How can zip be used to chunk data into equal tied groups?-CodePudding

>>> n = 3
>>> x = range(n ** 2),
>>> xn = list(zip(*[iter(x)] * n))

In PEP 618, the author gives this example of how zip can be used to chunk data into equal sized groups.

How does it work?

I think that it relies on an implementation detail of zip such that if it takes the first element of each of the elements of the list [iter(x)] * n that equates to the first n elements because of the changing state of iter(x) as each of the elements are taken.

This is because the following code replicates the above behavior:

n = 3
x = range(n ** 2)
xn = [iter(x)] * n

res = []

while True:    
        try:    
                col = []    
                for element in xn:    
                        col.append(next(element))    
                res.append(col)    
        except:    
                break

However, I would like to make sure that this is indeed the case and that this is a reliable behavior that can be used to chunk elements of an iterable.

CodePudding user response：

It's not an implementation of zip. It's how iterables work in Python - they always "consume" and move forward.

eg:

whatever = iter([1, 2, 3])
next(whatever)
# 1
next(whatever)
# 2

What zip does is "advance" each object it's provided with and given the example you've provided [iter(x)] * n... this becomes basically zip(whatever, whatever, whatever)

Since zip works in sequence - it takes the first next from whatever - then the next from whatever which has already moved on from the first next, so it's the value of 2. Which means the next one is 3. etc...

It's behaviour by design and the language guarantees it.

CodePudding user response：

It's not really specific to zip, but you basically have that right. In effect, it's zipping 3 references to the same iterator, causing it to round-robin between them. During each iteration, one more element is consumed from the iterator.

Effectively, it's the same as doing this:

>>> n = 3
>>> x = range(n ** 2)
>>> a = b = c = iter(x)
>>> list(zip(a, b, c))
[(0, 1, 2), (3, 4, 5), (6, 7, 8)]

Note that it only produces equal sized groups and may drop elements (that part is a characteristic of zip, because it's limited by the smallest iterable, though you could use itertools.zip_longest if you want):

>>> n = 4
>>> x = range(n ** 2)
>>> a = b = c = iter(x)
>>> list(zip(a, b, c))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 14)]