how to rank values in a vector and give them corresponding values?

Now I'm doing it by looping trhough a sorted vector, but maybe there is a faster way using internal R functions, and maybe I don't even need to sort.

vect = c(41,42,5,6,3,12,10,15,2,3,4,13,2,33,4,1,1)
vect = sort(vect)
print(vect)
outvect = mat.or.vec(length(vect),1)
outvect[1] = counter = 1
for(i in 2:leng开发者_C百科th(vect)) {
    if (vect[i] != vect[i-1]) { counter = counter + 1 }
    outvect[i] = counter
}

    print(cbind(vect,outvect))

 vect outvect
 [1,]    1       1
 [2,]    1       1
 [3,]    2       2
 [4,]    2       2
 [5,]    3       3
 [6,]    3       3
 [7,]    4       4
 [8,]    4       4
 [9,]    5       5
[10,]    6       6
[11,]   10       7
[12,]   12       8
[13,]   13       9
[14,]   15      10
[15,]   33      11
[16,]   41      12
[17,]   42      13

The code is used to make charts with integers on the X axis instead of real data because for me distance between the X values is not important. So in my case the smallest x value is always 1. and the largest is always equal to how many X values are there.

-- edit: due to some misuderstanding about my question I added self sufficient code with output.

---

That's more clear. Hence :

> vect = c(41,42,5,6,3,12,10,15,2,3,4,13,2,33,4,1,1)
> cbind(vect,as.numeric(factor(vect)))
 [1,]   41 12
 [2,]   42 13
 [3,]    5  5
 [4,]    6  6
 [5,]    3  3
 [6,]   12  8
 [7,]   10  7
 [8,]   15 10
 [9,]    2  2
[10,]    3  3
[11,]    4  4
[12,]   13  9
[13,]    2  2
[14,]   33 11
[15,]    4  4
[16,]    1  1
[17,]    1  1

No sort needed. And as said, see also ?factor

and if you want to preserve the order, then:

> cbind(vect,as.numeric(factor(vect,levels=unique(vect))))
      vect   
 [1,]   41  1
 [2,]   42  2
 [3,]    5  3
 [4,]    6  4
 [5,]    3  5
 [6,]   12  6
 [7,]   10  7
 [8,]   15  8
 [9,]    2  9
[10,]    3  5
[11,]    4 10
[12,]   13 11
[13,]    2  9
[14,]   33 12
[15,]    4 10
[16,]    1 13
[17,]    1 13

---

Joris solution is right on, but if you have a long vectors, it is a bit (3x) more efficient to use match and unique:

> x=sample(1e5, 1e6, replace=TRUE)
> # preserve order:
> system.time( a<-cbind(x, match(x, unique(x))) )
   user  system elapsed 
   0.20    0.00    0.22 
> system.time( b<-cbind(x, as.numeric(factor(x,levels=unique(x)))) )
   user  system elapsed 
   0.70    0.00    0.72 
> all.equal(a,b)
[1] TRUE
> 
> # sorted solution:
> system.time( a<-cbind(x, match(x, sort(unique(x)))) )
   user  system elapsed 
   0.25    0.00    0.25 
> system.time( b<-cbind(x, as.numeric(factor(x))) )
   user  system elapsed 
   0.72    0.00    0.72 
> all.equal(a,b)
[1] TRUE

---

You can try this : (Note that you may want a different behaviour for repeated values. This will give each value a unique rank)

> x <- sample(size=10, replace=T, x=1:100)
> x1 <- vector(length=length(x))
> x1[order(x)] <- 1:length(x)
> cbind(x, x1)
       x x1
 [1,] 40  1
 [2,] 46  4
 [3,] 43  3
 [4,] 41  2
 [5,] 47  5
 [6,] 84 10
 [7,] 75  8
 [8,] 60  7
 [9,] 59  6
[10,] 80  9

---

It looks like you are counting runs in the data, if that is the case, look at the rle function.

---

You apparently want the results of something like table() but lined up next to the values: Try using the ave() function:

csvdata$counts <- ave(csvdata[, "X"], factor(csvdata[["X"]]), FUN=length)

The trick here is that the syntax of ave is a bit different than tapply because you put in an arbitrarily long set of factor arrguments and you need to put in the FUN= in front of the function because the arguments after triple dots are not process by order. They need to be named.