r - how to calculate the median on grouped dataset?

Question

My dataset is as following:

salary  number
1500-1600   110
1600-1700   180
1700-1800   320
1800-1900   460
1900-2000   850
2000-2100   250
2100-2200   130
2200-2300   70
2300-2400   20
2400-2500   10

How can I calculate the median of this dataset? Here's what I have tried:

x <- c(110, 180, 320, 460, 850, 250, 130, 70, 20, 10)
colnames <- "numbers"
rownames <- c("[1500-1600]", "(1600-1700]", "(1700-1800]", "(1800-1900]", 
              "(1900-2000]", "(2000,2100]", "(2100-2200]", "(2200-2300]",
              "(2300-2400]", "(2400-2500]")
y <- matrix(x, nrow=length(x), dimnames=list(rownames, colnames))
data.frame(y, "cumsum"=cumsum(y))

            numbers cumsum
[1500-1600]     110    110
(1600-1700]     180    290
(1700-1800]     320    610
(1800-1900]     460   1070
(1900-2000]     850   1920
(2000,2100]     250   2170
(2100-2200]     130   2300
(2200-2300]      70   2370
(2300-2400]      20   2390
(2400-2500]      10   2400

Here, you can see the half-way frequency is 2400/2=1200. It is between 1070 and 1920. Thus the median class is the (1900-2000] group. You can use the formula below to get this result:

Median = L + h/f (n/2 - c)

where:

L is the lower class boundary of median class
h is the size of the median class i.e. difference between upper and lower class boundaries of median class
f is the frequency of median class
c is previous cumulative frequency of the median class
n/2 is total no. of observations divided by 2 (i.e. sum f / 2)

Alternatively, median class is defined by the following method:

Locate n/2 in the column of cumulative frequency.

Get the class in which this lies.

And in code:

> 1900 + (1200 - 1070) / (1920 - 1070) * (2000 - 1900)    
[1] 1915.294

Now what I want to do is to make the above expression more elegant - i.e. 1900+(1200-1070)/(1920-1070)*(2000-1900). How can I achieve this?

Answer 1

Since you already know the formula, it should be easy enough to create a function to do the calculation for you.

Here, I've created a basic function to get you started. The function takes four arguments:

• frequencies: A vector of frequencies ("number" in your first example)
• intervals: A 2-row matrix with the same number of columns as the length of frequencies, with the first row being the lower class boundary, and the second row being the upper class boundary. Alternatively, "intervals" may be a column in your data.frame, and you may specify sep (and possibly, trim) to have the function automatically create the required matrix for you.
• sep: The separator character in your "intervals" column in your data.frame.
• trim: A regular expression of characters that need to be removed before trying to coerce to a numeric matrix. One pattern is built into the function: trim = "cut". This sets the regular expression pattern to remove (, ), [, and ] from the input.

Here's the function (with comments showing how I used your instructions to put it together):

GroupedMedian <- function(frequencies, intervals, sep = NULL, trim = NULL) {
  # If "sep" is specified, the function will try to create the 
  #   required "intervals" matrix. "trim" removes any unwanted 
  #   characters before attempting to convert the ranges to numeric.
  if (!is.null(sep)) {
    if (is.null(trim)) pattern <- ""
    else if (trim == "cut") pattern <- "\[|\]|\(|\)"
    else pattern <- trim
    intervals <- sapply(strsplit(gsub(pattern, "", intervals), sep), as.numeric)
  }

  Midpoints <- rowMeans(intervals)
  cf <- cumsum(frequencies)
  Midrow <- findInterval(max(cf)/2, cf) + 1
  L <- intervals[1, Midrow]      # lower class boundary of median class
  h <- diff(intervals[, Midrow]) # size of median class
  f <- frequencies[Midrow]       # frequency of median class
  cf2 <- cf[Midrow - 1]          # cumulative frequency class before median class
  n_2 <- max(cf)/2               # total observations divided by 2

  unname(L + (n_2 - cf2)/f * h)
}

---

Here's a sample data.frame to work with:

mydf <- structure(list(salary = c("1500-1600", "1600-1700", "1700-1800", 
    "1800-1900", "1900-2000", "2000-2100", "2100-2200", "2200-2300", 
    "2300-2400", "2400-2500"), number = c(110L, 180L, 320L, 460L, 
    850L, 250L, 130L, 70L, 20L, 10L)), .Names = c("salary", "number"), 
    class = "data.frame", row.names = c(NA, -10L))
mydf
#       salary number
# 1  1500-1600    110
# 2  1600-1700    180
# 3  1700-1800    320
# 4  1800-1900    460
# 5  1900-2000    850
# 6  2000-2100    250
# 7  2100-2200    130
# 8  2200-2300     70
# 9  2300-2400     20
# 10 2400-2500     10

---

Now, we can simply do:

GroupedMedian(mydf$number, mydf$salary, sep = "-")
# [1] 1915.294

---

Here's an example of the function in action on some made up data:

set.seed(1)
x <- sample(100, 100, replace = TRUE)
y <- data.frame(table(cut(x, 10)))
y
#           Var1 Freq
# 1   (1.9,11.7]    8
# 2  (11.7,21.5]    8
# 3  (21.5,31.4]    8
# 4  (31.4,41.2]   15
# 5    (41.2,51]   13
# 6    (51,60.8]    5
# 7  (60.8,70.6]   11
# 8  (70.6,80.5]   15
# 9  (80.5,90.3]   11
# 10  (90.3,100]    6

### Here's GroupedMedian's output on the grouped data.frame...
GroupedMedian(y$Freq, y$Var1, sep = ",", trim = "cut")
# [1] 49.49231

### ... and the output of median on the original vector
median(x)
# [1] 49.5

---

By the way, with the sample data that you provided, where I think there was a mistake in one of your ranges (all were separated by dashes except one, which was separated by a comma), since strsplit uses a regular expression by default to split on, you can use the function like this:

x<-c(110,180,320,460,850,250,130,70,20,10)
colnames<-c("numbers")
rownames<-c("[1500-1600]","(1600-1700]","(1700-1800]","(1800-1900]",
            "(1900-2000]"," (2000,2100]","(2100-2200]","(2200-2300]",
            "(2300-2400]","(2400-2500]")
y<-matrix(x,nrow=length(x),dimnames=list(rownames,colnames))
GroupedMedian(y[, "numbers"], rownames(y), sep="-|,", trim="cut")
# [1] 1915.294