我需要在一组八聚体（8个字母的集合）上运行正交编码函数，并将它们作为nx160数字的矩阵返回（其中n是数据上的八聚体数）。
正交编码功能是：
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
reload(sys)
sys.setdefaultencoding("utf-8")
有些人问过，这是一个例子，即使这不是不起作用的部分：
orthocode <- function(octamer){
matcode <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
octamer_char <- as.character(octamer)
octamer_split <- strsplit(octamer_char,"")[[1]]
for (letter in octamer_split){
ifelse (letter == "A", (matcode = rbind(matcode,c(1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "R", (matcode = rbind(matcode,c(0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "N", (matcode = rbind(matcode,c(0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "D", (matcode = rbind(matcode,c(0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "C", (matcode = rbind(matcode,c(0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "Q", (matcode = rbind(matcode,c(0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "E", (matcode = rbind(matcode,c(0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "G", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "H", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "I", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "L", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0))),
ifelse (letter == "K", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0))),
ifelse (letter == "M", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0))),
ifelse (letter == "F", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0))),
ifelse (letter == "P", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0))),
ifelse (letter == "S", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0))),
ifelse (letter == "T", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0))),
ifelse (letter == "W", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0))),
ifelse (letter == "Y", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0))),
ifelse (letter == "V", (matcode = rbind(matcode,c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1)))
))))))))))))))))))))
}
matcode <- matcode[-1,]
matcode <- c(matcode)
return(matcode)
}
该函数正在处理单个八度音，但当我尝试使用lapply时，结果只是一个160数字向量，这次代码改变了（并且没有意义）。
orthocode("ARNDCQEG")
[1] 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[81] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
结果如下：
lapply(data[1], orthocode)
正交码功能实际上正在运行。我需要知道的是如何从数据帧中获取八聚体，对它们进行操作，最终得到一个类似于此的矩阵：
$V1
[1] 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[81] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
输出必须是n行，160列的矩阵。在我必须运行的数据上，结果矩阵应该是947x160。
有什么想法吗？
答案 0 :(得分：2)
switch具有CASE构造的语义，存在于其他语言中。在没有好的例子的情况下进行轻微测试，但试试这个：
orthocode <- function(octamer){
matcode <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
octamer_char <- as.character(octamer)
octamer_split <- strsplit(octamer_char,"")[[1]]
for (letter in octamer_split){
val <- switch( letter,
"A" = c(1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
"R" = c(0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
"N" = c(0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
"D" = c(0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
"C" = c(0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
"Q" = c(0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0),
"E" = c(0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0),
"G" = c(0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0),
"H" = c(0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),
"I" = c(0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0),
"L" = c(0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0),
"K" = c(0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0),
"M" = c(0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0),
"F" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0),
"P" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0),
"S" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0),
"T" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0),
"W" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0),
"Y" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0),
"V" = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1)
)
matcode=c(matcode,val)
}
matcode
}
请注意，我删除了使用此：matcode <- c(matcode)的行，因为它具有破坏矩阵结构的副作用。
dat <- list("ARNDE", "CQEGD")
我明白了：
t( sapply(dat, orthocode) )
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31] [,32]
[1,] 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
[,33] [,34] [,35] [,36] [,37] [,38] [,39] [,40] [,41] [,42] [,43] [,44] [,45] [,46] [,47]
[1,] 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
[,48] [,49] [,50] [,51] [,52] [,53] [,54] [,55] [,56] [,57] [,58] [,59] [,60] [,61] [,62]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[,63] [,64] [,65] [,66] [,67] [,68] [,69] [,70] [,71] [,72] [,73] [,74] [,75] [,76] [,77]
[1,] 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
[,78] [,79] [,80] [,81] [,82] [,83] [,84] [,85] [,86] [,87] [,88] [,89] [,90] [,91] [,92]
[1,] 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
[,93] [,94] [,95] [,96] [,97] [,98] [,99] [,100] [,101] [,102] [,103] [,104] [,105] [,106]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 1 0 0
[,107] [,108] [,109] [,110] [,111] [,112] [,113] [,114] [,115] [,116] [,117] [,118] [,119]
[1,] 1 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0 0
[,120]
[1,] 0
[2,] 0
如果我最后使用它，我更喜欢结果（但它不是你所说的）：
matcode <- matcode[-1, ,drop=FALSE]
rownames(matcode) <- octamer_split
return(matcode) # here the return call is needed.
答案 1 :(得分：2)
我们可以使用ifelse简化match，然后删除forloop：
orthocode <- function(octamer){
matcode <- rep(0, 20)
octamer_char <- as.character(octamer)
octamer_split <- strsplit(octamer_char,"")[[1]]
t(sapply(octamer_split, function(letter){
res <- matcode
res[ match(letter, c("A","R","N","D","C","Q","E","G","H","I",
"L","K","M","F","P","S","T","W","Y","V"))] <- 1
res
}))
}
答案 2 :(得分：2)
R是矢量化的。忘记为每个案例运行一个单独的代码块。不要在循环中生长物体。我只想跟
一起去orthocode <- function(octamer) {
# Predifine identity matrix
m <- diag(20)
# Predefine values vector (no "J" or "B" here btw)
rownames(m) <- c("A", "R", "N", "D", "C", "Q", "E", "G", "H", "I", "L",
"K", "M", "F", "P", "S", "T", "W", "Y", "V")
# Create a character vector for each input
octamer_split <- strsplit(as.character(octamer), "", fixed = TRUE)
# match values for each value
t(sapply(octamer_split, function(x) m[match(x, rownames(m)),]))
}
此功能适用于单个输入或矢量。您可以使用
进行测试orthocode(c("ARNDCQEG", "NGJKAEPS", "ABGSWKLA"))
或者仅使用
orthocode(data[, 1])
P.S。
你的矢量中没有J或B，因此不确定如何处理你的例子。在这种情况下，它返回NA s