I am relatively new to R. I am iterating over a vector in R by using for() loop. However, based on a certain condition, I need to skip some values in the vector. The first thought that comes to mind is to change the loop index within the loop. I have tried that but somehow its not changing it. There must be some what to achieve this in R.
Thanks in advance.
Sami
You can change the loop index within a for loop, but it will not affect the execution of the loop; see the Details section of ?"for":
The ‘seq’ in a ‘for’ loop is evaluated at the start of the loop;
changing it subsequently does not affect the loop. If ‘seq’ has
length zero the body of the loop is skipped. Otherwise the
variable ‘var’ is assigned in turn the value of each element of
‘seq’. You can assign to ‘var’ within the body of the loop, but
this will not affect the next iteration. When the loop terminates,
‘var’ remains as a variable containing its latest value.
Use a while loop instead and index it manually:
i <- 1
while(i < 100) {
# do stuff
if(condition) {
i <- i+3
} else {
i <- i+1
}
}
Look at
?"next"
The next command will skip the rest of the current iteration of the loop and begin the next one. That may accomplish what you want.
Without an example it is hard to see what you want to do, but you can always use an if-statement inside a for-loop:
foo <- 1:10*5
for (i in seq(length(foo)))
{
if (foo[i] != 15) print(foo[i])
}
In R, local alterations in the index variable are "corrected" with the next pass:
for (i in 1:10){
if ( i==5 ) {i<-10000; print(i)} else{print(i)}
}
#-----
[1] 1
[1] 2
[1] 3
[1] 4
[1] 10000
[1] 6
[1] 7
[1] 8
[1] 9
[1] 10
Since you have some criterion for skipping you should apply the criterion to the loop vector inside the for-parentheses. E.g:
for( i in (1:10)[-c(3,4,6,8,9)] ) {
print(i)}
#----
[1] 1
[1] 2
[1] 5
[1] 7
[1] 10
Related
I am relatively new to R and am trying to create a for loop with a conditional that references the previous row for equivalence. In order to learn to write this code for my own data, I created a simpler, representative data frame:
df<- c(1,1,2,0,0,0,0,1,1,2)
My goal would be to print a 1 for every value in df that is different from the previous value. For this df, this should look like:
[1] 0,0,1,1,0,0,0,1,0,1.
Here is what I have tried thus far:
for(i in df){
if(x[i] != x[i-1]){
print(1)
}else{
print(0)
}
}
From the above code, I consistently get the error "argument is of length zero". Very possible that I am making a simple mistake, but I appreciate any suggestions!
Maybe try this. You are using x which does not exist. Instead use df inside the loop. Also as you need comparison from one element to next, it would be better to start the loop from second position because the first element does not have any previous value to be compared. Here the code:
#Data
df<- c(1,1,2,0,0,0,0,1,1,2)
#Loop
for(i in 2:length(df)){
if(df[i] != df[i-1]){
print(1)
}else{
print(0)
}
}
Output:
[1] 0
[1] 1
[1] 1
[1] 0
[1] 0
[1] 0
[1] 1
[1] 0
[1] 1
I am trying to understand the for and if-statement in r, so I run a code where I am saying that if the sum of rows are bigger than 3 then return 1 else zero:
Here is the code
set.seed(2)
x = rnorm(20)
y = 2*x
a = cbind(x,y)
hold = c()
Now comes the if-statement
for (i in nrow(a)) {
if ([i,1]+ [i,2] > 3) hold[i,] == 1
else ([i,1]+ [i,2]) <- hold[i,] == 0
return (cbind(a,hold)
}
I know that maybe combining for and if may not be ideal, but I just want to understand what is going wrong. Please keep the explanation at a dummy level:) Thanks
You've got some issues. #mnel covered a better way to go about doing this, I'll focus on understanding what went wrong in this attempt (but don't do it this way at all, use a vectorized solution).
Line 1
for (i in nrow(a)) {
a has 20 rows. nrow(a) is 20. Thus your code is equivalent to for (i in 20), which means i will only ever be 20.
Fix:
for (i in 1:nrow(a)) {
Line 2
if ([i,1]+ [i,2] > 3) hold[i,] == 1
[i,1] isn't anything, it's the ith row and first column of... nothing. You need to reference your data: a[i,1]
You initialized hold as a vector, c(), so it only has one dimension, not rows and columns. So we want to assign to hold[i], not hold[i,].
