如何连接两个数据帧没有重复？

我想将两个数据帧 A、 B连接到一个没有重复行的新数据帧(如果 B中的行已经存在于 A中，不要添加) :

数据帧 A:

   I    II
0  1    2
1  3    1

数据帧 B:

   I    II
0  5    6
1  3    1

新数据框架:

     I    II
0  1    2
1  3    1
2  5    6

我怎么能这么做？

151752 次浏览
小开
最佳答案

The simplest way is to just do the concatenation, and then drop duplicates.

>>> df1
A  B
0  1  2
1  3  1
>>> df2
A  B
0  5  6
1  3  1
>>> pandas.concat([df1,df2]).drop_duplicates().reset_index(drop=True)
A  B
0  1  2
1  3  1
2  5  6

The reset_index(drop=True) is to fix up the index after the concat() and drop_duplicates(). Without it you will have an index of [0,1,0] instead of [0,1,2]. This could cause problems for further operations on this dataframe down the road if it isn't reset right away.

小开

In case you have a duplicate row already in DataFrame A, then concatenating and then dropping duplicate rows, will remove rows from DataFrame A that you might want to keep.

In this case, you will need to create a new column with a cumulative count, and then drop duplicates, it all depends on your use case, but this is common in time-series data

Here is an example:

df_1 = pd.DataFrame([
{'date':'11/20/2015', 'id':4, 'value':24},
{'date':'11/20/2015', 'id':4, 'value':24},
{'date':'11/20/2015', 'id':6, 'value':34},])


df_2 = pd.DataFrame([
{'date':'11/20/2015', 'id':4, 'value':24},
{'date':'11/20/2015', 'id':6, 'value':14},
])




df_1['count'] = df_1.groupby(['date','id','value']).cumcount()
df_2['count'] = df_2.groupby(['date','id','value']).cumcount()


df_tot = pd.concat([df_1,df_2], ignore_index=False)
df_tot = df_tot.drop_duplicates()
df_tot = df_tot.drop(['count'], axis=1)
>>> df_tot


date    id  value
0   11/20/2015  4   24
1   11/20/2015  4   24
2   11/20/2015  6   34
1   11/20/2015  6   14
小开

I'm surprised that pandas doesn't offer a native solution for this task. I don't think that it's efficient to just drop the duplicates if you work with large datasets (as Rian G suggested).

It is probably most efficient to use sets to find the non-overlapping indices. Then use list comprehension to translate from index to 'row location' (boolean), which you need to access rows using iloc[,]. Below you find a function that performs the task. If you don't choose a specific column (col) to check for duplicates, then indexes will be used, as you requested. If you chose a specific column, be aware that existing duplicate entries in 'a' will remain in the result.

import pandas as pd


def append_non_duplicates(a, b, col=None):
if ((a is not None and type(a) is not pd.core.frame.DataFrame) or (b is not None and type(b) is not pd.core.frame.DataFrame)):
raise ValueError('a and b must be of type pandas.core.frame.DataFrame.')
if (a is None):
return(b)
if (b is None):
return(a)
if(col is not None):
aind = a.iloc[:,col].values
bind = b.iloc[:,col].values
else:
aind = a.index.values
bind = b.index.values
take_rows = list(set(bind)-set(aind))
take_rows = [i in take_rows for i in bind]
return(a.append( b.iloc[take_rows,:] ))


# Usage
a = pd.DataFrame([[1,2,3],[1,5,6],[1,12,13]], index=[1000,2000,5000])
b = pd.DataFrame([[1,2,3],[4,5,6],[7,8,9]], index=[1000,2000,3000])


append_non_duplicates(a,b)
#        0   1   2
# 1000   1   2   3    <- from a
# 2000   1   5   6    <- from a
# 5000   1  12  13    <- from a
# 3000   7   8   9    <- from b


append_non_duplicates(a,b,0)
#       0   1   2
# 1000  1   2   3    <- from a
# 2000  1   5   6    <- from a
# 5000  1  12  13    <- from a
# 2000  4   5   6    <- from b
# 3000  7   8   9    <- from b
小开

Another option:

concatenation = pd.concat([
dfA,
dfB[dfB['I'].isin(dfA['I']) == False], # <-- get all the data in dfB that doesn't show up in dfB (based on values in column 'I')
])

The object concatenation will be:

     I    II
0  1    2
1  3    1
2  5    6

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