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

正如这里提到的 那

df1[~df1.apply(tuple,1).isin(df2.apply(tuple,1))]

是正确的解决方案，但它会产生错误的输出如果

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

在这种情况下，上面的溶液会给出 空数据帧，相反，你应该使用concat方法后，从每个数据帧删除重复。

使用concate和drop_duplicate

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

对称差分

如果你只对其中一个数据帧中的行感兴趣，而不是两个数据帧中的行，你在寻找集的差异:

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

⚠️只有在两个数据帧都不包含任何重复的情况下才有效。

设置差分/关系代数差分

如果你对关系代数差异/集差异感兴趣，即df1-df2或df1\df2:

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

⚠️只有在两个数据帧都不包含任何重复的情况下才有效。

试试这个: Df_new = df1。merge(df2, how='outer'， indicator=True)。查询('_merge == "left_only"')。下降(_merge, 1)

它将产生一个新的数据框架，其差异是:df1中存在的值，而df2中不存在。

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

通过索引查找差异。假设df1是df2的一个子集，并且在进行子集设置时将索引前移

df1.loc[set(df1.index).symmetric_difference(set(df2.index))].dropna()

# Example

df1 = pd.DataFrame({"gender":np.random.choice(['m','f'],size=5), "subject":np.random.choice(["bio","phy","chem"],size=5)}, index = [1,2,3,4,5])

df2 =  df1.loc[[1,3,5]]

df1

 gender subject
1      f     bio
2      m    chem
3      f     phy
4      m     bio
5      f     bio

df2

  gender subject
1      f     bio
3      f     phy
5      f     bio

df3 = df1.loc[set(df1.index).symmetric_difference(set(df2.index))].dropna()

df3

  gender subject
2      m    chem
4      m     bio