基于条件语句的矩阵元素的条件更新

Question

我相信我只是犯了一个简单的错误。我有一个大的矩阵3307592x9，我需要迭代，如果列8（字符串/字符串）== 9（字符/字符串）（不区分大小写）然后列3-7（数字0-1）需要是1自我。我写的代码是：

for (i in 1:3307592){ 
    if(grepl(chr2SnpFreqNorm[i,8], chr2SnpFreqNorm[i,9], ignore.case=TRUE)){ 
     chr2SnpFreqNorm[i,3] <- 1 - chr2SnpFreqNorm[i,3] 
     chr2SnpFreqNorm[i,4] <- 1 - chr2SnpFreqNorm[i,4] 
     chr2SnpFreqNorm[i,5] <- 1 - chr2SnpFreqNorm[i,5] 
     chr2SnpFreqNorm[i,6] <- 1 - chr2SnpFreqNorm[i,6] 
     chr2SnpFreqNorm[i,7] <- 1 - chr2SnpFreqNorm[i,7] 
    } 
} 

当我尝试执行我的[R客户端只是挂了半小时以上我取消命令。我不确定我做错了什么，因为代码看起来对我来说是正确的。

/编辑 实例数据

> chr2SnpFreqNorm[1:10,] 
     ID pos ceuChr2SnpFreq chsChr2SnpFreq lwkChr2SnpFreq 
1 rs187078949 10133 0.070588235   0.000 0.030927835 
2 rs191522553 10140 0.005882353   0.000 0.005154639 
3 rs149483862 10286 0.100000000   0.135 0.226804124 
4 rs150919307 10297 0.147058824   0.070 0.113402062 
5 rs186644623 10315 0.000000000   0.000 0.000000000 
6 rs193294418 10345 0.017647059   0.000 0.036082474 
7 rs185496709 10386 0.082352941   0.020 0.087628866 
8 rs188771313 10419 0.229411765   0.085 0.056701031 
9 rs192945962 10425 0.100000000   0.020 0.015463918 
10 rs184397180 10431 0.064705882   0.005 0.036082474 
    tsiChr2SnpFreq yriChr2SnpFreq ALT AA 
1  0.035714286 0.045454545 A a 
2  0.005102041 0.005681818 A C 
3  0.239795918 0.170454545 A t 
4  0.168367347 0.130681818 T t 
5  0.000000000 0.005681818 G C 
6  0.030612245 0.028409091 A G 
7  0.035714286 0.113636364 T t 
8  0.147959184 0.090909091 G G 
9  0.091836735 0.034090909 G c 
10 0.015306122 0.045454545 T a 

> 

Answer 1

在基础R你可以做简单的

flip <- Vectorize(grepl)(chr2SnpFreqNorm[,8], chr2SnpFreqNorm[,9], ignore.case=TRUE) 

chr2SnpFreqNorm[flip,3:7] <- 1 - chr2SnpFreqNorm[filp,3:7] 

这可能是有点慢因为Vectorize隐藏了一个循环。但是，如果你需要的是翻转行，其中，列8和9准确匹配（除的情况下），然后使用此过滤器来代替：

flip <- tolower(chr2SnpFreqNorm[,8])==tolower(chr2SnpFreqNorm[,9]) 

Answer 2

for是不是你在R的朋友，这里有一个解决方案使用apply和条件索引

## create some toy data  
matrix(ncol=5, nrow = 100, c(runif(300), sample(c('A','G','C','T','a','c','g','t'), replace=T, 200))) -> data 

flip_allele_freqs <- function(x) { 
## function will return 1-x on any x that looks like a number less than 1 
    n = as.numeric(x) 
    if (is.na(n)) { ## cant convert to numeric, must be str 
     return(x) 
    } 
    if (n < 1) { 
     return(1 - n) 
    } else { 
     return x 
    } 
} 

## apply the flip alleles function to the rows where the two last columns are equal 
##fold the new data back into the old matrix 

data[toupper(data[,5]) == toupper(data[,4]),] <- 
    apply(data[toupper(data[,5]) == toupper(data[,4]),], c(1,2), flip_allele_freqs) 

与GWAS祝你好运！

Answer 3

首先你的数据：

DF <- structure(list(ID = c("rs187078949", "rs191522553", "rs149483862", 
"rs150919307", "rs186644623", "rs193294418", "rs185496709", "rs188771313", 
"rs192945962", "rs184397180"), pos = c(10133L, 10140L, 10286L, 
10297L, 10315L, 10345L, 10386L, 10419L, 10425L, 10431L), ceuChr2SnpFreq = c(0.070588235, 
0.005882353, 0.1, 0.147058824, 0, 0.017647059, 0.082352941, 0.229411765, 
0.1, 0.064705882), chsChr2SnpFreq = c(0, 0, 0.135, 0.07, 0, 0, 
0.02, 0.085, 0.02, 0.005), lwkChr2SnpFreq = c(0.030927835, 0.005154639, 
0.226804124, 0.113402062, 0, 0.036082474, 0.087628866, 0.056701031, 
0.015463918, 0.036082474), tsiChr2SnpFreq = c(0.035714286, 0.005102041, 
0.239795918, 0.168367347, 0, 0.030612245, 0.035714286, 0.147959184, 
0.091836735, 0.015306122), yriChr2SnpFreq = c(0.045454545, 0.005681818, 
0.170454545, 0.130681818, 0.005681818, 0.028409091, 0.113636364, 
0.090909091, 0.034090909, 0.045454545), ALT = c("A", "A", "A", 
"T", "G", "A", "T", "G", "G", "T"), AA = c("a", "C", "t", "t", 
"C", "G", "t", "G", "c", "a")), .Names = c("ID", "pos", "ceuChr2SnpFreq", 
"chsChr2SnpFreq", "lwkChr2SnpFreq", "tsiChr2SnpFreq", "yriChr2SnpFreq", 
"ALT", "AA"), row.names = c("1", "2", "3", "4", "5", "6", "7", 
"8", "9", "10"), class = "data.frame") 

而且现在data.table解决方案：

#use data.table for excellent efficiency 
library(data.table) 
DT <- data.table(DF) 

#subtract 1 from columns 3 to 7 if columns ALT and AA are equal (case insensitive) 
DT[tolower(ALT)==tolower(AA), 3:7 := lapply(.SD, `-`, e2 = 1), .SDcols=3:7] 

#    ID pos ceuChr2SnpFreq chsChr2SnpFreq lwkChr2SnpFreq tsiChr2SnpFreq yriChr2SnpFreq ALT AA 
# 1: rs187078949 10133 -0.929411765   -1.000 -0.969072165 -0.964285714 -0.954545455 A a 
# 2: rs191522553 10140 0.005882353   0.000 0.005154639 0.005102041 0.005681818 A C 
# 3: rs149483862 10286 0.100000000   0.135 0.226804124 0.239795918 0.170454545 A t 
# 4: rs150919307 10297 -0.852941176   -0.930 -0.886597938 -0.831632653 -0.869318182 T t 
# 5: rs186644623 10315 0.000000000   0.000 0.000000000 0.000000000 0.005681818 G C 
# 6: rs193294418 10345 0.017647059   0.000 0.036082474 0.030612245 0.028409091 A G 
# 7: rs185496709 10386 -0.917647059   -0.980 -0.912371134 -0.964285714 -0.886363636 T t 
# 8: rs188771313 10419 -0.770588235   -0.915 -0.943298969 -0.852040816 -0.909090909 G G 
# 9: rs192945962 10425 0.100000000   0.020 0.015463918 0.091836735 0.034090909 G c 
# 10: rs184397180 10431 0.064705882   0.005 0.036082474 0.015306122 0.045454545 T a