Iterables¶
cartes¶
Returns the cartesian product of sequences as a generator.
 Examples::
>>> from sympy.utilities.iterables import cartes >>> list(cartes([1,2,3], 'ab')) [(1, 'a'), (1, 'b'), (2, 'a'), (2, 'b'), (3, 'a'), (3, 'b')]
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 lowlevel 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 partions.
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']
Docstring¶

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
 R831
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
graph : 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.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 https://kogswww.informatik.unihamburg.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
 R834
 R835
https://stackoverflow.com/questions/4856615/recursivepermutation/4857018
 R836
 R837
https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm
 R838
Generating involutions, derangements, and relatives by ECO Vincent Vajnovszki, DMTCS vol 1 issue 12, 2010

sympy.utilities.iterables.
generate_derangements
(perm)[source]¶ 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])) []

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
 R840
T. Beyer and S.M. Hedetniemi: constant time generation of rooted trees, SIAM J. Computing Vol. 9, No. 4, November 1980
 R841
https://stackoverflow.com/questions/1633833/orientedforesttaocpalgorithminpython

sympy.utilities.iterables.
group
(seq, multiple=True)[source]¶ 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

sympy.utilities.iterables.
has_dups
(seq)[source]¶ 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

sympy.utilities.iterables.
has_variety
(seq)[source]¶ 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

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 bitslength lists starting from
[0, 0,..., 0]
through[1, 1, ..., 1]
are desired, pass a noninteger 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 noninteger 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.
interactive_traversal
(expr)[source]¶ Traverse a tree asking a user which branch to choose.

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.
kbins
(l, k, ordered=None)[source]¶ Return sequence
l
partitioned intok
bins.Examples
>>> from __future__ import print_function
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)[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, is_set=False, small=None)[source]¶ Return a tuple representing the rotation of the sequence in which the lexically smallest elements appear first, e.g. \(cba >acb\).
If
directed
is False then the smaller of the sequence and the reversed sequence is returned, e.g. \(cba > abc\).For more efficient processing,
is_set
can be set to True if there are no duplicates in the sequence.If the smallest element is known at the time of calling, it can be passed as
small
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'

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_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

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.
 Returns
sym : Symbol
The subscripted symbols.

sympy.utilities.iterables.
ordered_partitions
(n, m=None, sort=True)[source]¶ Generates ordered partitions of integer
n
. Parameters
m : integer (default None)
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 mn 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
 R844
Generating Integer Partitions, [online], Available: https://jeromekelleher.net/generatingintegerpartitions.html
 R845
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.
 Parameters
m : integer (default gives partitions of all sizes)
limits number of parts in partition (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 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}
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})]
References
 R846
modified from Tim Peter’s version to allow for k and m values: http://code.activestate.com/recipes/218332generatorforintegerpartitions/

sympy.utilities.iterables.
permute_signs
(t)[source]¶ Return iterator in which the signs of nonzero 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.
postorder_traversal
(node, keys=None)[source]¶ 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 depthfirst 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 ofordered
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)) [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)]

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.
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.utilities.iterables 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/2/library/itertools.html#recipes
roundrobin(‘ABC’, ‘D’, ‘EF’) –> A D E B F C
Recipe credited to George Sakkis

sympy.utilities.iterables.
runs
(seq, op=<builtin 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.
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 nonbinary output
The nonbinary 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 nonzero 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
graph : 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.utilities.iterables 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 a generator of the nsized 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.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