我有下面的数据框：
country<- c("CA","IN","US")
text <- c("paint red green", "painting red", "painting blue")
word <- c("green, red, blue", "red", "red, blue")
df <- data.frame(country, text, word)
对于每一行，我想在文本列的文本中找到单词列中的单词，并将它们呈现在新列中，因此将在文本中显示已创建的单词，以逗号分隔。所以新列应该是：
df$new_col <- c("green,red","red","blue")
我正在使用这些代码行，但运行甚至崩溃需要很多时间。
setDT(df)[, new_col:= paste(df$word[unlist(lapply(df$word,function(x) grepl(x, df$text,
ignore.case = T)))], collapse = ","), by = 1:nrow(df)]
有没有办法更改代码以提高效率？
非常感谢！
试试这个
mapply(function(x,y){paste(intersect(x,y),collapse=", ")},
strsplit(as.character(df$text),"\\, | "),
strsplit(as.character(df$word),"\\, | "))
[1] "red, green" "red" "blue"
另一个使用mapplicationgrep注册表的基本R解决方案，即，
df <- within(df, newcol <- mapply(function(x,y) toString(grep(x,y,value = TRUE)),
gsub("\\W+","|",word),
regmatches(text,gregexpr("\\w+",text))))
这样
> df
country text word newcol
1 CA paint red green green, red, blue red, green
2 IN painting red red red
3 US painting blue red, blue blue
library(tidyverse)
df %>%
mutate(newcol = stringr::str_extract_all(text,gsub(", +","|",word)))
country text word newcol
1 CA paint red green green, red, blue red, green
2 IN painting red red red
3 US painting blue red, blue blue
在这种情况下，newcol是一个列表。要使其成为字符串，我们可以这样做：
df%>%
mutate(newcol = text %>%
str_extract_all(gsub(", +", "|", word)) %>%
invoke(toString, .))
使用data. table，您可以执行：
df[,id := .I][,newcol := do.call(toString,str_extract_all(text,gsub(', +',"|",word))),
by = id][, id := NULL][]
country text word newcol
1: CA paint red green green, red, blue red, green
2: IN painting red red red
3: US painting blue red, blue blue
