## Monday, August 31, 2009

### Example 7.11: Plot an empirical cumulative distribution function from scratch

In example 7.8, we used built-in functions to produce an empirical CDF plot. But the empirical cumulative distribution function (CDF) is simple to calculate directly, and it might be useful to have more control over its appearance than is aﬀorded by the direct method employed in example 7.8. In a later post, we'll use that control to improve upon example 7.8.

SAS

In SAS, we first read the data from a locally stored Stata data set. Then, we calculate the empirical CDF for pcs. We do this by ﬁrst sorting (1.5.6) on pcs. Then the order statistic of the observation divided by the total number of obervations is the empirical probability that X ≤ x. We ﬁnd the number of observations using the nobs data set option (A.6.1) and the observation number using the _n_ implied variable (1.4.15). Then we can plot the empirical CDF using proc gplot (section 5.1.1) to make a scatterplot and a symbol statement with the j interpolation (as in section 1.13.5) to connect the points.

`proc import datafile="C:\book\help.dta"   out=help dbms=dta;run;proc sort data=help;   by pcs;run;data help_a;set help nobs=totalobs;   ecdf_pcs = _n_ / totalobs;run;symbol1 i=j v=none c=blue;proc gplot data=help_a;   plot ecdf_pcs * pcs;run;quit;`

R
Now, R. First, make the ECDF values in a similar fashion to SAS, then make an empty frame (5.1.1), then ﬁll the frame with lines connecting the calculated values (5.2.1).

`library(foreign)ds <- read.dta  ("http://www.math.smith.edu/sasr/datasets/help.dta")attach(ds)sortpcs <- sort(pcs)ecdfpcs <- (1:length(sortpcs))/length(sortpcs)plot(sortpcs, ecdfpcs, type="n")lines(sortpcs, ecdfpcs)` Robert said...

Is there a simple way to replace every value in ds with a corresponding value of the empirical distribution?

i.e.
ds #returns# 2 2 3 1 10 2
sort(ds) #returns# 1 2 2 2 3 10
ecdf #returns# 1/6 4/6 4/6 4/6 5/6 1
we want ecdfv #returns# 4/6 4/6 5/6 1/6 1 4/6

Robert said...

ohh cool.... figured out.... look:

ecdfv <- c(length(ds))
for(i in 1:length(ds)){ecdfv[i] <- length(ds[ds<=ds[i]&!is.na(ds)])/length(ds[!is.na(ds)])}

:DD

Ken Kleinman said...

Pretty cool, Robert. I can imagine another solution where you'd attach an ID vector in the original order, compute the ECDF as we outline above, then re-sort on the ID vector to find the value of the ECDF in the original order (or save it in place, if that's the point). Here's my ugly version of that approach:

myvec <- c(2,2,3,1,10,2)
vecid <- 1:length(myvec)
mydata <- data.frame(cbind(myvec,vecid))
mydata <- mydata[order(myvec),]
mydata\$myvec <- (1:length(mydata\$myvec))/length(mydata\$myvec)
mydata <- mydata[order(mydata\$vecid),]

But looking this over, I see that it does not handle ties correctly.(Nor does our code in the example above.)

Can you explain how your code handles ties correctly? I can't see it.

Anonymous said...

Thanks!

duyan said...

Could we use the same method in case of a bivariate empirical cumulative density function ?
In SAS, Is there any one knows another than using the procedure KDE ? Thank you for your help .

duyan said...

Could we use the same method in case of a bivariate empirical cumulative density function ?
In SAS, Is there any one knows another than using the procedure KDE ? Thank you for your help .

Jon K. said...

What about the ecdf generated by stats::ecdf?

x<-ecdf(pcs)
plot(x,pch=NA)