我在处理副本时遇到了问题，当一边有副本，另一边至少有一个副本时，所以我使用了Counter。集合做一个更好的差异，确保双方有相同的计数。这不会返回副本，但如果双方有相同的计数，则不会返回任何副本。

from collections import Counter

def diff(df1, df2, on=None):
    """
    :param on: same as pandas.df.merge(on) (a list of columns)
    """
    on = on if on else df1.columns
    df1on = df1[on]
    df2on = df2[on]
    c1 = Counter(df1on.apply(tuple, 'columns'))
    c2 = Counter(df2on.apply(tuple, 'columns'))
    c1c2 = c1-c2
    c2c1 = c2-c1
    df1ondf2on = pd.DataFrame(list(c1c2.elements()), columns=on)
    df2ondf1on = pd.DataFrame(list(c2c1.elements()), columns=on)
    df1df2 = df1.merge(df1ondf2on).drop_duplicates(subset=on)
    df2df1 = df2.merge(df2ondf1on).drop_duplicates(subset=on)
    return pd.concat([df1df2, df2df1])

> df1 = pd.DataFrame({'a': [1, 1, 3, 4, 4]})
> df2 = pd.DataFrame({'a': [1, 2, 3, 4, 4]})
> diff(df1, df2)
   a
0  1
0  2

我发现deepdiff库是一个很棒的工具，如果需要不同的细节或排序问题，它也可以很好地扩展到数据框架。你可以尝试不同的to_dict('records')， to_numpy()和其他导出:

import pandas as pd
from deepdiff import DeepDiff

df1 = pd.DataFrame({
    'Name':
        ['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa'],
    'Age':
        [23,45,12,34,27,44,28,39,40]
})

df2 = df1[df1.Name.isin(['John','Smith','Wale','Tom','Menda','Yuswa'])]

DeepDiff(df1.to_dict(), df2.to_dict())
# {'dictionary_item_removed': [root['Name'][1], root['Name'][4], root['Name'][7], root['Age'][1], root['Age'][4], root['Age'][7]]}

pandas DataFrame.compare中有一种新的方法，即比较2个不同的dataframe，并返回数据记录中每列中变化的值。

例子

第一个Dataframe

Id Customer Status      Date
1      ABC   Good  Mar 2023
2      BAC   Good  Feb 2024
3      CBA    Bad  Apr 2022

第二个Dataframe

Id Customer Status      Date
1      ABC    Bad  Mar 2023
2      BAC   Good  Feb 2024
5      CBA   Good  Apr 2024

比较Dataframes

print("Dataframe difference -- \n")
print(df1.compare(df2))

print("Dataframe difference keeping equal values -- \n")
print(df1.compare(df2, keep_equal=True))

print("Dataframe difference keeping same shape -- \n")
print(df1.compare(df2, keep_shape=True))

print("Dataframe difference keeping same shape and equal values -- \n")
print(df1.compare(df2, keep_shape=True, keep_equal=True))

结果

Dataframe difference -- 

    Id       Status            Date          
  self other   self other      self     other
0  NaN   NaN   Good   Bad       NaN       NaN
2  3.0   5.0    Bad  Good  Apr 2022  Apr 2024

Dataframe difference keeping equal values -- 

    Id       Status            Date          
  self other   self other      self     other
0    1     1   Good   Bad  Mar 2023  Mar 2023
2    3     5    Bad  Good  Apr 2022  Apr 2024

Dataframe difference keeping same shape -- 

    Id       Customer       Status            Date          
  self other     self other   self other      self     other
0  NaN   NaN      NaN   NaN   Good   Bad       NaN       NaN
1  NaN   NaN      NaN   NaN    NaN   NaN       NaN       NaN
2  3.0   5.0      NaN   NaN    Bad  Good  Apr 2022  Apr 2024

Dataframe difference keeping same shape and equal values -- 

    Id       Customer       Status            Date          
  self other     self other   self other      self     other
0    1     1      ABC   ABC   Good   Bad  Mar 2023  Mar 2023
1    2     2      BAC   BAC   Good  Good  Feb 2024  Feb 2024
2    3     5      CBA   CBA    Bad  Good  Apr 2022  Apr 2024

除了公认的答案，我想提出一个更广泛的解决方案，可以找到两个数据框架的2D集差异与任何索引/列(他们可能不符合两个数据框架)。此外，该方法允许设置浮动元素的容忍度，用于数据帧比较(它使用np.isclose)

import numpy as np
import pandas as pd

def get_dataframe_setdiff2d(df_new: pd.DataFrame, 
                            df_old: pd.DataFrame, 
                            rtol=1e-03, atol=1e-05) -> pd.DataFrame:
    """Returns set difference of two pandas DataFrames"""

    union_index = np.union1d(df_new.index, df_old.index)
    union_columns = np.union1d(df_new.columns, df_old.columns)

    new = df_new.reindex(index=union_index, columns=union_columns)
    old = df_old.reindex(index=union_index, columns=union_columns)

    mask_diff = ~np.isclose(new, old, rtol, atol)

    df_bool = pd.DataFrame(mask_diff, union_index, union_columns)

    df_diff = pd.concat([new[df_bool].stack(),
                         old[df_bool].stack()], axis=1)

    df_diff.columns = ["New", "Old"]

    return df_diff

例子:

In [1]

df1 = pd.DataFrame({'A':[2,1,2],'C':[2,1,2]})
df2 = pd.DataFrame({'A':[1,1],'B':[1,1]})

print("df1:\n", df1, "\n")

print("df2:\n", df2, "\n")

diff = get_dataframe_setdiff2d(df1, df2)

print("diff:\n", diff, "\n")

Out [1]

df1:
   A  C
0  2  2
1  1  1
2  2  2 

df2:
   A  B
0  1  1
1  1  1 

diff:
     New  Old
0 A  2.0  1.0
  B  NaN  1.0
  C  2.0  NaN
1 B  NaN  1.0
  C  1.0  NaN
2 A  2.0  NaN
  C  2.0  NaN 

通过使用drop_duplicate

pd.concat([df1,df2]).drop_duplicates(keep=False)

更新:

上面的方法只适用于那些本身没有副本的数据帧。例如:

df1=pd.DataFrame({'A':[1,2,3,3],'B':[2,3,4,4]})
df2=pd.DataFrame({'A':[1],'B':[2]})

它将输出如下所示，这是错误的

错误输出:

pd.concat([df1, df2]).drop_duplicates(keep=False)
Out[655]: 
   A  B
1  2  3

正确的输出

Out[656]: 
   A  B
1  2  3
2  3  4
3  3  4

如何实现这一目标?

方法一:将isin与tuple结合使用

df1[~df1.apply(tuple,1).isin(df2.apply(tuple,1))]
Out[657]: 
   A  B
1  2  3
2  3  4
3  3  4

方法二:与指标合并

df1.merge(df2,indicator = True, how='left').loc[lambda x : x['_merge']!='both']
Out[421]: 
   A  B     _merge
1  2  3  left_only
2  3  4  left_only
3  3  4  left_only

使用lambda函数，您可以过滤_merge值为“left_only”的行，以获得df1中df2中缺失的所有行

df3 = df1.merge(df2, how = 'outer' ,indicator=True).loc[lambda x :x['_merge']=='left_only']
df