r - 如何计算分组数据集的中位数？

Question

我的数据集如下:

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

如何计算此数据集的中位数？这是我尝试过的:

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

在这里，您可以看到中途频率为 2400/2 = 1200。它在 1070和 1920之间。因此， 中类是 (1900-2000]组。您可以使用以下公式获得此结果:

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

在哪里:

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)

或者，通过以下方法定义 中位数类:

Locate n/2 in the column of cumulative frequency.

Get the class in which this lies.

并在代码中:

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

现在我要做的是使上面的表达式更优雅-即 1900+(1200-1070)/(1920-1070)*(2000-1900)。我怎样才能做到这一点？

Answer 1

由于您已经知道公式，因此创建一个函数为您进行计算应该足够容易。

在这里，我创建了一个基本功能来帮助您入门。该函数有四个参数:

frequencies:频率向量(第一个示例中为“数字”)

intervals:2行matrix，其列数与频率的长度相同，第一行是下层边界，第二行是上层边界。另外，“intervals”可以是data.frame中的一列，并且您可以指定sep(可能还需要trim)以使该函数自动为您创建所需的矩阵。

sep:intervals中“data.frame”列中的分隔符。

trim:字符的正则表达式，在尝试强制转换为数字矩阵之前需要将其删除。函数中内置了一种模式:trim = "cut"。这将设置正则表达式模式以从输入中删除(，)，[和]。

这是功能(带有注释，显示了我如何使用您的说明将其组合在一起):

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)
}

以下是可使用的示例 data.frame:

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

现在，我们可以简单地执行以下操作:

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

这是一些作用于某些组合数据的函数的示例:

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

顺便说一句，在您提供的示例数据中，我认为您的一个范围内有一个错误(除破折号外，所有破折号均用破折号隔开，其中一个用逗号隔开)，因为 strsplit默认使用正则表达式来继续，您可以使用如下功能:

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