Intends to be fast and use memory efficiently, and also to be hooked together to express more complicated iteration-based algorithms

Takes several iterators as arguments and returns a single iterator

from itertools import *

a = [1, 2, 3]
b = ['a', 'b', 'c']
c = [10, 20, 30]

# chain
d = chain(a, b, c) # decorator
for i in d:
    print(i, end = ' ') # 1 2 3 a b c 10 20 30
print()

# chain.from_iterable
d = chain.from_iterable([a, b, c]) # decorator
for i in d:
    print(i, end = ' ') # 1 2 3 a b c 10 20 30
print()
        

# zip() stops when the first input iterator is exhausted
a = [1, 2, 3]
b = ['a', 'b', 'c', 'd']

for e in zip(a, b):
    print(e)

# (1, 'a')
# (2, 'b')
# (3, 'c')
        

# substitutes None for any missing values
from itertools import *

a = [1, 2, 3]
b = ['a', 'b', 'c', 'd']

for e in zip_longest(a, b):
    print(e)

# (1, 'a')
# (2, 'b')
# (3, 'c')
# (None, 'd')
        

# slice with parameters, start, stop, and step
from itertools import *

l = list(range(10))

for i in islice(l, 1, 5, 2):
    print(i, end=' ')
print('\n')

# 1 3
        

# return n independent iterators from a single iterable
# the new iterators created by tee() share their input, so the original iterator should not be used after the new ones are created
from itertools import *

l = list(range(10))

l1, l2, l3 = tee(l, 3)

print(list(l1))
print(list(l2))
print(list(l3))
        

def multiply(x, y):
    return (x, y, x * y)

r = map(multiply, range(5), range(5, 10))

for i in r:
    print(i) # tuple

# (0, 5, 0)
# (1, 6, 6)
# (2, 7, 14)
# (3, 8, 24)
# (4, 9, 36)
        

from itertools import *

def multiply(x, y):
    return (x, y, x * y)

values = [(0, 5), (1, 6), (2, 7), (3, 8), (4, 9)]

r = starmap(multiply, values)

for i in r:
    print(i)

# (0, 5, 0)
# (1, 6, 6)
# (2, 7, 14)
# (3, 8, 24)
# (4, 9, 36)
        

# generate infinite numbers
from itertools import *

c = count(10, 2)

for i in zip(c, [1, 2, 3]):
    print(i)

# (10, 1)
# (12, 2)
# (14, 3)
        

# repeats the contents of the arguments indefinitely
from itertools import *

for i in zip(range(6), cycle([1, 2, 3])):
    print(i)
        

# produces the same value repeatly
from itertools import *

for i in repeat('Hello World!', 3):
    print(i)
        

# returns an iterator containing elements of the input iterator after a condition becomes false
from itertools import *

def condition(x):
    return x < 1

for i in dropwhile(condition, [-2, -1, 0, 1, 2]):
    print(i)

# 1
# 2
        

# returns an iterator containing elements of the input iterator before a condition becomes false
from itertools import *

def should_drop(x):
    return x < 1

for i in takewhile(should_drop, [-2, -1, 0, 1, 2]):
    print(i)

# -2
# -1
# 0
        

# returns an iterator that includes only items for which the condition returns true
def should_drop(x):
    return x < 1

for i in filter(should_drop, [-2, -1, 0, 1, 2]):
    print(i)

# -2
# -1
# 0
        

# returns an iterator that includes only items for which the condition returns false
from itertools import *

def should_drop(x):
    return x < 1

for i in filterfalse(should_drop, [-2, -1, 0, 1, 2]):
    print(i)

# 1
# 2
        

# filter the contents of an iterable with the values in another iterable
# a true value causes the value to be produced, a false value causes it to be ignored
from itertools import *

conditions = [False, False, True, False, True]
data = range(5)

for i in compress(data, conditions):
    print(i, end=' ')
print()
        

# passing the nth and n+1st item to a function and producing the return value instead of either input
from itertools import *

r = accumulate(range(5))

print(list(r)) # [0, 1, 3, 6, 10]
        

from itertools import *

def f(a, b):
    return a*10 + b*10

r = accumulate(range(5), f)

print(list(r)) # [0, 10, 120, 1230, 12340]
        

# cartesian product
from itertools import *

l1 = [1, 2, 3]
l2 = ['A', 'B', 'C']

r = product(l1, l2)

for i in r:
    print(i)

# (1, 'A')
# (1, 'B')
# (1, 'C')
# (2, 'A')
# (2, 'B')
# (2, 'C')
# (3, 'A')
# (3, 'B')
# (3, 'C')
        

# all possible ordering tuple with r length
# permutations[l, r]
from itertools import *

l1 = [1, 2, 3]

r = permutations(l1, 2)

for i in r:
    print(i)

# (1, 2)
# (1, 3)
# (2, 1)
# (2, 3)
# (3, 1)
# (3, 2)
        

# in sorted order, no repeated elements
from itertools import *

l1 = [1, 2, 3]

r = combinations(l1, 2)

for i in r:
    print(i)

# (1, 2)
# (1, 3)
# (2, 3)
        

# in sorted order, with repeated elements
from itertools import *

l1 = [1, 2, 3]

r = combinations_with_replacement(l1, 2)

for i in r:
    print(i)

# (1, 1)
# (1, 2)
# (1, 3)
# (2, 2)
# (2, 3)
# (3, 3)
        

PyMOTW

Python STL