How to create a lambda function to return next cell value?

Question

For example :

the excepted output is the followwing :

What I did try is the following but it does not work and I am not sure if it is possible to do with lambda. ) )

DF["Output"] = DF.apply(lambda x: x[1].val2, axis=1)

The final goal is to apply thanks to a lambda function the following calculation :

(21/45) (45/54) (54/79) (79/45) ...

Wihtout having to create a new column with the shifted values as suggested in Ynjxsjmh Answer

score 1 · Accepted Answer · edited Jul 29 '22 at 08:24

1

Using Series.shift(-1)

DF["Output"] = DF["Val2"].shift(-1)

edited Jul 29 '22 at 08:24

TourEiffel

answered Jul 29 '22 at 08:22

Ynjxsjmh

So I can't use lambda function to do so ? – TourEiffel Jul 29 '22 at 08:23
why would you need a lambda? – mozway Jul 29 '22 at 08:25
@mozway because the real goal is to calculate the following formula : (21/45) (45/54) etc.. without having to create a new column. – TourEiffel Jul 29 '22 at 08:26
then provide a complete example, you don't need a new column (and even if you did this would still be better than apply) – mozway Jul 29 '22 at 08:28
@TourEiffel You can use `DF["Val2"].div(DF["Val2"].shift(-1))` – Ynjxsjmh Jul 29 '22 at 08:28
1

@TourEiffel If your index is like 0,1,2, you can try `df.apply(lambda row: df['Val2'].iloc[row.name+1], axis=1)` but you need to special handle the integer out of bounds. – Ynjxsjmh Jul 29 '22 at 08:31

1 Answers1