## itertools ##
itertools.permutations(iterable[, r]) Return successive r length permutations of elements in the iterable.
If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated.
Permutations are emitted in lexicographic sort order. So, if the input iterable is sorted, the permutation tuples will be produced in sorted order.
Elements are treated as unique based on their position, not on their value. So if the input elements are unique, there will be no repeat values in each permutation.
def permutations(iterable, r=None): # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC # permutations(range(3)) --> 012 021 102 120 201 210 pool = tuple(iterable) n = len(pool) r = n if r is None else r if r > n: return indices = range(n) cycles = range(n, n-r, -1) yield tuple(pool[i] for i in indices[:r]) while n: for i in reversed(range(r)): cycles[i] -= 1 if cycles[i] == 0: indices[i:] = indices[i+1:] + indices[i:i+1] cycles[i] = n - i else: j = cycles[i] indices[i], indices[-j] = indices[-j], indices[i] yield tuple(pool[i] for i in indices[:r]) break else: return
The code for permutations() can be also expressed as a subsequence of product(), filtered to exclude entries with repeated elements (those from the same position in the input pool):
def permutations(iterable, r=None): pool = tuple(iterable) n = len(pool) r = n if r is None else r for indices in product(range(n), repeat=r): if len(set(indices)) == r: yield tuple(pool[i] for i in indices)
The number of items returned is n! / (n-r)! when 0 <= r <= n or zero when r > n.
New in version 2.6.
itertools.product(*iterables[, repeat]) Cartesian product of input iterables.
Equivalent to nested for-loops in a generator expression. For example, product(A, B) returns the same as ((x,y) for x in A for y in B).
The nested loops cycle like an odometer with the rightmost element advancing on every iteration. This pattern creates a lexicographic ordering so that if the input’s iterables are sorted, the product tuples are emitted in sorted order.
To compute the product of an iterable with itself, specify the number of repetitions with the optional repeat keyword argument. For example, product(A, repeat=4) means the same as product(A, A, A, A).
This function is equivalent to the following code, except that the actual implementation does not build up intermediate results in memory:
def product(*args, **kwds): # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111 pools = map(tuple, args) * kwds.get('repeat', 1) result = [] for pool in pools: result = [x+[y] for x in result for y in pool] for prod in result: yield tuple(prod)
## set ##
len(s) Return the cardinality of set s.
x in s Test x for membership in s.
x not in s Test x for non-membership in s.
add(elem) Add element elem to the set.
remove(elem) Remove element elem from the set. Raises KeyError if elem is not contained in the set.
discard(elem) Remove element elem from the set if it is present.
pop() Remove and return an arbitrary element from the set. Raises KeyError if the set is empty.
clear() Remove all elements from the set.