Combining several binary variables

Question

I have 5 variables, var1, var2 etc which are all coded as such:

Factor w/ 2 levels "no","yes": 2 1 1 2 1 2 1 1 1 1 ...

I would like to combine them into one. So far I have only used:

comb_drug <- with(dt1,interaction(var1, var2, var2, var4, var5))

which gives a variable with 32 levels. I would now like to create a variable with the following 3 levels:

all 5 are yes
any 4 are yes
less than 4 are yes

What is the best way to do this ? Here is some example data:

var1 <- as.factor(c(2,2,1,2,2,1,2,1,2,2))
var2 <- as.factor(c(2,1,2,2,2,1,2,2,2,2))
var3 <- as.factor(c(2,2,1,2,2,2,2,2,1,2))
var4 <- as.factor(c(2,2,1,2,2,2,2,2,1,2))
var5 <- as.factor(c(2,2,2,1,2,1,2,1,1,2))

dt <- data.frame(var1,var2,var3,var4,var5)

for ( i in 1:5) {
    levels(dt[,i]) <- c("no","yes")
}

   var1 var2 var3 var4 var5
1   yes  yes  yes  yes  yes
2   yes   no  yes  yes  yes
3    no  yes   no   no  yes
4   yes  yes  yes  yes   no
5   yes  yes  yes  yes  yes
6    no   no  yes  yes   no
7   yes  yes  yes  yes  yes
8    no   no  yes  yes   no
9   yes  yes   no   no   no
10  yes  yes  yes  yes  yes

I would instead like

    newvar
1   allyes
2   4yes
3   lessthan4yes
4   4yes
5   allyes
6   lessthan4yes
7   allyes
8   lessthan4yes
9   lessthan4yes
10  allyes

David Arenburg · Accepted Answer · 2016-12-14 08:54:45Z

3

An alternative that might be slightly faster than apply(x,1,sum) (rowSums)

dt$nYes <- rep(c('<4','4','all'),times = c(3,1,1))[rowSums(dt=='yes')]

edited Dec 14, 2016 at 8:54

David Arenburg

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answered Apr 24, 2013 at 11:12

mnel

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A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2013-04-24 11:09:56Z

2

This should get you on your way... Just add up the number of "yes" values per row:

dt$newvar <- apply(dt, 1, function(x) sum(x == "yes"))
dt$newvar
#  [1] 5 4 2 4 5 2 5 3 2 5

From there, you can do some clever factoring to get what you need... or this might be good enough for your purposes.

Actually, rowSums would be a lot faster probably:

dt$newvar <- rowSums(dt == "yes")

answered Apr 24, 2013 at 11:09

A5C1D2H2I1M1N2O1R2T1

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Comments

Carl Witthoft · Accepted Answer · 2013-04-24 13:04:53Z

1

If you subtract 1 from all your data, you'll have zeroes and ones, which is directly interpretable as TRUE/FALSE, which makes software jocks happier :-) . As an added bonus, for some vector of T/F (or 1 and 0), sum(myvector) gives you the number of TRUE directly. At that point, you could even have a look-up matrix like

sum  label
0    allno
1     one_no
2    lessthan4yes
3    lessthan4yes
4    4yes
5    yes

and do a direct replacement as newvec <- lutmat[lutmat[,1]==sums,2] .

answered Apr 24, 2013 at 13:04

Carl Witthoft

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