R 根据第二个数据集和拆分列重新编码
我有一个矩阵“mat”,012个编码的SNP作为列,人作为行。例如:R 根据第二个数据集和拆分列重新编码,r,R,我有一个矩阵“mat”,012个编码的SNP作为列,人作为行。例如: > mat<-matrix(c("0","1","0","1","2","0","1","1","2"),3,byrow=T) > rownames(mat)<-c("ID1","ID2","ID3") > colnames(mat)<-c("rs123","rs333","rs9000") > mat rs123 rs333 rs9000 ID1 "0" "1"
> mat<-matrix(c("0","1","0","1","2","0","1","1","2"),3,byrow=T)
> rownames(mat)<-c("ID1","ID2","ID3")
> colnames(mat)<-c("rs123","rs333","rs9000")
> mat
rs123 rs333 rs9000
ID1 "0" "1" "0"
ID2 "1" "2" "0"
ID3 "1" "1" "2"
>mat行名称(mat)列名称(mat)mat
rs123 rs333 rs9000
ID1“0”1“0”
ID2“1”2“0”
ID3“1”“1”“2”
在不同的矩阵“mat2”中,我将各自的等位基因分为两列(即次要和主要等位基因),SNP分为两行
> mat2<-matrix(c("A","T","C","T","T","G"),3,byrow=T)
> rownames(mat2)<-c("rs123","rs333","rs9000")
> colnames(mat2)<-c("Allele_A","Allele_B")
> mat2
Allele_A Allele_B
rs123 "A" "T"
rs333 "C" "T"
rs9000 "T" "G"
>mat2行名(mat2)列名(mat2)mat2
等位基因A等位基因B
rs123“A”“T”
rs333“C”“T”
rs9000“T”“G”
现在我想把第一个矩阵中的012个编码的SNP重新编码为两列:如果它们的编码为零,它们应该是两个新列中各自的等位基因A,如果是1,它们应该是A/B,如果是2,它们应该是B/B。在我的示例中,我希望得到以下信息:
> mat3<-matrix(c("A","C","T","A","T","T","A","T","T","T","T","T","A","C","G","T","T","G" ),3,byrow=T)
> rownames(mat3)<-c("ID1","ID2","ID3")
> colnames(mat3)<-c("rs123_1","rs333_1","rs9000_1","rs123_2","rs333_2","rs9000_2")
> mat3
rs123_1 rs333_1 rs9000_1 rs123_2 rs333_2 rs9000_2
ID1 "A" "C" "T" "A" "T" "T"
ID2 "A" "T" "T" "T" "T" "T"
ID3 "A" "C" "G" "T" "T" "G"
>mat3行名(mat3)列名(mat3)mat3
rs123_1 rs333_1 rs9000_1 rs123_2 rs333_2 rs9000_2
ID1“A”“C”“T”“A”“T”“T”
ID2“A”“T”“T”“T”“T”
ID3“A”“C”“G”“T”“T”“G”
你能帮我做到这一点吗?提前谢谢你 这里是另一个data.table解决方案
library(reshape2)
library(data.table)
mat<-data.table(matrix(c("ID1","0","1","0","ID2","1","2","0","ID3","1","1","2"),3,byrow=T))
setnames(mat, c("ID","rs123","rs333","rs9000"))
mat2<-data.table(matrix(c("rs123","A","T","rs333","C","T","rs9000","T","G"),3,byrow=T))
setnames(mat2, c("rs","Allele_A","Allele_B"))
mat <- data.table(melt(mat, id.vars = 'ID'))
setnames(mat,'variable','rs')
mat <- merge(mat,mat2, by = 'rs', all.x = TRUE)
mat[,a1 := Allele_A]
mat[value == 2,a1 := Allele_B]
mat[,a2 := Allele_B]
mat[value == 0,a2 := Allele_A]
mat1 <- dcast(mat, ID~rs, value.var = 'a1')
mat2 <- dcast(mat, ID~rs, value.var = 'a2')
mat <- merge(mat1,mat2,by = 'ID', suffixes = c('_1','_2'))
library(data.table)
DT.mat1 <- as.data.table(mat)
DT.mat2 <- as.data.table(mat2, keep.rownames=TRUE)
## Make sure all characters
DT.mat2[, names(DT.mat2) := lapply(.SD, as.character)]
# set key to the row names
setkey(DT.mat2, rn)
您可以使用
microbenchmark
(来自microbenchmark包)来计时。我不确定您的实际数据有多大,但我猜随着数据的增长,这将是相当快的不如Codoremifa建议的漂亮,作为一个函数:
