Error in if statement in r - r

This might be a silly question, but since I could not find the way to search it, I wanted to ask. I have the code below:
(m is a numeric value defined earlier)
if (0 < m <= 1.00){
print("a")
} else if (1.00 < m <= 2.00){
print("likely benign")
} else if (2.00 < m <= 3.00){
print("b")
} else if (3.00 < m <= 4.00){
print("c")
} else if (4.00 < m <= 5.00){
print("d")
} else if (5.00 < m <= 6.00){
print("e")
} else{
print("f")
}
It gives the error:
Error: unexpected '}' in " }"
but I used if in exactly the same way on the upper part of the code and there is no error from that part. How can I fix this? Thanks in advance.

We need to change the syntax to &
if(0 < m & m <= 1.00){
---
and also in all the else if conditions.
It is not clear whether m is of length 1 or >1. If it is length 1, use the & and if greater than 1, it may be better to use ifelse instead of if/else or if we are using if/else, use && instead of & as #ZheyuanLi suggested in the comments.

Related

R: How to use current function output's within this function

I write a R function using if & else if in it. See the code below:
i_ti_12 <- function(x){
if (x <= 44)
{ti = exp(-13.2238 + 0.152568*x)}
else if (x >= 49)
{ti = -0.01245109 + 0.000315605*x}
else (x > 44 & x < 49)
{ti = (x-44)*(i_ti_12(49))/(49-44) + (49-x)*(i_ti_12(44))/(49-44)}
return(ti)
}
I want to use this function's output, i_ti_12(49) within this function, but the above code doesn't work. The output is:
> i_ti_12(49)
Error: C stack usage 7974292 is too close to the limit
The simple solution is just replace i_ti_12(49) by -0.01245109 + 0.000315605*49, but its not a clear way to solve it and might not work in complex cases.
So I really want to know and to learn if there are clear methods to do this? I mean, like above simple example, write a conditional function using one condition's output in this function. Any help is highly appreciate.
Your last else is followed by a condition (x > 44 & x < 49), which actually is not correct. If you have (x > 44 & x < 49) there, that means you will execute that statement, and ti = (x-44)*(i_ti_12(49))/(49-44) + (49-x)*(i_ti_12(44))/(49-44) is something independent with your if-else structure.
In that case, when you call i_ti_12(49), your function does not know when the recursion should be terminated since you did not define that case.
You can try the code below:
i_ti_12 <- function(x){
if (x <= 44)
{ti = exp(-13.2238 + 0.152568*x)}
else if (x >= 49)
{ti = -0.01245109 + 0.000315605*x}
else
{ti = (x-44)*(i_ti_12(49))/(49-44) + (49-x)*(i_ti_12(44))/(49-44)}
return(ti)
}
such that
> i_ti_12(49)
[1] 0.003013555

if else statement concatenation - R

This is a very common question: 1, 2, 3, 4, 5, and still I cannot find even an answer to my problem.
If a == 1, then do X.
If a == 0, then do Y.
If a == 0 and b == 1, then do Z.
Just to explain: the if else statements has to do Y if a==0 no matter the value of b. But if b == 1 and a == 0, Z will do additional changes to those already done by Y.
My current code and its error:
if (a == 1){
X
} else if(a == 0){
Y
} else if (a == 0 & b == 1){
Z}
Error in !criterion : invalid argument type
An else only happens if a previous if hasn't happened.
When you say
But if b == 1 and a == 0, Z will do additional changes to those already done by Y
Then you have two options:
## Option 1: nest Z inside Y
if (a == 1){
X
} else if(a == 0){
Y
if (b == 1){
Z
}
}
## Option 2: just use `if` again (not `else if`):
if (a == 1) {
X
} else if(a == 0) {
Y
}
if (a == 0 & b == 1) {
Z
}
Really, you don't need any else here at all.
## This will work just as well
## (assuming that `X` can't change the value of a from 1 to 0
if (a == 1) {
X
}
if (a == 0) {
Y
if (b == 1){
Z
}
}
Typically else is needed when you want to have a "final" action that is done only if none of the previous if options were used, for example:
# try to guess my number between 1 and 10
if (your_guess == 8) {
print("Congratulations, you guessed my number!")
} else if (your_guess == 7 | your_guess = 9) {
print("Close, but not quite")
} else {
print("Wrong. Not even close!")
}
In the above, else is useful because I don't want to have enumerate all the other possible guesses (or even bad inputs) that a user might enter. If they guess 8, they win. If they guess 7 or 9, I tell them they were close. Anything else, no matter what it is, I just say "wrong".
Note: this is true for programming languages in general. It is not unique to R.
However, since this is in the R tag, I should mention that R has if{}else{} and ifelse(), and they are different.
if{} (and optionally else{}) evaluates a single condition, and you can run code to do anything in {} depending on that condition.
ifelse() is a vectorized function, it's arguments are test, yes, no. The test evaluates to a boolean vector of TRUE and FALSE values. The yes and no arguments must be vectors of the same length as test. The result will be a vector of the same length as test, with the corresponding values of yes (when test is TRUE) and no (when test is FALSE).
I believe you want to include Z in the second condition like this:
if (a == 1){X}
else if(a == 0){
Y
if (b == 1){Z}
}

Error in if (num < 0) { : missing value where TRUE/FALSE needed

y <- as.integer(readline(prompt ="Enter a number: "))
factorial = 1
if (y< 0){
print("Error")
} else if (y== 0)
{
print("1")
} else
{
for(i in 1:y) {
factorial = factorial * i
}
return(factorial)
}
wondering why this is giving:
Error in if (y< 0) { : missing value where TRUE/FALSE needed
is it cause the first line has data type NA_integer?
There are three possible ways to pass values to the if statement.
y <- 1
if (y > 0) print("more")
This one works as expected.
y <- 1:3
if (y > 0) print("ignores all but 1st element")
As the warning message will tell you, only the first element was used to evaluate it. You could use any or all to make this right.
y <- NA
if (y > 0) print("your error")
This case actually gives you your error. I would wager a bet that y is somehow NA. You will probably need to provide a reproducible example (with data and the whole shebang) if you'll want more assistance. Note also that it helps visually structure your code to improve readability.

