R匹配列表中的单词
我有一个字符向量R匹配列表中的单词,r,list,string-matching,R,List,String Matching,我有一个字符向量 var1 <- c("pine tree", "forest", "fruits", "water") 与var1有2个匹配项 [[1]] [1] "tree" "house" "star" [[3]] [1] "apple" "orange" "grapes" 1与var1匹配 [[1]] [1] "tree" "house" "star" [[3]] [1] "apple" "orange" "grapes" 0与var1匹配 [[1]]
var1 <- c("pine tree", "forest", "fruits", "water")
与var1有2个匹配项
[[1]]
[1] "tree" "house" "star"
[[3]]
[1] "apple" "orange" "grapes"
1与var1匹配
[[1]]
[1] "tree" "house" "star"
[[3]]
[1] "apple" "orange" "grapes"
0与var1匹配
[[1]]
[1] "tree" "house" "star"
[[3]]
[1] "apple" "orange" "grapes"
所需输出为以下等级:
[1] "house" "tree" "dense forest"
[2] "tree" "house" "star"
[3] "apple" "orange" "grapes"
我试过了
sapply(var1, grep, var2, ignore.case=T, value=T)
没有得到想要的输出
如何解决?若能提供代码片段,将不胜感激。
谢谢
编辑:
该问题已按上述短语从一个单词匹配到另一个单词匹配进行了编辑。您可以尝试
var2[[which.max(lapply(var2, function(x) sum(var1 %in% x)))]]
[1] "house" "tree" "forest"
根据OP和@franks评论的最后修改
var2[order(-sapply(var2, function(x) sum(var1 %in% x)))]
[[1]]
[1] "house" "tree" "forest"
[[2]]
[1] "tree" "house" "star"
[[3]]
[1] "apple" "orange" "grapes"
OP更改了所需的输出,我建议
var2[顺序(-sappy(var2,函数(x)sum(var1%in%x))]
此外,在最新的R中,有一个length
函数,因此您可以执行length(sappy(var2,intersect,var1))
谢谢@Frank,我不知道我可以像在vectors中一样访问列表的元素。@Frank就是这样elegant@john在这个网站上,发布一个新的问题通常比改变一个已经有答案的问题(基本上是否定答案)要好。“我接受当x”也有点不好,我想;我们大多数人这样做是为了帮助和学习,而不是为了互联网积分。不管怎样,只要让你知道;没什么大不了的。