这看起来很简单,但无论我使用哪个 *apply 函数,正确的答案都让我困惑。我没有尝试过任何其他包,因为看起来 *apply 绝对应该能够做到这一点。
My data:
data = list(foo=c("first", "m", "last"), bar=c("first", "m", "last"))
我真正认为应该起作用的是:
lapply(data, FUN=paste)
但它给了我:
$foo
[1] "first" "m" "last"
$bar
[1] "first" "m" "last"
I want:
$foo
[1] "first m last"
$bar
[1] "first m last"
当然,我尝试过很多其他的东西:
> paste(data)
[1] "c(\"first\", \"m\", \"last\")" "c(\"first\", \"m\", \"last\")"
> paste(data, collapse = "")
[1] "c(\"first\", \"m\", \"last\")c(\"first\", \"m\", \"last\")"
> paste(data, sep = "")
[1] "c(\"first\", \"m\", \"last\")" "c(\"first\", \"m\", \"last\")"
> paste(data, collapse = "", sep="")
[1] "c(\"first\", \"m\", \"last\")c(\"first\", \"m\", \"last\")"
> paste(as.vector(data), collapse = "", sep="")
[1] "c(\"first\", \"m\", \"last\")c(\"first\", \"m\", \"last\")"
> paste(c(data), collapse = "", sep="")
[1] "c(\"first\", \"m\", \"last\")c(\"first\", \"m\", \"last\")"
> paste(c(data, recursive = T), collapse = "", sep="")
[1] "firstmlastfirstmlast"
我不明白这个引用的“c”废话是从哪里来的。