pandas dataframe split and get last element of list

Question

I have a pandas dataframe and in one column I have a string where words are separated by '_', I would like to extract the last element of this string (which is a number) and make a new column with this. I tried the following

df = pd.DataFrame({'strings':['some_string_25','a_different_one_13','and_a_last_one_40']})
df.assign(number = lambda x: x.strings.str.split('_')[0])

but it gives me this in my last column

number
some
string
25

but I would like to get this

number
25
13
40

How can I do this?

Answer 1

Use Series.str.split for split and select last value of list by indexing or use Series.str.extract by last integer of strings - (\\d+) is for match int and $ for end of string:

df['last'] = df['strings'].str.split('_').str[-1]
df['last1'] = df['strings'].str.extract('(\d+)$')

print (df)
              strings last last1
0      some_string_25   25    25
1  a_different_one_13   13    13
2   and_a_last_one_40   40    40

Difference is possible see in changed data:

df = pd.DataFrame({'strings':['some_string_25','a_different_one_13','and_a_last_one_40', 
                              'aaaa', 'sss58']})

df['last'] = df['strings'].str.split('_').str[-1]

df['last1'] = df['strings'].str.extract('(\d+)$')

print (df)
              strings   last last1
0      some_string_25     25    25
1  a_different_one_13     13    13
2   and_a_last_one_40     40    40
3                aaaa   aaaa   NaN
4               sss58  sss58    58

Answer 2

Can do:

df['number']=df['strings'].apply(lambda row: row.split('_')[-1])

or:

df['number']=[row[-1] for row in df['strings'].str.split('_')]

Answer 3

Please try this

df = pd.DataFrame({'strings':['some_string_25','a_different_one_13','and_a_last_one_40']})
df['number'] = df.strings.apply(lambda x: x.split('_')[-1])
df