== is used for equality testing. = or <- are for assignment. Right now, if the >3 condition is met, then you check if hold[i,] is equal to 1. (And do nothing with the result).
Fix:
if (a[i,1]+ a[i,2] > 3) hold[i] <- 1
Line 3
else ([i,1]+ [i,2]) <- hold[i,] == 0
As above for assignment vs equality testing. (Here you used an arrow assignment, but put it in the wrong place - as if you're trying to assign to the else)
else happens whenever the if condition isn't met, you don't need to try to repeat the condition
Fix:
else hold[i] <- 0
Fixed code together:
for (i in 1:nrow(a)) {
if (a[i,1] + a[i,2] > 3) hold[i] <- 1
else hold[i] <- 0
}
You aren't using curly braces for your if and else expressions. They are not required for single-line expressions (if something do this one line). They are are required for multi-line (if something do a bunch of stuff), but I think they're a good idea to use. Also, in R, it's good practice to put the else on the same line as a } from the preceding if (inside the for loop or a function it doesn't matter, but otherwise it would, so it's good to get in the habit of always doing it). I would recommend this reformatted code:
for (i in 1:nrow(a)) {
if (a[i, 1] + a[i, 2] > 3) {
hold[i] <- 1
} else {
hold[i] <- 0
}
}
Using ifelse
ifelse() is a vectorized if-else statement in R. It is appropriate when you want to test a vector of conditions and get a result out for each one. In this case you could use it like this:
hold <- ifelse(a[, 1] + a[, 2] > 3, 1, 0)
ifelse will take care of the looping for you. If you want it as a column in your data, assign it directly (no need to initialize first)
a$hold <- ifelse(a[, 1] + a[, 2] > 3, 1, 0)
Such operations in R are nicely vectorised.
You haven't included a reference to the dataset you wish to index with your call to [ (eg a[i,1])
using rowSums
h <- rowSums(a) > 3
I am going to assume that you are new to R and trying to learn about the basic function of the for loop itself. R has fancy functions called "apply" functions that are specifically for doing basic math on each row of a data frame. I am not going to talk about these.
You want to do the following on each row of the array.
Sum the elements of the row.
Test that the sum is greater than 3.
Return a value of 1 or 0 representing the result of 2.
For 1, luckily "sum" is a built in function. It pays off to check out the built in functions within every programming language because they save you time. To sum the elements of a row, just use sum(a[row_number,]).
For 2, you are evaluating a logical statement "is x >3?" where x is the result from 1. The ">3" statement returns a value of true or false. The logical expression is a fancy "if then" statement without the "if then".
> 4>3
[1] TRUE
> 2>3
[1] FALSE
For 3, a true or false value is a data structure called a "logical" value in R. A 1 or 0 value is a data structure called a "numeric" value in R. By converting the "logical" into a "numeric", you can change the TRUE to 1's and FALSE to 0's.
> class(4>3)
[1] "logical"
> as.numeric(4>3)
[1] 1
> class(as.numeric(4>3))
[1] "numeric"
A for loop has a min, a max, a counter, and an executable. The counter starts at the min, and increments until it goes to the max. The executable will run for each run of the counter. You are starting at the first row and going to the last row. Putting all the elements together looks like this.
for (i in 1:nrow(a)){
hold[i] <- as.numeric(sum(a[i,])>3)
}
So I'm currently trying to get a random initial pathway between nodes. I've tried the following code, but at times it 'skips' a node i,e sometimes the same node is visited twice rather than it traversing each one. But since I've defined a visited node's 'column' as all 0 I don't see why this should happen when using the which(>0) command. Any advice?
A<-matrix(sample(1:15,25,replace=TRUE), ncol=5)
n=nrow(A)
b=c()
a=c(1:nrow(A))
b[1]=sample(a,1)
for(i in 2:n){
A[,b[i-1]]<-rep(0,n)
d=which(A[b[i-1],]>0)
b[i]=sample(d,1)
}
print(b)
The problem is that sample behaves differently when you pass it a vector of length 1. Observe
set.seed(14)
x<-c(5,3)
sample(x, 1)
# [1] 5
x<-5
sample(x, 1)
# [1] 4
you see that sample returned 4. When you pass in a vector of length one, it draws from 1:x. You can write your own wrapper if you like
Sample<-function(x,n ) {
if(length(x)>1)
sample(x,n)
else if (length(x)==1 & n==1) {
x
} else {
stop("error")
}
}
and then use this function instead.
