r - Recode dataframe based on one column

Question

I have a 5845*1095 (rows*columns) data frame that looks like this:

 9  286593   C     C/C     C/A     A/A
 9  334337   A     A/A     G/A     A/A
 9  390512   C     C/C     C/C     C/C

c <-  c("9", "286593", "C", "C/C", "C/A", "A/A") 
d <-  c("9", "334337", "A", "A/A", "G/A", "A/A")
e <-   c("9", "390512", "C", "C/C", "C/C", "C/C")
dat <- data.frame(rbind(c,d,e))

I want the values in the third column to be used to change the columns to its right so if (per row 1) column 3 is "C", then column 4 is turned from "C/C" to "0" as it has the same letter. One letter match is "1" (can be first or second letter) and no letter match is "2" .

9 286593  C  0  1  2
9 334337  A  0  1  0
9 390512  C  0  0  0 

c <-  c("9", "286593", "C", "0", "1", "2") 
d <-  c("9", "334337", "A", "0", " 1", "0")
e <-   c("9", "390512", "C", "0", "0", "0")
dat <- data.frame(rbind(c,d,e))

I am interested to see the best way to do this as I want to get out of the habit of using nested For loops in R.

score 5 · Accepted Answer

首先你的数据：

c <-  c("9", "286593", "C", "C/C", "C/A", "A/A")
# Note: In your original data, you had a space in "G/A", which I did remove. 
# If this was no mistake, we would also have to deal with the space.
d <-  c("9", "334337", "A", "A/A", "G/A", "A/A")
e <-   c("9", "390512", "C", "C/C", "C/C", "C/C")
dat <- data.frame(rbind(c,d,e))

现在我们生成一个包含所有可能字母的向量。

values <- c("A", "C", "G", "T")
dat$X3 <- factor(dat$X3, levels=values) # This way we just ensure that it will later on be possible to compare the reference values to our generated data. 

# Generate all possible combinations of two letters
combinations <- expand.grid(f=values, s=values)
combinations <- cbind(combinations, v=with(combinations, paste(f, s, sep='/')))

主函数找到每列的每个组合的正确列，然后将其与参考列 3 进行比较。

compare <- function(col, val) {
    m <- match(col, combinations$v)
    2 - (combinations$f[m] == val) - (combinations$s[m] == val)
}

最后，我们使用 apply 在所有需要更改的列上运行该函数。您可能希望将 6 更改为您的实际列数。

dat[,4:6] <- apply(dat[,4:6], 2, compare, val=dat[,3])

请注意，与迄今为止的其他解决方案相比，此解决方案不使用字符串比较，而是纯粹基于因子级别的方法。看看哪个表现更好会很有趣。

编辑

我刚刚做了一些基准测试：

    test replications elapsed relative user.self sys.self user.child sys.child
1   arun      1000000   2.881    1.116     2.864    0.024          0         0
2  fabio      1000000   2.593    1.005     2.558    0.030          0         0
3 roland      1000000   2.727    1.057     2.687    0.048          0         0
5  thilo      1000000   2.581    1.000     2.540    0.036          0         0
4  tyler      1000000   2.663    1.032     2.626    0.042          0         0

这让我的版本稍微快一点。但是，差异几乎为零，因此您可能对每种方法都很好。公平地说：我没有对添加额外因子水平的部分进行基准测试。这样做也可能会排除我的版本。

score 4 · Accepted Answer

这是一种方法：

FUN <- function(x) {
    a <- strsplit(as.character(unlist(x[-1])), "/")
    b <- sapply(a, function(y) sum(y %in% as.character(unlist(x[1]))))
    2 - b
}

dat[4:6] <-  t(apply(dat[, 3:6], 1, FUN))

## > dat
##   X1     X2 X3 X4 X5 X6
## c  9 286593  C  0  1  2
## d  9 334337  A  0  1  0
## e  9 390512  C  0  0  0

score 4 · Accepted Answer

这是一种使用方法apply：

out <- apply(dat[, -(1:2)], 1, function(x) 
        2 - grepl(x[1], x[-1]) -  
        x[-1] %in% paste(x[1], x[1], sep="/"))
cbind(dat[, (1:3)], t(out))

score 3 · Accepted Answer

这个解决方案不是很有效：

dat <-  cbind(dat[,-(4:6)],
              t(sapply(seq_len(nrow(dat)),function(i){
                res <- dat[i,]
                res[,4:6] <- lapply(res[,4:6],function(x) 2-sum(gregexpr(res[,3],x)[[1]]>0))
              })))

#  X1     X2 X3 X4 X5 X6
#c  9 286593  C  0  1  2
#d  9 334337  A  0  1  0
#e  9 390512  C  0  0  0

score 2 · Accepted Answer

丑陋，但它有效！

fff<-apply(dat[,4:ncol(dat)],2,substr,1,1)!=dat[,3]
ggg<-apply(dat[,4:ncol(dat)],2,substr,3,3)!=dat[,3]
final<-fff+ggg
cbind(dat,final)
X1     X2 X3  X4  X5  X6 X4 X5 X6
c  9 286593  C C/C C/A A/A  0  1  2
d  9 334337  A A/A G/A A/A  0  1  0
e  9 390512  C C/C C/C C/C  0  0  0

score 2 · Accepted Answer

对 R-golf 的另一个贡献：

cbind(dat[, 1:3],
      apply(dat[, -(1:3)], 2, function(x) {
        2 - (dat[[3]] == gsub('..$', '', x)) - (dat[[3]] == gsub('^..', '', x))
      }))

r - Recode dataframe based on one column

6 回答 6

编辑

Related

Reference