Python/Pandas：如何跨多个列对匹配项执行联接

Question

问题：

我想用更复杂的连接标准将两个表连接在一起。一个表有3个不同的可能电话号码，另一个表格有2个不同的可能电话号码。我不知道哪一个是每一行的主要电话号码。因此，我希望根据以下条件加入：第一个表中number、Phone1、Phone2列中的电话号码可以在第二个表的ANI或{}列中找到。在

样本数据：

一个数据帧看起来像这样。。。在

                           application_uuid      number      Phone1      Phone2
0      b7754a2e-84be-4aec-a04e-0eba93dca5d8  5196942368         NaN         NaN
1      6ca3f0c3-0c83-4ebd-afe3-23977f1c6608  6475219092         NaN         NaN
2      3b5a083e-7765-4f27-941d-d2b4cbd6f26a  6476256563         NaN         NaN
3      229fee54-437f-4812-abec-7034fcb9a655        None         NaN         NaN
4      866a2cd2-5628-4e6b-b649-d92e2f0585ce  7092164418  7096391545  7092164977
5      8259410d-8d3d-4381-a0b3-6d6ce67b0917  6476387217  6476387217  6475313526
6      c359b03b-5e5f-4d4e-a5b0-ee37ac90c292        None         NaN         NaN
7      d70414a9-8fd9-4d1d-a77d-17f06743fd00  7054987969         NaN         NaN
8      0452edf9-2d58-4ad5-b1e2-0621ac517104  6136219401         NaN         NaN
9      cb3ab85c-fd42-4aff-a9b8-1743565b31e6        None         NaN         NaN
10     563e3e4d-e59a-4afc-b804-91aa14de919d  7056582202  7056582202  7056584200
11     3dd1df61-a36f-490b-ac15-225a83a21551        None  7096899998  7096899998
12     6bc42df3-e869-4794-a595-e3238ccf5284  5873415009         NaN         NaN
13     8bf11117-038f-4d2d-b4c6-9b2c6423d626  6473435642         NaN         NaN
14     0a854fe5-af66-40b0-b202-3e9367dc5a75  6478594204         NaN         NaN
15     b5884de8-2e0c-4b38-a3fd-7911cf4840b1  7787075288  7787075288  7787075288
16     f74cf212-cff0-48cc-b210-539dcdcccf72  7802676838  7806678567        None
17     9bffe5bf-b5d8-4e74-b4c9-9f1b5b238af3        None         NaN         NaN
18     dce91c00-a1ea-4111-a6ee-5ff5fd0cfb5f  6476093140         NaN         NaN
19     29cd024e-2c51-4682-b274-809c3cfb2b2b        None         NaN         NaN
20     ec55317b-fc20-416a-b26d-e95300f89c79        None         NaN         NaN
21     b3d00cd8-9d8e-415e-99b1-d8944e7b31e1        None         NaN         NaN
22     b3328787-edb7-4e08-a76c-370a74135fba        None         NaN         NaN
23     c8baf235-e702-41db-b4f8-8c2bf38109bf        None         NaN         NaN
24     cd9179bc-0594-4d25-9d7f-ddf6671777e2  7802428155         NaN         NaN
25     370855c0-b3fa-4d87-8d54-b84d34e7f35f        None         NaN         NaN
26     82244e78-3802-4890-96f6-e5267172f0e9        None         NaN         NaN
27     c7b0054c-29ac-4c76-bc5d-8cdbc93f5157  7052093358  7055268791  7052093358
28     d90e6e87-f7ef-43e1-9c85-35572fae838c  4039696044         NaN         NaN
29     bdd2474f-f4be-402b-8672-d73da90d7066        None         NaN         NaN

另一个数据帧看起来像这样。。。在

^{pr2}$

我的方法是将每行中的电话号码追加为列表中的一列。然后我创建了一个搜索函数。这种方式不实用，不雅观，而且太慢了。在

def f(row):
    phone_numbers_59 = phone_data['Number'].tolist()
    callid = phone_data['CALL ID'].tolist()

    get_callid = []
    for i in range(0, len(phone_numbers_59)):
        if any([x in phone_numbers_59[i] for x in row['Numbers']]):
            get_callid.append(callid[i])

    if len(get_callid) > 0:
        return get_callid
    else:
        return "NA"

s = data.apply(f, axis=1)

Answer 1

number, Phone1, Phone2 can be found in either ANI or DNIS

如果一次只接受一个条件，就更简单了（不要在Python中编写大的for循环，正如您所看到的那样，这是很慢的）：

for col in ('ANI', 'DNIS'):
    right = df2.set_index(col, drop=False)
    df1 = df1.join(right, 'number', rsuffix='_num_'+col)
    df1 = df1.join(right, 'Phone1', rsuffix='_p1_'+col)
    df1 = df1.join(right, 'Phone2', rsuffix='_p2_'+col)

它的作用是将列添加到df1六次：每个组合一次。rsuffix用于消除列名的歧义。您可能最终得到多个匹配项（可能是Phone1匹配ANI和{}匹配DNIS），在这种情况下，如何解决或组合它们取决于您（也许使用groupby()）。在