How to create a consecutive index based on a group

问题:

I have a data frame (all_data) in which I have a list of sites (1... to n) and their scores e.g.

  site  score
     1    10
     1    11  
     1    12
     4    10 
     4    11
     4    11
     8    9
     8    8
     8    7

I want create a column that numbers each level of site in numerical order, like a counter. In the example, the sites (1, 4, and 8) would be have a corresponding counter from 1 to 3 in the \'number\' column:

site  score number
     1    10    1
     1    11    1 
     1    12    1 
     4    10    2
     4    11    2
     4    11    2
     8    9     3
     8    8     3 
     8    7     3

I am sure this must be easily solved, but I have not found a way yet.

回答1:

Try Data$number <- as.numeric(as.factor(Data$site))

On a sidenote : the difference between the solution of me and @Chase on one hand, and the one of @DWin on the other, is the ordering of the numbers. Both as.factor and factor will automatically sort the levels, whereas that doesn\'t happen in the solution of @DWin :

Dat <- data.frame(site = rep(c(1,8,4), each = 3), score = runif(9))

Dat$number <- as.numeric(factor(Dat$site))
Dat$sitenum <- match(Dat$site, unique(Dat$site) ) 

Gives

> Dat
  site     score number sitenum
1    1 0.7377561      1       1
2    1 0.3131139      1       1
3    1 0.7862290      1       1
4    8 0.4480387      3       2
5    8 0.3873210      3       2
6    8 0.8778102      3       2
7    4 0.6916340      2       3
8    4 0.3033787      2       3
9    4 0.6552808      2       3

回答2:

Two other options:

1) Using the .GRP function from the data.table package:

library(data.table)
setDT(dat)[, num := .GRP, by = site]

with the example dataset from below this results in:

> dat
    site      score num
 1:    1 0.14945795   1
 2:    1 0.60035697   1
 3:    1 0.94643075   1
 4:    8 0.68835336   2
 5:    8 0.50553372   2
 6:    8 0.37293624   2
 7:    4 0.33580504   3
 8:    4 0.04825135   3
 9:    4 0.61894754   3
10:    8 0.96144729   2
11:    8 0.65496051   2
12:    8 0.51029199   2

2) Using the group_indices function from dplyr:

dat$num <- group_indices(dat, site)

or when you want to work around non-standard evaluation:

library(dplyr)
dat %>% 
  mutate(num = group_indices_(dat, .dots = c(\'site\')))

which results in:

   site      score num
1     1 0.42480366   1
2     1 0.98736177   1
3     1 0.35766187   1
4     8 0.06243182   3
5     8 0.55617002   3
6     8 0.20304632   3
7     4 0.90855921   2
8     4 0.25215078   2
9     4 0.44981251   2
10    8 0.60288270   3
11    8 0.46946587   3
12    8 0.44941782   3

As can be seen, dplyr gives a different order of the group numbers.

---

If you want another number every time the group changes, there are several other options:

1) with base R:

# option 1:
dat$num <- cumsum(c(TRUE, head(dat$site, -1) != tail(dat$site, -1)))

# option 2:
x <- rle(dat$site)$lengths
dat$num <- rep(seq_along(x), times=x)

2) with the data.table package:

library(data.table)
setDT(dat)[, num := rleid(site)]

which all result in:

> dat
   site      score num
1     1 0.80817855   1
2     1 0.07881334   1
3     1 0.60092828   1
4     8 0.71477988   2
5     8 0.51384565   2
6     8 0.72011650   2
7     4 0.74994627   3
8     4 0.09564052   3
9     4 0.39782587   3
10    8 0.29446540   4
11    8 0.61725367   4
12    8 0.97427413   4

---

Used data:

dat <- data.frame(site = rep(c(1,8,4,8), each = 3), score = runif(12))

回答3:

This should be fairly efficient and understandable:

Dat$sitenum <- match(Dat$site, unique(Dat$site))  

回答4:

You can turn site into a factor and then return the numeric or integer values of that factor:

dat <- data.frame(site = rep(c(1,4,8), each = 3), score = runif(9))
dat$number <- as.integer(factor(dat$site))
dat

  site     score number
1    1 0.5305773      1
2    1 0.9367732      1
3    1 0.1831554      1
4    4 0.4068128      2
5    4 0.3438962      2
6    4 0.8123883      2
7    8 0.9122846      3
8    8 0.2949260      3
9    8 0.6771526      3

回答5:

Another solution using the data.table package.

Example with the more complete datset provided by Jaap:

setDT(dat)[, number := frank(site, ties.method = \"dense\")]
dat
    site     score number
 1:    1 0.3107920      1
 2:    1 0.3640102      1
 3:    1 0.1715318      1
 4:    8 0.7247535      3
 5:    8 0.1263025      3
 6:    8 0.4657868      3
 7:    4 0.6915818      2
 8:    4 0.3558270      2
 9:    4 0.3376173      2
10:    8 0.7934963      3
11:    8 0.9641918      3
12:    8 0.9832120      3