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I have two vectors with species names following two different methods. Some names are the same, others are different and both are sorted in different ways. An example: list 1: c(Homo sapiens sapiens, Homo sapiens neanderthalensis, Homo erectus,...,n) List 2: c(Homo erectus, Homo sapiens, Homo neanderthalensis,...,n+1)

I write n and n+1 to denote that these lists have different lengths.

I would like to create a new list that consists out of two values: in the case that there is a match between the two vectors (e.g. Homo erectus) I would like to have the name of list 2 at the location the name has in List 1, or in case there is a mismatch a "0" at the location in List 1. So in this case this new list would be newlist: c(0,0, Homo erectus,...)

For this I have written the following code, but it does not work.

data<-read.table("species.txt",sep="\t",header=TRUE)
list1<-as.vector(data$Species1)
list2<-as.vector(data$Species2)
newlist<-as.character(rep(0,length(list1)))

for (i in 1:length(list1)){
for (j in 1:length(list2)){
if(list1[i] == list2[j]){newlist[i]<- list2[j]}else {newlist[i]= 0}
}
}

I hope this is clear.

Thanks for any help!

mnicolai
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    Please share a [minimal reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) and show expected output. This will make it much easier for others to help you. – markus Oct 29 '18 at 10:47

3 Answers3

0

Take this reproducible example:

set.seed(1)
list1 <- letters[1:10]
list1names
list2 <- letters[sample(1:10, 10)]

You can avoid a loop using ifelse:

newlist <- ifelse(list1==list2, list2, 0)

The issue is that you did not declare newname, did you mean newlist ?

If you want to use a loop you can use only one loop and not 2 because length(list1) = length(list2):

for (i in 1:length(list1)){
    if(list1[i] == list2[i]){newlist[i]<- list2[i]}else {newlist[i]= 0}
}

In general if you want to match elements in vectors you can use match like this:

> list1
 [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"
> list2
 [1] "c" "d" "e" "g" "b" "h" "i" "f" "j" "a"
> match(list1, list2)
 [1] 10  5  1  2  3  8  4  6  7  9

As you can see match gets the indexes of the elements in list2 which are equal to the elements in list1. This is useful in case you have another table data2, and you would like to fetch the column in data2 for corresponding elements from data$list1 in data2$list3, you would use:

data <- data.frame(list1, list2)
list3 <- list2
columntoget <- 1:length(list2)
data2 <- data.frame(list3, columntoget)
data$mynewcolumn <- data2$columntoget[match(data$list1, data2$list3)]
> data$mynewcolumn
 [1] 10  5  1  2  3  8  4  6  7  9
gaut
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  • Dear Gpier, thanks for the quick reply. Some remarks. Yes, newname=newlist (I used different names in my original code and forgot to change this). I also forgot to mention that both lists have different lengths. – mnicolai Oct 29 '18 at 11:09
  • Or please give us example data and what you expect as a result – gaut Oct 29 '18 at 11:18
  • While I did not formulate my question as good as it should be, "match" seems do to the trick.Thanks. – mnicolai Nov 06 '18 at 09:17
  • Welcome, please consider accepting this answer in this case – gaut Nov 06 '18 at 09:21
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I'm not completely certain that I understand what you're trying to achieve, but I think this does what you're after. 

list1 <- c("Homo sapiens sapiens","Homo sapiens neanderthalensis","Homo erectus")
list2 <- c("Homo erectus","Homo sapiens","Homo neanderthalensis")

sapply(list1, function(x) { ifelse(x %in% list2, list2[which(list1 == x)], 0) } )
abarnsey
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The inner for loop uses newname[i] where it should be newlist[i]. Using your code, you overwrite the newlist[i] entries j times with either 0 or a species name. This is probably not what you want.

Strawberryshrub
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Frank Ruge
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