根据条件python3将变量追加到列表

Question

I have a array

arr=['a','b','c']

and a variable

var='a'

I am trying to remove the variable from array and append the resultant array to a new array.

newarray = []

if var in arr:
    newarray.append(arr.remove(var))
    print(newarray)

This does not print anything. However when I run only arr.remove(var) it works...I am not able to append the resultant smaller array to a new variable.

Answer 1

Looks like your code got a bit mangled. Here's how to fix it:

arr = ['a','b','c']
var = 'a'
newarray = []

if var in arr:
    newarray.append(var)
    arr.remove(var)
    print(newarray)

To understand why your original code printed [None] , you first need to understand array.remove() . array.remove is a void function : it does not return a value, only performs tasks. If you try to get or call its return value in a function such as array.append() , Python doesn't know how to react and returns None . The None value was then appended to the array properly by the array.append() function.

Answer 2

From your description, this is what you may be looking for:

arr = ['a','b','c']
var = 'a'
newarray = []
if var in arr:
    arr.remove(var)
    newarray.append(arr)
    print(newarray)

Output:

[['b', 'c']]

Note the output of following:

if var in arr:
    print(arr.remove(var))
    print(newarray.append(arr))
    print(newarray)

Output:

None
None
[['b', 'c']]

arr.remove(var) and newarray.append(arr) work on list in place but do not return anything.

Hence newarray.append(arr.remove(var)) means newarray.append(None)

Answer 3

use pop() function instead. Change:

newarray.append(var)
print(arr.remove(var))

to:

newarray.append(arr.pop(0))
print(newarr)