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SOLVED

MOD function

PNC97
8 - Asteroid

I'm working on math where  

 

if [Running Sum] >[sch-work] then (if [sch-work] = 0 then [Running Sum] else Number.Mod([Running Sum],[#"sch-work"])) else 0),

 

I tried by this below formula

 

IF [Running Sum] >[sch-work]
THEN
If [sch-work] = 0 THEN [Running Sum]
ELSE Mod([Running Sum],[sch-work]) ENDIF

ELSE 0 ENDIF

 

But I didn't got the expected output 

 

Prudhvinath_Chowdary_0-1623095183334.png

 

 

If possible can you help with the workflow or logic to be used

 

5 REPLIES 5
apathetichell
20 - Arcturus

I'm fairly sure mod returns an integer. So all of those are going to be rounded to zero. You can test this by adding a number greater than 7 and seeing that the result is 1.

 

if you want to get the mod of that try this:

[Running Sum]-(floor([Running Sum]/[sch-work])*6)

PNC97
8 - Asteroid

Yes it is rounding to 0.

 

But is there any luck to get the double (type) output same as excel 

apathetichell
20 - Arcturus

see my formula above - I'd use that as a workaround.

 

this is what R says about Modulo btw:

Error in mod(6.11116, 6) :
Arguments 'n', 'm' must be integers or vectors of integers.

Qiu
21 - Polaris
21 - Polaris

@PNC97 

Looks like you have already an answer and just to add and try to be more dynamic.

Qiu_0-1623105923806.png

Capture1A.PNG

 

sherinamahtani
7 - Meteor

Check out my solution!

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