CODEWITHSUNDEEP
pythonpandas
ali bakhtiari
ali bakhtiari

Reputation: 1097

adding values to pandas dataframe columns based on another dataframe

I have a dataframe that looks like this(df):

HOUSEID    PERSONID      WHY_TRP
20000017      1            1
20000017      1            1
20000017      1            1
20000017      2            1
20000017      2            3
20000231      1            11
20000231      1            11
20000231      2            11
20000521      1            11
20000521      2            11
20000521      2            3

Each row describes a trip made by a person. I have another dataframe of the same kind in which each row describes a person(df_p):

    HOUSEID   PERSONID   
    20000017      1      
    20000017      2     
    20000231      1    
    20000231      2    
    20000521      1    
    20000521      2

I want to make three new columns in the second dataframe to show the frequency of 1, 3 and 11 for each person. basically I already have a second dataframe (df_p) with other features so I shouldn't use groupby. for some reason the first and second dataframe don't have the same number of people. that's why I needed the strategy below. This is the code I tried but took hours to complete(1 million iterations):

df_p.insert(2, 'WHY_TRP_1', 0)
df_p.insert(3, 'WHY_TRP_2', 0)
df_p.insert(4, 'WHY_TRP_3', 0)

def trip_counter(i, r):
  if r[2] == 1:
    df_p.loc[(df_p['HOUSEID'] == r[0]) & (df_p['PERSONID'] == r[1]), ['WHY_TRP_1']] += 1 
  elif r[2] == 3:
    df_p.loc[(df_p['HOUSEID'] == r[0]) & (df_p['PERSONID'] ==  r[1]), ['WHY_TRP_3']] += 1 
  elif r[2] == 11:
    df_p.loc[(df_p['HOUSEID'] == r[0]) & (df_p['PERSONID'] ==  r[1]), ['WHY_TRP_11']] += 1


for i ,r in df.iterrows():
  trip_counter(i ,r)

output:

     HOUSEID   PERSONID   WHY_TRP_1     WHY_TRP_3      WHY_TRP_11
    20000017      1            3            0            0
    20000017      2            1            1            0
    20000231      1            0            0            2
    20000231      2            0            0            1
    20000521      1            0            0            1
    20000521      2            0            1            1

Is there a faster way to do this?

thank you

Upvotes: 1

Views: 71

Answers (2)

anky
anky

Reputation: 75080

You could also do a pivot_table and then merge:

m = df.pivot_table(index=['HOUSEID','PERSONID'],
                   columns='WHY_TRP',aggfunc=len,fill_value=0)

out= df_p.merge(m.add_prefix('WHY_TRP'),left_on=['HOUSEID','PERSONID'],right_index=True)

print(out)

    HOUSEID  PERSONID  WHY_TRP1  WHY_TRP3  WHY_TRP11
0  20000017         1         3         0          0
1  20000017         2         1         1          0
2  20000231         1         0         0          2
3  20000231         2         0         0          1
4  20000521         1         0         0          1
5  20000521         2         0         1          1

Upvotes: 1

Randy
Randy

Reputation: 14849

You can get a table of the counts by doing a groupby on the first dataframe and unstacking WHY_TRP, and then you can just merge it to the second:

counts = df.groupby(["HOUSEID", "PERSONID", "WHY_TRP"]).apply(len).unstack(fill_value=0)

counts.columns = counts.columns.map(lambda x: f"WHY_TRP_{x}")

counts

WHY_TRP            WHY_TRP_1  WHY_TRP_3  WHY_TRP_11
HOUSEID  PERSONID
20000017 1                 3          0           0
         2                 1          1           0
20000231 1                 0          0           2
         2                 0          0           1
20000521 1                 0          0           1
         2                 0          1           1

df2.merge(counts, how="left", left_on=["HOUSEID", "PERSONID"], right_index=True)

    HOUSEID  PERSONID  WHY_TRP_1  WHY_TRP_3  WHY_TRP_11
0  20000017         1          3          0           0
1  20000017         2          1          1           0
2  20000231         1          0          0           2
3  20000231         2          0          0           1
4  20000521         1          0          0           1
5  20000521         2          0          1           1

Upvotes: 1

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