# [R] ifelse reformulation

arun smartpink111 at yahoo.com
Thu Oct 25 21:27:14 CEST 2012

```Hi,

Did you mean this?
group1<-c(40,50,"60",70)
#or
group2<-c(50,"var1","var2",60)

In either of the above cases, when you check
str(group1) # all are converted to character.
# chr [1:4] "40" "50" "60" "70"

str(group2)
# chr [1:4] "50" "var1" "var2" "60"

Suppose, I am comparing the test1 dataset (x4: x6 columns) with group2
apply(test1,1,function(x) ifelse(x[5:7]%in%group2,0,1)) # 2nd row is 50 for x4 column
#  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
#[1,] 1 0 1 1 1 1 1 1 1  1  1  1  1  1  1
#[2,] 1 1 1 1 1 1 1 1 1  1  1  1  1  1  1
#[3,] 1 1 1 1 1 1 1 1 1  1  1  1  1  1  1

# final result
as.vector(apply(test1,1,function(x) ifelse(any(x[5:7]%in%group2),0,1)))
# [1] 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1

I guess it works.

as.vector(apply(test1,1,function(x) ifelse(any(x[4:6]%in%group),0,1)))
^^^^^^^
It should be 5:7.
group<-c(40,50,60,70)
as.vector(apply(test1,1,function(x) ifelse(any(x[5:7]%in%group),0,1)))
#[1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1

ifelse(test1\$x4==40|test1\$x4==50|test1\$x4==60|test1\$x4==70|test1\$x5==40|test1\$x5==50|test1\$x5==60|test1\$x5==70|test1\$x6==40|test1\$x6==50|test1\$x6==60|test1\$x7==70,0,1)
# [1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1

as.vector(apply(test1,1,function(x) ifelse(any(x[5:7]==40|x[5:7]==50|x[5:7]==60|x[5:7]==70),0,1)))
#[1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1

A.K.

----- Original Message -----
From: brunosm <brunosm87 at gmail.com>
To: r-help at r-project.org
Cc:
Sent: Thursday, October 25, 2012 12:55 PM
Subject: Re: [R] ifelse reformulation

Arun, thank you very much... but a new problem arose...

What if... in the variable group that i want to compare, there is numeric
and non numeric types?

Or, if you think its better, can i have two variables, one numeric and one
non numeric, and made the comparision?

Best regards,

Bruno

2012/10/12 arun kirshna [via R] <ml-node+s789695n4645988h70 at n4.nabble.com>

> HI,
> Try this:
>    id x1 x2 x3 x4 x5 x6 x7
> 1   1 36 26 21 32 31 27 31
> 2   2 45 21 46 50 22 36 29
> 3   3 49 47 35 44 33 31 46
> 4   4 42 32 38 28 39 45 32
> 5   5 29 42 39 48 25 35 34
> 6   6 39 31 30 37 46 43 44
> 7   7 41 40 25 23 42 40 24
> 8   8 27 29 47 34 26 38 28
> 9   9 25 35 29 36 43 34 23
> 10 10 24 44 37 26 27 46 22
> 11 11 38 50 32 49 37 24 40
> 12 12 20 34 48 25 30 41 36
> 13 13 26 46 20 40 29 20 43
> 14 14 33 37 49 31 47 30 30
> 15 15 43 39 27 35 48 47 27
> count40<-
> ifelse(test1\$x1==40|test1\$x2==40|test1\$x3==40|test1\$x4==40|test1\$x5==40|test1\$x6==40|test1\$x7==40,0,1)
>
> count40
>  #[1] 1 1 1 1 1 1 0 1 1 1 0 1 0 1 1
> #if you want to get the same result,
>  apply(test1,1,function(x) ifelse(any(x==40),0,1))
> # 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
>  #1  1  1  1  1  1  0  1  1  1  0  1  0  1  1
> A.K.
>
>
>
>
>
> ------------------------------
> below:
> http://r.789695.n4.nabble.com/ifelse-reformulation-tp4645981p4645988.html
> .
>

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