Python Pandas：如何比较单元格和两列的值以及如何使用 If...Else 语句创建具有新值的另一列

Question

I'm trying and researching a lot how to do this, but I'm having trouble mixing pandas with if, else and/or get values by index and compare it with if, else and assign a column with codes/values. Explanation: I have this table below, I want to compare the cells in the ID column, and if the value of the posterior cell is equal to the previous one AND if in the COD column the posterior cell is equal to the previous one, THEN Result column = "no", otherwise "pass" and if neither then "unknown"

This is the formula in excel that I made: =IF(B3=B2,IF(C3=C2,"NO","PASS"),"UNKNOWN").

Below I have also posted some code attempts. I can even create two columns with the first test (from the ID column cells) and the second test (from the COD column cells), and return with Boolean results, but I can't get the If, Else to join it all together and generate the values I want in another column. Would I appreciate it if someone could help me?

df = df.sort_values(by=['ID'])
df['matchesID'] = df['ID'].shift(1) == df['ID']
df['matchesCod']= df['Cod'].shift(1) == df['Cod']

or

df = df.sort_values(by=['ID'])
test = (df['SWENo'].shift(1) == df['SWENo']) & (df['Cod'].shift(1) == df['Cod'])

I was trying something like this below

if df['ID'].shift(1) == df['ID'] and df['Cod'].shift(1) == df['Cod']:
    listProg.append('not')
elif df['ID'].shift(1) == df['ID'] and df['Cod'].shift(1) != df['Cod']:
    listProg.append('pass')
else:
    listProg.append('Unknown')

But the result is: "ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()".

If you can help me I appreciate it, it can be with pandas or not or mixing. I just need it to work. Thank you guys.

Answer 1

Similar approach in pandas will be to use numpy.where function.

With this code:

import numpy as np

df['Result'] = np.where(df['ID'] == df['ID'].shift(), np.where(df['Cod'] == df['Cod'].shift(), 'NO', 'PASS'), 'UNKNOWN')

I get below results:

   ID  Cod   Result
0   1    1  UNKNOWN
1   2    1  UNKNOWN
2   2    1       NO
3   3    1  UNKNOWN
4   4    1  UNKNOWN
5   4    2     PASS
6   4    2       NO
7   5    1  UNKNOWN
8   6    1  UNKNOWN

which seems more inline with your description of how Result value is derived.