[R] APPLY as alternate to FOR loop?
Bill.Venables at csiro.au
Bill.Venables at csiro.au
Mon Mar 31 07:02:25 CEST 2008
As far as I know there is no function called 'APPLY'
There is one called 'apply', but why are you determined to use it here?
It is essentially concealed looping.
Here is an alternative way of solving your problem:
___
x <- rbind(x1 = c(1, NA, NA, NA, NA, 1, 2, 2),
x2 = c(1, NA, NA, NA, NA, 1, 2, 1),
x3 = c(1, NA, 1, 1, 1, 1, 1, 5),
x4 = c(1, NA, NA, NA, NA, NA, NA, 5))
badRows <- which(rowSums(is.na(x)) > 3)
is.na(x[as.matrix(expand.grid(badRows, 1:ncol(x)))]) <- TRUE
___
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
x1 NA NA NA NA NA NA NA NA
x2 NA NA NA NA NA NA NA NA
x3 1 NA 1 1 1 1 1 5
x4 NA NA NA NA NA NA NA NA
>
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of zack holden
Sent: Monday, 31 March 2008 2:32 PM
To: r-help at r-project.org
Subject: [R] APPLY as alternate to FOR loop?
Dear list,
Below I've written a clunky for loop that counts NA's in a row,
replacing all with NA if there are
more than 3 missing values, or keeping the values if 4 or more are
present. This is sample code from a very large
dataframe I'm trying to clean up.
I know there are many simpler more elegant solutions to this little
problem.
Would someone be willing to show me how to create a function that I can
APPLY to
each row rather than looping? I've tried and can't get it.
Thank you,
Zack
################################################# Count NA's in each
row. IF > 3 NA's in a row, make all
NA###########################################################
## test dataframe #######x1 <- c(1,NA,NA,NA,NA,1,2,2)x2 <-
c(1,NA,NA,NA,NA,1,2,1)x3 <- c(1,NA,1,1,1,1,1,5)x4 <-
c(1,NA,NA,NA,NA,NA,NA,5)x <- rbind(x1,x2,x3,x4)test <- rowSums(is.na(x))
## count numer of NA's in rowx <- cbind(x, test) ## add
row NA count to datax <- data.frame(x) ## make
dataframe
# FOR LOOP to apply across all rows of dataframe -- for(i in 1:nrow(x))
{if(x[i,9] > 4) {x[i,1:7] <- NA} else { x[i,1:7] <- x[i,1:7]i = i+1}}
#####################################################
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