来自forcats的as_factor可以返回有序因子吗?这似乎是一个缺失的功能,即使不是,尽管我还没有在GitHub page上看到它被报告为一个问题。
我试过了:
y <- forcats::as_factor(c("a", "z", "g"), ordered = TRUE)
is.ordered(y)
# FALSE如果我做不到,那么这样做有没有潜在的危险:
y <- ordered(forcats::as_factor(c("a", "z", "g")))或者这样做会更好:
y <- factor(c("a", "z", "g"), levels = unique(c("a", "z", "g")), ordered = TRUE))发布于 2017-08-31 12:59:49
看起来这确实是一种意想不到的行为。forcats::as_factor强制它按出现的顺序排序,但不知何故没有设置标志。但是将它与base::factor结合使用,不需要显式地指定顺序,只需设置标志就可以了。
y <- forcats::as_factor(c("a", "z", "g"))
y
[1] a z g
Levels: a z g
is.ordered(y)
[1] FALSE
k <- factor(y, c("a","z","g"), ordered = TRUE)
k
[1] a z g
Levels: a < z < g
is.ordered(k)
[1] TRUE
k2 <- factor( y, ordered = TRUE)
k2
[1] a z g
Levels: a < z < g
is.ordered(k2)
[1] TRUE
k3 <- factor(forcats::as_factor(c("a","g","z")), ordered = TRUE)
k3
[1] a g z
Levels: a < g < z
is.ordered(k3)
[1] TRUEhttps://stackoverflow.com/questions/45972983
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