6

我想将列表转换如下：

    list(structure(c(16L, 17L, 71L, 87L, 113L, 120L, 127L, 128L, 
      144L, 177L, 207L, 213L), .Names = c("246653_at", "246897_at", 
      "251347_at", "252988_at", "255528_at", "256535_at", "257203_at", 
      "257582_at", "258807_at", "261509_at", "265050_at", "265672_at")))

变成一个字符对象：

         c("246653_at", "246897_at", "251347_at", "252988_at", "255528_at", 
         "256535_at", "257203_at", "257582_at", "258807_at", "261509_at", 
         "265050_at", "265672_at")

我尝试使用

    as.character(fin[1])

这给了我

   [1] "c(16, 17, 71, 87, 113, 120, 127, 128, 144, 177, 207, 213)"

我提到了这个堆栈溢出帖子，但无法解决。

4

2 回答 2

9

如果您的对象是x：

> names(unlist(x))
 [1] "246653_at" "246897_at" "251347_at" "252988_at" "255528_at" "256535_at"
 [7] "257203_at" "257582_at" "258807_at" "261509_at" "265050_at" "265672_at"
于 2013-08-06T11:14:59.233 回答
3

要不就...

names(fin[[1]])
 [1] "246653_at" "246897_at" "251347_at" "252988_at" "255528_at" "256535_at"
 [7] "257203_at" "257582_at" "258807_at" "261509_at" "265050_at" "265672_at"
于 2013-08-06T11:19:41.640 回答