CODEWITHSUNDEEP
rfor-loopcount
user2630096
user2630096

Reputation: 21

count of records within levels of a factor

I am trying to populate a field in a table (or create a separate vector altogether, whichever is easier) with consecutive numbers from 1 to n, where n is the total number of records that share the same factor level, and then back to 1 for the next level, etc. That is, for a table like this

data<-matrix(c(rep('A',4),rep('B',3),rep('C',4),rep('D',2)),ncol=1)

the result should be a new column (e.g. "sample") as follows:

sample<-c(1,2,3,4,1,2,3,1,2,3,4,1,2)

Upvotes: 2

Views: 1477

Answers (5)

digEmAll
digEmAll

Reputation: 57210

You can use rle function together with lapply :

sample <- unlist(lapply(rle(data[,1])$lengths,FUN=function(x){1:x}))

data <- cbind(data,sample)

Or even better, you can combine rle and sequence in the following one-liner (thanks to @Arun suggestion)

data <- cbind(data,sequence(rle(data[,1])$lengths))

> data
      [,1] [,2]
 [1,] "A"  "1" 
 [2,] "A"  "2" 
 [3,] "A"  "3" 
 [4,] "A"  "4" 
 [5,] "B"  "1" 
 [6,] "B"  "2" 
 [7,] "B"  "3" 
 [8,] "C"  "1" 
 [9,] "C"  "2" 
[10,] "C"  "3" 
[11,] "C"  "4" 
[12,] "D"  "1" 
[13,] "D"  "2"

Upvotes: 2

Thomas
Thomas

Reputation: 44535

You can get it as follows, using ave:

data <- data.frame(data)
new <- ave(rep(1,nrow(data)),data$data,FUN=cumsum)
all.equal(new,sample) # check if it's right.

Upvotes: 2

Frank P.
Frank P.

Reputation: 513

    factors <- unique(data)
    f1 <- length(which(data == factors[1]))
    ...
    fn <- length(which(data == factors[length(factors)]))

You can use a for loop or 'apply' family to speed that part up.

Then,

    sample <- c(1:f1, 1:f2, ..., 1:fn)

Once again you can use a for loop for that part. Here is the full script you can use:

    data<-matrix(c(rep('A',4),rep('B',3),rep('C',4),rep('D',2)),ncol=1)

    factors <- unique(data)
    f <- c()

    for(i in 1:length(factors)) {
      f[i] <- length(which(data == factors[i]))
    }

    sample <- c()

   for(i in 1:length(f)) {
      sample <- c(sample, 1:f[i])
    }

    > sample
     [1] 1 2 3 4 1 2 3 1 2 3 4 1 2

Upvotes: 0

user1981275
user1981275

Reputation: 13372

My answer:

sample <- unlist(lapply(levels(factor(data)), function(x)seq_len(sum(factor(data)==x))))

Upvotes: 0

alexwhan
alexwhan

Reputation: 16026

There are lots of different ways of achieving this, but I prefer to use ddply() from plyr because the logic seems very consistent to me. I think it makes more sense to be working with a data.frame (your title talks about levels of a factor):

dat <- data.frame(ID = c(rep('A',4),rep('B',3),rep('C',4),rep('D',2)))
library(plyr)
ddply(dat, .(ID), summarise, sample = 1:length(ID))
#    ID sample
# 1   A      1
# 2   A      2
# 3   A      3
# 4   A      4
# 5   B      1
# 6   B      2
# 7   B      3
# 8   C      1
# 9   C      2
# 10  C      3
# 11  C      4
# 12  D      1
# 13  D      2

Upvotes: 1

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