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Mapping values inside pandas column

I used the code below to map the 2 values inside S column to 0 but it didn't work. Any suggestion on how to solve this?
N.B : I want to implement an external function inside the map.

df = pd.DataFrame({'Age':[30,40,50,60,70,80],'Sex':
['F','M','M','F','M','F'],'S':
[1,1,2,2,1,2]})
def app(value):
for n in df['S']:
if n == 1:
return 1
if n == 2:
return 0
df["S"] = df.S.map(app)


8 Answers
8

Use eq to create a boolean series and conver that boolean series to int with astype:

eq

astype

df['S'] = df['S'].eq(1).astype(int)


OR

df['S'] = (df['S'] == 1).astype(int)


Output:

Age Sex S
0 30 F 1
1 40 M 1
2 50 M 0
3 60 F 0
4 70 M 1
5 80 F 0


loc

df.S.mask(df.S>1,0)

mask

Don't use apply, simply use loc to assign the values:

apply

loc

df.loc[df.S.eq(2), 'S'] = 0

Age Sex S
0 30 F 1
1 40 M 1
2 50 M 0
3 60 F 0
4 70 M 1
5 80 F 0


If you need a more performant option, use np.select. This is also more scalable, as you can always add more conditions:

np.select

df['S'] = np.select([df.S.eq(2)], [0], 1)


You're close but you need a few corrections. Since you want to use a function, remove the for loop and replace n with value. Additionally, use apply instead of map. Apply operates on the entire column at once. See this answer for how to properly use apply vs applymap vs map

for

n

value

apply

map

Apply

apply

applymap

map

def app(value):
if value == 1:
return 1
elif value == 2:
return 0
df['S'] = df.S.apply(app)
Age Sex S
0 30 F 1
1 40 M 1
2 50 M 0
3 60 F 0
4 70 M 1
5 80 F 0


If you only wish to change values equal to 2, you can use pd.DataFrame.loc:

pd.DataFrame.loc

df.loc[df['S'] == 0, 'S'] = 0


pd.Series.apply is not recommend and this is just a thinly veiled, inefficient loop.

pd.Series.apply

You could use .replace as follows:
df["S"] = df["S"].replace([2], 0)
This will replace all of 2 values to 0 in one line

Go with vectorize numpy operation:

df['S'] = np.abs(df['S'] - 2)


and stand yourself out from competitions in interviews and SO answers :)

>>>df = pd.DataFrame({'Age':[30,40,50,60,70,80],'Sex':
['F','M','M','F','M','F'],'S':
[1,1,2,2,1,2]})


>>> def app(value):
return 1 if value == 1 else 0
# or app = lambda value : 1 if value == 1 else 0

>>> df["S"] = df["S"].map(app)

>>> df
Age S Sex
Age S Sex
0 30 1 F
1 40 1 M
2 50 0 M
3 60 0 F
4 70 1 M
5 80 0 F


You can do:

import numpy as np

df['S'] = np.where(df['S'] == 2, 0, df['S'])


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