我想从x中减去y,这意味着删除一个" A",三个" B"和一个" E"来自x,因此xNew
将为c("A", "C", "A","B","D")
。这也意味着
length(xNew)=length(x) - length(y)
x <- c("A","A","C","A","B","B","B","B","D","E")
y <- c("A","B","B","B","E")
setdiff不起作用,因为
xNew <- setdiff(x,y)
xNew
[1] "C" "D"
匹配也不起作用
xNew <- x[-match(y,x)]
xNew
[1] "A" "C" "A" "B" "B" "B" "D"
它删除了&#34; B&#34;在第五个位置3次,所以还有三个&#34; B&#34;左
有人知道如何做到这一点,R中是否有可用的功能,或者我们应该写一个私有函数? 非常感谢。
答案 0 :(得分:4)
您可以使用pmatch
:
x[-pmatch(y,x)]
#[1] "A" "C" "A" "B" "D"
修改强>
如果您的数据可以是超过1个字符的字符串,则可以选择获取所需内容:
xNew <- unlist(sapply(x[!duplicated(x)],
function(item, tab1, tab2) {
rep(item,
tab1[item] - ifelse(item %in% names(tab2), tab2[item], 0))
}, tab1=table(x), tab2=table(y)))
实施例
x <- c("AB","BA","C","CA","B","B","B","B","D","E")
y <- c("A","B","B","B","E")
xNew
# AB BA C CA B D
#"AB" "BA" "C" "CA" "B" "D"