itertools in Python (2)

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*Memo:

• 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().
• My post explains an iterator (1).

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itertools has the functions to create iterators.

*more-itertools has more functions by installing with pip install more-itertools.

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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:

*Memo:

• The 1st argument is iterable(Required-Type:Iterable).
• The 2nd argument is func(Optional-Default:None-Type:Callable or NoneType):
  • It can be operator.
• The 3rd argument is initial(Optional-Default:None-Type:Any or NoneType):
  • It's the 1st element.

from itertools import accumulate

v = accumulate(iterable=[])
v = accumulate(iterable=[], func=None, initial=None)

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=lambda a, b: a+b)
v = accumulate(iterable=[1, 2, 3, 4, 5], func=add)
v = accumulate(iterable=[2, 3, 4, 5], initial=1)
v = accumulate(iterable=[2, 3, 4, 5], func=lambda a, b: a+b, 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=lambda a, b: a*b):
# for x in accumulate([1, 2, 3, 4, 5], func=mul):
    print(x)
# 1
# 2
# 6
# 24
# 120

from itertools import accumulate
from operator import mul

for x in accumulate([1, 2, 3, 4, 5], func=lambda a, b: a*b, initial=10):
# for x in accumulate([1, 2, 3, 4, 5], func=mul, initial=10):
    print(x)
# 10
# 10
# 20
# 60
# 240
# 1200

batched() can return the iterator which batches the elements of iterable one by one to return the batches one by one as shown below:

*Memo:

• The 1st argument is iterable(Required-Type:Iterable).
• The 2nd argument is n(Required-Type:int):
  • It's the length of a batch.
  • It must be 1 <= x.
• The 3rd argument is strict(Optional-Default:False-Type:bool):
  • If it's True, error occurs if the last batch is shorter than n.
  • strict= must be used.

from itertools import batched

v = batched(iterable='', n=1)
v = batched(iterable='', n=1, strict=False)
v = batched(iterable=[], n=1)

print(v)
# <itertools.batched object at 0x0000026905D0CE80>

print(next(v))
# StopIteration:

from itertools import batched

v = batched(iterable='ABCDEFGHIJ', n=4)
v = batched(iterable=['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'], n=4)

print(next(v)) # ('A', 'B', 'C', 'D')
print(next(v)) # ('E', 'F', 'G', 'H')
print(next(v)) # ('I', 'J')

from itertools import batched

v = batched(iterable='ABCDEFGHIJ', n=4, strict=True)
v = batched(iterable=['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'], n=4,
            strict=True)

print(next(v)) # ('A', 'B', 'C', 'D')
print(next(v)) # ('E', 'F', 'G', 'H')
print(next(v)) # ValueError: batched(): incomplete batch

from itertools import batched

for x in batched(iterable='ABCDEFGHIJ', n=4):
# for x in batched(iterable=['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'],
#                  n=4):
    print(x)
# ('A', 'B', 'C', 'D')
# ('E', 'F', 'G', 'H')
# ('I', 'J')

from itertools import batched

for x in batched(iterable='ABCDEFGHIJ', n=4, strict=True):
# for x in batched(iterable=['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'],
#                  n=4, strict=True):
    print(x)
# ('A', 'B', 'C', 'D')
# ('E', 'F', 'G', 'H')
# ValueError: batched(): incomplete batch

chain() and chain.from_iterable() can return the iterator which returns the elements of chained *iterables and iterable respectively one by one as shown below:

*Memo for chain():

• The 1st arguments are *iterables(Optional-Default:()-Type:Iterable):
  • Don't use any keywords like *iterables=, iterables=, etc.

*Memo for chain.from_iterable():

• The 1st argument is iterable(Required-Type:Iterable):
  • Don't use iterable=.

from itertools import chain

v = chain()
v = 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('ABCDEF')
v = chain('ABC', 'DE', 'F')
v = chain(['A', 'B', 'C', 'D', 'E', '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:

from itertools import chain

for x in chain('ABCDEF'):
# for x in chain('ABC', 'DE', 'F'):
# for x in chain(['A', 'B', 'C', 'D', 'E', 'F']):
# for x in chain(['A', 'B', 'C'], ['D', 'E'], ['F']):
# for x in chain.from_iterable('ABCDEF'):
# for x in chain.from_iterable(['A', 'B', 'C', 'D', 'E', 'F']):
    print(x)
# A
# B
# C
# D
# E
# F