How to compare two rows in pandas (across all columns), and sum the output?

Question

I have data like this-

    User1  User2  User3  User4  User5  User6  User7  User8
w1      1      1      1      1      0      1      1      1
w2      0      1      0      0      1      1      1      1
w3      0      0      1      1      1      1      1      1
w4      1      1      1      0      0      0      0      1
w5      1      0      1      0      1      1      1      0
w6      1      1      1      1      1      1      1      1

Now what I want to do is compare every two consecutive weeks, and find all cases where the change is 1->0 .

So output for the above data would be something like this-

    Column
w1      n/a    
w2      3   
w3      1
w4      4   
w5      2 
w6      0

I am not understanding, are you counting the position of the change or number of changes in this output? — Richard K Yu, Jul 18 '20 at 19:23

score 2 · Accepted Answer · answered Jul 18 '20 at 19:29

2

could also do it like this:

(df > df.shift(-1)).sum(axis=1)

answered Jul 18 '20 at 19:29

Derek Eden

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This works, but the final df was one row shifted up as compared to the one that I wanted. Used shift to get it the way I wanted, thanks! – Vivek Singh Jul 18 '20 at 19:43

Reza · Answer 2 · 2020-07-18T20:16:10.703

0

Let say the dataframe name is df, so you can achieve it by

abs(df.diff()).sum(axis=1)

edited Jul 18 '20 at 20:16

answered Jul 18 '20 at 19:27

Reza

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How to compare two rows in pandas (across all columns), and sum the output?

2 Answers2