我想按如下方式转换列表:
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")))
进入角色对象:
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])
给了我
[1] "c(16, 17, 71, 87, 113, 120, 127, 128, 144, 177, 207, 213)"
我提到this stack overflow post但无法解决。
答案 0 :(得分:6)
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"
答案 1 :(得分:3)
或者只是......
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"