df1中所有不在df2中的行 [英] all rows in df1 that are NOT in df2

问题描述

我有一个看起来像的df(df1)

I have a df (df1) that looks like:

df1 = pd.DataFrame([
        ['YYZ', 'SFO', 1],
        ['YYZ', 'YYD', 1],
        ['YYZ', 'EWR', 1],
        ['YYZ', 'DFW', 1],
        ['YYZ', 'LAX', 1],
        ['YYZ', 'YYC', 1]
    ], columns=['city1', 'city2', 'val'])

我还有另一个df(df2)，它是df1的子集:

I have another df (df2) that is a subset of df1:

df2 = pd.DataFrame([
        ['YYZ', 'SFO', 1],
        ['YYZ', 'YYD', 1]
    ], columns=['city1', 'city2', 'val'])

我想要df1中所有不在df2中的行.

I want all rows in df1 that are NOT in df2.

我尝试了本文基于条件切片的各种选项2列的值，但是我无法使其正常工作.

I've tried various options described in this post conditional slicing based on values of 2 columns, however I haven't been able to get it to work.

您的帮助将不胜感激.

推荐答案

• 将merge与indicator=True一起使用
• 然后使用query仅删除带有'left_only'的内容
• Use merge with indicator=True
• Then use query to strip out only those with 'left_only'

df1.merge(
    df2, how='outer', indicator=True
).query('_merge == "left_only"').drop('_merge', 1)

  city1 city2  val
2   YYZ   EWR    1
3   YYZ   DFW    1
4   YYZ   LAX    1
5   YYZ   YYC    1

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