当我写作
dt[a>0, {...}, by=...]
Run Code Online (Sandbox Code Playgroud)
在{...}之前或之后处理的a>0过滤?(似乎答案是在之前).
我可以想象这两个订单都很有用,所以正确的问题是,我想,我该如何控制订单或过滤与处理?
这个i=论点是(非常明智地)首先处理的,因为你可以用以下的东西来确认.
library(data.table)
dt <- data.table(a=c(0,1,0,1), grp=c("a", "a", "b", "b"))
# a grp
# 1: 0 a
# 2: 1 a
# 3: 0 b
# 4: 1 b
## Show that filtering op in i= is performed before processing in j=
dt[a>0, if(any(a<=0)) stop("a<=0 must've been passed on to j") else a, by=grp]
# grp V1
# 1: a 1
# 2: b 1
## Check that error _is_ thrown when when verboten elements make it past filter
dt[a<=0, if(any(a<=0)) stop("a<=0 must've been passed on to j") else a, by=grp]
# Error in `[.data.table`(dt, a <= 0, if (any(a <= 0)) \\
# stop("a<=0 must've been passed on to j") else a, :
# a<=0 must've been passed on to j
Run Code Online (Sandbox Code Playgroud)
要执行第二次过滤操作,只需将其置于第二次调用中[.data.table():
dt[,tot:=sum(a),by=grp][a>0,]
# a grp tot
# 1: 1 a 1
# 2: 1 b 1
Run Code Online (Sandbox Code Playgroud)