基于条件语句的矩阵元素条件更新答案

【问题标题】：Conditional updating of matrix elements based on conditional statement基于条件语句的矩阵元素条件更新
【发布时间】：2013-09-01 03:53:45
【问题描述】：

我确信我只是犯了一个简单的错误。

我有一个需要迭代的大矩阵 3307592x9，如果第 8 列（字符/字符串）== 9（字符/字符串）（不区分大小写），则需要第 3-7 列（数字 0-1） 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 客户端时，我会在取消命令之前挂起半个多小时。我不确定我做错了什么，因为代码对我来说是正确的。

/编辑示例数据

> 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

>

【问题讨论】：

您的主要错误是使用for 循环而不是矢量化操作。（我会根据您的数据大小推荐打包 data.table。）我也不清楚为什么使用grepl。 tolower 和 == 的组合就足够了。如果你 gave example data 向你展示如何做到这一点会更容易。
类似于fixAncestor
您的样本数据没有用，因为其中的条件从未满足。
修复了样本数据
我已经修改了我的答案以使用它。

标签： r

【解决方案1】：

首先是你的数据：

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")

现在是 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

【讨论】：

我要查一下 :=
只需阅读data.table intro 和FAQ。

【解决方案2】：

在base R中你可以简单地做

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

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

这可能有点慢，因为Vectorize 隐藏了一个循环。但是，如果您只需要翻转第 8 列和第 9 列完全匹配的行（大小写除外），请改用此过滤器：

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

【讨论】：

【解决方案3】：

for 不是你在R 中的朋友，这是使用apply 和条件索引的解决方案

## 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)

祝 GWAS 好运！

【讨论】：

我很确定它做到了。谢谢！
如果您使用ifelse 而不是if 和else，则无需为此使用apply。