6 messages · Dimitri Liakhovitski, Weidong Gu, Bert Gunter +1 more
Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer!
Dimitri Liakhovitski marketfusionanalytics.com
You can do it by subsetting or indexing r<-c(0,0,0,rnorm(10,10,5))
mean(r)
[1] 8.052215
mean(r[r!=0])
[1] 10.46788 Weidong Gu On Tue, Jul 19, 2011 at 4:36 PM, Dimitri Liakhovitski
Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer! -- Dimitri Liakhovitski marketfusionanalytics.com
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
In the more general case, that approach is prone to machine precision error (FAQ 7.31). Here's a clunky but safer alternative:
set.seed(1234) testvec <- sample(0:10, 100, replace=TRUE) mean(testvec)
[1] 4.31
mean(testvec[testvec != 0])
[1] 4.842697
mean(testvec[!sapply(testvec, function(x)isTRUE(all.equal(x, 0)))])
[1] 4.842697
(Is there an elementwise equivalent to all.equal() that I'm missing?) Sarah
You can do it by subsetting or indexing ?r<-c(0,0,0,rnorm(10,10,5))
mean(r)
[1] 8.052215
mean(r[r!=0])
[1] 10.46788 Weidong Gu On Tue, Jul 19, 2011 at 4:36 PM, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote:
Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer! -- Dimitri Liakhovitski marketfusionanalytics.com
Sarah Goslee http://www.functionaldiversity.org
Thanks a lot, Sarah.
I assume, if the values against which I am comparing are REALLY zero
("0") - then even the first one (mean(testvec[testvec != 0])) should
work, right?
Dimitri
In the more general case, that approach is prone to machine precision error (FAQ 7.31). Here's a clunky but safer alternative:
set.seed(1234) testvec <- sample(0:10, 100, replace=TRUE) mean(testvec)
[1] 4.31
mean(testvec[testvec != 0])
[1] 4.842697
mean(testvec[!sapply(testvec, function(x)isTRUE(all.equal(x, 0)))])
[1] 4.842697
(Is there an elementwise equivalent to all.equal() that I'm missing?) Sarah On Tue, Jul 19, 2011 at 4:48 PM, Weidong Gu <anopheles123 at gmail.com> wrote:
You can do it by subsetting or indexing ?r<-c(0,0,0,rnorm(10,10,5))
mean(r)
[1] 8.052215
mean(r[r!=0])
[1] 10.46788 Weidong Gu On Tue, Jul 19, 2011 at 4:36 PM, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote:
Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer! -- Dimitri Liakhovitski marketfusionanalytics.com
-- Sarah Goslee http://www.functionaldiversity.org
Dimitri Liakhovitski marketfusionanalytics.com
Sarah et. al:
In the more general case, that approach is prone to machine precision error (FAQ 7.31). Here's a clunky but safer alternative:
Perhaps ?zapsmall . However, I would agree with your sentiments that it may depend on context. Finite precision can be a tricky thing. -- Bert
set.seed(1234) testvec <- sample(0:10, 100, replace=TRUE) mean(testvec)
[1] 4.31
mean(testvec[testvec != 0])
[1] 4.842697
mean(testvec[!sapply(testvec, function(x)isTRUE(all.equal(x, 0)))])
[1] 4.842697
(Is there an elementwise equivalent to all.equal() that I'm missing?) Sarah On Tue, Jul 19, 2011 at 4:48 PM, Weidong Gu <anopheles123 at gmail.com> wrote:
You can do it by subsetting or indexing ?r<-c(0,0,0,rnorm(10,10,5))
mean(r)
[1] 8.052215
mean(r[r!=0])
[1] 10.46788 Weidong Gu On Tue, Jul 19, 2011 at 4:36 PM, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote:
Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer! -- Dimitri Liakhovitski marketfusionanalytics.com
-- Sarah Goslee http://www.functionaldiversity.org
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
"Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions." -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics
On Tue, Jul 19, 2011 at 5:20 PM, Dimitri Liakhovitski
Thanks a lot, Sarah.
I assume, if the values against which I am comparing are REALLY zero
("0") - then even the first one (mean(testvec[testvec != 0])) should
work, right?
Dimitri
Well, yes. But what's "really" zero?
((.2 + .1) - .3) == 0
[1] FALSE
all.equal(((.2 + .1) - .3), 0)
[1] TRUE "0" is a string, and string comparison is a different issue, not subject to machine precision. Sarah
On Tue, Jul 19, 2011 at 4:56 PM, Sarah Goslee <sarah.goslee at gmail.com> wrote:
In the more general case, that approach is prone to machine precision error (FAQ 7.31). Here's a clunky but safer alternative:
set.seed(1234) testvec <- sample(0:10, 100, replace=TRUE) mean(testvec)
[1] 4.31
mean(testvec[testvec != 0])
[1] 4.842697
mean(testvec[!sapply(testvec, function(x)isTRUE(all.equal(x, 0)))])
[1] 4.842697
(Is there an elementwise equivalent to all.equal() that I'm missing?) Sarah On Tue, Jul 19, 2011 at 4:48 PM, Weidong Gu <anopheles123 at gmail.com> wrote:
You can do it by subsetting or indexing ?r<-c(0,0,0,rnorm(10,10,5))
mean(r)
[1] 8.052215
mean(r[r!=0])
[1] 10.46788 Weidong Gu On Tue, Jul 19, 2011 at 4:36 PM, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote:
Sorry if it's been discussed before - don't seem to find it. I'd like to calculate a mean while ignoring zeros. "mean" doesn't seem to have an option for that. Any other function/package that could do it? Thanks for a pointer!
Sarah Goslee http://www.functionaldiversity.org