Converting Simple Python Function to Lambda Function

Question

I have a simple quantile function that I was going to use to select some data. I am trying to convert that function to a lambda function, but it does not appear to be working the way I expected. I want the lambda function to be able to set values above the 90th percentile as 'yes' and those below as 'no'. Need help figuring out what I am doing wrong.

Below is the code with a small set of data for illustrative purposes.

import pandas as pd
import numpy as np

# the simple function
def selector(x):
    return np.quantile(x, 0.90)

# the lambda function (not working properly.)
df.apply(lambda x: 'yes' if (x >= np.quantile(x, 0.90)) else 'no'))

df = pd.DataFrame({'tm': [263, 257, 263, 268, 268,259, 265,
               263, 256, 263, 264, 268, 268, 263,
               262, 260, 262, 235, 263, 264, 264]})

jezrael · Accepted Answer · 2022-06-08 07:13:19Z

3

Use numpy.where if performance is important:

df['new'] = np.where(df['tm'] >= np.quantile(df['tm'], 0.90), 'yes', 'no')
print (df)
     tm  new
0   263   no
1   257   no
2   263   no
3   268  yes
4   268  yes
5   259   no
6   265   no
7   263   no
8   256   no
9   263   no
10  264   no
11  268  yes
12  268  yes
13  263   no
14  262   no
15  260   no
16  262   no
17  235   no
18  263   no
19  264   no
20  264   no

Your solution with lambda function is necessary change by test quantile of original column tm:

df['new'] = df['tm'].apply(lambda x: 'yes' if (x >= np.quantile(df['tm'], 0.90)) else 'no')

edited Jun 8, 2022 at 7:13

answered Jun 8, 2022 at 6:39

jezrael

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Comments

bekirduran · Accepted Answer · 2022-06-08 07:11:33Z

0

can you try this lambda code :

df2 = df.applymap(lambda  x : "yes" if x >= selector(x)  else "no")

answered Jun 8, 2022 at 7:11

bekirduran

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