Question

我有一个如下所示的DataFrame：

Market | Status | Team | Member |
-------|--------|------|--------|
Chicago|   1    |  ENG |  None  |
Chicago|   1    |  ENG |  None  |
SF Bay |   3    |  ENG |  Julia |

用户及其电子邮件的字典：

TeamMembers = {
    "Julia": "julia@email.com", "Tyler": "tyler@email.com", "Kyle": "kyle@email.com"
}

在我的DataFrame中，如果没有成员，我想随机分配一个成员，但如果市场价值相同，那么成员也需要相同。

我想用

name, email = random.choice(list(TeamMembers.items()))

获取具体的姓名和电子邮件地址，但我不确定如何根据市场相同的值来操纵DataFrame。

Answer 1

您可以将transform与fillna一起使用，也可以通过将name更改为item来生成key：

df['Member'] = (df.groupby('Market')['Member']
                  .transform(lambda x: x.fillna(random.choice(list(TeamMembers.keys())))))
print (df)
    Market  Status Team Member
0  Chicago       1  ENG   Kyle
1  Chicago       1  ENG   Kyle
2   SF Bay       3  ENG  Julia

Answer 2

这是另一种解决方案。这个的好处是，如果芝加哥已被映射到一个成员，其他实例将被映射到同一成员，即使当前None。

import pandas as pd
import random

df = pd.DataFrame([['Chicago', 1, 'ENG', None],
                   ['Chicago', 1, 'ENG', None],
                   ['SF Bay', 3, 'ENG', 'Julia'],
                   ['SF Bay', 2, 'ENG', None],
                   ['NY', 1, 'ENG', None],
                   ['NY', 2, 'ENG', None]],
                  columns=['Market', 'Status', 'Team', 'Member'])

TeamMembers = {"Julia": "julia@email.com", "Tyler": "tyler@email.com", "Kyle": "kyle@email.com"}

existing_map = df.dropna(subset=['Member']).set_index('Market')['Member'].to_dict()
unmapped = list(set(df.loc[pd.isnull(df['Member']), 'Market']) - set(existing_map))

MemberChoices = list(TeamMembers.keys())
random.shuffle(unmapped)
random.shuffle(MemberChoices)

additional_map = {k: MemberChoices[i % len(MemberChoices)] for i, k in enumerate(unmapped)}

new_map = {**existing_map, **additional_map}

df['Member'] = df['Member'].fillna(df['Market'].map(new_map))

#     Market  Status Team Member
# 0  Chicago       1  ENG  Tyler
# 1  Chicago       1  ENG  Tyler
# 2   SF Bay       3  ENG  Julia
# 3   SF Bay       2  ENG  Julia
# 4       NY       1  ENG   Kyle
# 5       NY       2  ENG   Kyle

Answer 3

没有groupby

k=df.Market.unique().tolist()
list(TeamMembers.keys())
Out[31]: ['Julia', 'Tyler', 'Kyle']

d=dict(zip(k,random.sample(set(list(TeamMembers.keys())), 2)))
df.Member=df.Member.fillna(df.Market.map(d))

根据另一列中的相同值将值分配给列

3 个答案: