DataFrame String Manipulation

问题:

I have a dataframe that has one column with data that looks like this:

AAH.
AAH.
AAR.UN
AAR.UN
AAR.UN
AAR.UN
AAV.
AAV.
AAV.

I think I need to use the apply method to trim the column data. So if there is anything after the period keep the data unchanged but if there is nothing after the period then return just the letters without the period at the end. I know I can probably use a lambda function and maybe a string split or something to do this but have not much of an idea to make it happen.

This is kind of what I have so far:

df.apply(lambda x: string.split('.'))

I am not sure if I can use an if statement or something with the lambda function this way?

Any guidance appreciated.

回答1:

Since there's only one column, you can take advantage of vectorized string operations via .str (docs):

>>> df
        0
0    AAH.
1    AAH.
2  AAR.UN
3  AAR.UN
4  AAR.UN
5  AAR.UN
6    AAV.
7    AAV.
8    AAV.
>>> df[0] = df[0].str.rstrip('.')
>>> df
        0
0     AAH
1     AAH
2  AAR.UN
3  AAR.UN
4  AAR.UN
5  AAR.UN
6     AAV
7     AAV
8     AAV

Otherwise you'd have to do something like df.applymap(lambda x: x.rstrip(".")), or drop down to numpy char methods.

回答2:

You can also use lambda function to do this:

>>> L = [['AAH.'],
         ['AAR.UN'],
         ['AAR.UN'],
         ['AAV.'],
         ['AAV.']]

>>> df = pd.DataFrame(L)
>>> M = lambda x: x[0][:-1] if x[0][-1]=='.' else x[0][:]
>>> df = df.apply(M, axis=1)

>>> df
0       AAH
1    AAR.UN
2    AAR.UN
3       AAV
4       AAV

回答3:

def change_to_date(string):
    seq = (string[:2],string[2:5],string[5:])
    return '-'.join(seq)

pt['DATE'] = pt['DATE'].apply(change_to_date)

I applied a simple function to the column to manipulate all string values, for somewhat similar problem.