Replace multiple values in Pandas column without loop

Question

I have Pandas dataframe and want to substitute a lot of values on specific column of df.

How to implement the code below without loop?

for i in range(len(data.loc[:, 'CityID'])):
    if data.loc[:, 'CityID'][i] == 1:
        data.loc[:, 'CityID'][i] = 1
    elif data.loc[:, 'CityID'][i] in (2, 3, 4, 21):
        data.loc[:, 'CityID'][i] = 2
    elif data.loc[:, 'CityID'][i] in (33, 34):
        data.loc[:, 'CityID'][i] = 4
    else:
        data.loc[:, 'CityID'][i] = 3

score 4 · Answer 1 · answered Sep 03 '18 at 11:15

4

Use:

m1 = data['CityID'] == 1
m2 = data['CityID'].isin([2, 3, 4, 21])
m3 = data['CityID'].isin([33, 34])

data['CityID'] = np.select([m1, m2, m3], [1,2,4], default=3)

answered Sep 03 '18 at 11:15

jezrael

822,522
95
1,334
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score 4 · Accepted Answer · answered Sep 03 '18 at 11:16

You can use a dictionary with pd.Series.map:

d = {1: 1, 2: 2, 3: 2, 4: 2, 21: 2, 33: 4, 34: 4}

data['CityID'] = data['CityID'].map(d).fillna(3)

If the dictionary construction feels laborious, you can use unpacking:

d = {1: 1, **dict.fromkeys((2, 3, 4, 21), 2), **dict.fromkeys((33, 34), 4}

Replace multiple values in Pandas column without loop

2 Answers2