[英]How two panda data frames with same column values can be merged to form the third data frame that shows the difference of the values
dataframe:df1 数据框:DF1
name age id salary
0 Smith 30 2 2000
1 Ron 24 3 30000
2 Mike 35 4 40000
3 Jack 21 5 5000
4 Roshan 20 6 60000
5 Steve 45 8 8000
6 Peter 32 1 1000
dataframe:df2 数据框:DF2
name age salary id
0 Peter 28 10000 1
1 Smith 30 1500 2
2 Ron 24 7000 3
3 Mike 35 20000 4
4 Jack 21 5000 5
5 Cathy 20 9000 6
6 Steve 45 56000 8
df1 and df2 To be merged on ID's.Please not that ID's are same in both df1 and df2 but id order is different.df3 needs to be created like below- df1和df2要在ID上合并。请不要让df1和df2中的ID相同,但ID顺序不同。df3需要如下创建:
name age id salary
0 Smith 30 2 2000|1500
1 Ron 24 3 30000|7000
2 Mike 35 4 40000 |20000
3 Jack 21 5 5000
4 Roshan|Cathy 20 6 60000|9000
5 Steve 45 8 8000|56000
6 Peter 32|28 1 1000|10000
I am planning to put the above output to excel sheet using to_excel functionality. 我打算使用to_excel功能将上述输出放到excel工作表中。 Before that i want to add one more extra column to this data frame which says 'match and 'mismatch' . 在此之前,我想在此数据帧中再添加一列,即“匹配和不匹配”。 Logic would be if any one of the row showing minimum of one difference value result should be mismatch else match.I am mocking the output below something like this- 逻辑上,如果显示最小一个差值结果的行中的任何一个应该不匹配否则匹配。我正在模拟这样的输出-
id age name salary Result 0 2 30 Smith 2000|1500 Mismatch 1 3 24 Ron 30000|7000 Mismatch 3 5 21 Jack 5000 Match 4 6 20 Roshan|Cathy 60000|9000 MisMatch 5 8 45 Steve 8000|56000 MisMatch 6 1 32|28 Peter 1000|10000 MisMatch id年龄名称薪水结果0 2 30 Smith 2000 | 1500不匹配1 3 24 Ron 30000 | 7000不匹配3 5 21 Jack 5000匹配4 6 20 Roshan | Cathy 60000 | 9000不匹配5 5 45 Steve 8000 | 56000不匹配6 1 32 | 28彼得1000 | 10000错误配对
What can i use for achieving such result 我可以用什么来达到这样的结果
Use merge
first and then join columns by condition with numpy.where
, last filter only columns by df1.columns
: 首先使用merge
,然后按条件与numpy.where
列,最后按df1.columns
仅过滤列:
cols = df1.columns.difference(['id'])
df = df1.merge(df2, on='id', suffixes=('','_'))
s = df[cols].astype(str) + '|' + df[cols + '_'].astype(str).values
mask = df[cols].values != df[cols + '_'].values
arr = np.where(mask, s, df[cols].astype(str))
df = df1[['id']].join(pd.DataFrame(arr, columns=cols))
print (df)
id age name salary
0 2 30 Smith 2000|1500
1 3 24 Ron 30000|7000
2 4 35 Mike 40000|20000
3 5 21 Jack 5000
4 6 20 Roshan|Cathy 60000|9000
5 8 45 Steve 8000|56000
6 1 32|28 Peter 1000|10000
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