Python: Conditional Row Values From a New Column Using dictionary, function, and lambda-CodePudding

I have a dataframe:

id
id1
id2
id3
id8
id9

I want to add a new column new with conditional row values as follows:

If a row from id == id1, then the new row is id1 is cat 1

If a row from id == id2, then the new row is id2 is cat 2

If a row from id == id3, then the new row is id3 is cat 3

else, idx is cat 0, where x is the id that is not one of id1, id2, or id3

This is what I tried so far. I think the solution should be to wrap the for loop inside a function and use that function with apply and/or lambda.

import pandas as pd

df = pd.DataFrame({
'id': ['id1', 'id2', 'id3', 'id8', 'id9']
}) 
df

dict = {'id1': '1', 'id2': '2', 'id3': '3'}

for k, val in dict.items():
    if k == "id1" or k == "id2" or k == "id3" in df['state']:
        print(str(k)   " is cat "   str(val))
    else:
        print(str(k)   " is cat 0")

Desired result:

id     new
id1   id1 is cat 1
id2   id2 is cat 2
id3   id3 is cat 3 
id8   id8 is cat 0
id9   id9 is cat 0

CodePudding user response：

You can get number and save in m then use numpy.where and if number m.isin(['1','2','3']) use number of m else use 0.

import numpy as np
m = df['id'].str[2:]
tf = m.isin(['1','2','3'])
df['new'] = np.where(tf, 
                     df['id']   " is cat "   m , 
                     df['id']   " is cat 0")
print(df)

Output:

    id           new
0  id1  id1 is cat 1
1  id2  id2 is cat 2
2  id3  id3 is cat 3
3  id8  id8 is cat 0
4  id9  id9 is cat 0

CodePudding user response：

It's a simple dictionary lookup.

import pandas as pd

df = pd.DataFrame({
'id': ['id1', 'id2', 'id3', 'id8', 'id9']
}) 

subst = {'id1': '1', 'id2': '2', 'id3': '3'}

def fix(row):
    val = subst.get(row['id'],0)
    return f"{row['id']} is cat {val}"

df['new'] = df.apply(fix,axis=1)
print(df)

Output:

    id          new
0  id1  id1 is cat 1
1  id2  id2 is cat 2
2  id3  id3 is cat 3
3  id8  id8 is cat 0
4  id9  id9 is cat 0