r - 尽管谓词为真，但对命名向量的名称使用过滤器会返回一个空的命名向量？

Question

考虑out<-setNames(sample(26,replace = T),letters)。通过构造，输出的元素之一必须命名为a。但是，当我尝试时Filter(function(x) names(x)=="a",out)，我得到一个大小为 0 的命名数字。为什么会这样？

标签： rnames

Answer 1

The function Filter loses the naming information of the vector (because it use lapply internally on out which also loses this information).

However, you can filter the names using: out[names(out) == "a"]. The result is a vector containing 6 with the name a.