CODEWITHSUNDEEP
rmoduloarithmetic-expressions
Pradeep Chaghran
Pradeep Chaghran

Reputation: 296

R - Addition after modulo operation

Running the snippet D=1;for (j in 1:5){D=(D-1)%%4;print(D)} I get the expected outcome:

[1] 0
[1] 3
[1] 2
[1] 1
[1] 0

But when I add 1 to the result D=1;for (j in 1:5){D=((D-1)%%4)+1;print(D)}, it changes to:

[1] 1
[1] 1
[1] 1
[1] 1
[1] 1

Why does this happen?

This is what I would expect:

D=1;for (j in 1:5){D=(D-1)%%4;print(D+1)}
[1] 1
[1] 4
[1] 3
[1] 2
[1] 1

Edit:

The code chunks above were written in a compressed form in order to avoid a lengthy post. A properly styled R code should look like this:

D <- 1
for (j in 1:5) {
D <- (D - 1) %% 4
print(D)
}

Upvotes: 4

Views: 80

Answers (3)

Lajos Arpad
Lajos Arpad

Reputation: 76414

You initialize D with 1. When j is 0 and later, then you compute

D=((D-1)%%4)+1
  • D-1 is 1-1, which is 0
  • 0%%4 is 0
  • 0 + 1 is 1

therefore, when D is 1, then the formula above will yield 1, independently of j's value, since D only depends on its own previous value and whenever it was 1, the formula yields 1, hence D will never change. Instead, try this formula:

D=(D%%4)+1
  • if D was 0, then the result will be 1
  • if D was 1, then the result will be 2
  • if D was 2, then the result will be 3
  • if D was 3, then the result will be 4
  • if D was 4, the result will be 1
  • and so on
  • and so on

Since you want

[1] 1
[1] 4
[1] 3
[1] 2
[1] 1

you will need to initialize D with 0, so the first result will be 1:

D=0;for (j in 1:5){D=(D%%4)+1;print(D)}

EDIT

The solution above provides 1 2 3 4 ... If, instead 1 4 3 2 1 ... is needed, then one can use (4-D)%%4 instead of D%%4 and initialize D with 2.

Upvotes: 5

ThomasIsCoding
ThomasIsCoding

Reputation: 101014

You just need to add two more lines, say D <- D + 1 and D <- D - 1 like below

D <- 1
for (j in 1:5) {
    D <- ((D - 1) %% 4)
    D <- D + 1
    print(D)
    D <- D - 1
}

and you will obtain

[1] 1
[1] 4
[1] 3
[1] 2
[1] 1

Upvotes: 1

Axel Elias
Axel Elias

Reputation: 23

It returns you 1 every time because what you are doing is : D = 1

In your loop :

j = 1 : D = ((D - 1) %% 4 ) + 1 wich do D = ((1 - 1) %% 4 ) + 1 wich is equal to 0 %% 4 + 1 = 1

So at the end of this execution you get the same value of D as in the beginning and you do that 4 other times.

So your value of D doesn't change.

(English isn't my first language but if you need more details I would be happy to give you more)

Upvotes: 2

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