How to "peek" at iterator, while leaving the items on?

Question

Let's say I want to see the fifth item of an iterator, but I want to leave all the items on. What is the best way to do that? Is there a module? I could probably hack together a function that makes a new iterator, but I want to know if its already been done?

Answer 1

I think what you want is itertools.tee, which essentially makes two iterators under the hood. Use one to peek, expending the items therein, while the contents "remain" in the other one.

In [10]: g = (i for i in range(5))

In [11]: a,b = itertools.tee(g)

In [12]: next(a)
Out[12]: 0

In [13]: next(a)
Out[13]: 1

In [14]: next(a)
Out[14]: 2

In [15]: next(b)
Out[15]: 0

Notice that next(b) returns 0 instead of 3, which is the behavior you are looking for

Answer 2

The more_itertools third-party library has a more_itertools.spy tool that allows you to lookahead:

import more_itertools as mit

iterator = iter("abcdefg")
lookahead, iterable = mit.spy(iterator, 6) 
lookahead[5]
# 'f'

From the source, this tool essentially implements what @inspectorG4dget described. See also more_itertools.peekable.