R:如何基于模式从字符串中提取文本片段

Question

我有一个数据集，其中每一行包含一个这种类型的文本字符串。

1)list(text = \"incredible hosts\", relevance = 0.87518, count = 1), list(text = \"Japan\", relevance = 0.675236, count = 1), list(text = \"support\", relevance = 0.625663, count = 1), list(text = \"result\", relevance = 0.359757, count = 1)


2)list(text = \"British fleet\", relevance = 0.912888, count = 1), list(text = \"worst maritime disasters\", relevance = 0.904047, count = 1), list(text = \"British history\", relevance = 0.755491, count = 1), list(text = \"Scilly Isles\", relevance = 0.716508, count = 1), list(text = \"sailors\", relevance = 0.691141, count = 1), list(text = \"evening\", relevance = 0.597375, count = 1), list(text = \"Tragedy\", relevance = 0.577141, count = 1), list(text = \"prize\", relevance = 0.565035, count = 1), list(text = \"rocks\", relevance = 0.543257, count = 1), list(text = \"innovation\", relevance = 0.529463, count = 1), list(text = \"longitude\", relevance = 0.335207, count = 1)

基本上，我只想提取\“和\”之间的文本字符串。

得到这样的东西

1) "incredible hosts, Japan, support , result"
2) "British fleet, worst maritime disasters, British history, scilly Isles, sailors, evening, etc..."

此外，我希望创建一个数据框架，帮助le跟踪文本中包含的每一段文本的相关性评分(考虑到不同的raws可能有不同数量的文本)，这样可以得到如下内容：

 col1                 col2.   col3.    col4.   col5.     col6.....  colA1    colA2.  .....
 incredible hosts     Japon  support  result    NA.      NA        0.87518.   0.675236....
 british fleet.       worst marit.......

基本上，一个列的数量等于一行中的最大文本块数，对于对应于分数的列是相同的(每个相关性评分都是指一段文本，因此它们是相同的数目)。

如果我能找到一种方法，首先提取文本片段，然后用逗号将它们分开，然后对相关分数进行同样的处理，我想我可以很容易地将两者合并在一个数据格式中。所以问题主要是从文本中提取这两样东西。

谢谢你的帮助，

卡洛

Answer 1

上面显示的字符串几乎是正确的R代码。因此，只要进行最小的修改，我们就可以直接将数据读入R：

txt1 <- 'list(text = \"incredible hosts\", relevance = 0.87518, count = 1), list(text = \"Japan\", relevance = 0.675236, count = 1), list(text = \"support\", relevance = 0.625663, count = 1), list(text = \"result\", relevance = 0.359757, count = 1)'
txt2 <- 'list(text = \"British fleet\", relevance = 0.912888, count = 1), list(text = \"worst maritime disasters\", relevance = 0.904047, count = 1), list(text = \"British history\", relevance = 0.755491, count = 1), list(text = \"Scilly Isles\", relevance = 0.716508, count = 1), list(text = \"sailors\", relevance = 0.691141, count = 1), list(text = \"evening\", relevance = 0.597375, count = 1), list(text = \"Tragedy\", relevance = 0.577141, count = 1), list(text = \"prize\", relevance = 0.565035, count = 1), list(text = \"rocks\", relevance = 0.543257, count = 1), list(text = \"innovation\", relevance = 0.529463, count = 1), list(text = \"longitude\", relevance = 0.335207, count = 1)'

txt1 <- gsub("text = ", "id = 1, text = ", txt1) # this is just if you want to have an ID later on
txt2 <- gsub("text = ", "id = 2, text = ", txt2)

list1 <- eval(parse(text = paste0("list(", txt1, ")")))
list2 <- eval(parse(text = paste0("list(", txt2, ")")))


df <- dplyr::bind_rows(list1, list2)
df
#> # A tibble: 15 x 4
#>       id text                     relevance count
#>    <dbl> <chr>                        <dbl> <dbl>
#>  1     1 incredible hosts             0.875     1
#>  2     1 Japan                        0.675     1
#>  3     1 support                      0.626     1
#>  4     1 result                       0.360     1
#>  5     2 British fleet                0.913     1
#>  6     2 worst maritime disasters     0.904     1
#>  7     2 British history              0.755     1
#>  8     2 Scilly Isles                 0.717     1
#>  9     2 sailors                      0.691     1
#> 10     2 evening                      0.597     1
#> 11     2 Tragedy                      0.577     1
#> 12     2 prize                        0.565     1
#> 13     2 rocks                        0.543     1
#> 14     2 innovation                   0.529     1
#> 15     2 longitude                    0.335     1

从这里开始，你可以更容易地把它们变成你想要的任何形式。

更新

在您的评论之后，我更改了我的回答，向您展示了如何在更大的数据集中使用这种方法，并将其放入quanteda：

假设您读取了数据，现在每个文本都是一个向量中的值：

txt <- c('list(text = \"incredible hosts\", relevance = 0.87518, count = 1), list(text = \"Japan\", relevance = 0.675236, count = 1), list(text = \"support\", relevance = 0.625663, count = 1), list(text = \"result\", relevance = 0.359757, count = 1)',
         'list(text = \"British fleet\", relevance = 0.912888, count = 1), list(text = \"worst maritime disasters\", relevance = 0.904047, count = 1), list(text = \"British history\", relevance = 0.755491, count = 1), list(text = \"Scilly Isles\", relevance = 0.716508, count = 1), list(text = \"sailors\", relevance = 0.691141, count = 1), list(text = \"evening\", relevance = 0.597375, count = 1), list(text = \"Tragedy\", relevance = 0.577141, count = 1), list(text = \"prize\", relevance = 0.565035, count = 1), list(text = \"rocks\", relevance = 0.543257, count = 1), list(text = \"innovation\", relevance = 0.529463, count = 1), list(text = \"longitude\", relevance = 0.335207, count = 1)')

与其更改每个对象，不如循环遍历每个元素：

txt <- lapply(seq_along(txt), function(i) { # this is just if you want to have an ID later on
  gsub("text = ", paste0("id = ", i, ", text = "), txt[i])
})

list <- lapply(txt, function(x) {
  dplyr::bind_rows(eval(parse(text = paste0("list(", x, ")"))))
})

df <- dplyr::bind_rows(list)

一旦您有了一个data.frame，在您可以使用quanteda工作之前，只需要做一点数据争论：

library(dplyr)
df_wide <- df %>% 
  group_by(id) %>% 
  summarise(text = paste(text, collapse = " "), relevance = list(relevance))

library(quanteda)

corp <- corpus(df_wide, docid_field = "id", text_field = "text")
corp
#> Corpus consisting of 2 documents and 1 docvar.
corp$documents$relevance
#> [[1]]
#> [1] 0.875180 0.675236 0.625663 0.359757
#> 
#> [[2]]
#>  [1] 0.912888 0.904047 0.755491 0.716508 0.691141 0.597375 0.577141
#>  [8] 0.565035 0.543257 0.529463 0.335207

Answer 2

下面是一个基本R方法，它至少输出所有匹配的引号：

x <- "list(text = \"incredible hosts\", relevance = 0.87518, count = 1), list(text = \"Japan\", relevance = 0.675236, count = 1), list(text = \"support\", relevance = 0.625663, count = 1), list(text = \"result\", relevance = 0.359757, count = 1)"
m <- gregexpr("\"(.*?)\"", x)
regmatches(x, m)[[1]]

[1] "\"incredible hosts\"" "\"Japan\""            "\"support\""
[4] "\"result\""