://doi.org/10.1016/0196-6774(92)90047-G
- sympy.utilities.iterables.numbered_symbols(prefix='x', cls=None, start=0, exclude=(), *args, **assumptions)[source]#
Generate an infinite stream of Symbols consisting of a prefix and increasing subscripts provided that they do not occur in
exclude
.- 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 useWild
orDummy
.start : int, optional
The start number. By default, it is 0.
exclude : list, tuple, set of cls, optional
Symbols to be excluded.
*args, **kwargs
Additional positional and keyword arguments are passed to the cls class.
- Returns:
sym : Symbol
The subscripted symbols.
- sympy.utilities.iterables.ordered_partitions(n, m=None, sort=True)[source]#
Generates ordered partitions of integer n.
- Parameters:
n : int
m : int, optional
The default value gives partitions of all sizes else only those with size m. In addition, if m is not None then partitions are generated in place (see examples).
sort : bool, default: True
Controls whether partitions are returned in sorted order when m is not None; when False, the partitions are returned as fast as possible with elements sorted, but when m|n the partitions will not be in ascending lexicographical order.
Examples
>>> from sympy.utilities.iterables import ordered_partitions
All partitions of 5 in ascending lexicographical:
>>> for p in ordered_partitions(5): ... print(p) [1, 1, 1, 1, 1] [1, 1, 1, 2] [1, 1, 3] [1, 2, 2] [1, 4] [2, 3] [5]
Only partitions of 5 with two parts:
>>> for p in ordered_partitions(5, 2): ... print(p) [1, 4] [2, 3]
When
m
is given, a given list objects will be used more than once for speed reasons so you will not see the correct partitions unless you make a copy of each as it is generated:>>> [p for p in ordered_partitions(7, 3)] [[1, 1, 1], [1, 1, 1], [1, 1, 1], [2, 2, 2]] >>> [list(p) for p in ordered_partitions(7, 3)] [[1, 1, 5], [1, 2, 4], [1, 3, 3], [2, 2, 3]]
When
n
is a multiple ofm
, the elements are still sorted but the partitions themselves will be unordered if sort is False; the default is to return them in ascending lexicographical order.>>> for p in ordered_partitions(6, 2): ... print(p) [1, 5] [2, 4] [3, 3]
But if speed is more important than ordering, sort can be set to False:
>>> for p in ordered_partitions(6, 2, sort=False): ... print(p) [1, 5] [3, 3] [2, 4]
References
[R1046]Generating Integer Partitions, [online], Available: https://jeromekelleher.net/generating-integer-partitions.html
[R1047]Jerome Kelleher and Barry O’Sullivan, “Generating All Partitions: A Comparison Of Two Encodings”, [online], Available: https://arxiv.org/pdf/0909.2331v2.pdf
- sympy.utilities.iterables.partitions(n, m=None, k=None, size=False)[source]#
Generate all partitions of positive integer, n.
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”.
- Parameters:
n : int
m : int, optional
limits number of parts in partition (mnemonic: m, maximum parts)
k : int, optional
limits the numbers that are kept in the partition (mnemonic: k, keys)
size : bool, default: False
If
True
, (M, P) is returned where M is the sum of the multiplicities and P is the generated partition. IfFalse
, only the generated partition is returned.
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 partition (the sum of the values in the returned dict) are limited with m (default value, None, gives partitions from 1 through n):
>>> for p in partitions(6, m=2): ... print(p) ... {6: 1} {1: 1, 5: 1} {2: 1, 4: 1} {3: 2}
References
[R1048]modified from Tim Peter’s version to allow for k and m values: https://code.activestate.com/recipes/218332-generator-for-integer-partitions/
- sympy.utilities.iterables.permute_signs(t)[source]#
Return iterator in which the signs of non-zero elements of t are permuted.
Examples
>>> from sympy.utilities.iterables import permute_signs >>> list(permute_signs((0, 1, 2))) [(0, 1, 2), (0, -1, 2), (0, 1, -2), (0, -1, -2)]
- sympy.utilities.iterables.postfixes(seq)[source]#
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]]
- sympy.utilities.iterables.prefixes(seq)[source]#
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]]
- sympy.utilities.iterables.random_derangement(t, choice=None, strict=True)[source]#
Return a list of elements in which none are in the same positions as they were originally. If an element fills more than half of the positions then an error will be raised since no derangement is possible. To obtain a derangement of as many items as possible–with some of the most numerous remaining in their original positions–pass \(strict=False\). To produce a pseudorandom derangment, pass a pseudorandom selector like \(choice\) (see below).
