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579

answers:

2

I'm running the following script:

cause = c(1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2); time =  c(1, 1, 2, 3, 3, 2, 2, 1, 1, 2, 2); table(cause, time)

And I get the following:
          time
cause 1 2 3
       1 2 2 2
       2 2 3 0

What I want is this:
               time
     cause 1 2 3
  Maltreat 2 2 2
Non-Maltr 2 3 0

So, my question is: how do you rename the rows of a table in R?

In the same vein, how would you rename the columns of that table?

+3  A: 

One way to do it is to use factors or lists of strings instead of indexes. So:

cause1 <- c("Maltreat", "Non-malt")[cause]

> print(cause1)
 [1] "Maltreat" "Maltreat" "Maltreat" "Maltreat" "Maltreat" "Non-malt"
 [7] "Maltreat" "Non-malt" "Non-malt" "Non-malt" "Non-malt"

> table(cause1, time)
          time
cause1     1 2 3
  Maltreat 2 2 2
  Non-malt 2 3 0

And, in case you're worried about memory or speed, R is pretty good at representing this sort of thing efficiently internally, with only a single instance of the whole string stored, and the rest done with indexes.

Incidentally, you'll be happier in the long run with data frames:

> df <- data.frame(cause=as.factor(c("Maltreat", "Non-malt")[cause]), time=time)
> summary(df)
      cause        time      
 Maltreat:6   Min.   :1.000  
 Non-malt:5   1st Qu.:1.000  
              Median :2.000  
              Mean   :1.818  
              3rd Qu.:2.000  
              Max.   :3.000  
> table(df)
          time
cause      1 2 3
  Maltreat 2 2 2
  Non-malt 2 3 0
Harlan
+4  A: 

There are two easy ways to do this:

z <- table(cause, time)

Use the colnames/rownames functions:

> colnames(z)
[1] "1" "2" "3"
> rownames(z)
[1] "1" "2"

Or use dimnames:

> dimnames(z)
$cause
[1] "1" "2"
$time
[1] "1" "2" "3"
> dimnames(z)$cause
[1] "1" "2"

In any case, choose your names as a vector and assign them:

> dimnames(z)$cause <- c("Maltreat","Non-malt")
> z
          time
cause      1 2 3
  Maltreat 2 2 2
  Non-malt 2 3 0
Shane