R loop not functional

Question

[vinayjrao]

I am writing an R-script to check for a particular expression value in a narrow range of values. For example, expression count of Gene A = 0.8. I want the flexibility to be 0.2, therefore if Gene X (the gene under test) = 0.5 to 1.1, I'll consider the expressions to be equal. The R-script I have written has an if-statement nested within a for-loop, and the loop is functioning at times, and sometimes it isn't.

I am pasting the script below to avoid any confusion -

a = 0.8
b= 0.68

a_vec = seq(a-0.3,a+0.3, by=0.001)

for(i in a_vec) {
    if(i == b) {
            print(b)
    }
}

When b = 0.68, 0.3, 0.35 there is no output, but when b = 0.8, 1, 1.1, 1.03 it prints 0.8, 1, 1.1, 1.03

[Benn]

Hi, one tip. If you want people to read your code, please make your code human readable. We are not machines.

[vinayjrao]

I'm sorry, but what is the problem?

[Devon Ryan]

Nothing any more, I modified your post :)

[vinayjrao]

Thanks for the modification, Devon :)

[Benn]

This looks more legible!

[VHahaut]

I am not sure I understand what you want to do. Is it "if your value(gene X) is comprised between value(gene A) +/- 0.3 you print something like 'TRUE' if not 'FALSE' " ?

If so it would be either to go for a statement like the following one: if(b < a+0.3 | b > a-0.3) print(TRUE) else { print(FALSE)}

[vinayjrao]

I want the value printed, not TRUE or FALSE. I need to use these values further. I need the values printed mainly because I will have a huge collection, and need to know which are the ones present.

[russhh]

You'd be better using Filter() than trying to use print() - especially if you have a huge collection of things to check

[novice]

If you're using for-loops in R, you're doing it wrong!

[Benn]

I disagree. I know that R fanatics want you to use the apply family of functions, but if R is not the only language you use, the for loop is the most important loop in bioinformatics.

Answer 1

[Devon Ryan]

Welcome to the joys of floating point precision. In general, you should not check i == b, but rather abs(i-b) < 1e-6 or some other reasonable tolerance.

[vinayjrao]

Is abs() for absolute? If yes, then I'll need something else, because I have negative counts too.

[cpad0112]

if ((abs(b-a)<=0.3)==T) {print (b)} # if script
ifelse ((abs(b-a)<=0.3),b,NA) # ifelse script
b[sapply(b, function(x) abs(x-a)<=0.3)] # sapply script
b[sapply((abs(b-a)<=0.3), isTRUE)] # sapply and istrue
b[(abs(b-a)<=0.3)] # shortening further

Code:

>a = 0.8
>b=c( 0.68, 0.3, 0.35, 0.8, 1, 1.1, 1.03 )
>ifelse((abs(b-a)<=0.3),as.numeric(b),NA)  # script 2
>b[sapply(b, function(x) abs(x-a)<=0.3)] # script 3
>b[(abs(b-a)<=0.3)]

Result

> ifelse((abs(b-a)<=0.3),as.numeric(b),NA)
[1] 0.68   NA   NA 0.80 1.00   NA 1.03

> b[sapply(b, function(x) abs(x-a)<=0.3)]
[1] 0.68 0.80 1.00 1.03

> b[sapply((abs(b-a)<=0.3), isTRUE)]
[1] 0.68 0.80 1.00 1.03

> b[(abs(b-a)<=0.3)]
[1] 0.68 0.80 1.00 1.03

[Devon Ryan]

Yes, abs() is the absolute value. In general, though, I would suggest that what cpad0112 showed is generally a more reasonable approach.

Answer 2

[lakhujanivijay]

A solution to this in R is not to use ==, but rather the all.equal function. Since all.equal gives lots of detail about the differences if there are any, isTRUE(all.equal(...)).

a = 0.8
b = 0.68

a_vec = seq(a-0.3,a+0.3, by=0.001)

for(i in a_vec){
if(isTRUE(all.equal(i,0.68))) cat(b)  
}

[steve]

+1 for isTRUE(), I too learned about that one the hard way