如何通过R dataframe中每个id的百分位数对有序数据进行二进制[r]

Question

I have dataframe that contains 70-80 rows of ordered response time (rt) data for each of 228 people each with a unique id# (everyone doesn't have the same amount of rows). I want to bin each person's RTs into 5 bins. I want the 1st bin to be their fastest 20 percent of RTs, 2nd bin to be their next fastest 20 percent RTs, etc., etc. Each bin should have the same amount of trials in it (unless the total # of trial is odd).

My current dataframe looks like this:

id     RT
7000   225
7000   250
7000   253
7001   189
7001   201
7001   225

I'd like my new dataframe to look like this:

id   RT    Bin
7000  225    1
7000  250    1

After getting my data to look like this, I will aggregate by id and bin

The only way I can think of to do this is to split the data into a list (using the split command), loop through each person, use the quantile command to get break points for the different bins, assign a bin value (1-5) to every response time. This feels very convoluted (and would be difficult for me). I'm in a bit of a jam and I would greatly appreciate any help in how to streamline this process. Thanks.

Answer 1

The answer @Chase gave split the range into 5 groups of equal length (difference of endpoints). What you seem to want is pentiles (5 groups with equal number in each group). For that, you need the cut2 function in Hmisc

library("plyr")
library("Hmisc")

dat <- data.frame(id = rep(1:10, each = 10), value = rnorm(100))

tmp <- ddply(dat, "id", transform, hists = as.numeric(cut2(value, g = 5)))

tmp now has what you want

> tmp
    id       value hists
1    1  0.19016791     3
2    1  0.27795226     4
3    1  0.74350982     5
4    1  0.43459571     4
5    1 -2.72263322     1
....
95  10 -0.10111905     3
96  10 -0.28251991     2
97  10 -0.19308950     2
98  10  0.32827137     4
99  10 -0.01993215     4
100 10 -1.04100991     1

With the same number in each hists for each id

> table(tmp$id, tmp$hists)

     1 2 3 4 5
  1  2 2 2 2 2
  2  2 2 2 2 2
  3  2 2 2 2 2
  4  2 2 2 2 2
  5  2 2 2 2 2
  6  2 2 2 2 2
  7  2 2 2 2 2
  8  2 2 2 2 2
  9  2 2 2 2 2
  10 2 2 2 2 2

Answer 2

Here's a reproducible example using package plyr and the cut function:

dat <- data.frame(id = rep(1:10, each = 10), value = rnorm(100))

ddply(dat, "id", transform, hists = cut(value, breaks = 5))

    id       value             hists
1    1 -1.82080027     (-1.94,-1.41]
2    1  0.11035796     (-0.36,0.166]
3    1 -0.57487134    (-0.886,-0.36]
4    1 -0.99455189    (-1.41,-0.886]
....
96  10 -0.03376074    (-0.233,0.386]
97  10 -0.71879488   (-0.853,-0.233]
98  10 -0.17533570    (-0.233,0.386]
99  10 -1.07668282    (-1.47,-0.853]
100 10 -1.45170078    (-1.47,-0.853]

Pass in labels = FALSE to cut if you want simple integer values returned instead of the bins.

Answer 3

Here's an answer in plain old R.

#make up some data
df <- data.frame(rt = rnorm(60), id = rep(letters[1:3], rep(20)) )

#and this is all there is to it
df <- df[order(df$id, df$rt),]
df$bin <- rep( unlist( tapply( df$rt, df$id, quantile )), each = 4)

You'll note that quantile command used can be set to use any quantiles. The defaults are for quintiles but if you want deciles then use

quantile(x, seq(0, 1, 0.1))

in the function above.

The answer above is a bit fragile. It requires equal numbers of RTs/id and I didn't tell you how to get to the magic number 4. But, it also will run very fast on a large dataset. If you want a more robust solution in base R.

library('Hmisc')
df <- df[order(df$id),]
df$bin <- unlist(lapply( unique(df$id), function(x) cut2(df$rt[df$id==x], g = 5) ))

This is much more robust than the first solution but it isn't as fast. For small datasets you won't notice.