从列值 pandas 替换所有值的更好方法

Question

I am looking for a better way to replace all column values with some other value.

What I currently have is this:

gender_text = ['undefined', 'male', 'female']

df.loc[df['gender'] == 0, 'gender'] = gender_text[0]
df.loc[df['gender'] == 1, 'gender'] = gender_text[1]
df.loc[df['gender'] == 2, 'gender'] = gender_text[2]

df.head()

I was hoping for something a bit more elegant and use the gender value (0, 1 or 2) as the index to choose from in gender_text to have everything fit in one line.

Answer 1

You can use a dictionary.

import pandas as pd
df = pd.DataFrame({'gender':[0,0,2,1,1,2]})

gender_text = {0:'undefined', 1:'male', 2:'female'}
df['gender'].map(gender_text)

# Out[33]: 
# 0    undefined
# 1    undefined
# 2       female
# 3         male
# 4         male
# 5       female
# Name: gender, dtype: object

Alternatively, you can also pd.merge , which might be better for larger datasets.

import pandas as pd
df = pd.DataFrame({'gender':[0,0,2,1,1,2]})
df_map = pd.DataFrame({'gender': [0, 1, 2], 'gender_new': ['undefined', 'male', 'female']})

df['gender'] = df.merge(df_map, on=['gender'])['gender_new']

Answer 2

You can define a dict

replace_values = {0 :'undefined', 1 : 'male', 2 : 'female'}

And replace multiple values using replace

df = df.replace({"gender": replace_values}) 

Alternatively, replace each value in the column using

df.gender = df.gender.replace(0, 'undefined')
df.gender = df.gender.replace(1, 'male')
df.gender = df.gender.replace(2, 'female')

Answer 3

This is one of the usecase of the map function (use np.select for much faster performance)-

gender_text  = {0 :'undefined', 1 : 'male', 2 : 'female'}
df['gender'] = df['gender'].map(gender_text)

Or you can use apply -

df['gender'] =  df['gender'].apply(lambda x :  gender_text[x])

Or you can use np.select

condlist = [df['gender'] == 0,
            df['gender'] == 1,
            df['gender'] == 2]

choicelist = ['undefined',
              'male',
              'female']
df['gender'] = np.select(condlist, choicelist)

Performace Comparison . — >

%timeit df['gender'] = df['gender'].map(gender_text)
411 µs ± 10.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df['gender'] = np.select(condlist,choicelist)
101 µs ± 322 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)