Why named expression is necessary in this Numpy function chain?

user_219153 0
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This snippet:

a = np.arange(12, 0, -1).reshape(3, 4)
(a := a.flatten()).sort()
print(a)

produces

[ 1  2  3  4  5  6  7  8  9 10 11 12]

as expected. Without named expression:

a = np.arange(12, 0, -1).reshape(3, 4)
a = a.flatten().sort()
print(a)

I'm getting None. Why is that?

method-chaining numpy python
2 Answers

Quoting from PEP 572 – Assignment Expressions

Syntax and semantics

In most contexts where arbitrary Python expressions can be used, a named expression can appear. This is of the form NAME := expr where expr is any valid Python expression other than an unparenthesized tuple, and NAME is an identifier.

The value of such a named expression is the same as the incorporated expression, with the additional side-effect that the target is assigned that value:

So this code,

(a := a.flatten()).sort()
print(a)

is roughly equivalent to

a = a.flatten()
a.sort()
print(a)

This means that you are not assigning the result of sort anywhere. You are letting a.sort() to complete and print a later(which will result in expected output)

Remember sort is an in-place operation and it returns None, So in your second code you are assigning return value of sort(which is None).

a = a.flatten().sort()
user_66994470

ndarray.sort performs sorting in-place without returning the value. In your first example you create a, then assign the result of a.flatten() to it. After that the sort function executes sorting in-place on the array which a points to. Then you print a not result of sort().

In second example you print the result of sort which is None, instead of sorted array a.

user_121179680
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