Question

考虑到我在下面这些列旁边还有另一列，我想通过这 3 列创建一个新列，这些列定义了每行的最终状态。

status_1                        status_2       status_3
a_accepted_with_comment         a_revised     c_approved
a_accepted_with_comment         c_rejected       nan
a_rejected                      a_approved       nan
a_rejected                         nan           nan

从第 3 列开始，如果具有值的最后一列显示 c_approved 则新列将给予批准

从第 3 列开始，如果具有值的最后一列显示 c_rejected，则新列将被拒绝

从第 3 列开始，如果最后一列有值显示 a_approved 则新列将给出修正

从第 3 列开始，如果具有值的最后一列显示 a_rejected，则新列将被拒绝

决赛桌是这样的：

status_1                        status_2       status_3       final_status
a_accepted_with _comment         a_revised     c_approved       approved
a_accepted_with_comment         c_rejected       nan           rejected
b_rejected                      a_approved       nan           revised
a_rejected                       nan             nan           rejected

如何在 python 中创建具有如此多条件的新列？

提前致谢。

Answer 1

让我们用 ffill 试试 np.select

s = df.ffill(1).iloc[:,-1]
c1 = s=='c_approved'
c2 = s.isin(['c_rejected','a_rejected'])
c3 = s=='a_approved'
df['new'] = np.select([c1,c2,c3],['approve','rejected','revised'])
df
Out[210]: 
                  status_1    status_2    status_3       new
0  a_accepted_with_comment   a_revised  c_approved   approve
1  a_accepted_with_comment  c_rejected         NaN  rejected
2               a_rejected  a_approved         NaN   revised
3               a_rejected         NaN         NaN  rejected

Answer 2

您可以使用 ffill 和 map 来跟踪您的每个条件及其结果。

response_rules = {
    "c_approved": "approved",
    "c_rejected": "rejected",
    "a_approved": "revised",
    "a_rejected": "rejected"
}

df["final_status"] = df.ffill(axis=1)["status_3"].map(response_rules)
print(df)
                  status_1    status_2    status_3 final_status
0  a_accepted_with_comment   a_revised  c_approved     approved
1  a_accepted_with_comment  c_rejected         NaN     rejected
2               a_rejected  a_approved         NaN      revised
3               a_rejected         NaN         NaN     rejected

如果您有很多规则，更好的设计模式可能是保留一个易于阅读/可编辑的字典，将结果映射到每个标准，然后在调用 .map

之前将其反转

response_rules = {
    "approved": ["c_approved"],
    "rejected": ["c_rejected", "a_rejected"],
    "revised": ["a_approved"]
}
# invert dictionary
inverted_rules = {vv: k for k, v in response_rules.items() for vv in v}

# same as before
df["final_status"] = df.ffill(axis=1)["status_3"].map(inverted_rules)

print(df)
                  status_1    status_2    status_3 final_status
0  a_accepted_with_comment   a_revised  c_approved     approved
1  a_accepted_with_comment  c_rejected         NaN     rejected
2               a_rejected  a_approved         NaN      revised
3               a_rejected         NaN         NaN     rejected



# Just so you can see:
print(inverted_rules) 
{'a_approved': 'revised',
 'a_rejected': 'rejected',
 'c_approved': 'approved',
 'c_rejected': 'rejected'}

基于多个条件和多列熊猫创建新列

2 个答案: