I have a dataframe with 6 columns I need to compare each 3 columns with other three columns another. The 6 columns are same data but values of first 3 are from one method and other three are other method. So I need to compare them for differences or variations.

Df.head()



  A    B    C  A-1  B-1  C-1

190  289  300  190  287  267

And my conditions are,

conditions = [(combined_min['A'] == combined_min['A-1']) & (combined_min['B'] == combined_min['B-1'] & combined_min['C'] == combined_min['C-1']),

              (combined_min['A'] > combined_min['A-1']) & (combined_min['B'] > combined_min['B-1'] & combined_min['C'] > combined_min['C-1']),

              (combined_min['A'] < combined_min['A-1']) & (combined_min['B'] < combined_min['B-1'] & combined_min['C'] < combined_min['C-1'])]

And my choices are,

choices     = [ "same", 'kj_greater', 'mi_greater' ]

Then I tried,

combined_min['que'] = np.select(conditions,choices, default=np.nan)

But it is throwing error message,

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

In the end I need a dataframe like this,

  A    B    C  A-1  B-1  C-1         que

190  289  300  190  287  267  kj_greater

The if the columns A, B, and C, are higher then kj_greater otherwise mi_greater, if all 6 are same then same.

Edit

After a bit of digging/reflection, I realized that I was wrong: it turns out that & is a logical operator in Pandas. & implements pairwise logical and between pd.Series and pd.DataFrame objects. Unfortunately, & has different operator precedence than and, so you have to be careful with it (in this case, & has higher precedence than ==, >, or <). The bug in the OP's code just comes down to a lack of parentheses in the right places.

So to get the kind of labeling that the OP was originally after, the code would be:

import numpy as np

import pandas as pd



data= [

    [191, 289, 300, 190, 287, 267],

    [191, 289, 300, 200, 312, 400],

    [191, 289, 300, 191, 289, 300],

    [191, 289, 300, 200, 287, 400],

]

combined_min = pd.DataFrame(data=data, columns=['A', 'B','C','A-1','B-1','C-1'])



cond = lambda x: [(x['A'] == x['A-1']) & (x['B'] == x['B-1']) & (x['C'] == x['C-1']),

                  (x['A'] > x['A-1'])  & (x['B'] > x['B-1'])  & (x['C'] > x['C-1']),

                  (x['A'] < x['A-1'])  & (x['B'] < x['B-1'])  & (x['C'] < x['C-1'])]

choices = ['same', 'kj_greater', 'mi_greater']



combined_min['que'] = np.select(cond(combined_min), choices, default=np.nan)

print(combined_min)

This outputs:

     A    B    C  A-1  B-1  C-1         que

0  191  289  300  190  287  267  kj_greater

1  191  289  300  200  312  400  mi_greater

2  191  289  300  191  289  300        same

3  191  289  300  200  287  400         nan

Optionally, cond can be boiled down to a one-liner:

from functools import reduce

from operator import eq, gt, lt, and_



cond = lambda x: [reduce(and_, (op(x[c], x['{}-1'.format(c)]) for c in 'ABC')) for op in (eq, gt, lt)]

Though this reduces readability somewhat.

The problem is that you are missing parenthesis on conditions. Each conditions has to be surrounded by parenthesis.

conditions = [(combined_min['A'] == combined_min['A-1']) & (combined_min['B'] == combined_min['B-1']) & (combined_min['C'] == combined_min['C-1']),

          (combined_min['A'] > combined_min['A-1']) & (combined_min['B'] > combined_min['B-1']) & (combined_min['C'] > combined_min['C-1']),

          (combined_min['A'] < combined_min['A-1']) & (combined_min['B'] < combined_min['B-1']) & (combined_min['C'] < combined_min['C-1'])]

You're error is in your conditions. The problem is that you are not directly comparing booleans, but rather a set of pd.Series containing a boolean, which connot be directly compared as you do.

So:

df['A'] == df['A-1']

Returns:

0    True

dtype: bool

So when you do:

df['A'] == df['A-1'] & df['A'] == df['A-1']

You get the error you mentioned. Try separating each term using parenthesis, and using any() to get the boolean from the pd.Series:

((df['A'] == df['A-1']) & (df['A'] == df['A-1'])).any()