Examples
>>> from sympy.utilities.iterables import random_derangement >>> t = 'SymPy: a CAS in pure Python' >>> d = random_derangement(t) >>> all(i != j for i, j in zip(d, t)) True
A predictable result can be obtained by using a pseudorandom generator for the choice:
>>> from sympy.core.random import seed, choice as c >>> seed(1) >>> d = [''.join(random_derangement(t, c)) for i in range(5)] >>> assert len(set(d)) != 1 # we got different values
By reseeding, the same sequence can be obtained:
>>> seed(1) >>> d2 = [''.join(random_derangement(t, c)) for i in range(5)] >>> assert d == d2
- sympy.utilities.iterables.reshape(seq, how)[source]#
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], {2}, (1, (3,), 1)]) [[0, 1, [2, 3, 4], {5, 6}, (7, (8, 9, 10), 11)]]
- sympy.utilities.iterables.rotate_left(x, y)[source]#
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]
- sympy.utilities.iterables.rotate_right(x, y)[source]#
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]
- sympy.utilities.iterables.rotations(s, dir=1)[source]#
Return a generator giving the items in s as list where each subsequent list has the items rotated to the left (default) or right (
dir=-1
) relative to the previous list.Examples
>>> from sympy import rotations >>> list(rotations([1,2,3])) [[1, 2, 3], [2, 3, 1], [3, 1, 2]] >>> list(rotations([1,2,3], -1)) [[1, 2, 3], [3, 1, 2], [2, 3, 1]]
- sympy.utilities.iterables.roundrobin(*iterables)[source]#
roundrobin recipe taken from itertools documentation: https://docs.python.org/3/library/itertools.html#itertools-recipes
roundrobin(‘ABC’, ‘D’, ‘EF’) –> A D E B F C
Recipe credited to George Sakkis
- sympy.utilities.iterables.runs(seq, op=<built-in function gt>)[source]#
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]]
- sympy.utilities.iterables.sequence_partitions(l, n, /)[source]#
Returns the partition of sequence \(l\) into \(n\) bins
- Parameters:
l : Sequence[T]
A nonempty sequence of any Python objects
n : int
A positive integer
- Yields:
out : list[Sequence[T]]
A list of sequences with concatenation equals \(l\). This should conform with the type of \(l\).
Explanation
Given the sequence \(l_1 \cdots l_m \in V^+\) where \(V^+\) is the Kleene plus of \(V\)
The set of \(n\) partitions of \(l\) is defined as:
\[\{(s_1, \cdots, s_n) | s_1 \in V^+, \cdots, s_n \in V^+, s_1 \cdots s_n = l_1 \cdots l_m\}\]Examples
>>> from sympy.utilities.iterables import sequence_partitions >>> for out in sequence_partitions([1, 2, 3, 4], 2): ... print(out) [[1], [2, 3, 4]] [[1, 2], [3, 4]] [[1, 2, 3], [4]]
Notes
This is modified version of EnricoGiampieri’s partition generator from https://stackoverflow.com/questions/13131491/partition-n-items-into-k-bins-in-python-lazily
See also
- sympy.utilities.iterables.sequence_partitions_empty(l, n, /)[source]#
Returns the partition of sequence \(l\) into \(n\) bins with empty sequence
- Parameters:
l : Sequence[T]
A sequence of any Python objects (can be possibly empty)
n : int
A positive integer
- Yields:
out : list[Sequence[T]]
A list of sequences with concatenation equals \(l\). This should conform with the type of \(l\).
