Python：在数据帧中，创建一个新列，并使用从另一列的值中切出的字符串

Question

I have a pretty simple problem that i can't get around.

I create a dataframe and i would like to generate a new column with a sliced string from one column with a slice value from another column.

Ex: From this:

dftest = pd.DataFrame({'string' : ['EXAMPLE']*5, 'position' : [1, 2, 3, 4, 5]})

   position   string
0         1  EXAMPLE
1         2  EXAMPLE
2         3  EXAMPLE
3         4  EXAMPLE
4         5  EXAMPLE
5         1  OTHER
6         2  OTHER
7         3  OTHER

I want this:

   position   string    new
0         1  EXAMPLE  E
1         2  EXAMPLE  EX
2         3  EXAMPLE  EXA
3         4  EXAMPLE  EXAM
4         5  EXAMPLE  EXAMP
5         1  OTHER    O
6         2  OTHER    OT
7         3  OTHER    OTH

I tried:

dftest['new'] = dftest.string.str[:dftest.position]
dftest['new'] = dftest.string.str[:dftest['position']]
dftest['new'] = dftest.string[:dftest.position]

as well as different row iteration methods but each time I end up with Nan values.

Any help would be greatly appreciated

Answer 1

You can do the following

dftest['new'] = [dftest.iloc[i]['string'][0:dftest.iloc[i]['position']] for i in range(0,len(dftest))]

This will check the position.

Answer 2

One method is to enumerate the string using a list comprehension.

dftest['new'] = [s[:n] for s, n in zip(dftest.string, dftest.position)]

>>> dftest
   position   string    new
0         1  EXAMPLE      E
1         2  EXAMPLE     EX
2         3  EXAMPLE    EXA
3         4  EXAMPLE   EXAM
4         5  EXAMPLE  EXAMP
5         1    OTHER      O
6         2    OTHER     OT
7         3    OTHER    OTH

Answer 3

You could use iterrows method:

for i, row in df.iterrows():
    df.loc[i, 'new'] = row['string'][:row['position']]

Example:

In [60]: dftest
Out[60]:
   position   string
0         1  EXAMPLE
1         2  EXAMPLE
2         3  EXAMPLE
3         4  EXAMPLE
4         5  EXAMPLE
5         1    OTHER
6         2    OTHER
7         3    OTHER

for i, row in dftest.iterrows():
    dftest.loc[i, 'new'] = row['string'][:row['position']]


In [62]: dftest
Out[62]:
   position   string    new
0         1  EXAMPLE      E
1         2  EXAMPLE     EX
2         3  EXAMPLE    EXA
3         4  EXAMPLE   EXAM
4         5  EXAMPLE  EXAMP
5         1    OTHER      O
6         2    OTHER     OT
7         3    OTHER    OTH

EDIT

Or you could use apply which is more convenient:

dftest.apply(lambda x: x['string'][:x['position']], axis=1)