示例代码：

Question

我试图编写一个R函数，它将根据数据帧每行中原始字符串的长度对可变数量的5元素子串进行采样。我首先计算了我希望每次绘制重复的次数，并希望将其添加到函数中，以便每行采样的数量基于＆＃34; num_draws＆＃34;该行的列。我的想法是使用一个通用实例，然后使用函数外部的apply语句来操作每一行，但是我无法弄清楚如何设置函数来调用col 3作为一个通用实例（没有调用第一行的值或所有行的值）。

示例数据框：

  BP                             TF                                  num_draws
1 CGGCGCATGTTCGGTAATGA           TFTTTFTTTFFTTFTTTTTF                6
2 ATAAGATGCCCAGAGCCTTTTCATGTACTA TFTFTFTFFFFFFTTFTTTTFTTTTFFTTT      9
3 TCTTAGGAAGGATTC                FTTTTTTTTTFFFFF                     4

期望的输出：

[1]GGCGC FTTTF 
   AATGA TTTTF 
   TTFFT TGTTC 
   TAATG TTTTT
   AATGA TTTTF   
   CGGCG TFTTT

[2]AGATG FTFTF
   ATAAG TFTFT
   ATGCC FTFFF
   GCCCA FFFFF
   ATAAG TFTFT
   GTACT TFFTT
   GCCCA FFFFF
   TGCCC TFFFF
   AGATG FTFTF

[3]TTAGG TTTTT
   CTTAG TTTTT
   GGAAG TTTTT
   GGATT TTFFF

示例代码：

#make example data frame
BaseP1 <- paste(sample(size = 20, x = c("A","C","T","G"), replace = TRUE), collapse = "")
BaseP2 <- paste(sample(size = 30, x = c("A","C","T","G"), replace = TRUE), collapse = "")
BaseP3 <- paste(sample(size = 15, x = c("A","C","T","G"), replace = TRUE), collapse = "")
TrueFalse1 <- paste(sample(size = 20, x = c("T","F"), replace = TRUE), collapse = "")
TrueFalse2 <- paste(sample(size = 30, x = c("T","F"), replace = TRUE), collapse = "")
TrueFalse3 <- paste(sample(size = 15, x = c("T","F"), replace = TRUE), collapse = "")
my_df <- data.frame(c(BaseP1,BaseP2,BaseP3), c(TrueFalse1, TrueFalse2, TrueFalse3)) 


#calculate number of draws by length 
frag_length<- 5 
my_df<- cbind(my_df, (round((nchar(my_df[,1]) / frag_length) * 1.5, digits = 0)))
colnames(my_df) <- c("BP", "TF", "num_draws")

#function to sample x number of draws per row (this does not work)
Fragment = function(string) {
  nStart = sample(1:(nchar(string) -5), 1)
  samp<- substr(string, nStart, nStart + 4)
replicate(n= string[,3], expr = samp)
  }


apply(my_df[,1:2], c(1,2), Fragment)

Answer 1

一种选择是将函数更改为另一个参数n并在nStart调用中创建replicate

Fragment = function(string, n) {
   replicate(n= n,  {nStart <- sample(1:(nchar(string) -5), 1)
                  samp <- substr(string, nStart, nStart + 4)
              })   

}

apply(my_df, 1, function(x) data.frame(lapply(x[1:2], Fragment, n = x[3])))
$`1`
#     BP    TF
#1 GGCGC FFTTF
#2 GGTAA TFFTT
#3 GCGCA TTFTT
#4 CGCAT TFFTT
#5 GGCGC FTTTF
#6 TGTTC FTTFT

#$`2`
#     BP    TF
#1 GTACT TTTTF
#2 ATAAG FTTFT
#3 GTACT TFTFF
#4 TAAGA TTTTF
#5 CCTTT FFTTF
#6 TCATG TTTTF
#7 CCAGA TFTFT
#8 TTCAT TFTFT
#9 CCCAG FTFTF

#$`3`
#     BP    TF
#1 AAGGA TTTFF
#2 AGGAT TTTTT
#3 CTTAG TFFFF
#4 TAGGA TTTFF

函数来给出字符串长度

示例代码：

1 个答案: