定义数据框架:

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'])

df1

    Name  Age
0   John   23
1   Mike   45
2  Smith   12
3   Wale   34
4  Marry   27
5    Tom   44
6  Menda   28
7   Bolt   39
8  Yuswa   40

df2

    Name  Age
0   John   23
2  Smith   12
3   Wale   34
5    Tom   44
6  Menda   28
8  Yuswa   40

两者之间的区别是:

df1[~df1.isin(df2)].dropna()

    Name   Age
1   Mike  45.0
4  Marry  27.0
7   Bolt  39.0

地点:

isin(df2)返回df1中也在df2中的行。 ~(元素逻辑NOT)在表达式前面对结果求反，因此我们得到df1中不在df2中的元素——两者之间的差值。 .dropna()删除NaN显示所需输出的行

注意:这只适用于len(df1) >= len(df2)。如果df2比df1长，可以反转表达式:df2[~df2.isin(df1)].dropna()

import pandas as pd
# given
df1 = pd.DataFrame({'Name':['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa',],
    'Age':[23,45,12,34,27,44,28,39,40]})
df2 = pd.DataFrame({'Name':['John','Smith','Wale','Tom','Menda','Yuswa',],
    'Age':[23,12,34,44,28,40]})

# find elements in df1 that are not in df2
df_1notin2 = df1[~(df1['Name'].isin(df2['Name']) & df1['Age'].isin(df2['Age']))].reset_index(drop=True)

# output:
print('df1\n', df1)
print('df2\n', df2)
print('df_1notin2\n', df_1notin2)

# df1
#     Age   Name
# 0   23   John
# 1   45   Mike
# 2   12  Smith
# 3   34   Wale
# 4   27  Marry
# 5   44    Tom
# 6   28  Menda
# 7   39   Bolt
# 8   40  Yuswa
# df2
#     Age   Name
# 0   23   John
# 1   12  Smith
# 2   34   Wale
# 3   44    Tom
# 4   28  Menda
# 5   40  Yuswa
# df_1notin2
#     Age   Name
# 0   45   Mike
# 1   27  Marry
# 2   39   Bolt

nice @liangli的解决方案略有变化，不需要改变现有数据框架的索引:

newdf = df1.drop(df1.join(df2.set_index('Name').index))

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

另一个可能的解决方案是使用numpy广播:

df1[np.all(~np.all(df1.values == df2.values[:, None], axis=2), axis=0)]

输出:

    Name  Age
1   Mike   45
4  Marry   27
7   Bolt   39

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