R help - Jul 2010 - information reduction-database management question

If you redefine your NAs as below to be detected as some arbitrary large
number, then the code should work through.  Any 5's left in your dataset can
be replaced just as easily by NAs again.  Not elegant, but effective.

site <- c("s1", "s1", "s1",
"s2","s2", "s2")
pref <- c(1, 2, 3, 1, 2, 3)
R1 <- c(NA, NA, 1, NA,NA,NA)
R2 <- c(NA, 0, 1, 1, NA, 1)
R3 <- c(NA, 1, 1, NA, 1, 1)
R4 <- c(0, NA, 0, 1, NA, 0)
R5 <- c(NA, 0, 1, NA, 1, 1)
datum <- data.frame(site, pref, R1, R2, R3, R4, R5)

## For 1 column;
datum$R1[is.na(datum$R1)==T]<-5
tapply(datum$R1, datum$site, min, na.rm=T)

## Can loop this over all columns;

new<-matrix(0,5,2)
for (i in 3:7)
{datum[,i][is.na(datum[,i])==T]<-5
new[i-2,]<-tapply(datum[,i], datum$site, min, na.rm=T)}

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I don't think the approach would change much with text.  You would have
to write a function which picks the 'min' or whatever that means to you
with text and then it should work ok,

 

Paul

 

From: Brad Patrick Schneid [via R]
[mailto:ml-node+2279677-1095983982-120784@n4.nabble.com] 
Sent: 06 July 2010 15:48
To: Paul Chatfield
Subject: Re: information reduction-database management question

 

Thanks Paul, 
I appreciate your time and this is an interesting approach.
Unfortunately, I need it to work for all types of information, including
character data (i.e. text).   
Again.. thanks for your suggestion! 
Brad 

Paul Chatfield wrote:

If you redefine your NAs as below to be detected as some arbitrary large
number, then the code should work through.  Any 5's left in your dataset
can be replaced just as easily by NAs again.  Not elegant, but
effective. 

site <- c("s1", "s1", "s1",
"s2","s2", "s2")
pref <- c(1, 2, 3, 1, 2, 3) 
R1 <- c(NA, NA, 1, NA,NA,NA) 
R2 <- c(NA, 0, 1, 1, NA, 1) 
R3 <- c(NA, 1, 1, NA, 1, 1) 
R4 <- c(0, NA, 0, 1, NA, 0) 
R5 <- c(NA, 0, 1, NA, 1, 1) 
datum <- data.frame(site, pref, R1, R2, R3, R4, R5) 

## For 1 column; 
datum$R1[is.na(datum$R1)==T]<-5 
tapply(datum$R1, datum$site, min, na.rm=T) 

## Can loop this over all columns; 

new<-matrix(0,5,2) 
for (i in 3:7) 
{datum[,i][is.na(datum[,i])==T]<-5 
new[i-2,]<-tapply(datum[,i], datum$site, min, na.rm=T)} 

 

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