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Я пытаюсь заполнить нулевое значение job_industry_category из фрейма данных поиска. Например:

 df = pd.DataFrame()
df['job_title'] = ['Executive Secretary', 'Administrative Officer' , 'Recruiting Manager' , 'Senior Editor', 'Media Manager I']
df['job_industry_category'] = ['Health', 'Financial Services' , 'Property', NaN, NaN]
df
             job_title           job_industry_category
0       Executive Secretary              Health
1       Administrative Officer    Financial Services
2       Recruiting Manager              Property
3       Senior Editor                     NaN
4       Media Manager I                   NaN

lookup = pd.DataFrame()
lookup['job_title'] = ['Executive Secretary', 'Senior Editor', 'Media Manager I']
lookup['job_industry_category'] = ['Retail', 'Manufacturing', 'Health']
lookup
             job_title           job_industry_category
0       Executive Secretary              Health
1       Senior Editor                 Manufacturing
2       Media Manager I                  Health
  

И результат, который я ожидаю, будет:

 df
             job_title           job_industry_category
0       Executive Secretary              Health
1       Administrative Officer    Financial Services
2       Recruiting Manager              Property
3       Senior Editor                Manufacturing
4       Media Manager I                  Health
  

Я пытался использовать map , как это:
df.loc[df['job_industry_category'].isnull(), 'job_industry_category'] = lookup['job_title'].map(lookup) А также удаление na из другого сообщения:

 def remove_na(x):
    if pd.isnull(x['job_industry_category']):
        return freq_job_ind[x['job_title']]
    else:
        return x['job_industry_category']

df['job_industry_category'] = df.apply(remove_na, axis=1)
  

Но оба не сработали, и я не уверен, есть ли лучший способ сделать это?
Заранее благодарю вас!

Получите недостающие местоположения с помощью isna(), а затем используйте map с set_index .

 % ipython
Python 3.8.5 (default, Sep  4 2020, 07:30:14) 
Type 'copyright', 'credits' or 'license' for more information
IPython 7.18.1 -- An enhanced Interactive Python. Type '?' for help.

In [1]: 
import pandas as pd
import numpy as np
df = pd.DataFrame({
    'job_title': ['Executive Secretary', 'Administrative Officer',
                  'Recruiting Manager', 'Senior Editor', 'Media Manager I'],
    'job_industry_category': ['Health', 'Financial Services',
                              'Property', np.nan, np.nan]})
df
Out[1]: 
                job_title job_industry_category
0     Executive Secretary                Health
1  Administrative Officer    Financial Services
2      Recruiting Manager              Property
3           Senior Editor                   NaN
4         Media Manager I                   NaN

In [2]: 
lookup = pd.DataFrame({
    'job_title': ['Executive Secretary', 'Senior Editor', 'Media Manager I'],
    'job_industry_category': ['Retail', 'Manufacturing', 'Health']})
lookup
Out[2]: 
             job_title job_industry_category
0  Executive Secretary                Retail
1        Senior Editor         Manufacturing
2      Media Manager I                Health

In [3]: 
missing = df['job_industry_category'].isna()

In [4]: 
df.loc[missing, 'job_industry_category'] = df.loc[missing, 'job_title'].map(
    lookup.set_index('job_title')['job_industry_category'])
df
Out[4]: 
                job_title job_industry_category
0     Executive Secretary                Health
1  Administrative Officer    Financial Services
2      Recruiting Manager              Property
3           Senior Editor         Manufacturing
4         Media Manager I                Health
  

 #Boolean select NaN
m=df.job_industry_category.isna()
#Mask the NaNs and map across values using a dict of lookup['job_title']:lookup['job_industry_category']   df.loc[m,'job_industry_category']=df.loc[m,'job_title'].map(dict(zip(lookup.job_title,lookup.job_industry_category)))

 

    job_title         job_industry_category
0     Executive Secretary                Health
1  Administrative Officer    Financial Services
2      Recruiting Manager              Property
3           Senior Editor         Manufacturing
4         Media Manager I                Health