Explanation
Given the sequence \(l_1 \cdots l_m \in V^*\) where \(V^*\) is the Kleene star of \(V\)
The set of \(n\) partitions of \(l\) is defined as:
\[\{(s_1, \cdots, s_n) | s_1 \in V^*, \cdots, s_n \in V^*, s_1 \cdots s_n = l_1 \cdots l_m\}\]There are more combinations than
sequence_partitions()
because empty sequence can fill everywhere, so we try to provide different utility for this.Examples
>>> from sympy.utilities.iterables import sequence_partitions_empty >>> for out in sequence_partitions_empty([1, 2, 3, 4], 2): ... print(out) [[], [1, 2, 3, 4]] [[1], [2, 3, 4]] [[1, 2], [3, 4]] [[1, 2, 3], [4]] [[1, 2, 3, 4], []]
See also
- sympy.utilities.iterables.sift(seq, keyfunc, binary=False)[source]#
Sift the sequence,
seq
according tokeyfunc
.- Returns:
When
binary
isFalse
(default), the output is a dictionarywhere elements of
seq
are stored in a list keyed to the valueof keyfunc for that element. If
binary
is True then a tuplewith lists
T
andF
are returned whereT
is a listcontaining elements of seq for which
keyfunc
wasTrue
andF
containing those elements for whichkeyfunc
wasFalse
;a ValueError is raised if the
keyfunc
is not binary.
Examples
>>> from sympy.utilities import sift >>> from sympy.abc import x, y >>> from sympy import sqrt, exp, pi, Tuple
>>> 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 will not 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)]}
Sometimes you expect the results to be binary; the results can be unpacked by setting
binary
to True:>>> sift(range(4), lambda x: x % 2, binary=True) ([1, 3], [0, 2]) >>> sift(Tuple(1, pi), lambda x: x.is_rational, binary=True) ([1], [pi])
A ValueError is raised if the predicate was not actually binary (which is a good test for the logic where sifting is used and binary results were expected):
>>> unknown = exp(1) - pi # the rationality of this is unknown >>> args = Tuple(1, pi, unknown) >>> sift(args, lambda x: x.is_rational, binary=True) Traceback (most recent call last): ... ValueError: keyfunc gave non-binary output
The non-binary sifting shows that there were 3 keys generated:
>>> set(sift(args, lambda x: x.is_rational).keys()) {None, False, True}
If you need to sort the sifted items it might be better to use
ordered
which can economically apply multiple sort keys to a sequence while sorting.See also
- sympy.utilities.iterables.signed_permutations(t)[source]#
Return iterator in which the signs of non-zero elements of t and the order of the elements are permuted.
Examples
>>> from sympy.utilities.iterables import signed_permutations >>> list(signed_permutations((0, 1, 2))) [(0, 1, 2), (0, -1, 2), (0, 1, -2), (0, -1, -2), (0, 2, 1), (0, -2, 1), (0, 2, -1), (0, -2, -1), (1, 0, 2), (-1, 0, 2), (1, 0, -2), (-1, 0, -2), (1, 2, 0), (-1, 2, 0), (1, -2, 0), (-1, -2, 0), (2, 0, 1), (-2, 0, 1), (2, 0, -1), (-2, 0, -1), (2, 1, 0), (-2, 1, 0), (2, -1, 0), (-2, -1, 0)]
- sympy.utilities.iterables.strongly_connected_components(G)[source]#
Strongly connected components of a directed graph in reverse topological order.
- Parameters:
G : tuple[list, list[tuple[T, T]]
A tuple consisting of a list of vertices and a list of edges of a graph whose strongly connected components are to be found.
Examples
Consider a directed graph (in dot notation):
digraph { A -> B A -> C B -> C C -> B B -> D }
where vertices are the letters A, B, C and D. This graph can be encoded using Python’s elementary data structures as follows:
>>> V = ['A', 'B', 'C', 'D'] >>> E = [('A', 'B'), ('A', 'C'), ('B', 'C'), ('C', 'B'), ('B', 'D')]
The strongly connected components of this graph can be computed as
>>> from sympy.utilities.iterables import strongly_connected_components
>>> strongly_connected_components((V, E)) [['D'], ['B', 'C'], ['A']]
This also gives the components in reverse topological order.
Since the subgraph containing B and C has a cycle they must be together in a strongly connected component. A and D are connected to the rest of the graph but not in a cyclic manner so they appear as their own strongly connected components.
Notes
The vertices of the graph must be hashable for the data structures used. If the vertices are unhashable replace them with integer indices.
This function uses Tarjan’s algorithm to compute the strongly connected components in \(O(|V|+|E|)\) (linear) time.
