why zip(*k) can't work when k is a iterator?

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Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.

Note that by unpacking, you are essentially doing the same as:

zip(1,2,3,4)

Which also raises:

TypeError: zip argument #1 must support iteration

Given that the first element it encounters is not an iterable, and as specified in the docs, zip is used to:

Make an iterator that aggregates elements from each of the iterables

So if you want to use zip, make sure you're feeding it iterables.

Python zip(), The * operator can be used in conjunction with zip() to unzip the list. zip(* zippedList). Example 3: Unzipping the Value Using zip(). coordinate = ['x', 'y',� The function takes in iterables as arguments and returns an iterator. This iterator generates a series of tuples containing elements from each iterable. zip () can accept any type of iterable, such as files, lists, tuples, dictionaries, sets, and so on.

argument #1 in your case is the integer 1- which does not support iteration.

this on the other hand will work:

c = [range(3), range(3)]
k = iter(c)
a = zip(*k)

now argument #1 is a range object which supports iteration.

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To see what is happening, lets try to print *k

In [1]: c = [1,2,3,4] 
   ...: k = iter(c) 
   ...: print(*k)                                                                                                              
1 2 3 4

What did we see here, we got 4 integers in our case, but that is not an expected. (Also equivalent to zip(1,2,3,4) as per yatu's answer

From the docs: https://docs.python.org/3/library/functions.html#zip

zip(*iterables) . Make an iterator that aggregates elements from each of the iterables.

So we need to pass it an iterator, which we are not doing as we see here

Hence we will pass an iterator to zip as the docs say

In [10]: c = [1,2,3,4]                                                                                                         
In [12]: list(zip(c))                                                                                                          
Out[12]: [(1,), (2,), (3,), (4,)]

Also zip is useful when we have atleast 2 iterators, with one iterator it doesn't work, but if we have a list of lists, then the unpacking would work

In [13]: c = [[1,2,3,4],[5,6,7,8]]                                                                                             

In [14]: a = zip(*c)                                                                                                           

In [15]: list(a)                                                                                                               
Out[15]: [(1, 5), (2, 6), (3, 7), (4, 8)]

Because unpacking c gave the zip two iterators, which were the sublists inside the bigger list!

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LLTTP (little late to the party).

One part of these problems is to make it work with the minimal effort and provide meaningful results. Here is what I would do:

# Python 3.x
c = [1,2,3,4]
k = iter(c)
a = zip(k,k)
list(a) #    [(1, 2), (3, 4)]

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Zip in Python. zip takes n number of iterables and…, To unzip a list of tuples we zip(*listP_of_tuples) . Unzip creates We can loop over the zip object or the iterator to get the actual list. Consider� The zip() function takes iterables (can be zero or more), aggregates them in a tuple, and return it. In this tutorial, we will learn about Python zip() in detail with the help of examples.

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