类型错误：| 不支持的操作数类型：“str”和“bool”

Question

我有不同列的数据框users。我的目标是添加列 [uses_name] 当密码与每个用户的名字或姓氏相同时，它应该是 True。

例如，十二行中的 [user_name] 包含 milford.hubbard。然后在 [uses_name] 中将是 True，因为 [password] 和 [last_name] 是相同的。

为此，我使用正则表达式创建了两列 [first_name] 和 [last_name]。当创建 [uses_name] 时，| 运算符有问题。我在 pandas doc 中阅读了有关布尔索引的更多信息，但没有找到答案。

我的代码：

import pandas as pd

users = pd.read_csv('datasets/users.csv')

# Extracting first and last names into their own columns

users['first_name'] = users['user_name'].str.extract(r'(^\w+)', expand=False)

users['last_name'] = users['user_name'].str.extract(r'(\w+$)', expand=False)

# Flagging the users with passwords that matches their names

users['uses_name'] = users['password'].isin(users['first_name'] | users['last_name'])

# Counting and printing the number of users using names as passwords

print(users['uses_name'].count())

# Taking a look at the 12 first rows

print(users.head(12))

当我尝试编译这个时，我给出了一个错误：

TypeError: unsupported operand type(s) for |: 'str' and 'bool'

users 数据框中的前 12 行，创建了 first_name 和 last_name 列：

id          user_name            password   first_name  last_name
0    1    vance.jennings          joobheco      vance    jennings
1    2    consuelo.eaton        0869347314   consuelo       eaton
2    3   mitchel.perkins        fabypotter    mitchel     perkins
3    4    odessa.vaughan         aharney88     odessa     vaughan
2    3   mitchel.perkins        fabypotter    mitchel     perkins
3    4    odessa.vaughan         aharney88     odessa     vaughan
4    5    araceli.wilder        acecdn3000    araceli      wilder
5    6  shawn.harrington           5278049      shawn  harrington
6    7        evelyn.gay            master     evelyn         gay
7    8       noreen.hale            murphy     noreen        hale
8    9       gladys.ward           lwsves2     gladys        ward
9   10   brant.zimmerman  1190KAREN5572497      brant   zimmerman
10  11     leanna.abbott          aivlys24     leanna      abbott
11  12   milford.hubbard           hubbard    milford     hubbard

Answer 1

这行得通：

users['uses_name']= (users['password']==users['first_name'] )| (users['password']==users['last_name'])

Answer 2

你可以 concat ，因为两者都是系列

users['password'].isin(pd.concat([users['first_name'],users['last_name']]))

既然你改了问题，那就更新一个

df[['first_name','last_name']].eq(df.password,axis=0).any(1)

Answer 3

使用numpy.union1d:

val = np.union1d(users['first_name'], users['last_name'])
users['uses_name'] = users['password'].isin(val)
print (users)
    id         user_name          password first_name   last_name  uses_name
0    1    vance.jennings          joobheco      vance    jennings      False
1    2    consuelo.eaton        0869347314   consuelo       eaton      False
2    3   mitchel.perkins        fabypotter    mitchel     perkins      False
3    4    odessa.vaughan         aharney88     odessa     vaughan      False
2    3   mitchel.perkins        fabypotter    mitchel     perkins      False
3    4    odessa.vaughan         aharney88     odessa     vaughan      False
4    5    araceli.wilder        acecdn3000    araceli      wilder      False
5    6  shawn.harrington           5278049      shawn  harrington      False
6    7        evelyn.gay            master     evelyn         gay      False
7    8       noreen.hale            murphy     noreen        hale      False
8    9       gladys.ward           lwsves2     gladys        ward      False
9   10   brant.zimmerman  1190KAREN5572497      brant   zimmerman      False
10  11     leanna.abbott          aivlys24     leanna      abbott      False
11  12   milford.hubbard           hubbard    milford     hubbard       True

Answer 4

我认为最好的办法是执行 set 联合并将其传递给 isin：

users['uses_name'] = users['password'].isin(
   set(users['first_name']).union(users['last_name'])
)

users 

    id         user_name          password first_name   last_name  uses_name
0    1    vance.jennings          joobheco      vance    jennings      False
1    2    consuelo.eaton        0869347314   consuelo       eaton      False
2    3   mitchel.perkins        fabypotter    mitchel     perkins      False
3    4    odessa.vaughan         aharney88     odessa     vaughan      False
2    3   mitchel.perkins        fabypotter    mitchel     perkins      False
3    4    odessa.vaughan         aharney88     odessa     vaughan      False
4    5    araceli.wilder        acecdn3000    araceli      wilder      False
5    6  shawn.harrington           5278049      shawn  harrington      False
6    7        evelyn.gay            master     evelyn         gay      False
7    8       noreen.hale            murphy     noreen        hale      False
8    9       gladys.ward           lwsves2     gladys        ward      False
9   10   brant.zimmerman  1190KAREN5572497      brant   zimmerman      False
10  11     leanna.abbott          aivlys24     leanna      abbott      False
11  12   milford.hubbard           hubbard    milford     hubbard       True

注意| 是逻辑或，它对pandas 中的字符串列没有意义。