Print the nth step of a Generator in an easy way

Question

I want to know if there is a better and cleaner way of printing the 3rd step of a generator function.
Currently I have written the following code

def imparesgen():
  n = 0
  while n<200: 
    n=n+2
    yield n

gen = imparesgen()

y = 0
for x in gen:
  y+=1
  if y == 3:
    print(x)

This worked, but, is there maybe a simpler way of doing this? Without the use of a list.

Asked By: Fire Frekox

||

Source

Answer 1

You could iterate over the generator three times. Since zip stops at the shortest input, you’ll only get three iterations provided gen yields at least three elements.

gen = imparesgen()
for _, item in zip(range(3), gen):
    pass

# Now, item is the third element
print(item)

Alternatively, use the next() function to get the next element of the generator:

gen = imparesgen()
next(gen)
next(gen)
item = next(gen)
print(item)

Or, if you don’t want to write out those next lines multiple times:

gen = imparesgen()
for _ in range(3):
    item = next(gen)

print(item)

Answered By: Pranav Hosangadi

Answer 2

From Itertools recipes:

def nth(iterable, n, default=None):
    "Returns the nth item or a default value"
    return next(islice(iterable, n, None), default)

Applied to your example:

import itertools

def imparesgen():
  n = 0
  while n<200:
    n=n+2
    yield n

gen = imparesgen()

print(next(itertools.islice(gen, 3, None)))

Answered By: treuss

Answer 3

Since a generator acts like an iterator, you can call next on it :

third = next(next(next(gen)))

I don’t think you can go much faster than that in pure-Python. But I think that without benchmarking the code, the speedup won’t be noticed.

Answered By: Lenormju

Answer 4

Is this what you’re looking for?:

from itertools import islice

def imparesgen():
  n = 0
  while n<200: 
    n=n+2
    yield n

gen = imparesgen()

third = list(islice(gen, 2, 3))[0]  # -> 6

Answered By: Tim-Schaeffer_Crown

Print the nth step of a Generator in an easy way

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