Question

我有dataframe（约昏暗的446664 X 234），名为mydf（dput已提供）。此dataframe包含REF列和ALT。

REF每行只有一个字母，但ALT可以有一个，两个甚至三个字母，用逗号（“，”）分隔，其余列（示例列）是我需要完成所有工作的专栏。

考虑到REF中的任何字母为0且ALT中的第一个字母分别为1，第二个字母2和第三个字母3，我需要创建一个函数，以便：

我可以用字母替换所有样本列中的数字（即REF和ALT除外），如果有“./.";
用NA / NA填充它们并折叠“/”以获得每个单元格中的成对字母。

最后，我需要反映行（transpose）中的所有示例列，如result所示。谢谢！

mydf<-
structure(list(REF = structure(c(1L, 4L, 3L, 2L, 3L), .Label = c("A", 
"C", "G", "T"), class = "factor"), ALT = structure(c(6L, 6L, 
1L, 9L, 1L), .Label = c("A", "A,C", "A,G", "A,T", "C", "C,G", 
"C,T", "G", "G,T", "T"), class = "factor"), X860 = structure(c(1L, 
3L, 2L, 1L, 1L), .Label = c("./.", "0/0", "0/1", "0/2", "1/1"
), class = "factor"), X861 = structure(c(1L, 6L, 2L, 1L, 1L), .Label = c("./.", 
"0/0", "0/1", "0/2", "1/1", "1/2"), class = "factor"), X862 = structure(c(6L, 
3L, 1L, 2L, 1L), .Label = c("./.", "0/0", "0/1", "0/2", "1/1", 
"2/2"), class = "factor")), .Names = c("REF", "ALT", "X860", 
"X861", "X862"), row.names = c(NA, -5L), class = "data.frame")

预期产出：

X860 NANA TC GG NANA NANA
X861 NANA CG GG NANA NANA 
X862 GG TC NANA CC NANA

Answer 1

得到了这个，但我很不确定它的表现：

letters <- strsplit(paste(mydf$REF,mydf$ALT,sep=","),",") # concatenate the letters to have an index to work on from the numbers
values <- t(mydf[,3:ncol(mydf)]) # let's work on each column needing values
nbval <- ncol(values) # Save time for later and save the length of values 

#Prepare the two temp vectors used later
chars <- vector("character",2) 
ret <- vector("character",nbval)

#Loop over the rows (and transpose the result)
t(sapply(rownames(values),
   function(x) { 
     indexes <- strsplit(values[x,],"/") # Get a list with pairs of indexes

     for(i in 1:nbval) { # Loop over the number of columns :/
       for (j in 1:2) { # Loop over the pair 
         chars[j] <- ifelse(indexes[[i]][j] == ".", "NA",letters[[i]][as.integer(indexes[[i]][j])+1]) # Get NA if . or the letter with the correct index at this postion
       }
       ret[i] <- paste0(chars[1],chars[2]) # concatenate the two chars
     }
     return(ret) # return this for this row
   }
))

带样本数据的输出：

     [,1]   [,2] [,3]   [,4]   [,5]  
X860 "NANA" "TC" "GG"   "NANA" "NANA"
X861 "NANA" "CG" "GG"   "NANA" "NANA"
X862 "GG"   "TC" "NANA" "CC"   "NANA"

从评论中更新了该函数的版本（因为其余代码没有改变）：

#Loop over the rows (and transpose the result)
t(sapply(rownames(values),
   function(x) {
     indexes <- strsplit(values[x,],"/") # Get a list with pairs of indexes
     for(i in 1:nbval) { # Loop over the number of columns :/
       if (values[x,i] == "./.") { # test if we have ./. and if yes, set to NA
         ret[i] <- "NA"
       } else { # if it's not ./. then try to find the corresponding letters
         for (j in 1:2) { # Loop over the pair 
           chars[j] <- ifelse(indexes[[i]][j] == ".", "NA",letters[[i]][as.integer(indexes[[i]][j])+1]) # Get NA if . or the letter with the correct index at this postion
         }
         ret[i] <- paste0(chars[1],chars[2]) # concatenate the two chars
       }
     }
     return(ret) # return this for this row
   }
))

输出：

     [,1] [,2] [,3] [,4] [,5]
X860 "NA" "TC" "GG" "NA" "NA"
X861 "NA" "CG" "GG" "NA" "NA"
X862 "GG" "TC" "NA" "CC" "NA"

Answer 2

这是我的尝试：

#ref alt together, we will access by index of each allele
refalt <- paste(mydf$REF,mydf$ALT,sep=",")

#loop over for each SNP column
t(
  sapply(3:5, function(snp){
    #take one SNPs
    x <- as.character(mydf[,snp])
    sapply(1:length(x), function(ix){
      #get genotype and make numeric alleles
      geno <- unlist(strsplit(x[ix],"/",fixed=TRUE))
      a1 <- as.numeric(geno[1])+1
      a2 <- as.numeric(geno[2])+1
      #match with refalt on ix
      allele <- unlist(strsplit(refalt[ix],",",fixed=TRUE))
      #output - here I would prefer to output "NANA" as NA but it is up to you.
      paste0(allele[a1],allele[a2])
    })
  })
)

#output
#       [,1]   [,2] [,3]   [,4]   [,5]  
# [1,] "NANA" "TC" "GG"   "NANA" "NANA"
# [2,] "NANA" "CG" "GG"   "NANA" "NANA"
# [3,] "GG"   "TC" "NANA" "CC"   "NANA"

如何转换基因分型数据

2 个答案: