Analyze if the value of a column is less than another and this another is less than another and so o-CodePudding

Currently I do it this way:

import pandas as pd

dt = pd.DataFrame({
    '1st':[1,0,1,0,1],
    '2nd':[2,1,2,1,2],
    '3rd':[3,0,3,2,3],
    '4th':[4,3,4,3,4],
    '5th':[5,0,5,4,5],
    'minute_traded':[6,5,6,5,6]
})

dt = dt[
    (dt['1st'] < dt['2nd']) & 
    (dt['2nd'] < dt['3rd']) & 
    (dt['3rd'] < dt['4th']) & 
    (dt['4th'] < dt['5th']) & 
    (dt['5th'] < dt['minute_traded'])
]

print(dt)

Result:

   1st  2nd  3rd  4th  5th  minute_traded
0    1    2    3    4    5              6
2    1    2    3    4    5              6
3    0    1    2    3    4              5
4    1    2    3    4    5              6

Is there a more correct method for an analysis like this that always uses the same pattern and only changes the columns to be analyzed?

CodePudding user response：

You can take column-wise differences from left to right and see if all of them are less than 0 to determine the mask:

dt.loc[dt.diff(-1, axis="columns").iloc[:, :-1].lt(0).all(axis="columns")]

(.iloc[:, :-1] is to drop the rightmost difference result which is NaNs since there's no right column to it.)

to get

   1st  2nd  3rd  4th  5th  minute_traded
0    1    2    3    4    5              6
2    1    2    3    4    5              6
3    0    1    2    3    4              5
4    1    2    3    4    5              6

CodePudding user response：

Using shift to perform the comparison and all to aggregate as single boolean for boolean indexing:

out = dt[dt.shift(axis=1).lt(dt).iloc[:, 1:].all(axis=1)]

Output:

   1st  2nd  3rd  4th  5th  minute_traded
0    1    2    3    4    5              6
2    1    2    3    4    5              6
3    0    1    2    3    4              5
4    1    2    3    4    5              6