Question

我正在尝试使用 R编程语言将特定的字符串模式转换为三个不同列的二进制列。

这是我所拥有的：

have <- structure(list(rep1 = c("china", "na", "bay", "eng", "giad", 
"china", "sing", "giad", "na", "china", "china, camp", "guat,camp", 
"na", "na", "cis", "trans", "stron, mon"), rep2 = c("china", 
"na", "bay", "eng", "giad", "china", "sing", "giad", "na", "china", 
"china, camp", "camp", "na", "na", "cis", "trans", "stron, mon"
), rep3 = c("na", "na", "bay", "eng", "giad", "china", "sing", 
"giad", "china", "china", "china, camp", "camp", "na", "na", 
"cis", "trans", "stron, mon")), row.names = c(NA, -17L), class = c("data.table", 
"data.frame"))

这就是我想要的：

    want <- structure(list(rep1 = c("china", "na", "bay", "eng", "giad", 
"china", "sing", "giad", "na", "china", "china, camp", "guat,camp", 
"na", "na", "cis", "trans", "stron, mon"), rep2 = c("china", 
"na", "bay", "eng", "giad", "china", "sing", "giad", "na", "china", 
"china, camp", "camp", "na", "na", "cis", "trans", "stron, mon"
), rep3 = c("na", "na", "bay", "eng", "giad", "china", "sing", 
"giad", "china", "china", "china, camp", "camp", "na", "na", 
"cis", "trans", "stron, mon"), rep1_chi = c(1, 0, 0, 0, 0, 1, 
0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0), rep2_chi = c(1, 0, 0, 0, 0, 
1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0), rep3_chi = c(0, 0, 0, 0, 
0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0), rep1_bay = c(0, 0, 1, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep2_bay = c(0, 0, 
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep3_bay = c(0, 
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep1_gia = c(0, 
0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep2_gia = c(0, 
0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep3_gia = c(0, 
0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep1_sin = c(0, 
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep2_sin = c(0, 
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), rep3_sin = c(0, 
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), class = "data.frame", row.names = c(NA, 
-17L))

我能够使用ifelse和stringr::str_detect创建一个可行的解决方案，如下所示：

want <- have %>% dplyr::select(rep1, rep2, rep3) %>% mutate(
      rep1_chi = ifelse(str_detect(rep1,"chi") == T,1,0),
      rep2_chi = ifelse(str_detect(rep2,"chi") == T,1,0),
      rep3_chi = ifelse(str_detect(rep3,"chi") == T,1,0),
      rep1_bay = ifelse(str_detect(rep1,"bay") == T,1,0),
      rep2_bay = ifelse(str_detect(rep2,"bay") == T,1,0),
      rep3_bay = ifelse(str_detect(rep3,"bay") == T,1,0),          
      rep1_gia = ifelse(str_detect(rep1,"gia") == T,1,0),
      rep2_gia = ifelse(str_detect(rep2,"gia") == T,1,0),
      rep3_gia = ifelse(str_detect(rep3,"gia") == T,1,0),           
      rep1_sin = ifelse(str_detect(rep1,"sin") == T,1,0),
      rep2_sin = ifelse(str_detect(rep2,"sin") == T,1,0),
      rep3_sin = ifelse(str_detect(rep3,"sin") == T,1,0))

我最大的问题是，它似乎很重复。 我想知道是否有更优雅的解决方案？考虑到“ rep”列的编号是1-3，我认为可能会有更好的编程方法。

通过SO，我发现使用model.matrix的{{3}}似乎在需要每种模式并且只对单个列感兴趣的情况下效果很好。我尝试将其转换为一个函数，以便可以选择多个列-但我仍然必须删除不感兴趣的模式的字符串。

Answer 1

这是使用mutate_all的方法。如果您只想对特定列执行此操作，则只需使用mutate_at并指定列即可。

library(dplyr)
library(stringr)

mutate_all(have, funs(chi = as.numeric(str_detect(., "chi")),
                  bay = as.numeric(str_detect(., "bay")),
                  gia = as.numeric(str_detect(., "gia")),
                  sin = as.numeric(str_detect(., "sin"))))

mutate_at示例，其中包含vars：

want <- have %>% mutate_at(vars(rep1,rep2,rep3), funs( 
                           tox = as.numeric(str_detect(., "chi")), 
                           bay = as.numeric(str_detect(., "bay")), 
                           gia = as.numeric(str_detect(., "gia")), 
                           iso = as.numeric(str_detect(., "sin"))))

Answer 2

这里有一些丑陋且效率低下的（性能方面的）基本代码，您不必自己构造colname：

want_new <- have
colold <- colnames(want_new)
for (p in pattern) {
  cname <- paste0(
    colold, 
    "_",
    p
  )
  for (col in cname) {
    want_new[, col] <- as.numeric(str_detect(
      want_new[, gsub(paste0("_", p), "", col, fixed)],
      p
    ))
  }
}

可以肯定，可以通过进一步调整来改善这一点。

将多个字符串模式转换为二进制列

2 个答案: