R convert string to variable name and replace the variable

Question

I read all the 'string to variable name' posts but none of them covered my particular problem. I have a list of vectors (DNA sequence data) made using the seqinr package 'read.fasta'. I have a data frame of variants and their location and I want to convert the list vector elements at the locations specified in the data frame to their alternate values. On an individual basis this can be done using

list$name[number] <- alternate.character

# I tried
 for (i in 1:length(df$CHROM))
  if (is.na(df$Call[i])) {next} else {get(paste("test$",df$CHROM[i],"[",df$POS[i],"]",sep=""))  <- df$Call[i]}

# example data
 test <- list("One" = c("a","t","a","g","c"),
              "Two" = c("g","a","t","t","a","c","a"))
 df <- data.frame("CHROM"=c(rep("One",2),rep("Two",3)),
             "POS" = c(2,4,1,3,6),
             "REF" = c("t","g","g","t","c"),
             "ALT" = c("a","a","t","g","t"),
             "Call" = c("T","A","G",NA,"T"))

But 'get' returns the vector element from the list and doesn't allow me to assign it as the variant in the parent list.

What I want is the list to go from

$One
[1] "a" "t" "a" "g" "c"

$Two
[1] "g" "a" "t" "t" "a" "c" "a"

to

$One
[1] "a" "T" "a" "A" "c"

$Two
[1] "G" "a" "t" "t" "a" "T" "a"

For the test data this isn't a problem because you can just do it individually, but the real data is over 10,000 sequences and over 100,000 variants. Bonus points if you can vectorize it, I don't have enough experience nesting apply functions to get it to work with information from a list and a data frame at the same time.

sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-pc-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_GB.UTF-8       LC_NUMERIC=C              
 [3] LC_TIME=en_GB.UTF-8        LC_COLLATE=en_GB.UTF-8    
 [5] LC_MONETARY=en_GB.UTF-8    LC_MESSAGES=en_GB.UTF-8   
 [7] LC_PAPER=en_GB.UTF-8       LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C            
 [11] LC_MEASUREMENT=en_GB.UTF-8 LC_IDENTIFICATION=C       

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] seqinr_3.0-7

loaded via a namespace (and not attached):
[1] tools_3.0.2

Actually, yes they are irrelevant for this part sorry, they were used to get the 'Call' character earlier. — JeremyS
– JeremyS, Commented Jan 13, 2014 at 8:20
Continuing your approach see something like for(i in seq_len(nrow(df))) {if(!is.na(as.character(df$Call[i]))) test[[as.character(df$CHROM[i])]][as.numeric(as.character(df$POS[i]))] <- as.character(df$Call[i])} ; test — alexis_laz
– alexis_laz, Commented Jan 13, 2014 at 14:26

Sven Hohenstein · Accepted Answer · 2014-01-13 08:05:04Z

1

You can use sapply for this task:

res <- sapply(names(test), function(x) {
  tmp <- df[df$CHROM == x & !is.na(df$Call), ]
  replace(test[[x]], tmp$POS, as.character(tmp$Call))
})


res
# $One
# [1] "a" "T" "a" "A" "c"
#
# $Two
# [1] "G" "a" "t" "t" "a" "T" "a"

edited Jan 13, 2014 at 8:05

answered Jan 13, 2014 at 7:59

Sven Hohenstein

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1 Comment

JeremyS Over a year ago

That does exactly what I want. And it looks so simple. Thank you.

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