Python 等效于 excel 嵌套 if 过滤 Pandas DataFrame 行的条件

Question

Selecting specific excel rows using python. So in excel I would do

If(And(Or(A<>({"Closed",""}),Or(B<>({"Closed",""})))

For obtaining those columns in a data frame that is neither Closed or blank. Tried using

df = df[(~df.A.isin([Closed","No Data"])) &(~df.B.isin([Closed","No Data"]))]

The problem is python is removing columns which are for example:

A                        B
Approved       Closed
No Data          Restrict
Restrict           No Data

Which I don't want As suggested in one of the links also tried

df.loc[(df[A] != "Closed") & (df[B] != "Closed") & (df[A] != "No data") & (df[B] != "No data")

Got the same result as when I tried .isin

Answer 1

I will use this sample data:

           A         B  ~df.A.isin  ~df.B.isin  ~A & ~B  ~A | ~B
0     Closed    Closed       False       False    False    False
1     Closed   No Data       False       False    False    False
2   Approved    Closed        True       False    False     True
3    No Data   No Data       False       False    False    False
4     Closed  Approved       False        True    False     True
5    No Data  Restrict       False        True    False     True
6   Approved   No Data        True       False    False     True
7     Closed  Restrict       False        True    False     True
8   Approved  Approved        True        True     True     True
9    No Data  Approved       False        True    False     True
10  Restrict   No Data        True       False    False     True
11  Restrict  Approved        True        True     True     True

~df.A.isin column shows the value of ~df.A.isin(["Closed","No Data"]) , which is True for rows where A contains neither Closed nor No Data

~df.B.isin column shows the value of ~df.B.isin(["Closed","No Data"]) , which is True for rows where B contains neither Closed nor No Data

~A & ~B column shows the value of (~df.A.isin(["Closed","No Data"])) &(~df.B.isin(["Closed","No Data"]))

~A | ~B ~A | ~B column shows the value of (~df.A.isin(["Closed","No Data"])) |(~df.B.isin(["Closed","No Data"]))

You first attempt lacks a " at the beginning of Closed" . Adding it we have

df[(~df.A.isin(["Closed","No Data"])) &(~df.B.isin(["Closed","No Data"]))]

which gives us:

           A         B  ~df.A.isin  ~df.B.isin  ~A & ~B  ~A | ~B
8   Approved  Approved        True        True     True     True
11  Restrict  Approved        True        True     True     True

The result shows only those rows that are completely without Closed and without No Data .

The suggestion in comments by Wen-Ben:

df[(~df.A.isin(["Closed","No Data"])) |(~df.B.isin(["Closed","No Data"]))]

gives us:

           A         B  ~df.A.isin  ~df.B.isin  ~A & ~B  ~A | ~B
2   Approved    Closed        True       False    False     True
4     Closed  Approved       False        True    False     True
5    No Data  Restrict       False        True    False     True
6   Approved   No Data        True       False    False     True
7     Closed  Restrict       False        True    False     True
8   Approved  Approved        True        True     True     True
9    No Data  Approved       False        True    False     True
10  Restrict   No Data        True       False    False     True
11  Restrict  Approved        True        True     True     True

Here we have | ( or ) instead of & ( and ), so the rows can contain Closed or No Data , but not in both A and B. This means the rows that have:

       A         B
Approved    Closed
 No Data  Restrict
Restrict   No Data

will be included, but not rows that have:

     A         B
Closed    Closed
Closed   No Data

---

Your second attempt:

df.loc[(df[A] != "Closed") & (df[B] != "Closed") &
       (df[A] != "No data") & (df[B] != "No data")

needs quotes around column labels. You can either use df.A or df['A'] , but not df[A]

Also, you spelled data in No data with lowercase d , while in other places you have it with uppercase D - No Data . In python, that's not the same. If we fix that:

df.loc[(df['A'] != "Closed") & (df['B'] != "Closed") &
       (df['A'] != "No Data") & (df['B'] != "No Data")]

which gives us the same thing as the first attempt:

           A         B  ~df.A.isin  ~df.B.isin  ~A & ~B  ~A | ~B
8   Approved  Approved  True  True  True  True  True  True  True
11  Restrict  Approved  True  True  True  True  True  True  True

If you rearrange this expression a little, use parentheses and | ( or ):

df.loc[((df['A'] != "Closed") & (df['A'] != "No Data")) | 
       ((df['B'] != "Closed") & (df['B'] != "No Data"))]

we get:

           A         B  ~df.A.isin  ~df.B.isin  ~A & ~B  ~A | ~B
2   Approved    Closed        True       False    False     True
4     Closed  Approved       False        True    False     True
5    No Data  Restrict       False        True    False     True
6   Approved   No Data        True       False    False     True
7     Closed  Restrict       False        True    False     True
8   Approved  Approved        True        True     True     True
9    No Data  Approved       False        True    False     True
10  Restrict   No Data        True       False    False     True
11  Restrict  Approved        True        True     True     True