Print the nth step of a Generator in an easy way-CodePudding

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.

CodePudding user response：

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)))

CodePudding user response：

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)

CodePudding user response：

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

CodePudding user response：

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.