r 根据方括号拆分数据框中的列

Question

我有一个数据框：

x <- data.frame(a = letters[1:7], b = letters[2:8], 
   c = c("bla bla    [ text1 ]", "bla bla  [text2]", "how how [text3  ]",
   "wow wow   [ text4a ] [ text4b  ]", "ba ba [ text5a  ][  text5b]", 
    "my text A", "my text B"), stringsAsFactors = FALSE)
x

我想根据其中两个方括号 [...] 之间的内容拆分 c 列。如果 c 列仅包含一组方括号，我希望字符串转到下一列。如果 c 列包含两组由 [ 和 ] 包围的字符串，我只想将最后一个 [ ] 之间的字符串转到新列。

这是我的做法。看起来很复杂，我正在使用循环。有没有可能以更简约的方式做到这一点？

library(stringr)

# Counting number of square brackets "[" in column c:
sqrbrack_count <- str_count(x$c, pattern = '\[')

# Creating a new column:
x$newcolumn <- NA

for(i in 1:nrow(x)){                 # looping through rows of x
  if(sqrbrack_count[i] == 0) next    # do nothing of 0 square brackets
  minilist <- str_split_fixed(x[i, "c"], pattern = '\[', n = Inf)  # split string
  if(sqrbrack_count[i] == 1) {       # if there is only one square bracket "["
    x[i, "c"] <- minilist[1]
    x[i, "newcolumn"] <- minilist[2]
  } else {                           # if there are >1 square bracket "["
    x[i, "c"] <- paste(minilist[1:2], collapse = "+")
    x[i, "newcolumn"] <- minilist[3]
  }
}
# Replacing renmaning square brackets we don't need anymore:
x$c <- str_replace(x$c, pattern = " \]", replacement =  "")
x$c <- str_replace(x$c, pattern = "\]", replacement =  "")
x$newcolumn <- str_replace(x$newcolumn, pattern = " \]", replacement =  "")
x$newcolumn <- str_replace(x$newcolumn, pattern = "\]", replacement =  "")
x

Answer 1

下面的代码更短一些，可能更容易理解，因为大部分复杂的逻辑都发生在两行中。我已经在这两行上面添加了注释，我认为其余部分是不言自明的。

library(plyr)
# find all strings between characters '[' and ']'
strmatches = lapply(1:nrow(x), function(y) {regmatches(x$c[y], gregexpr("(?<=\[).*?(?=\])", x$c[y], perl=T))[[1]]})
# parse these to a dataframe called 'new_cols'
new_cols = rbind.fill(lapply(strmatches, function(x) {as.data.frame(t(x),stringsAsFactors = F)}))
df = cbind(x,new_cols)
df$c = gsub("\[.*$", "", x$c) # only keep everything before '['
df$c[!is.na(df$V2)] = paste0(df$c[!is.na(df$V2)], '+',df$V1[!is.na(df$V2)])
df$V1[!is.na(df$V2)] = df$V2[!is.na(df$V2)]
df$V2=NULL
colnames(df)[colnames(df)=="V1"]="newcolumn"

输出：

  a b                   c        V1
1 a b         bla bla        text1 
2 b c           bla bla       text2
3 c d            how how    text3  
4 d e wow wow   + text4a   text4b  
5 e f    ba ba + text5a      text5b
6 f g           my text A      <NA>
7 g h           my text B      <NA>

希望对您有所帮助！

PS：这符合您的预期输出，但您可能想在其中添加一些 str_trim。

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