Reputation: 539
I have a dataframe like this:
import pandas as pd
import numpy as np
df=pd.DataFrame({'c1':[1,2,4,5],
'c2':[3,'P','N','T'],
'c3':np.nan})
the df
:
c1 c2 c3
0 1 3 NaN
1 2 P NaN
2 4 N NaN
3 5 T NaN
I want to change the c3
value based on c2
columns:
the result I wanted:
c1 c2 c3
0 1 3 NaN
1 2 P 1.0
2 4 N 3.0
3 5 T 5.0
I use the concat
to get this result:
df1=df[df.c2 == 'P']
df1['c3'] =1
df2=df[df.c2 == 'N']
df2['c3'] =3
df3=df[df.c2 == 'T']
df3['c3'] =5
df4=df[(df.c2 != 'N') & (df.c2 != 'P') & (df.c2 != 'T')]
new_df=pandas.concat([df1,df2,df3,df4]).reset_index()
new_df[['c1','c2','c3']]
I want to use apply
function to get the same result. I always replace the whole c3
columns when I use apply
function:
def new_col(x,df):
if x== 'P':
df['c3'] = 1
elif x == 'N':
df['c3'] = 3
elif x == 'T':
df['c3'] =5
else:
df['c3']=np.nan
df.c2.apply(new_col,df=df)
df
How I change the new_col
function ?
Upvotes: 1
Views: 296
Reputation: 862641
You can use:
def new_col(x):
a = np.nan
if x == 'P':
a = 1
elif x == 'N':
a = 3
elif x == 'T':
a = 5
return a
df['c3'] = df.c2.apply(new_col)
print (df)
c1 c2 c3
0 1 3 NaN
1 2 P 1.0
2 4 N 3.0
3 5 T 5.0
Another solution:
df.loc[df.c2 == 'P', 'C3'] = 1
df.loc[df.c2 == 'N', 'C3'] = 3
df.loc[df.c2 == 'T', 'C3'] = 5
print (df)
c1 c2 c3 C3
0 1 3 NaN NaN
1 2 P NaN 1.0
2 4 N NaN 3.0
3 5 T NaN 5.0
Upvotes: 1