Jua*_*tro 49 join r data.table
我正在使用data.table 1.9.3,也许我错了,但我不记得之前的预期.
我构建了2个data.tables,dta和dtb
> dta
idx vala fdx
1: 1 2 a
2: 2 4 a
3: 3 6 b
> dtb
idx valb
1: 1 3
2: 4 6
> dput(x = dta)
structure(list(idx = c(1, 2, 3), vala = c(2, 4, 6), fdx = c("a",
"a", "b")), .Names = c("idx", "vala", "fdx"), row.names = c(NA,
-3L), class = c("data.table", "data.frame"), .internal.selfref =
<pointer: 0x0000000000110788>, sorted = "idx")
> dput(x = dtb)
structure(list(idx = c(1, 4), valb = c(3, 6)), .Names = c("idx",
"valb"), row.names = c(NA, -2L), class = c("data.table", "data.frame"
), .internal.selfref = <pointer: 0x0000000000110788>, sorted = "idx")
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在这两种情况下,关键都是idx.
当然,以下工作
> dta[dtb, sum(valb)]
[1] 9
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但事实并非如此
> dta[dtb, sum(valb), by = fdx]
Error in `[.data.table`(dta, dtb, sum(valb), by = fdx) :
object 'valb' not found
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但这样做
> dta[dtb][, sum(valb), by = fdx]
fdx V1
1: a 3
2: NA 6
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如果我们看到中间步骤
> dta[dtb]
idx vala fdx valb
1: 1 2 a 3
2: 4 NA NA 6
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我原以为是
dta[dtb, sum(valb), by = fdx] == dta[dtb][, sum(valb), by = fdx]
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我哪里出错了?
只是一个猜测
library(data.table)
dta <- data.frame(idx=c(1,2,3),
vala=c(2,4,6),
fdx=c('a','a','b'))
dta <- data.table(dta)
dtb <- data.frame(idx=c(1,4),
valb=c(3,6))
dtb <- data.table(dtb)
setkey(dta,idx)
setkey(dtb,idx)
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所以当你打电话时
dta[dtb, sum(valb)]
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这有点像打电话
tmp <- dta[dtb]
attach(tmp)
sum(valb)
detach(tmp)
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但是,如果您致电
dta[dtb, sum(valb), by=fdx]
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那么这就有点像打电话
tmp <- dta[dtb]
# attach(tmp) : attach doesn't happen
sum(valb)
# detach(tmp) : detach doesn't happen
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该函数不知道如何处理附加参数。例如,这也会引发错误:
dta[dtb, class(fdx), sum(valb)]
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然而,这有效
dta[dtb][, sum(valb), by=fdx]
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这有点像
tmp <- dta[dtb]
tmp[, sum(valb), by=fdx]
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就像我说的,这只是对为什么该功能可能无法按预期工作的猜测。