Question

我想拆分字符。虽然我有一个大型数据框可以工作，但下面的小例子展示了需要做什么。

  mydf <- data.frame (name = c("L1", "L2", "L3"), 
    M1 = c("AC", "AT", NA), M2 = c("CC", "--", "TC"), M3 = c("AT", "TT", "AG"))

我想将变量M1到M3的字符分割（在真实数据集中我有> 6000个变量）

  name  M1a M1b   M2a M2b  M3a  M3b 
   L1   A    C    C    C    A     T
   L2   A    T    -    -    T     T
   L3   NA   NA   T     C    A     G

我尝试了以下代码：

func<- function(x) {sapply( strsplit(x, ""),
                     match, table= c("A","C","T","G", "--", NA))}

odataframe <- data.frame(apply(mydf, 1, func) )
colnames(odataframe) <-  paste(rep(names(mydf), each = 2), c("a", "b"), sep = "")
odataframe

Answer 1

你走了：

splitCol <- function(x){
  x <- as.character(x)
  x[is.na(x)] <- "$$"
  z <- matrix(unlist(strsplit(x, split="")), ncol=2, byrow=TRUE)
  z[z=="$"] <- NA
  z
}


newdf <- as.data.frame(do.call(cbind, lapply(mydf[, -1], splitCol)))
names(newdf) <- paste(rep(names(mydf[, -1]), each=2), c("a", "b"), sep="")
newdf <- data.frame(mydf[, 1, drop=FALSE], newdf)

newdf
  name  M1a  M1b M2a M2b M3a M3b
1   L1    A    C   C   C   A   T
2   L2    A    T   -   -   T   T
3   L3 <NA>  <NA   T   C   A   G

Answer 2

Andrie的代码作为可复制函数

splitCol <- function(dataframe, splitVars=names(dataframe)){
split.DF <- dataframe[,splitVars]
keep.DF <- dataframe[, !names(dataframe) %in% c(splitVars)]

X <- function(x)matrix(unlist(strsplit(as.character(x), split="")), ncol=2, byrow=TRUE)

newdf <- as.data.frame(do.call(cbind, suppressWarnings(lapply(split.DF, X))) )
names(newdf) <- paste(rep(names(split.DF), each=2), c(".a", ".b"), sep="") 
data.frame(keep.DF,newdf)
}

测试出来

splitCol(mydf)
splitCol(mydf, c('M1','M2'))

请不要将此作为正确答案投票。安德烈的答案显然是第一个正确的答案。这只是他的代码扩展到更多情况。感谢问题并感谢Andrie代码。

在列和名称中拆分字符

2 个答案: