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r - 函数对给定字符串长度的可变数量的子字符串进行采样

转载 作者:行者123 更新时间:2023-12-01 22:35:00 26 4
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我正在尝试编写一个 R 函数,该函数将根据数据帧每行中原始字符串的长度对可变数量的 5 元素子字符串进行采样。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,以便为每行获取的样本数基于该行的“num_draws”列。我的想法是使用一个通用实例,然后在函数外部使用 apply 语句来作用于每一行,但我不知道如何设置函数来调用 col 3 作为通用实例(不调用仅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)

最佳答案

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

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

关于r - 函数对给定字符串长度的可变数量的子字符串进行采样,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45002224/

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