问题
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?
回答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-rowmatrixwith 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 yourdata.frame, and you may specifysep(and possibly,trim) to have the function automatically create the required matrix for you.sep: The separator character in your "intervals" column in yourdata.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
回答2:
I've written it like this to clearly explain how it's being worked out. A more compact version is appended.
library(data.table)
#constructing the dataset with the salary range split into low and high
salarydata <- data.table(
salaries_low = 100*c(15:24),
salaries_high = 100*c(16:25),
numbers = c(110,180,320,460,850,250,130,70,20,10)
)
#calculating cumulative number of observations
salarydata <- salarydata[,cumnumbers := cumsum(numbers)]
salarydata
# salaries_low salaries_high numbers cumnumbers
# 1: 1500 1600 110 110
# 2: 1600 1700 180 290
# 3: 1700 1800 320 610
# 4: 1800 1900 460 1070
# 5: 1900 2000 850 1920
# 6: 2000 2100 250 2170
# 7: 2100 2200 130 2300
# 8: 2200 2300 70 2370
# 9: 2300 2400 20 2390
# 10: 2400 2500 10 2400
#identifying median group
mediangroup <- salarydata[
(cumnumbers - numbers) <= (max(cumnumbers)/2) &
cumnumbers >= (max(cumnumbers)/2)]
mediangroup
# salaries_low salaries_high numbers cumnumbers
# 1: 1900 2000 850 1920
#creating the variables needed to calculate median
mediangroup[,l := salaries_low]
mediangroup[,h := salaries_high - salaries_low]
mediangroup[,f := numbers]
mediangroup[,c := cumnumbers- numbers]
n = salarydata[,sum(numbers)]
#calculating median
median <- mediangroup[,l + ((h/f)*((n/2)-c))]
median
# [1] 1915.294
The compact version -
EDIT: Changed to a function at @AnandaMahto's suggestion. Also, using more general variable names.
library(data.table)
#Creating function
CalculateMedian <- function(
LowerBound,
UpperBound,
Obs
)
{
#calculating cumulative number of observations and n
dataset <- data.table(UpperBound, LowerBound, Obs)
dataset <- dataset[,cumObs := cumsum(Obs)]
n = dataset[,max(cumObs)]
#identifying mediangroup and dynamically calculating l,h,f,c. We already have n.
median <- dataset[
(cumObs - Obs) <= (max(cumObs)/2) &
cumObs >= (max(cumObs)/2),
LowerBound + ((UpperBound - LowerBound)/Obs) * ((n/2) - (cumObs- Obs))
]
return(median)
}
# Using function
CalculateMedian(
LowerBound = 100*c(15:24),
UpperBound = 100*c(16:25),
Obs = c(110,180,320,460,850,250,130,70,20,10)
)
# [1] 1915.294
回答3:
(Sal <- sapply( strsplit(as.character(dat[[1]]), "-"),
function(x) mean( as.numeric(x) ) ) )
[1] 1550 1650 1750 1850 1950 2050 2150 2250 2350 2450
require(Hmisc)
wtd.mean(Sal, weights = dat[[2]])
[1] 1898.75
wtd.quantile(Sal, weights=dat[[2]], probs=0.5)
Generalization to a weighed median might require looking for a package that has such.
回答4:
Have you tried median or apply(yourobject,2,median) if it is a matrix or data.frame ?
回答5:
What about this way? Create vectors for each salary bracket, assuming an even spread over each band. Then make one big vector from those vectors, and take the median. Similar to you, but a slightly different result. I'm not a mathematician, so the method could be incorrect.
dat <- matrix(c(seq(1500, 2400, 100), seq(1600, 2500, 100), c(110, 180, 320, 460, 850, 250, 130, 70, 20, 10)), ncol=3)
median(unlist(apply(dat, 1, function(x) { ((1:x[3])/x[3])*(x[2]-x[1])+x[1] })))
Returns 1915.353
回答6:
I think this concept should work you.
$salaries = array(
array("1500","1600"),
array("1600","1700"),
array("1700","1800"),
array("1800","1900"),
array("1900","2000"),
array("2000","2100"),
array("2100","2200"),
array("2200","2300"),
array("2300","2400"),
array("2400","2500"),
);
$numbers = array("110","180","320","460","850","250","130","70","20","10");
$cumsum = array();
$n = 0;
$count = 0;
foreach($numbers as $key=>$number){
$cumsum[$key] = $number;
$n += $number;
if($count > 0){
$cumsum[$key] += $cumsum[$key-1];
}
++$count;
}
$classIndex = 0;
foreach($cumsum as $key=>$cum){
if($cum < ($n/2)){
$classIndex = $key+1;
}
}
$classRange = $salaries[$classIndex];
$L = $classRange[0];
$h = (float) $classRange[1] - $classRange[0];
$f = $numbers[$classIndex];
$c = $numbers[$classIndex-1];
$Median = $L + ($h/$f)*(($n/2)-$c);
echo $Median;
来源：https://stackoverflow.com/questions/18887382/how-to-calculate-the-median-on-grouped-dataset