根据条件过滤出行

Question

我有一个数据框架:

| from id | from group | to id | to group   |
|    1    |      A     |   3   |      B     |
|    4    |      B     |   4   |      X     | 
|    5    |      F     |   5   |      J     |
|    2    |      B     |   3   |      A     |

查看'from group'和'to group'列。我想删除'A和B'和'F和J'同时出现在两列中的行。

预期输出:

| from id | from group | to id | to group   |
|    4    |      B     |   4   |      X     | 

我正在寻找一个灵活的解决方案。这意味着如果添加了第三个条件，它不会改变

Answer 1

假设您正在使用pandas，请尝试如下操作:

df.loc[~((df['from group'].isin(['A','B'])) & (df['to group'].isin(['A','B'])))] 

第一个括号前的~将定义即将到来的过滤器

的否定

Answer 2

看起来很长但很简单

df.query("not ((`from group` == 'A' and `to group` == 'B') or (`from group` == 'B' and `to group` == 'A') or (`from group` == 'F' and `to group` == 'J') or (`from group` == 'J' and `to group` == 'F'))")

-=- EDIT -=-

这里有一个更复杂但更可扩展的解决方案。只需将l1定义为要排除的对

l1 = [['A', 'B'], ['F', 'J']]
l2 = list(itertools.chain.from_iterable([itertools.permutations(i, r=2) for i in l1]))

df[[j not in l2 for j in list(zip(df['from group'], df['to group']))]]