按行过滤数据框

Question

Hi I am a beginner python user and I need some help. I am trying to filter one dataframe against another.

Df1

 date          emp#   sku     transaction#   
 2017-01-01    10     200     399              
 2017-01-01    10     201     399             
 2017-01-01    10     202     399             
 2017-01-01    11     203     399             
 2017-01-01    11     200     399            

Df2

 date          emp#   sku     transaction#
 2017-01-01    10     200     301
 2017-01-01    11     200     301

Desired Df1

 date          emp#   sku     transaction#
 2017-01-01    10     200     399
 2017-01-01    11     200     399

I know this can work with an inner join (one emp# and sku) but I would have erroneous columns, how can I do this as a filter?

Answer 1

Use merge and the on parameter:

Df1.merge(Df2, on=['date','emp#','sku'], suffixes=('','_y'))\
   .drop('transaction#_y', axis=1)

Output:

         date  emp#  sku  transaction#
0  2017-01-01    10  200           399
1  2017-01-01    11  200           399

Answer 2

Here is one way without pd.merge . The benefit of this method is you don't have to play around with column names.

df2 = df2.set_index(['emp#', 'sku'])
df2['transaction#'] = df1.set_index(['emp#', 'sku'])['transaction#']
df2 = df2.reset_index()

#    emp#  sku        date  transaction#
# 0    10  200  2017-01-01           399
# 1    11  200  2017-01-01           399

Answer 3

You can do a filter from df2 by converting the desired columns into a dictionary, with orientation set to list , and then check in the values exist using isin . Lastly, take the min of each row to ensure both conditions are met ie

1. False + False = False
2. False + True = False
3. True + False = False
4. True + True = True

---

cols = ['emp#','sku']
df1[df1[cols].isin(df2[cols].to_dict(orient='list')).min(1)]

         date  emp#  sku  transaction#
0  2017-01-01    10  200           399
4  2017-01-01    11  200           399

Answer 4

您需要一个内部联接，它看起来像：保留仅在两个目录中都存在的行：

df1.join(df2, how='inner')