I need to compare columns within a same dataframe and rank them

dataframe rank by column
pandas rank multiple columns
pandas rank within group
pandas rank over partition by
pandas sort by column
pandas rank same value
pandas rank unique
dask dataframe rank

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.

pandas.DataFrame.rank, How to rank the group of records that have the same value (i.e. ties):. average: average rank For DataFrame objects, rank only numeric columns if set to True. Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great

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'])]

Python, Pandas Dataframe.rank() method returns a rank of every respective index of a series passed. 'min', 'max', 'first', 'dense') which tells pandas what to do with same values. value and the rank function works on non-numeric value only if it's False. All the values in Name column are unique and hence there is no need to  Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Learn more Comparing two columns in pandas dataframe to create a third one

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()

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Comments
  • this is due to operator precedence. You should enclose each condition under a parenthesis .
  • With your above data, I think the result will be combined_min['que'] == np.nan, since combined_min['A'] == combined_min['A-1'] == 190. Maybe make combined_min['A'] == 191 in your example?
  • Thanks , it is printing NAH for all rows
  • @user1017373 I realized that I was wrong about & not being a logical operator, at least for Pandas objects. I posted a corrected answer with code that should label each row according to your original intention.
  • Thanks with this new editing it complains TypeError: object of type 'function' has no len()
  • @user1017373 I think I know what might be causing that TypeError. Is there a line in your code that looks like: np.select(cond, choices, default=np.nan)? You can't pass the cond lambda directly to np.select, you have to call it and pass the result. The line should instead look like np.select(cond(combined_min), choices, default=np.nan). If you're having trouble, first try just copy/pasting the code from my answer and see if you can get that to run as is.