*Memos:
- My post explains itertools about count(), cycle() and repeat().
- My post explains itertools about compress(), filterfalse(), takewhile() and dropwhile().
- My post explains itertools about groupby() and islice().
- My post explains itertools about pairwise(), starmap(), tee() and zip_longest().
- My post explains itertools about product() and permutations().
- My post explains itertools about combinations() and combinations_with_replacement().
itertools has the functions to create iterators.
*more-itertools has more functions by installing with pip install more-itertools.
accumulate() can return the iterator which accumulates the elements of iterable one by one to return the accumulated elements one by one as shown below:
*Memos:
- The 1st argument is
iterable(Required-Type:iterable). - The 2nd argument is
func(Optional-Default:None-Type:callable). *It can be operator. - The 3rd argument is
initial(Optional-Default:None-Type:object).
from itertools import accumulate
v = accumulate(iterable=[])
print(v)
# <itertools.accumulate object at 0x0000026906CF9850>
print(next(v)) # StopIteration:
from itertools import accumulate
from operator import add
v = accumulate(iterable=[1, 2, 3, 4, 5])
v = accumulate(iterable=[1, 2, 3, 4, 5], func=None, initial=None)
v = accumulate(iterable=[1, 2, 3, 4, 5], func=add)
v = accumulate(iterable=[1, 2, 3, 4, 5], func=lambda a, b: a+b)
v = accumulate(iterable=[2, 3, 4, 5], initial=1)
print(next(v)) # 1
print(next(v)) # 3
print(next(v)) # 6
print(next(v)) # 10
print(next(v)) # 15
print(next(v)) # StopIteration:
from itertools import accumulate
for x in accumulate([1, 2, 3, 4, 5]):
print(x)
print(next(v)) # 1
print(next(v)) # 3
print(next(v)) # 6
print(next(v)) # 10
print(next(v)) # 15
print(next(v)) # StopIteration:
from itertools import accumulate
for x in accumulate([1, 2, 3, 4, 5], initial=10):
# for x in accumulate([10, 1, 2, 3, 4, 5]):
print(x)
# 10
# 11
# 13
# 16
# 20
# 25
from itertools import accumulate
from operator import mul
for x in accumulate([1, 2, 3, 4, 5], func=mul):
# for x in accumulate([1, 2, 3, 4, 5], func=lambda a, b: a*b):
print(x)
# 1
# 2
# 6
# 24
# 120
batched() can return the iterator which batches the one or more elements of iterable one by one to return the batches one by one as shown below:
*Memos:
- The 1st argument is
iterable(Required-Type:iterable). - The 2nd argument is
n(Required-Type:int). *It's the number of batches. *It must be1 <= x. - The 3rd argument is
strict(Optional-Default:False): *Memos:- If it's
True, error occurs if the final batch is shorter thann. -
strict=must be used.
- If it's
from itertools import batched
v = batched(iterable='', n=1)
v = batched(iterable=[], n=1)
print(v)
# <itertools.batched object at 0x0000026905D0CE80>
print(next(v)) # StopIteration:
from itertools import batched
v = batched(iterable='ABCDEFGH', n=3)
v = batched(iterable='ABCDEFGH', n=3, strict=False)
v = batched(iterable=['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H'], n=3)
print(next(v)) # ('A', 'B', 'C')
print(next(v)) # ('D', 'E', 'F')
print(next(v)) # ('G', 'H')
from itertools import batched
v = batched(iterable='ABCDEFGH', n=3, strict=True)
print(next(v)) # ('A', 'B', 'C')
print(next(v)) # ('D', 'E', 'F')
print(next(v)) # ValueError: batched(): incomplete batch
from itertools import batched
for x in batched(iterable='ABCDEFGH', n=3):
print(x)
# ('A', 'B', 'C')
# ('D', 'E', 'F')
# ('G', 'H')
from itertools import batched
for x in batched(iterable='ABCDEFGH', n=3, strict=True):
print(x)
# ('A', 'B', 'C')
# ('D', 'E', 'F')
# ValueError: batched(): incomplete batch
You can use chain() and chain.from_iterable() as shown below:
*Memos:
-
chain()can return the iterator which chains*iterablesto return the elements one by one: *Memos:- The 1st or the later arguments are
*iterables(Optional-Type:iterable). - Don't use any keywords like
*iterables=,iterables=,*iterable=,iterable=, etc.
- The 1st or the later arguments are
-
chain.from_iterable()can return the iterator which returns the elements ofiterableone by one: *Memos:- The 1st argument is
iterable(Required-Type:iterable). - Don't use
iterable=.
- The 1st argument is
from itertools import chain
v = chain()
v = chain('')
v = chain([])
v = chain.from_iterable('')
v = chain.from_iterable([])
print(v)
# <itertools.chain object at 0x0000026906CEAD70>
print(next(v)) # StopIteration:
from itertools import chain
v = chain('ABC')
v = chain(['A', 'B', 'C'])
v = chain.from_iterable('ABC')
v = chain.from_iterable(['A', 'B', 'C'])
print(next(v)) # A
print(next(v)) # B
print(next(v)) # C
print(next(v)) # StopIteration:
from itertools import chain
v = chain('ABC', 'DE')
v = chain(['A', 'B', 'C'], ['D', 'E'])
v = chain.from_iterable('ABCDE')
v = chain.from_iterable(['A', 'B', 'C', 'D', 'E'])
print(next(v)) # A
print(next(v)) # B
print(next(v)) # C
print(next(v)) # D
print(next(v)) # E
print(next(v)) # StopIteration:
from itertools import chain
v = chain('ABC', 'DE', 'F')
v = chain(['A', 'B', 'C'], ['D', 'E'], ['F'])
v = chain.from_iterable('ABCDEF')
v = chain.from_iterable(['A', 'B', 'C', 'D', 'E', 'F'])
print(next(v)) # A
print(next(v)) # B
print(next(v)) # C
print(next(v)) # D
print(next(v)) # E
print(next(v)) # F
print(next(v)) # StopIteration:
for x in chain('ABC', 'DE', 'F'):
# for x in chain.from_iterable('ABCDEF'):
print(x)
# A
# B
# C
# D
# E
# F