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]))
Related
I want to associate points in these coordinates and associate them with the correspondent number. However every time I tried to run it I got this: the condition has length > 1
for (i in dados) {
if (dados[dados$latitude>="55.84" & dados$latitude<= "55.95" & dados$longitude>="-3.444" & dados$longitude<="-3.198", ]){
dados$neighbourhood <- 1
} else if (dados[dados$latitude>="55.84" & dados$latitude<= "55.95" & dados$longitude>="-3.198" & dados$longitude<="-3.058", ]){
dados$neighbourhood <- 2
} else if (dados[dados$latitude>="55.95" & dados$latitude<="56.01" & dados$longitude>="-3.444" & dados$longitude<="-3.198", ]){
dados$neighbourhood <- 3
} else if (dados[dados$latitude>="55.95" & dados$latitude<="56.01" & dados$longitude>="-3.189" & dados$longitude<="-3.058", ]){
dados$neigbourhod <- 4
}
}
Your help would be much appreciated thanks in advance.
Here is a vectorized way. Use the fact that binary numbers are written with 0/1 times powers of two. And add 1 to have one-based results.
i2 <- dados$latitude >= 55.95 & dados$latitude<= 56.01
i1 <- dados$longitude >= -3.198 & dados$longitude <= -3.058
dados$neigbourhod <- 1 + i1 + i2*2
Often, you need to execute some statements only when some condition is met. You can use following conditional statements in your code to do this.
if Statement: use it to execute a block of code, if a specified condition is true
else Statement: use it to execute a block of code, if the same condition is false
else if Statement: use it to specify a new condition to test, if the first condition is false
ifelse() Function: use it when to check the condition for every element of a vector
It is clearly that the documentation of R clearly goes against having a break line between "}" and "else". However, it is odd that the first piece of codes works but the second one does not work (syntax error)
First program
x = 1
stupid_function = function(x){
if(x != 1){
print("haha")
}
else if( x == 1){
print("hihi")
}
}
stupid_function(x)
[1] "hihi"
Second program
x = 1
if(x != 1){
print("haha")
}
else if( x == 1){
print("hihi")
}
Error in source("~/.active-rstudio-document", echo = TRUE) :
~/.active-rstudio-document:6:3: unexpected 'else'
5: }
6: else
In the second program it sees a line at a time as it is typed in so at the point that the line with the } is typed in it cannot know that there will be further lines with an else so it assumes the statement is finished.
In the first case it can see all the code before it is run because it can see all the code in the function so it knows that the } has not finished the statement.
This line of argument works for an if/else but does not work in general. For example, this will produce an error when the function is
defined.
f <- function(x) {
x
* 2
}
Note that the if else need not be in a function for it to work. It just need to be in a continuous form or in a way to be expressed as a continuous block of code. One way is being in a function. The other is to write it in one line, or even ensure that there is no line break between the if block and the else block:
x <- 1
if(x != 1) print('haha') else print('hihi')
[1] "hihi"
More blocks of statements:
x <- 1
if(x != 1){
print("haha")
} else if( x == 1){ # Note how else begins immediatley after }
print("hihi")
}
[1] "hihi"
Note that you need to know when to put the line breaks whether in a function or outside of a function. Otherwise the code might fail or even give incorrect results.
Using subtraction:
x <- 1
x -
2
[1] -1
x <- 1
x
- 2
[1] -2
You need to know when/where to have the line breaks. Its always safe to have else follow the closing brace } of the previous if statement. ie:
if(...){
....
} else if(...){
....
} else {
...
}
I made a function to to compute the sum of I(Xi
my.ecdf<- function(x,y) {
if(!is.null(dim(y)))
stop("y has more than one dimension")
n<-length(x)
i<-1:n
p<-if(x[i]<y) 1 else {
0
}
(sum(p))/n
}
But when I run it with input (rnorm(11),6), I get this error:
Warning message:
In if (x[i] < y) 1 else { :
the condition has length > 1 and only the first element will be used
Any ideas? I'm new to r so sorry if it's something obvious. (Also I don't want to use the for loop)
There are a number of issues in your code:
1) Whats the point of x[1:length(x)] in the if statement? Right now these are meaningless and can be dropped:
n<-length(x)
i<-1:n
x[i]
2) If statement accepts a logical argument not a vector of logical, you can consider adding all() any() etc like
if(all(x < y)) 1 else {0}
or use ifelse() statement for the assignment
3) Finally from what I can understand you overcomplicate things and the whole thing can be written as one-liner:
sum(x < y)/length(x)
This is a logical vector of the same length as y
is.null(dim(y))
You're using it as a logical test. An object with a length greater than 1 can't be unambiguously interpreted by the if statement. Consider if (TRUE FALSE FALSE TRUE) <do something>. When should you do that thing?
If you want to make sure y doesn't have more than one dimension, do
if(length(dim(y)) > 1){
stop("message")
}
Background
I'm developing a function that takes in a value for w between 1 and 3 and returns n values from one of 3 distributions.