But it seems like you are just shuffling your rows. Why not just permute the index with one call to sample:
sample(seq_len(nrow(A))
I have a command that generates a variable every 10 loops in R (index1, index2, index3... and so on). The command I have is functional, but I am thinking of a smarter way to write this command. Here's what my command looks like:
for (counter in 1:10){
for (i in 1:100){
if (counter == 1){
index1 <- data1 ## some really long command here, I just changed it to this simple command to illustrate the idea
}
if (counter == 2){
index2 <- data2
}
.
.
.
# until I reach index10
} indexing closure
} ## counter closure
Is there a way to write this without having to write the conditional if commands? I would like to generate index1, index2.... I am sure there is some easy way to do this but I just cannot think of it.
Thanks.
What you need is the modulo operator %%. inside the inner loop. Ex: 100%%10 returns 0 101%%10 returns 1 92%%10 returns 2 - in other words if it is multiple of 10 then you get 0. And the assign function.
Note: You no longer need the outer loop used in your example.
So to create a variable at every 10 iteration do something like this
for(i in 1:100){
#check if i is multiple of 10
if(i%%10==0){
myVar<-log(i)
assign(paste("index",i/10,sep=""), myVar)
}
}
ls() #shows that index1, index2, ...index10 objects have been created.
index1 #returns 2.302585
update:
Alternatively, you can store results in a vector
index<-vector(length=10)
for(i in 1:100){
#check if i is multiple of 10
if(i%%10==0){
index[i/10]<-log(i)
}
}
index #returns a vector with 10 elements, each a result at end of an iteration that is a multiple of 10.
I was wondering about how to write do-while-style loop?
I found this post:
you can use repeat{} and check conditions whereever using if() and
exit the loop with the "break" control word.
I am not sure what it exactly means. Can someone please elaborate if you understand it and/or if you have a different solution?
Pretty self explanatory.
repeat{
statements...
if(condition){
break
}
}
Or something like that I would think. To get the effect of the do while loop, simply check for your condition at the end of the group of statements.
See ?Control or the R Language Definition:
> y=0
> while(y <5){ print( y<-y+1) }
[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
So do_while does not exist as a separate construct in R, but you can fake it with:
repeat( { expressions}; if (! end_cond_expr ) {break} )
If you want to see the help page you cannot type ?while or ?repeat at the console but rather need to use ?'repeat' or ?'while'. All the "control-constructs" including if are on the same page and all need character quoting after the "?" so the interpreter doesn't see them as incomplete code and give you a continuation "+".
Building on the other answers, I wanted to share an example of using the while loop construct to achieve a do-while behaviour. By using a simple boolean variable in the while condition (initialized to TRUE), and then checking our actual condition later in the if statement. One could also use a break keyword instead of the continue <- FALSE inside the if statement (probably more efficient).
df <- data.frame(X=c(), R=c())
x <- x0
continue <- TRUE
while(continue)
{
xi <- (11 * x) %% 16
df <- rbind(df, data.frame(X=x, R=xi))
x <- xi
if(xi == x0)
{
continue <- FALSE
}
}
Noticing that user 42-'s perfect approach {
* "do while" = "repeat until not"
* The code equivalence:
do while (condition) # in other language
..statements..
endo
repeat{ # in R
..statements..
if(! condition){ break } # Negation is crucial here!
}
} did not receive enough attention from the others, I'll emphasize and bring forward his approach via a concrete example. If one does not negate the condition in do-while (via ! or by taking negation), then distorted situations (1. value persistence 2. infinite loop) exist depending on the course of the code.
In Gauss:
proc(0)=printvalues(y);
DO WHILE y < 5;
y+1;
y=y+1;
ENDO;
ENDP;
printvalues(0); # run selected code via F4 to get the following #
1.0000000
2.0000000
3.0000000
4.0000000
5.0000000
In R:
printvalues <- function(y) {
repeat {
y=y+1;
print(y)
if (! (y < 5) ) {break} # Negation is crucial here!
}
}
printvalues(0)
# [1] 1
# [1] 2
# [1] 3
# [1] 4
# [1] 5
I still insist that without the negation of the condition in do-while, Salcedo's answer is wrong. One can check this via removing negation symbol in the above code.