my dataframe is as below, I want to create a new column based on value of column "a" but the value is from "b" and "c"
a b c
1 0.1 2
0 3 0.2
1 0.4 5
I create a function as below, but it's not working.. any idea?
def proba_column(row):
if row["label"] == "1":
val = row["c"]
else:
val = row["b"]
return val
result['proba'] = result.apply(proba_column,axis=1)
expected result should be:
a b c proba
1 0.1 2 2
0 3 0.2 3
1 0.4 5 5
Your function compares the row to string 1 ("1") instead of integer 1 , if you replace "1" with 1 , apply will work as intended:
def proba_column(row):
if row["a"] == 1:
val = row["c"]
else:
val = row["b"]
return val
df['proba'] = df.apply(proba_column,axis=1)
However , you do not need to use apply for such cases , generally np.where as the above answer suggests suggests should do it . However adding in another method using df.lookup after using a series.map on column a with a dictionary:
df['proba'] = df.lookup(df.index,df['a'].map({1:"c",0:"b"}))
print(df)
a b c proba
0 1 0.1 2.0 2.0
1 0 3.0 0.2 3.0
2 1 0.4 5.0 5.0
You can use numpy.choose:
df["proba"] = np.choose(df["a"], [df["b"], df["c"]])
print(df)
a b c proba
0 1 0.1 2.0 2.0
1 0 3.0 0.2 3.0
2 1 0.4 5.0 5.0
From your code it looks like that your logic is to choose value from column c when a == 1, else choose b.
You can use numpy.where:
In[579]: import numpy as np
In [580]: df['proba'] = np.where(df.a.eq(1), df.c, df.b)
In [581]: df
Out[581]:
a b c proba
0 1 0.1 2.0 2.0
1 0 3.0 0.2 3.0
2 1 0.4 5.0 5.0
OR use df.where:
In [610]: df['proba'] = df.c.where(df.a.eq(1), df.b)
In [611]: df
Out[611]:
a b c proba
0 1 0.1 2.0 2.0
1 0 3.0 0.2 3.0
2 1 0.4 5.0 5.0