4
這看起來很簡單,但無論使用哪一種*應用函數,都無法正確回答。我沒有嘗試任何其他包,因爲它似乎*申請應該一定能夠做到這一點。將粘貼到矢量列表上以獲取字符串列表
我的數據:
data = list(foo=c("first", "m", "last"), bar=c("first", "m", "last"))
我真的覺得應該工作:
lapply(data, FUN=paste)
但它給我:
$foo
[1] "first" "m" "last"
$bar
[1] "first" "m" "last"
我想:
$foo
[1] "first m last"
$bar
[1] "first m last"
當然,我試過其他的東西整體轉換:
> 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"
我不明白的地方這引述「C」無稽之談的來源。
'lapply(數據,粘貼,崩潰=「「) ' –
呃,它的工作原理。作爲回答發佈? –