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.
Related
I'm trying to write a function that identifies if a number within a numerical vector is odd or even. The numerical vector has a length of 1000.
I know that the for loop works fine, and I just wanted to generalize it in the form of a function that takes a vector of any length
out<-vector()
f3<- function(arg){
for(i in 1:length(arg)){
if((arg[i]%%2==0)==TRUE){
out[i]<-1
}else{out[i]<-0
}
}
}
When run within a function, however, it just returns a NULL. Why is that, or what do I need to do to generalize the function work with any numerical vector?
As already mentioned by PKumar in the comments: Your function doesn't return anything, which means, the vector out exists only in the environment of your function.
To change this you can add return(out) to the end of your function. And you should also start your function with creating out before the loop. So your function would look like outlined below.
Note, that I assume you want to pass a vector of a certain length to your function, and get as a result a vector of the same length which contains 1 for even numbers and 0 for odd numbers. f3(c(1,1,2)) would return 0 0 1.
f3 <- function(arg){
out <- vector(length = length(arg), mode = "integer")
for(i in 1:length(arg)){
if((arg[i]%%2==0)==TRUE){ # note that arg[i]%%2==0 will suffice
out[i]<-1
} else {out[i]<-0
}
}
return(out) # calling out without return is enough and more inline with the tidyverse style guide
}
However, as also pointed out by sebastiann in the comments, some_vector %% 2 yields almost the same result. The difference is, that odd numbers yield 1 and even numbers 0. You can also put this into a function and subtract 1 from arg to reverse 0 and 1 :
f3 <- function(arg){
(arg-1) %% 2
}
A few thing to note about your code:
A function must return something
The logical if((arg[i]%%2==0)==TRUE) is redundant. if(arg[i]%%2==0) is enough, but wrong, because arg[i] does not exist.
the length(arg) is the length(1000) which, if ran, returns 1
You should change arg[i] with i and assign to i all the values from 1:1000, as follows:
R
out <-vector()
f3 <- function(arg){
for(i in 1:arg){
if(arg[i] %% 2 == 0){
out[i] <- 1
}
else{
out[i] <- 0
}
}
return(out)
}
f3(1000)
I would like to skip iterations in a for loop based on conditions. Intuitively I thought this would work:
for(i in 1:10){
if(i %in% c(1,2,3,4,5)){
print(i)
i <- i+2}
}
I would want it to return
1
4
but instead it returns
1
2
3
4
5
I am aware why this is happening.
Is there another way to skip (multiple) iterations based on a condition in a for loop in R?
It's not just bad practice to increment the counter inside the loop in R. It simply will not work. That's not the way the language is built. If you want to get 1 and 4 printed then try:
for(i in seq(1,10,by=3) ){
if(i %in% c(1,2,3,4,5)){
print(i)
}
}
Do also note that for-loops actually return NULL. There would be a side-effect of printing to the console, but no values of variables would change. If you want values to change you need to do assignment inside the loop.
The is a next control statement:
for(i in seq(1,10) ){
if( !(i %in% c(1,4)) ){ next }
print(i)
}
I'm trying to manually increment the i variable when a condition is met.
for(i in 1:x){
if(condition){
i <- i + 2
}
}
When debugging, the (i<-i+2) line is definitely being run, but i still only increments by 1, instead of 3. (+2 from the line and an additional +1 from the auto increment)
How can I increment while I'm within the loop?
So essentially you want to skip a few loop iterations based on a condition. It's a design choice that's rightfully frowned upon, but if you must, you need next. The following code skips the third, fifth and seventh iteration:
for(i in 1:10){
if(i %in% c(3,5,7)){
next
}
print(i)
}
Say you need to increment with 3 based on a certain condition, then you can use a temporary variable that helps you skip a number of steps. Note that this does go through every iteration, it just breaks out of the iteration in time:
skip <- 0 # the temporary variable helping us keeping track of the iterations
for(i in 1:10){
if(i == 5){ # the condition that causes the skip
skip <- 3
}
if(skip > 0){ # the control that skips as long as necessary
skip <- skip - 1
next
}
print(i)
}
When you run the loop, the value of the variable i is stored in tmp*. This means that whenever we reach the top the loop, i resets. For example
for(i in 1:2){
message(i)
i <- 10
message(i)
}
#1
#10
#2
#10
To get what you want, you could have something like:
k =1
for(i in 1:10){
if(condition) k <- k + 2
}
Once the sequence is created, you pretty much lose a lot of control over looping. In cases like this, I change it into a while loop and then do the conditional incrementing/decrementing at the end of the loop.
I'm agree with joris-meys, it's "frowned upon". But... A more simple approach is:
for(i in (0:3)*2+1){
cat(i," ")
}
or
for(i in (1:4)){
cat(i," ")
}
for(i in seq(0, 10, 2) ){
print(i)
}
you can do this..
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))