CODEWITHSUNDEEP
pythonpandas
Yale Newman
Yale Newman

Reputation: 1231

Pandas - Add values from series to dataframe column based on index of series matching some value in dataframe

Data

pb = {"mark_up_id":{"0":"123","1":"456","2":"789","3":"111","4":"222"},"mark_up":{"0":1.2987,"1":1.5625,"2":1.3698,"3":1.3333,"4":1.4589}}
data = {"id":{"0":"K69","1":"K70","2":"K71","3":"K72","4":"K73","5":"K74","6":"K75","7":"K79","8":"K86","9":"K100"},"cost":{"0":29.74,"1":9.42,"2":9.42,"3":9.42,"4":9.48,"5":9.48,"6":24.36,"7":5.16,"8":9.8,"9":3.28},"mark_up_id":{"0":"123","1":"456","2":"789","3":"111","4":"222","5":"333","6":"444","7":"555","8":"666","9":"777"}}
pb = pd.DataFrame(data=pb).set_index('mark_up_id')
df = pd.DataFrame(data=data)

Expected Output

test = df.join(pb, on='mark_up_id', how='left')
test['cost'].update(test['cost'] + test['mark_up'])
test.drop('mark_up',axis=1,inplace=True)

Or..

df['cost'].update(df['mark_up_id'].map(pb['mark_up']) + df['cost'])

Question

Is there a function that does the above, or is this the best way to go about this type of operation?

Upvotes: 1

Views: 77

Answers (1)

ansev
ansev

Reputation: 30940

I would use the second solution you propose or better this:

df['cost']=(df['mark_up_id'].map(pb['mark_up']) + df['cost']).fillna(df['cost'])

I think using update can be uncomfortable because it doesn't return anything.

Let's say Series.fillna is more flexible.

We can also use DataFrame.assign in order to continue working on the DataFrame that the assignment returns.

df.assign( Cost=(df['mark_up_id'].map(pb['mark_up']) + df['cost']).fillna(df['cost']) )

Time comparision with join method

%%timeit
df['cost']=(df['mark_up_id'].map(pb['mark_up']) + df['cost']).fillna(df['cost'])
#945 µs ± 46 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit
test = df.join(pb, on='mark_up_id', how='left')
test['cost'].update(test['cost'] + test['mark_up'])
test.drop('mark_up',axis=1,inplace=True)
#3.59 ms ± 137 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

slow..

%%timeit
df['cost'].update(df['mark_up_id'].map(pb['mark_up']) + df['cost'])
#985 µs ± 32.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Finally,I recommend you see: Underastanding inplace and When I should use apply

Upvotes: 2

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