# Re: [R] APPLY as alternate to FOR loop?

From: <Bill.Venables_at_csiro.au>
Date: Mon, 31 Mar 2008 15:02:25 +1000

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))

```

> 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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