CODEWITHSUNDEEP
pythonpandasdictionarydata-science
Fallag Tayeb
Fallag Tayeb

Reputation: 11

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)

Upvotes: 1

Views: 437

Answers (8)

dwdineen
dwdineen

Reputation: 1

You can do:

import numpy as np

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

Upvotes: 0

dimension
dimension

Reputation: 1030

>>>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

Upvotes: 0

user3483203
user3483203

Reputation: 51155

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

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:

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

Upvotes: 2

Scott Boston
Scott Boston

Reputation: 153460

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

Upvotes: 2

Patrick the Cat
Patrick the Cat

Reputation: 2158

Go with vectorize numpy operation:

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

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

Upvotes: 0

Anton Petrov
Anton Petrov

Reputation: 46

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

Upvotes: 0

jpp
jpp

Reputation: 164673

If you only wish to change values equal to 2, you can use 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.

Upvotes: 0

W Stokvis
W Stokvis

Reputation: 1439

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

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

Upvotes: 1

Related Questions