使用Python中的方法时，何时重新声明变量？

Question

While studying CS, I've noticed that some methods require you update and save the variable in order for the desired effect to kick in, while others just simply work when you execute the script.

An example with the set stop_words and list lst :

stop_words = stop_words.remove("whatever" ) # returns a variable of type none
stop_words.remove("whatever")  # has the desired effect of removing the word from the set
lst.append(2)  # doesn't update the list
lst = lst.append(2)  # updates the list.

What is the reason that I have to redeclare the lst variable when I use .append() but not when I want to use .remove() ?

Answer 1

TL;DR - only assign the result of an operation to a variable if the result is useful to you.

In python assignment, the expression on the right side of the equal sign is fully executed and resolved to a single anonymous object before the left hand side is evaluated for assignment. For example, suppose I perform a valid operation on the right hand side, but an invalid on one the left. The right side append worked even before python realized there was a problem on the left and raised an exception

>>> lst1 = []
>>> lst2 =[1]
>>> lst1[1] = lst2.append(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list assignment index out of range
>>> lst1
[]
>>> lst2
[1, 2]

In a similar manner, set.remove and list.append modify their set and list respectively and return None. They work exactly the same no matter what happens to the None later.

Both of your set examples remove "whatever" from the set. The difference is that in the first case you then rebind stop_words from the set to None . If there are no other references to the set, it is deleted. Same thing with the list.

stop_words = stop_words.remove("whatever") 
stop_words.remove("whatever")

This can be demonstrated by keeping a separate reference to the object to verify

>>> test_set = {"whatever", "whatever 2"}
>>> stop_words = test_set
>>> stop_words = stop_words.remove("whatever")
>>> repr(stop_words)
'None'
>>> test_set
{'whatever 2'}
>>> stop_words = test_set
>>> stop_words.remove("whatever 2")
>>> stop_words
set()
>>> test_set
set()

The same thing happens with lists

>>> test_list = []
>>> lst = test_list
>>> lst = lst.append(2)
>>> repr(lst)
'None'
>>> test_list
[2]
>>> lst = test_list
>>> lst.append(3)
>>> lst
[2, 3]
>>> test_list
[2, 3]

Answer 2

Lists are mutable, meaning that you calling a list method in itself will mutate it. For example, each of these will mutate the list without any re-assignments necessary.

lst = [1, 2, 3]
print(f"Original list: {lst}")
lst.remove(1)
print(f"Updated: {lst}")
lst.append(4)
print(f"Updated: {lst}")

Output:

original list: [1, 2, 3]
Updated: [2, 3]
Updated: [2, 3, 4]

The methods will return a None value (its only purpose is to mutate the list), so it is pointless to reassign it.

lst = [1, 2, 3]
lst = lst.remove(1)
print(lst)

Output:

None