Question

数据

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)

预期产量

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)

或..

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

问题

是否有执行上述操作的函数，或者这是进行此类操作的最佳方法？

Answer 1

我将使用您提出的第二种解决方案，或者更好的方法：

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

我认为使用更新可能会感到不舒服，因为它不会返回任何内容。

假设Series.fillna更灵活。

我们也可以使用DataFrame.assign 为了继续处理分配返回的DataFrame。

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

使用join方法的时间比较

%%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)

慢。

%%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)

最后，我建议您看到：Underastanding inplace和When I should use apply

熊猫-根据与数据框中某些值匹配的序列索引将序列中的值添加到数据框列

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