Removing the last element of a list: Why does ".pop() not work?

Question

I am working on a test dataset which is

print(df.head(10))

0                   NaN
1     93/2; 99/3; 05/4;
2                   NaN
3                   NaN
4                   NaN
5                   NaN

Now i want to convert the string "93/2; 99/3; 05/4;" to a more neat data structure for following analysis. Thus the first step would be so split on the ";"

df= df.apply(lambda x: x.split(';'))

which yields

0                        []
1    [93/2,  99/3,  05/4, ]
2                        []
3                        []
4                        []
5                        []
6                        []
7                        []
8                        []
9                        []

As you see the last element of the list is an empty value, which i want to delete. I was thinking about using the .pop() function but that yields

df = df.apply(lambda x: x.pop())
print(df.head(10))

0    
1    
2    
3    
4    
5    
6    
7    
8    
9 

if i am using slicing

df = df.apply(lambda x: x[:-1])

i get the expected output

0                      []
1    [93/2,  99/3,  05/4]
2                      []
3                      []
4                      []
5                      []
6                      []
7                      []
8                      []
9                      []

Could anyone please explain why the pop function is not working here as i expected?

Thank you in advance!

Answer 1

There is a vectorial function to split strings str.split , and one to strip characters:

Assuming a dataframe here, although your example might indicate you have a Series:

                # strip trailing ;   # split on ;
df['lst'] = df['col'].str.rstrip(';').str.split(';\s*')

if Series:

ser2 = ser.str.rstrip(';').str.split(';\s*')

output:

                 col                 lst
0                NaN                 NaN
1  93/2; 99/3; 05/4;  [93/2, 99/3, 05/4]
2                NaN                 NaN
3                NaN                 NaN
4                NaN                 NaN
5                NaN                 NaN

Answer 2

It works. You assign the return of pop() back into your df . pop() returns the element that gets popped. You assign that element to your df.