2014-01-13 125 views
1

我读取了所有'字符串到变量名称'的帖子,但没有一个涉及我的特殊问题。我有一个使用seqinr软件包'read.fasta'制作的载体列表(DNA序列数据)。我有变体的数据框及其位置,我想将数据框中指定位置的列表向量元素转换为其替代值。以个人为基础这可以使用R将字符串转换为变量名称并替换变量

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")) 

做到,但“得到”从列表返回向量元素,并且不允许我给它指定为在父列表中的变量。

我要的是单从

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

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

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

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

对于测试数据,因为你可以做到这一点单独这不是一个问题,但真实数据超过10,000个序列和超过100,000个变体。如果您可以对它进行矢量化,则可获得奖励点数,但我没有足够的经验嵌套应用函数,以使其能够同时处理来自列表和数据框的信息。

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 
+0

是REF和ALT列这里无关紧要? – Spacedman

+0

实际上,对于这部分对他们来说是不相干的,他们之前被用来获得'Call'字符。 – JeremyS

+0

继续你的方法看起来像'for(我在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])}'; '测试' –

回答

1

您可以使用sapply此任务:

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" 
+0

这正是我想要的。它看起来很简单。谢谢。 – JeremyS