Python zip object 'disappears' after iterating through?

Question

Total noob question here but I really want to know the answer.

I have no idea why the zip object simply "disappears" after I attempt to iterate through it in its list form: eg.

>>> A=[1,2,3]
>>> B=['A','B','C']
>>> Z=zip(A,B)
>>> list(Z)
>>> [('C', 3), ('B', 2), ('A', 1)]
>>> {p:q for (p,q) in Z}
{1: 'A', 2: 'B', 3: 'C'}
>>> {p:q for (p,q) in list(Z)}
{}
>>> list(Z)
[]

(this is in Python 3.4.2)

Can anybody help?

`zip` in Python2 returns a list while in Python3 a iterator. — Siyuan Ren, Feb 08 '15 at 04:08
Does this answer your question? [Performing len on list of a zip object clears zip](https://stackoverflow.com/questions/14637154/performing-len-on-list-of-a-zip-object-clears-zip) — Friedrich -- Слава Україні, Feb 25 '20 at 20:52

score 18 · Accepted Answer · answered Feb 08 '15 at 04:05

There was a change of behavior between Python2 to Python3:
in python2, zip returns a list of tuples while in python3 it returns an iterator.

The nature of iterator is that once it's done iterating the data - it points to an empty collection and that's the behavior you're experiencing.

Python2:

Python 2.7.9 (default, Jan 29 2015, 06:28:58)
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.56)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> A=[1,2,3]
>>> B=['A','B','C']
>>> Z=zip(A,B)
>>> Z
[(1, 'A'), (2, 'B'), (3, 'C')]
>>> list(Z)
[(1, 'A'), (2, 'B'), (3, 'C')]
>>> list(Z)
[(1, 'A'), (2, 'B'), (3, 'C')]
>>> list(Z)
[(1, 'A'), (2, 'B'), (3, 'C')]
>>> {p:q for (p,q) in Z}
{1: 'A', 2: 'B', 3: 'C'}
>>> Z
[(1, 'A'), (2, 'B'), (3, 'C')]
>>> Z
[(1, 'A'), (2, 'B'), (3, 'C')]

Python3:

Python 3.4.2 (v3.4.2:ab2c023a9432, Oct  5 2014, 20:42:22)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> A=[1,2,3]
>>> B=['A','B','C']
>>> Z=zip(A,B)
>>> list(Z)
[(1, 'A'), (2, 'B'), (3, 'C')]
>>> list(Z)
[]
>>>

i'd like to also ask if this 'exhaustion' behavior was designed simply for better memory efficiency? — Yibo Yang, Feb 08 '15 at 04:21
I believe so, yes: as Simeon Visser suggested in his answer, it's pretty easy to achieve the same behavior by wrapping `zip()` with `list()`. But when you don't need the whole sequence to be in-memory - it's more efficient to discard an item we've already iterated. — Nir Alfasi, Feb 08 '15 at 04:26

score 7 · Answer 2 · answered Feb 08 '15 at 03:57

7

zip creates an object for iterating once over the results. This also means it's exhausted after one iteration:

>>> a = [1,2,3]
>>> b = [4,5,6]
>>> z = zip(a,b)
>>> list(z)
[(1, 4), (2, 5), (3, 6)]
>>> list(z)
[]

You need to call zip(a,b) every time you wish to use it or store the list(zip(a,b)) result and use that repeatedly instead.

answered Feb 08 '15 at 03:57

Simeon Visser

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for migrating from python2 to python2 the list(zip(a,b)) worked well – zerocog Jun 06 '21 at 04:00

score 0 · Answer 3 · answered Feb 25 '20 at 20:48

0

For python 2 and 3, I use

xzip = zip
zip = lambda *x: list(xzip(*x))

Then zip returns always a list rather than an iterator.

IMO, xzip would have been a better name for zip in python 3.

answered Feb 25 '20 at 20:48

Friedrich -- Слава Україні

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Python zip object 'disappears' after iterating through?

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