Question

我正在尝试用* ply类型函数替换for循环。

我遇到的问题是我不确定如何重复更新相同的数据。

以下是一些示例数据（我知道这个具体示例可以通过其他方式完成，但这只是为了简单起见 - 我的真实示例要复杂得多）：

sample_pat_rep <-  data.frame(matrix(NA, ncol=2, nrow=3, dimnames=list(c(), c("Pattern","Replacement"))), stringsAsFactors=FALSE)
sample_pat_rep[1,] <-  c("a","A")
sample_pat_rep[2,] <-  c("b","B")
sample_pat_rep[3,] <-  c("c","C")

sample_strings <-  data.frame(matrix(NA, ncol=2, nrow=3, dimnames=list(c(), c("Original","Fixed"))), stringsAsFactors=FALSE)
sample_strings[1,] <-  c("aaaaaaaa bbbbbbbb cccccccc","aaaaaaaa bbbbbbbb cccccccc")
sample_strings[2,] <-  c("aAaAaAaA bBbBbBbB cCcCcCcC","aAaAaAaA bBbBbBbB cCcCcCcC")
sample_strings[3,] <-  c("AaAaAaAa BbBbBbBb CcCcCcCc","AaAaAaAa BbBbBbBb CcCcCcCc")

这是一个for循环版本：

sample_strings1 <- sample_strings
for (i in 1:nrow(sample_pat_rep))
{
  sample_strings1[,c("Fixed")] <- gsub(sample_pat_rep[i,c("Pattern")], sample_pat_rep[i,c("Replacement")], sample_strings1[,c("Fixed")],ignore.case = TRUE)
}

当我尝试用adply复制它时，它不会更新数据 - 它必须复制并重新绑定它。

sample_strings2 <- adply(.data=sample_pat_rep, .margins=1, .fun = function(x,data){

data[,c("Fixed")] <- gsub(x[,c("Pattern")], x[,c("Replacement")], data[,c("Fixed")],ignore.case = TRUE)
return(data)

}, data=sample_strings, .expand = FALSE, .progress = "none", .inform = FALSE, .parallel = FALSE, .paropts = NULL)

我确定有一个简单的解决方法。我看了拉普利，但不清楚这是解决方法。

也许写一个函数来调用？使用Rapply ??

提前致谢！

更新：新数据

这更接近实际情况。匹配是动态的，基于外部系统。我试图避免过于复杂的正则表达式或嵌套if ifses。

library(plyr)

sample_match <-  data.frame(matrix(NA, ncol=1, nrow=3, dimnames=list(c(), c("Match"))), stringsAsFactors=FALSE)
sample_match[1,] <-  c("dog")
sample_match[2,] <-  c("cat")
sample_match[3,] <-  c("bear")

sample_strings <-  data.frame(matrix(NA, ncol=2, nrow=3, dimnames=list(c(), c("Sentence","Has_Animal"))), stringsAsFactors=FALSE)
sample_strings[1,] <-  c("This person only has a cat",0)
sample_strings[2,] <-  c("This person has a cat and a dog",0)
sample_strings[3,] <-  c("This person has no animals",0)

sample_strings1 <- sample_strings
for (i in 1:nrow(sample_match))
{
 sample_strings1[,c("Has_Animal")] <- ifelse(grepl(sample_match[i,c("Match")], sample_strings1[,c("Sentence")]), 1,sample_strings1[,c("Has_Animal")])
} 


sample_strings2 <- adply(.data=sample_match, .margins=1, .fun = function(x,data){

 data[,c("Has_Animal")] <- ifelse(grepl(x[,c("Match")], data[,c("Sentence")]), 1,data[,c("Has_Animal")])
 return(data)

}, data=sample_strings, .expand = FALSE, .progress = "none", .inform = FALSE, .parallel = FALSE, .paropts = NULL)

Answer 1

更新：误解了问题，sample_strings2是必需的结果。更新了现在提供sample_strings1的答案，其中IIUC是必需的。

以下是使用base的解决方案：

pattern = paste(sample_match$Match, collapse="|")
transform(sample_strings, Has_Animal = grepl(pattern, Sentence)*1L)

#                          Sentence Has_Animal
# 1      This person only has a cat          1
# 2 This person has a cat and a dog          1
# 3      This person has no animals          0

如果您不想匹配包含该模式的字词，例如：concatenate包含cat，那么您可以将正则表达式\b用于字边界。

pattern = paste(paste("\\b", sample_match$Match, "\\b", sep=""), collapse="|")
grepl(pattern, c("cat", "concatenate"))
# [1] TRUE FALSE

Answer 2

这是问题的直接plyr方法：

ddply(sample_strings,.(Sentence),function(x,ref = sample_match) {
  any(unlist(strsplit(x[["Sentence"]]," ")) %in% ref[[1]])
  })

                         Sentence    V1
1 This person has a cat and a dog  TRUE
2      This person has no animals FALSE
3      This person only has a cat  TRUE

R：递归* ply / plyr函数;用于循环更换

2 个答案: