Pandas add a column whose value can be multiple different values

Question

I'm trying to add a new column to a pandas data set that take the numeric month from one column and stores the corresponding season into the new column.

So in essence,

if month == 12, season = 'Winter'
elif month == 2, season = 'Spring'
elif month == 5, season = 'Summer'
else month == 8, season = 'Fall'

I haven't seen a really clear solution for this. I've seen how to do it with 2 conditional values, but I'm still fairly new to pandas and python.

Edit: I was able to get it working with one of the solutions listed below (Thanks!), but I should have mentioned that I also need to include months 1,3,4,6,7,9,10,11

I am assuming you have a big df with many month entries. If you make a df with month-season (aka df.shape=(2,12)) then you can do a join on the month column. — julian
– julian, Commented Nov 29, 2019 at 19:04
What have you done to try to solve this? Have you tried anything? — AMC
– AMC, Commented Nov 30, 2019 at 2:56
Also, which solution did you manage to get working? I believe mine is now the simplest amongst them. — AMC
– AMC, Commented Nov 30, 2019 at 3:01

c0rias · Accepted Answer · 2019-11-29 19:07:58Z

1

You can create a dictionary with corresponding numbers and season names, that you can then map onto your new series:

season_dict = {12: "Winter",
              2: "Spring",
              5: "Summer",
              8: "Fall"}

df["season"] = df.month.replace(season_dict)

answered Nov 29, 2019 at 19:07

c0rias

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Comments

AMC · Accepted Answer · 2019-11-30 03:01:01Z

1

There are some wild solutions in here.

This obviously isn’t a complete example, but it should be sufficient to understand what you need to do:

# map of month number -> season name
months_dict = {12: 'Winter', 2: 'Spring', ...}

df['season'] = df['month_number'].map(months_dict)

Yes, that’s it.

Let me know if you have any questions :)

answered Nov 30, 2019 at 3:01

AMC

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Comments

RavinderSingh13 · Accepted Answer · 2019-12-01 05:15:08Z

1

Could you please try following. Fair warning, since no samples of Input and expected output are given so didn't test it.

import pandas as pd
import numpy as np
conditions=[df['month']==12,
            df['month']==2,
            df['month']==5,
            df['month']==8]
choices=['Winter','spring','Summer','fall']

df['season']=np.select(conditions,choices,default=np.nan)

edited Dec 1, 2019 at 5:15

answered Nov 29, 2019 at 19:04

RavinderSingh13

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8 Comments

Michael Over a year ago

That did work, yes. Just had to add duplicates of each choice to correspond to the months I didn't explicitly list.

AMC Over a year ago

Heads up, there’s a mismatched quote in there.

RavinderSingh13 Over a year ago

@AlexanderCécile, thanks buddy for letting know, I fixed it now, Thanks a TON.

RavinderSingh13 Over a year ago

@Michael, Glad that it helped you. Give it sometime and when you have few answers to your question you could select "anyone" of them as correct one too, cheers.

RavinderSingh13 Over a year ago

@AlexanderCécile, Hi Alex, apologies if you are not the down voter, in case you have given down vote then please do lemme know the reason for same, thank you.

|

mochsner · Accepted Answer · 2019-11-29 19:08:43Z

-1

Since you're already using numpy through pandas, the best way to do this would be through the np.where command.

df['season'] = np.where(df['month']==11|df['month']==12|df['month']==1, 'Winter')
df['season'] = np.where(df['month']==2|df['month']==3|df['month']==4, 'Spring')
df['season'] = np.where(df['month']==5|df['month']==6|df['month']==7, 'Summer')
df['season'] = np.where(df['month']==8|df['month']==9|df['month']==10, 'Fall')

Hope this helps!

answered Nov 29, 2019 at 19:08

mochsner

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3 Comments

AMC Over a year ago

Isn’t numpy.where() completely unnecessary here?

mochsner Over a year ago

I mean, it's an approach... yours is definitely a bit simpler though :) Wasn't familiar with the .map commands from my limited experience.

AMC Over a year ago

The general approach is fine, what I meant is that where() is designed to choose between two alternatives. That first line, for example, could be written df.loc[df['month'] == 1 | df['month'] == 11 | df['month'] == 12, 'season'] = 'Winter', or even simpler df.loc[df['month'].isin({1, 11, 12}), 'season'] = 'Winter'.

darthbhyrava · Accepted Answer · 2019-11-29 19:10:18Z

-1

season_map = {12: "Winter",
              2: "Spring",
              5: "Summer",
              8: "Fall"} # Add the rest to this dictionary

df['season'] = df['month_num'].apply(lambda x: season_map[x]) 
# 'season' is the new column, replace 'month_num' accordingly

answered Nov 29, 2019 at 19:10

darthbhyrava

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3 Comments

AMC Over a year ago

df['season'] = df['month_num'].apply(lambda x: season_map[x]) ---> df['season'] = df['month_num'].map(season_map)

darthbhyrava Over a year ago

I fail to understand why my answer has been downvoted. An accompanying explanation as to where my code is wrong would be welcome and helpful.

AMC Over a year ago

See my comment above.

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