熊猫根据索引向“数据框”列添加值，如果值存在，则追加

Question

I'm assigning value to column based on index value and if value exists then append the new value to old value

I have tried applying value to column based on index but couldn't append value to already existing value in column

Dataframe:-

    foo   bar 
0   1     1          
1   2     2     
2   3     3     
3   3     3     

Code :-

values = ['a', 'b']
for val in values:
   df.loc[df['foo'] == 1, foo2] = val

Outupt Dataframe:

    foo   bar   foo2  
0   1     1       b   
1   2     2     
2   3     3     
3   3     3     

Expected DataFrame:-

    foo   bar   foo2  
0   1     1     a,b  
1   2     2     
2   3     3     
3   3     3     

Answer 1

As you noticed the line

df.loc[df['foo'] == 1, foo2] = b

overwrites previous value so that you end up with only b . Is there a particular reason you need to solve your problem this way? You can write a custom function and use apply to append values but this is not how pandas is supposed to be used. I would recommend creating another column 'foo3' . You can always join the letters afterwards.

Answer 2

This should achieve what you are looking for. However, just as maow already pointed out, you should think about adding new columns and concatenating them when you need to.

import pandas as pd
df = pd.DataFrame({'foo': [1,2,3,3], 'bar': [1,2,3,3], 'foo2':[None, None, None, None]})
print(df)

a = 'a'
b = 'b'

# insert a
df.loc[df['foo'] == 1, 'foo2'] = a
print(df)

# concat b to a
df.loc[df['foo'] == 1, 'foo2'] += ',' + b
print(df)

Answer 3

This should solve your problem

import pandas as pd
df = pd.DataFrame({'foo': [1,2,3,3], 'bar': [1,2,3,3], 'foo2':[None, None, None, None]})
print(df)

print(df)
values = ['a', 'b']
for val in values:
    print(df.loc[df['foo'] == 1,'foo2'][0])
    if (df.loc[df['foo'] == 1,'foo2'][0]==None):
        df.loc[df['foo'] == 1, 'foo2']=val

    else:       
        df.loc[df['foo'] == 1, 'foo2']=df.loc[df['foo'] == 1:,'foo2'] + ',' +val

print(df)

output:

   foo  bar foo2
0    1    1  a,b
1    2    2
2    3    3
3    3    3