References
- sympy.utilities.iterables.subsets(seq, k=None, repetition=False)[source]#
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 \(k\) isNone
then all \(2^n\) subsets will be returned from shortest to longest.Examples
>>> from sympy import subsets
subsets(seq, k)
will return the \(\frac{n!}{k!(n - k)!}\) \(k\)-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 \(\frac{(n - 1 + k)!}{k!(n - 1)!}\) 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)]
- sympy.utilities.iterables.topological_sort(graph, key=None)[source]#
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
ValueError
will be raised:>>> topological_sort((V, E + [(10, 7)])) Traceback (most recent call last): ... ValueError: cycle detected
References
- sympy.utilities.iterables.unflatten(iter, n=2)[source]#
Group
iter
into tuples of lengthn
. Raise an error if the length ofiter
is not a multiple ofn
.
- sympy.utilities.iterables.uniq(seq, result=None)[source]#
Yield unique elements from
seq
as an iterator. The second parameterresult
is used internally; it is not necessary to pass anything for this.Note: changing the sequence during iteration will raise a RuntimeError if the size of the sequence is known; if you pass an iterator and advance the iterator you will change the output of this routine but there will be no warning.
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]]
- sympy.utilities.iterables.variations(seq, n, repetition=False)[source]#
Returns an iterator over the n-sized variations of
seq
(size N).repetition
controls whether items inseq
can appear more than once;Examples
variations(seq, n)
will return \(\frac{N!}{(N - n)!}\) permutations without repetition ofseq
’s elements:>>> from sympy 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
variations#
variations(seq, n) Returns all the variations of the list of size n.
Has an optional third argument. Must be a boolean value and makes the method return the variations with repetition if set to True, or the variations without repetition if set to False.
- Examples::
>>> from sympy.utilities.iterables import variations >>> list(variations([1,2,3], 2)) [(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)] >>> list(variations([1,2,3], 2, True)) [(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]
partitions#
Although the combinatorics module contains Partition and IntegerPartition
classes for investigation and manipulation of partitions, there are a few
functions to generate partitions that can be used as low-level tools for
routines: partitions
and multiset_partitions
. The former gives
integer partitions, and the latter gives enumerated partitions of elements.
There is also a routine kbins
that will give a variety of permutations
of partitions. And to obtain partitions as a list instead of a dictionary, there
is the ordered_partition
function which is quite fast. Finally, to simply
obtain a count of the number of partitions without enumerating them, there
is the nT
function.
See Also#
sympy.utilities.iterables.ordered_partitions, sympy.functions.combinatorial.numbers.nT
partitions:
>>> from sympy.utilities.iterables import partitions
>>> [p.copy() for s, p in partitions(7, m=2, size=True) if s == 2]
[{1: 1, 6: 1}, {2: 1, 5: 1}, {3: 1, 4: 1}]
multiset_partitions:
>>> from sympy.utilities.iterables import multiset_partitions
>>> [p for p in multiset_partitions(3, 2)]
[[[0, 1], [2]], [[0, 2], [1]], [[0], [1, 2]]]
>>> [p for p in multiset_partitions([1, 1, 1, 2], 2)]
[[[1, 1, 1], [2]], [[1, 1, 2], [1]], [[1, 1], [1, 2]]]
kbins:
>>> from sympy.utilities.iterables import kbins
>>> def show(k):
... rv = []
... for p in k:
... rv.append(','.join([''.join(j) for j in p]))
... return sorted(rv)
...
>>> show(kbins("ABCD", 2))
['A,BCD', 'AB,CD', 'ABC,D']
>>> show(kbins("ABC", 2))
['A,BC', 'AB,C']
>>> show(kbins("ABC", 2, ordered=0)) # same as multiset_partitions
['A,BC', 'AB,C', 'AC,B']
>>> show(kbins("ABC", 2, ordered=1))
['A,BC', 'A,CB',
'B,AC', 'B,CA',
'C,AB', 'C,BA']
>>> show(kbins("ABC", 2, ordered=10))
['A,BC', 'AB,C', 'AC,B',
'B,AC', 'BC,A',
'C,AB']
>>> show(kbins("ABC", 2, ordered=11))
['A,BC', 'A,CB', 'AB,C', 'AC,B',
'B,AC', 'B,CA', 'BA,C', 'BC,A',
'C,AB', 'C,BA', 'CA,B', 'CB,A']