Mapping a new column to a DataFrame by rows from another DataFrame





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I have a Pandas DataFrame stations with index as id:



id    station     lat     lng

1     Boston      45.343  -45.333

2     New York    56.444  -35.690


I have another DataFrame df1 that has the following:



duration   date       station   gender

NaN        20181118   NaN       M

9          20181009   2.0       F

8          20170605   1.0       F


I want to add to df1 so that it looks like the following DataFrame:



duration   date       station   gender  lat     lng 

NaN        20181118   NaN       M       nan     nan

9          20181009   New York  F       56.444  -35.690

8          20170605   Boston    F       45.343  -45.333


I tried doing this iteratively by referring to the station.iloc as shown in the following example but I have about 2 mil rows and it ended up taking a lot of time. 



stat_list =     

lng_list 

lat_list = 

for stat in df1:

  if not np.isnan(stat):

        ref = station.iloc[stat]

        stat_list.append(ref.station)

        lng_list.append(ref.lng)

        lat_list.append(ref.lat)

  else:

        stat_list.append(np.nan)

        lng_list.append(np.nan)

        lat_list.append(np.nan)


Is there a faster way to do this?






python pandas performance numpy dataframe







share|improve this question





























    1















    I have a Pandas DataFrame stations with index as id:



    id    station     lat     lng
    
1     Boston      45.343  -45.333
    
2     New York    56.444  -35.690


    I have another DataFrame df1 that has the following:



    duration   date       station   gender
    
NaN        20181118   NaN       M
    
9          20181009   2.0       F
    
8          20170605   1.0       F


    I want to add to df1 so that it looks like the following DataFrame:



    duration   date       station   gender  lat     lng 
    
NaN        20181118   NaN       M       nan     nan
    
9          20181009   New York  F       56.444  -35.690
    
8          20170605   Boston    F       45.343  -45.333


    I tried doing this iteratively by referring to the station.iloc as shown in the following example but I have about 2 mil rows and it ended up taking a lot of time. 



    stat_list =     
    
lng_list 
    
lat_list = 
    
for stat in df1:
    
  if not np.isnan(stat):
    
        ref = station.iloc[stat]
    
        stat_list.append(ref.station)
    
        lng_list.append(ref.lng)
    
        lat_list.append(ref.lat)
    
  else:
    
        stat_list.append(np.nan)
    
        lng_list.append(np.nan)
    
        lat_list.append(np.nan)


    Is there a faster way to do this?






    python pandas performance numpy dataframe







    share|improve this question

























      1












      1








      1








      I have a Pandas DataFrame stations with index as id:



      id    station     lat     lng
      
1     Boston      45.343  -45.333
      
2     New York    56.444  -35.690


      I have another DataFrame df1 that has the following:



      duration   date       station   gender
      
NaN        20181118   NaN       M
      
9          20181009   2.0       F
      
8          20170605   1.0       F


      I want to add to df1 so that it looks like the following DataFrame:



      duration   date       station   gender  lat     lng 
      
NaN        20181118   NaN       M       nan     nan
      
9          20181009   New York  F       56.444  -35.690
      
8          20170605   Boston    F       45.343  -45.333


      I tried doing this iteratively by referring to the station.iloc as shown in the following example but I have about 2 mil rows and it ended up taking a lot of time. 



      stat_list =     
      
lng_list 
      
lat_list = 
      
for stat in df1:
      
  if not np.isnan(stat):
      
        ref = station.iloc[stat]
      
        stat_list.append(ref.station)
      
        lng_list.append(ref.lng)
      
        lat_list.append(ref.lat)
      
  else:
      
        stat_list.append(np.nan)
      
        lng_list.append(np.nan)
      
        lat_list.append(np.nan)


      Is there a faster way to do this?






      python pandas performance numpy dataframe







      share|improve this question














      I have a Pandas DataFrame stations with index as id:



      id    station     lat     lng
      
1     Boston      45.343  -45.333
      
2     New York    56.444  -35.690


      I have another DataFrame df1 that has the following:



      duration   date       station   gender
      
NaN        20181118   NaN       M
      
9          20181009   2.0       F
      
8          20170605   1.0       F


      I want to add to df1 so that it looks like the following DataFrame:



      duration   date       station   gender  lat     lng 
      
NaN        20181118   NaN       M       nan     nan
      
9          20181009   New York  F       56.444  -35.690
      
8          20170605   Boston    F       45.343  -45.333


      I tried doing this iteratively by referring to the station.iloc as shown in the following example but I have about 2 mil rows and it ended up taking a lot of time. 



      stat_list =     
      
lng_list 
      
lat_list = 
      
for stat in df1:
      
  if not np.isnan(stat):
      
        ref = station.iloc[stat]
      
        stat_list.append(ref.station)
      
        lng_list.append(ref.lng)
      
        lat_list.append(ref.lat)
      
  else:
      
        stat_list.append(np.nan)
      
        lng_list.append(np.nan)
      
        lat_list.append(np.nan)


      Is there a faster way to do this?






      python pandas performance numpy dataframe




      python pandas performance numpy dataframe






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 25 '18 at 9:02









      Swopnil ShresthaSwopnil Shrestha

      636




      636
























          1 Answer
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          Looks like this would be best solved with a merge which should significantly boost performance:



          df1.merge(stations, left_on="station", right_index=True, how="left")


          This will leave you with two columns station_x and station_y if you only want the station column with the string names in you can do:



          df_merged = df1.merge(stations, left_on="station", right_index=True, how="left", suffixes=("_x", ""))
          
df_final = df_merged[df_merged.columns.difference(["station_x"])]


          (or just rename one of them before you merge)






          share|improve this answer


























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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Looks like this would be best solved with a merge which should significantly boost performance:



            df1.merge(stations, left_on="station", right_index=True, how="left")


            This will leave you with two columns station_x and station_y if you only want the station column with the string names in you can do:



            df_merged = df1.merge(stations, left_on="station", right_index=True, how="left", suffixes=("_x", ""))
            
df_final = df_merged[df_merged.columns.difference(["station_x"])]


            (or just rename one of them before you merge)






            share|improve this answer






























              1














              Looks like this would be best solved with a merge which should significantly boost performance:



              df1.merge(stations, left_on="station", right_index=True, how="left")


              This will leave you with two columns station_x and station_y if you only want the station column with the string names in you can do:



              df_merged = df1.merge(stations, left_on="station", right_index=True, how="left", suffixes=("_x", ""))
              
df_final = df_merged[df_merged.columns.difference(["station_x"])]


              (or just rename one of them before you merge)






              share|improve this answer




























                1












                1








                1







                Looks like this would be best solved with a merge which should significantly boost performance:



                df1.merge(stations, left_on="station", right_index=True, how="left")


                This will leave you with two columns station_x and station_y if you only want the station column with the string names in you can do:



                df_merged = df1.merge(stations, left_on="station", right_index=True, how="left", suffixes=("_x", ""))
                
df_final = df_merged[df_merged.columns.difference(["station_x"])]


                (or just rename one of them before you merge)






                share|improve this answer















                Looks like this would be best solved with a merge which should significantly boost performance:



                df1.merge(stations, left_on="station", right_index=True, how="left")


                This will leave you with two columns station_x and station_y if you only want the station column with the string names in you can do:



                df_merged = df1.merge(stations, left_on="station", right_index=True, how="left", suffixes=("_x", ""))
                
df_final = df_merged[df_merged.columns.difference(["station_x"])]


                (or just rename one of them before you merge)







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 25 '18 at 9:20

























                answered Nov 25 '18 at 9:11









                Sven HarrisSven Harris

                2,1961516




                2,1961516
































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