The problem I am having is when n or w are not of length 1. So I've added 2 parameters nIsList and wIsList to create the functionality I want. The way I want this to work is as follows:
(Works as needed)
If nIsList ex( c(1,2,3) ) return a list equivalent to running consume(w,1), consume(w,2), consume(w,3)
(Works as needed)
If wIsList ex( c(1,2,3) ) return a list equivalent to running consume(1,n), consume(2,n), consume(3,n)
(Doesn't work as needed)
If nIsList ex(1,2,3) and wIsList ex(1,2,3)
return a list equivalent to running consume(1,1), consume(2,2), consume(3,3). Instead, I get a list equivalent to running [consume(1,1), consume(1,2), consume(1,3)], [consume(2,1), consume(2,2), consume(2,3)], [consume(3,1),consume(3,2), consume(3,3)]
I understand why I am getting the results I am getting. I just can't seem to figure out how to get the result I want. (As explained above)
Question
I want the function to provide a list for each element in w and n that is consume(w[i], n[i]) when wIsList & nIsList are True. Is there a way to do that using lapply?
The code:
library("triangle")
consume <- function(w, n=1, nIsList=F, wIsList=F){
if(!nIsList & !wIsList){
if(w==1){
return(rtriangle(n,0.3,0.8))
}else if(w==2){
return(rtriangle(n,0.7,1))
}else if(w==3){
return(rtriangle(n,0.9,2,1.3))
}
}
else if(nIsList & !wIsList){
return(sapply(n, consume, w=w))
}
else if(nIsList & wIsList){
return(lapply(n, consume, w=w, wIsList=T))
}
else if(!nIsList & wIsList){
return(lapply(w, consume, n))
}
}
Note: I am having trouble summarizing this question. If you have any suggestions for renaming it please let me know and I will do so.
Thanks to JPC's comment, using mapply does the trick. The new code is as follows:
consume <- function(w, n=1){
nIsList <- length(n) > 1 # Change based on JPC's second comment
wIsList <- length(w) > 1 # Change based on JPC's second comment
if(!nIsList & !wIsList){
if(w==1){
return(rtriangle(n,0.3,0.8))
}else if(w==2){
return(rtriangle(n,0.7,1))
}else if(w==3){
return(rtriangle(n,0.9,2,1.3))
}
}
else if(nIsList & !wIsList){
return(sapply(n, consume, w=w))
}
else if(nIsList & wIsList){
return(mapply(consume,w,n)) ## Updated portion
}
else if(!nIsList & wIsList){
return(lapply(w, consume, n))
}
}
I was trying a tiny code with if statement, although it is very simple,but there is something I really confused here is the code
n<-857
while(n!=1){
if(n<=0)
print("please input a positive integer")
else if(n%%2==0)
n<-n/2
print(n)
else
n<-3*n+1
print(n)
}
as we see above,when running this code in R, there comes the error,but if I change the if statement like this
if(n<=0)
print("please input a positive integer")
else if(n%%2==0)
n<-n/2
else
n<-3*n+1
it is ok ,my question is that can we only write one line under each judgement? if I want to do something more after each judge, what should I do ,just like this case, I want to change the value of n,but also want to display it, what should I do? thank you very much
To be precise, this is not about lines but about statements. You can have the whole if else statement in one line:
> if (TRUE) 1 else 3
[1] 1
A statement will end at the end of the line (if complete), you can see that nicely in interactive mode if you enter the code line by line:
> if (TRUE)
+ 1
[1] 1
> else
Fehler: Unerwartete(s) 'else' in "else" # error: unexpected 'else' in "else"
> 3
[1] 3
if can come in form if (condition) statement or if (condition) statement else other.statement, the interpreter assumes the first version is meant if the statement is complete after line 2 - in interactive mode it cannot sensibly wait whether an else appears next. This is different in sourced code - there it is clear with the next line which form it is.
Semicolons end statements as well:
> if (TRUE) 1; else 3
[1] 1
Fehler: Unerwartete(s) 'else' in " else" # error: unexpected 'else' in "else"
But you can only have one statement in each branch of the condition.
> if (TRUE) 1; 2 else 3
[1] 1
Fehler: Unerwartete(s) 'else' in " 2 else" # error: unexpected 'else' in "2 else"
Curly braces group statements so they appear as one statement.
> if (TRUE) {1; 2} else 3
[1] 2
You have to use {} for allows the if statement to have more than one line. Try this:
n<-857
while(n!=1){
if(n<=0){
print("please input a positive integer")
}
else if(n%%2==0){
n<-n/2
print(n)
}
else {
n<-3*n+1
print(n)
}
}
To group statements, surround them with curly braces as you've done with the while loop:
if(n<=0) {
print("please input a positive integer")
} else if(n%%2==0) {
n<-n/2
print(n)
} else {
n<-3*n+1
print(n)
}
This will allow you to place multiple statements inside the if, the else if and the final else.
while the direct answer is, as has been noted, to use curly braces;
it is worth adding that you can integrate the <- assignment operator into many functions.
In your specific case:
print(n <- 3*n+1)
## instead of
# n <- 3*n+1
# print(n)
note that using = here will NOT work. It must be <-
Yes, you can use curly braces to contain more than statements for each conditional statement:
if (condition){
statement 1
print()
}
else{
statement 2
print()
}
Addition: if you want to use only if and else statement in one line you can use this:
ifelse(condition, statement1, statement2)
Ever heard of curly barces?
n<-857
while(n!=1){
if(n<=0) {
print("please input a positive integer")
} else if(n%%2==0) {
n<-n/2
print(n)
} else {
n<-3*n+1
print(n)
}
}