Multiple conditions in if statements in R

I am trying to cut down a list of gene names that I have been given. I'm trying to eliminate any repetitive names that may be present but I keep getting an error when running my code:
counter=0
i=0
j=0
geneNamesRevised=array(dim=length(geneNames))
for (i in 0:length(geneNamesRevised))
geneNamesRevised[i]=""
geneNamesRevised
for (i in 1:length(geneNames))
for (j in 1:length(geneNamesRevised))
if (geneNames[i]==geneNamesRevised[j])
{
break
}
else if ((j==length(geneNamesRevised)-1) &&
(geneNames[i]!=geneNamesRevised[j]))
{
geneNamesRevised[counter]=geneNames[i]
counter++
}
The error message is a repetitive string of :
the condition has length > 1 and only the first element will be usedthe condition has length > 1 and only the first element will be usedthe condition has length > 1 and only the first element will be used
and this error message is for the last "else if" statement that has the '&&'.
Thank you!
Why not just
geneNamesRevised <- unique( geneNames )
... which returns a shortened list. There is also a duplicated function that can be used to remove duplicates when negated.
There are a few problems in your code.
1) The else is incorrectly specified - or not :) thanks #Mohsen_Fatemi
2) & is usually what you need rather than &&
3) counter++ isn't R
Copy the code below and see if it runs
for (i in 1:length(geneNames)){
for (j in 1:length(geneNamesRevised)){
if (geneNames[i]==geneNamesRevised[j])
{
break
} else {
if ((j==length(geneNamesRevised)-1) & (geneNames[i]!=geneNamesRevised[j]))
{
geneNamesRevised[counter]=geneNames[i]
counter <- counter + 1
}
}
}
}
Edit
4) also you were missing braces for your fors
use & instead of && ,
else if ((j==length(geneNamesRevised)-1) & (geneNames[i]!=geneNamesRevised[j]))

How to get equivalent to R ifelse functionality, using logical operators

Please can anyone advise how I can turn the following statement into one that will do the same thing but NOT using ifelse please?
<-ifelse(y>=50, 0.2*x+0.8*y, ifelse(y<50 & x>70, y+10, ifelse(y<50 & x<70, y)))
x=80
y=60
So I the final code should give an answer of 64 - selecting the first condition. I will then test it to ensure the other 3 conditions give the correct result for varying values of x and y
Thanks a lot.
This should work:
finalmark <- (x * 0.2 + y * 0.8) * (y >= 50) + (y + 10 * (x > 70)) * (y < 50)
Something like this?
if(y>=50){
0.2*x+0.8*y
}else{
if(y<50 & x>70){
y+10
}else{
if(y<50 & x<70){
y
}else{
"OMG I did not expect this scenario"
}
}
}
try: y=45; x=70 to see why I have the last condition.
If y is a number then, once you've tested for y > = 50 then y must be less than 50 so don't keep testing for that. Similarly, once you've found x > 70 then you don't need the last ifelse. You don't have a return for x = 70. My guess is that you want to test for a <= or >= situation there.
ifelse(y>=50, 0.2*x+0.8*y, ifelse(x>70, y+10, y))
in scalar that's
if(y >= 50){
0.2*x+0.8*y
}else if(x > 70){
y+10
}else y
Given you seem to be having a hard time in general writing the logic I suggest you post a more complete question. It's possible (probable) that you're doing something here that you really don't want to do.
There are several approaches you can take. Below are a few examples of building a function 'f', so that 'f(x,y)' meets your criteria listed in the question using logic other than 'ifelse' statements.
Note: I'm also adding in one amendment to the original post, since 'x=70' would break the logic. I'm adding 'x>=70' to the second criterion.
Option 1: Use a standard 'if / else if / else' logic block. Personally, I like this option, because it's easily readable.
f <- function(x, y){
if (y>= 50){
return(0.2*x+0.8*y)
} else if (y < 50 & x >= 70){
return(y+10)
} else {
return(y)
}
}
Option 2: Combine your two logical tests (there are really only two) into a string, and use a switch. Note that the final and unnamed option is treated as an 'else'.
f <- function(x, y){
return(
switch(paste(x >= 70, y >= 50, sep=""),
TRUEFALSE = y + 10,
FALSEFALSE = y,
0.2*x+0.8*y
)
)
}
Option 3: Order your 'if' statements to reduce logical comparisons. This is the sort of thing to do if you have a large data set or very limited memory. This is slightly harder to troubleshoot, since you have to read the whole block to fully understand it. Option 1 is better if you don't have memory or cycle limitations.
f <- function(x, y){
if (y >= 50){
return(0.2*x+0.8*y)
} else {
if (x >=70){
return(y+10)
} else {
return(y)
}
}
}
There are other options, but these are the simplest that come readily to mind.

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