how to ignore NA when using cumsum WHILE retaining NAs?
-----Original Message-----
From: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org] On Behalf Of Tracy Bowerman
Sent: Saturday, November 21, 2009 12:19 PM
To: r-help
Subject: [R] how to ignore NA when using cumsum WHILE retaining NAs?
I would like to cumulatively sum rows in a matrix, in which
each row has 1
NA value, which I do NOT want to treat as 0s. The NAs are
placeholders
where there is actually no data, which is not the same as a
0. The usual
"na.rm=TRUE" does not seem to work with the command cumsum. Is there
another way to ignore the NAs or do I need to figure out a
different way to
do this?
Here's an example matrix of title "proportion":
Ntrail Strail NFJD Baldy Onion Crane
[1,] NA 0.2944937 0.1779969 0.1808015 0.2296511 0.11705683
[2,] 0.2882713 NA 0.1795668 0.1823961 0.2316766 0.11808925
[3,] 0.1716890 0.1769419 NA 0.3518116 0.2025204 0.09703714
[4,] 0.1726311 0.1779128 0.3482548 NA 0.2036317 0.09756961
[5,] 0.2252904 0.2321833 0.2059743 0.2092197 NA 0.12733223
[6,] 0.2052614 0.2115415 0.1764081 0.1791877 0.2276013 NA
I want cumulative sums for each row but maintain NAs. I do
NOT want to
treat NAs as 0s, otherwise the cumulative sums add 0 and
returns a number
where a NA should be.
It would really help if you supplied the expected
output for your example. Does the following do
what you want?
f <- function(mat) {
# use t() if you want output to be organized in rows like input
retval <- t(apply(replace(mat,is.na(mat),0), 1, cumsum))
replace(retval, is.na(mat), NA)
}
A slightly different approach that does the same thing is
g <- function(mat) {
for(i in seq_len(nrow(mat))) {
isNotNA <- !is.na(mat[i,])
mat[i,isNotNA] <- cumsum(mat[i,isNotNA])
}
mat
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Here's the code that works until it reaches an NA, then returns all subsequent values as NA. Adding na.rm=TRUE returns the error: Error in FUN(newX[, i], ...) : 2 arguments passed to 'cumsum' which requires 1 cumsums <- apply(proportion, 1, cumsum) Thank you for any suggestions, Tracy [[alternative HTML version deleted]]
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