CODEWITHSUNDEEP
pythonpandas
Michelle
Michelle

Reputation: 263

Replace column value based on value in other column

so far my dataframe looks like this:

ID   Area   Stage
1    P      X
2    Q      X
3    P      X
4    Q      Y

I would like to replace the area 'Q' with 'P' for every row where the Stage is equal to 'X'.

So the result should look like:

ID   Area   Stage
1    P      X
2    P      X
3    P      X
4    Q      Y

I tried:

data.query('Stage in ["X"]')['Area']=data.query('Stage in ["X"]')['Area'].replace('Q','P')

It does not work. Help is appreciated! :)

Upvotes: 8

Views: 3785

Answers (5)

bruno
bruno

Reputation: 32596

Note : this not an answer proposing a new way to do, but a comparison of the execution time each needs

All the proposals in the answers are quite 'magic' doing the job in one line of code thanks to pandas/numpy, anyway to do the job is good but to do it quickly is better, so I wanted to compare the execution time of each.

Here my program, in the loops I modify the dataframe two times to let it unchanged from a turn to the next ( I am not a Python programmer as you so sorry in advance if the way to do is 'poor') :

import pandas as pd
import numpy as np
import time

df=pd.DataFrame({'ID' : [i for i in range(1,1000)],
                 'Area' : ['P' if (i & 1) else 'Q' for i in range(1,1000)],
                 'Stage' : [ 'X' if (i & 2) else 'Y' for i in range(1,1000)]})

t0=time.process_time()
for i in range(1,100):
    df.loc[df['Stage']=='X', 'Area'] = df['Area'].replace('Q','q')
    df.loc[df['Stage']=='X', 'Area'] = df['Area'].replace('q','Q')

print("Quang Hoang", '%.2f' % (time.process_time() - t0))

t0=time.process_time()
for i in range(1,100):
    df.loc[df['Stage'] == 'X', 'Area'] = 'q'
    df.loc[df['Stage'] == 'X', 'Area'] = 'Q'

print("Joe Ferndz", '%.2f' % (time.process_time() - t0))

t0=time.process_time()
for i in range(1,100):
    df.loc[df['Area'].eq("Q") & df['Stage'].eq('X'),'Area']='q'
    df.loc[df['Area'].eq("q") & df['Stage'].eq('X'),'Area']='Q'

print("anky 1", '%.2f' % (time.process_time() - t0))

t0=time.process_time()
for i in range(1,100):
    df['Area'] = np.where(df['Area'].eq("Q") & df['Stage'].eq('X'),'q',df['Area'])
    df['Area'] = np.where(df['Area'].eq("q") & df['Stage'].eq('X'),'Q',df['Area'])

print("anky 2", '%.2f' % (time.process_time() - t0))

t0=time.process_time()
for i in range(1,100):
    df['Area']=np.where(df['Stage']=='X','q',df['Area'])
    df['Area']=np.where(df['Stage']=='X','Q',df['Area'])

print("RavinderSingh13", '%.2f' % (time.process_time() - t0))

On my PI 4 the result is :

Quang Hoang 1.60
Joe Ferndz 1.12
anky 1 1.55
anky 2 0.86
RavinderSingh13 0.38

if I use a dataframe having 100000 lines rather than 1000 the result is :

Quang Hoang 10.79
Joe Ferndz 6.61
anky 1 10.91
anky 2 9.64
RavinderSingh13 4.75

Note the proposals of Joe Ferndz and RavinderSingh13 suppose Area is only 'P' or 'Q'

Upvotes: 3

Joe Ferndz
Joe Ferndz

Reputation: 8508

To update a column using value from another column, use this option:

df.loc[df['Stage'] == 'X', 'Area'] = 'P'

This will check if value of 'Stage' is X. If True, then it will replace value of 'Area' to 'P'

Upvotes: 1

RavinderSingh13
RavinderSingh13

Reputation: 133750

Could you please try following.

import pandas as pd
import numpy as np
df['Area']=np.where(df['Stage']=='X','P',df['Area'])

Upvotes: 4

Quang Hoang
Quang Hoang

Reputation: 150805

You can use loc to specify where you want to replace, and pass the replaced series to the assignment:

df.loc[df['Stage']=='X', 'Area'] = df['Area'].replace('Q','P')

Output:

   ID Area Stage
0   1    P     X
1   2    P     X
2   3    P     X
3   4    Q     Y

Upvotes: 3

anky
anky

Reputation: 75130

you can use 2 boolean conditions and use loc:

df.loc[df['Area'].eq("Q") & df['Stage'].eq('X'),'Area']='P'
print(df)

   ID Area Stage
0   1    P     X
1   2    P     X
2   3    P     X
3   4    Q     Y

Or np.where

df['Area'] = np.where(df['Area'].eq("Q") & df['Stage'].eq('X'),'P',df['Area'])

Upvotes: 5

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