假设我有这个数据框:
df =data.frame(text=c("This is a very long sentence that I would like to trim because I might need to put it as a label somewhere",
"This is another very long sentence that I would also like to trim because I might need to put it as who knows what"),col2=c("1234","5678"))
在this帖子之后,我已经能够获得一个新专栏,它可以让我以完整的单词开始句子,这很好。
df$short_txt = sapply(strsplit(df$text, ' '), function(i) paste(i[cumsum(nchar(i)) <= 20], collapse = ' '))
> df$short_txt
[1] "This is a very long" "This is another very"
但是,我也有兴趣粘贴结尾前 20 个字符的完整单词的结尾,以获得与此输出接近的内容。
> df$short_txt
[1] "This is a very long...it as a label somewhere" "This is another very...it as who knows what"
我不知道如何完成
sapply
函数来达到这个结果。我尝试使用粘贴功能并将 cumsum
功能更改为 df$short_txt = sapply(strsplit(df$text, ' '), function(i) paste(i[cumsum(nchar(i)) <= 20],"...",i[cumsum(nchar(i)) >= (nchar(i)-20)], collapse = ' '))
但它没有返回我想要的。
感谢您的帮助。
也许我们可以正则表达式?
gsub("^(.{20}\\S*)\\b.*\\b(\\S*.{20})$", "\\1...\\2", df$text)
# [1] "This is a very long sentence...as a label somewhere" "This is another very...it as who knows what"
正则表达式解释:
^(.{20}\\S*)\\b.*\\b(\\S*.{20})$
^ $ beginning and end of string, respectively
(.........) (.........) first and second saved groups
.{20} .{20} exactly 20 characters of any kind
\\S* \\S* zero or more non-space characters
\\b \\b word boundaries
.* anything else (including nothing)
这不包括开头的
it
,因为没有它,子字符串的长度为 20。
我将查看
df$text[1]
以及前导/尾随完整单词子串的各种数字。
sapply(seq(10, 24, by = 2), function(len) gsub(sprintf("^(.{%d}\\S*)\\b.*\\b(\\S*.{%d})$", len, len), "\\1...\\2", df$text[1]))
# [1] "This is a very... somewhere"
# [2] "This is a very...label somewhere"
# [3] "This is a very...label somewhere"
# [4] "This is a very long... label somewhere"
# [5] "This is a very long... a label somewhere"
# [6] "This is a very long sentence...as a label somewhere"
# [7] "This is a very long sentence...it as a label somewhere"
# [8] "This is a very long sentence... it as a label somewhere"