Python 基于多个条件和多个列创建新列

考虑到下面这些列旁边还有其他列，我想通过这3列创建一个新列，定义每行的最终状态

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列中，如果最后一列的值显示c_approved，则新列将给出approved

从第3列中，如果具有值的最后一列显示c_rejected，则新列将给出rejected

从第3列中，如果具有值的最后一列显示已批准，则新列将给出修订的值

从第3列中，如果具有值的最后一列显示“已拒绝”，则新列将给出“已拒绝”

最后的表格如下：

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专栏

---

提前感谢。

让我们用

np尝试
ffill
。选择

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

让我们用
np尝试
ffill
。选择

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

您可以使用
ffill
和
map
跟踪您的每个标准及其结果

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


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

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'}

您可以使用
ffill
和
map
跟踪您的每个标准及其结果

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


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

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'}

谢谢你的回答！在我的例子中，列status_3并不总是为了给出最终的_状态而要读取的列。它可能来自状态2或状态1。我该怎么做呢？IIUC，调用
.ffill（axis=1）
就能解决这个问题。本质上，我们从左到右传播最后一个有效的
非NAN
响应，直到“status_3”列完全被最后一个有效条目填充。因此，
df.ffill（axis=1）[“status_3”]
的结果输出：
[“c_approved”、“c_reproved”、“a_approved”、“a_reproved”]
这就是我们在将调用链接到
.map（…）
时操作的内容，谢谢您的回答！在我的例子中，列status_3并不总是为了给出最终的_状态而要读取的列。它可能来自状态2或状态1。我该怎么做呢？IIUC，调用
.ffill（axis=1）
就能解决这个问题。本质上，我们从左到右传播最后一个有效的
非NAN
响应，直到“status_3”列完全被最后一个有效条目填充。因此，
df.ffill（axis=1）[“status_3”]
的结果输出：
[“c_approved”、“c_reproved”、“a_approved”、“a_reproved”]
这就是我们在将调用链接到
.map（…）