在 data.table 中对因子水平进行分组 [英] Grouping factor levels in a data.table

问题描述

我正在尝试在 data.table &想知道是否有 data.table-y 方法可以这样做.

I'm trying to combine factor levels in a data.table & wondering if there's a data.table-y way to do so.

例子:

DT = data.table(id = 1:20, ind = as.factor(sample(8, 20, replace = TRUE)))

我想说类型 1,3,8 属于 A 组；2和4在B组；和5,6,7在C组.

I want to say types 1,3,8 are in group A; 2 and 4 are in group B; and 5,6,7 are in group C.

这是我一直在做的，在问题的完整版中相当慢:

Here's what I've been doing, which has been quite slow in the full version of the problem:

DT[ind %in% c(1, 3, 8), grp := as.factor("A")]
DT[ind %in% c(2, 4), grp := as.factor("B")]
DT[ind %in% c(5, 6, 7), grp := as.factor("C")]

this 相关问题建议的另一种方法，我猜会这样翻译吗:

Another approach, suggested by this related question, would I guess translate like so:

DT[ , grp := ind]
levels(DT$grp) = c("A", "B", "A", "B", "C", "C", "C", "A")

或者也许(考虑到我有 65 个基础组和 18 个聚合组，这感觉更简洁一些)

Or perhaps (given I've got 65 underlying groups and 18 aggregated groups, this feels a little neater)

DT[ , grp := ind]
lev <- letters(1:8)
lev[c(1, 3, 8)] <- "A"
lev[c(2, 4)] <- "B"
lev[5:7] <- "C"
levels(DT$grp) <- lev

这两个看起来都很笨拙；这似乎是在 data.table 中执行此操作的适当方式吗?

Both of these seem unwieldy; does this seem like the appropriate way to do this in data.table?

作为参考，我用 10,000,000 个观察值和更多的子组/超组级别对此进行了增强版本的计时.我最初的方法是最慢的(必须运行所有这些逻辑检查的成本很高)，第二个最快，第三个紧随其后.但我更喜欢这种方法的可读性.

For reference, I timed a beefed up version of this with 10,000,000 observations and some more subgroup/supergroup levels. My original approach is slowest (having to run all those logic checks is costly), the second the fastest, and the third a close second. But I like the readability of that approach better.

(在搜索之前键入 DT 可以加快速度，但与后两种方法相比，它只会使差距减半)

(Keying DT before searching speeds things up, but it only halves the gap vis-a-vis the latter two methods)

推荐答案

更新:

我最近从 this 问题和对 ?levels 的仔细阅读.不需要合并、对应表等，只需将命名的 list 传递给 levels:

Update:

I recently learned of a much simpler way to re-associate factor levels from this question and a closer reading of ?levels. No merges, correspondence table, etc. necessary, just pass a named list to levels:

levels(DT$ind) = list(A = c(1, 3, 8), B = c(2, 4), C = 5:7)

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原答案:

正如@Arun 所建议的，我们可以选择将对应关系创建为单独的data.table，然后将其加入原始数据:

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Original Answer:

As suggested by @Arun we have the option of creating the correspondence as a separate data.table, then joining it to the original:

match_dt = data.table(ind = as.factor(1:12),
                      grp = as.factor(c("A", "B", "A", "B", "C", "C",
                                        "C", "A", "D", "E", "F", "D")))
setkey(DT, ind)
setkey(match_dt, ind)
DT = match_dt[DT]

我们也可以以(我认为的)更具可读性的方式来做到这一点(以边际速度成本):

We can also do this in (what I consider to be) the more readable fashion like so (with marginal speed costs):

levels <- letters[1:12]
levels[c(1, 3, 8)] <- "A"
levels[c(2, 4)] <- "B"
levels[5:7] <- "C"
levels[c(9, 12)] <- "D"
levels[10] <- "E"
levels[11] <- "F"
match_dt <- data.table(ind = as.factor(1:12),
                       grp = as.factor(levels))
setkey(DT, ind)
setkey(match_dt, ind)
DT = match_dt[DT]

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