itertools.repeat(object[, times]) Make an iterator that returns object over and over again. Runs indefinitely unless the times argument is specified. Used as argument to map() for invariant parameters to the called function. Also used with zip() to create an invariant part of a tuple record. Roughly equivalent to: def repeat(object, times=None): # repeat(10, 3) --> 10 10 10 if times is None: while True: yield object else: for i in range(times):

itertools.product(*iterables, repeat=1) Cartesian product of input iterables. Roughly 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

itertools.permutations(iterable, r=None) 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

itertools.islice(iterable, stop) itertools.islice(iterable, start, stop[, step]) Make an iterator that returns selected elements from the iterable. If start is non-zero, then elements from the iterable are skipped until start is reached. Afterward, elements are returned consecutively unless step is set higher than one which results in items being skipped. If stop is None, then iteration continues until the iterator is exhausted, if at all; otherwise, it stops at the specified position. Unlike

itertools.groupby(iterable, key=None) Make an iterator that returns consecutive keys and groups from the iterable. The key is a function computing a key value for each element. If not specified or is None, key defaults to an identity function and returns the element unchanged. Generally, the iterable needs to already be sorted on the same key function. The operation of groupby() is similar to the uniq filter in Unix. It generates a break or new group every time the value of the key function

itertools.filterfalse(predicate, iterable) Make an iterator that filters elements from iterable returning only those for which the predicate is False. If predicate is None, return the items that are false. Roughly equivalent to: def filterfalse(predicate, iterable): # filterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8 if predicate is None: predicate = bool for x in iterable: if not predicate(x): yield x

itertools.dropwhile(predicate, iterable) Make an iterator that drops elements from the iterable as long as the predicate is true; afterwards, returns every element. Note, the iterator does not produce any output until the predicate first becomes false, so it may have a lengthy start-up time. Roughly equivalent to: def dropwhile(predicate, iterable): # dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1 iterable = iter(iterable) for x in iterable: if not predicate(x):

itertools.cycle(iterable) Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely. Roughly equivalent to: def cycle(iterable): # cycle('ABCD') --> A B C D A B C D A B C D ... saved = [] for element in iterable: yield element saved.append(element) while saved: for element in saved: yield element Note, this member of the to

itertools.count(start=0, step=1) Make an iterator that returns evenly spaced values starting with number start. Often used as an argument to map() to generate consecutive data points. Also, used with zip() to add sequence numbers. Roughly equivalent to: def count(start=0, step=1): # count(10) --> 10 11 12 13 14 ... # count(2.5, 0.5) -> 2.5 3.0 3.5 ... n = start while True: yield n n += step When counting with floating point numbers, better accuracy can

itertools.compress(data, selectors) Make an iterator that filters elements from data returning only those that have a corresponding element in selectors that evaluates to True. Stops when either the data or selectors iterables has been exhausted. Roughly equivalent to: def compress(data, selectors): # compress('ABCDEF', [1,0,1,0,1,1]) --> A C E F return (d for d, s in zip(data, selectors) if s) New in version 3.1.