allele_count_to_genotype <- function(M_count, M_geno){
# assuming same order of SNPs in both matrices!
# generate empty target matrix with right col and row names
M_target <- matrix(ncol=ncol(M_count)*2, nrow=nrow(M_count))
rownames(M_target) <- rownames(M_count)
colnames(M_target) <- c(paste(colnames(M_count),"1",sep="_"),
paste(colnames(M_count),"2",sep="_"))
# fill the target matrix with counts
M_target[,1:ncol(M_count)] <- M_count
M_target[, (ncol(M_count)+1):ncol(M_target)] <- M_count
# substitute the counts
for(k in 1:ncol(M_count)){
M_target[,k] <- gsub("0", M_geno[k,1], M_target[,k])
M_target[,k] <- gsub("1", M_geno[k,1], M_target[,k])
M_target[,k] <- gsub("2", M_geno[k,2], M_target[,k])
M_target[,(ncol(M_count)+k)] <- gsub("0", M_geno[k,1], M_target[,(ncol(M_count)+k)])
M_target[,(ncol(M_count)+k)] <- gsub("1", M_geno[k,2], M_target[,(ncol(M_count)+k)])
M_target[,(ncol(M_count)+k)] <- gsub("2", M_geno[k,2], M_target[,(ncol(M_count)+k)])
}
return(M_target)
}
下面是一个使用
堆栈
、开关
和取消堆栈
的基本R方法:
s.mat <- stack(data.frame(mat))
left.allele.choice <- function(x) switch(x, "0"=1, "1"=1, "2"=2)
right.allele.choice <- function(x) switch(x, "0"=1, "1"=2, "2"=2)
get.allele.choice <- function(allele.choice) {
mapply(function(values, ind) mat2[ind, allele.choice(values)],
values=s.mat$values, ind=s.mat$ind)
}
left.side <- unstack(transform(s.mat,
values=get.allele.choice(left.allele.choice)))
right.side <- unstack(transform(s.mat,
values=get.allele.choice(right.allele.choice)))
result <- cbind(left.side, right.side)
colnames(result) <- make.unique(colnames(result))
# rs123 rs333 rs9000 rs123.1 rs333.1 rs9000.1
# 1 A C T A T T
# 2 A T T T T T
# 3 A C G T T G
s.mat它正在迭代.SD
(mat1列)的名称,并将它们用作mat2的键。然后是两个简单的ifelse
语句…显然+1代表聪明。缺乏可读性似乎是嵌套数据的一个常见特征。表解决方案——一些人可能会争辩说,这是速度提升的公平价格。哈哈哈!我确实认为理解data.table
语法就像理解嵌套的lapply
调用一样。明确的学习曲线,但一旦越过这一驼峰,可读性很强。但是,是的,你完全正确,速度的提高是值得的,这对我来说是完美的!
rs123 _ 1 rs333 _ 2 rs9000 _ 1 rs123 _ 2 rs333 _ 1 rs9000 _ 2
1: A T C C T G
2: A A T C T G
3: A A C C G T
allele_count_to_genotype <- function(M_count, M_geno){
# assuming same order of SNPs in both matrices!
# generate empty target matrix with right col and row names
M_target <- matrix(ncol=ncol(M_count)*2, nrow=nrow(M_count))
rownames(M_target) <- rownames(M_count)
colnames(M_target) <- c(paste(colnames(M_count),"1",sep="_"),
paste(colnames(M_count),"2",sep="_"))
# fill the target matrix with counts
M_target[,1:ncol(M_count)] <- M_count
M_target[, (ncol(M_count)+1):ncol(M_target)] <- M_count
# substitute the counts
for(k in 1:ncol(M_count)){
M_target[,k] <- gsub("0", M_geno[k,1], M_target[,k])
M_target[,k] <- gsub("1", M_geno[k,1], M_target[,k])
M_target[,k] <- gsub("2", M_geno[k,2], M_target[,k])
M_target[,(ncol(M_count)+k)] <- gsub("0", M_geno[k,1], M_target[,(ncol(M_count)+k)])
M_target[,(ncol(M_count)+k)] <- gsub("1", M_geno[k,2], M_target[,(ncol(M_count)+k)])
M_target[,(ncol(M_count)+k)] <- gsub("2", M_geno[k,2], M_target[,(ncol(M_count)+k)])
}
return(M_target)
}
> allele_count_to_genotype(M_count=mat, M_geno=mat2)
rs123_1 rs333_1 rs9000_1 rs123_2 rs333_2 rs9000_2
ID1 "A" "C" "T" "A" "T" "T"
ID2 "A" "T" "T" "T" "T" "T"
ID3 "A" "C" "G" "T" "T" "G"
s.mat <- stack(data.frame(mat))
left.allele.choice <- function(x) switch(x, "0"=1, "1"=1, "2"=2)
right.allele.choice <- function(x) switch(x, "0"=1, "1"=2, "2"=2)
get.allele.choice <- function(allele.choice) {
mapply(function(values, ind) mat2[ind, allele.choice(values)],
values=s.mat$values, ind=s.mat$ind)
}
left.side <- unstack(transform(s.mat,
values=get.allele.choice(left.allele.choice)))
right.side <- unstack(transform(s.mat,
values=get.allele.choice(right.allele.choice)))
result <- cbind(left.side, right.side)
colnames(result) <- make.unique(colnames(result))
# rs123 rs333 rs9000 rs123.1 rs333.1 rs9000.1
# 1 A C T A T T
# 2 A T T T T T
# 3 A C G T T G