我有一个data.table,我想总结一下。看起来是这样:
> DF
new_src action
1: cdn.adnxs.com 1
2: cdn.adnxs.com 1
3: cdn.adnxs.com 1
4: cdn.adnxs.com 3
5: s1.2mdn.net 1
6: cdn.adnxs.com 3
7: cdn.adnxs.com 3
8: cdn.adnxs.com 3
9: cdn.adnxs.com 3
10: cdn.adnxs.com 3我要用new_src聚合,用action找到最高的发生率,计算频率,打印这个action,打印总数。我可以在ddply中使用table并在ddply中重用变量,这样就不需要多次运行table了。我需要在data.table中这样做,但是我不能重用table结果,我必须运行table两次。
举例说明。这样做是可行的:
DF = structure(list(new_src = c("cdn.adnxs.com", "cdn.adnxs.com",
"cdn.adnxs.com", "cdn.adnxs.com", "s1.2mdn.net", "cdn.adnxs.com",
"cdn.adnxs.com", "cdn.adnxs.com", "cdn.adnxs.com", "cdn.adnxs.com"), action = c("1", "1", "1", "3", "1", "3", "3", "3", "3", "3")), .Names = c("new_src", "action"), class = c("data.table", "data.frame"), row.names = c(NA, -10L))
dt = DF[1:10,by=list(new_src),list(tb = sort(table(action),decreasing=T)[1], nm = names(sort(table(action),decreasing=T)[1]),tot = .N)]
View(dt)
ddpl = ddply(DF,.(new_src),summarize,tb = sort(table(action),decreasing=T)[1], nm = names(tb), tot = length(new_src))
View(ddpl)这不可能。
dt = DF[1:10,by=list(new_src),list(tb = sort(table(action),decreasing=T)[1], nm = names(tb),tot = .N)]有可能用data.table吗?谢谢
发布于 2016-11-11 16:21:18
我想你想让.N在这里:
DF[, .N, by=.(new_src, action)][
order(-N), .(topv = action[1], topn = N[1], n = sum(N)), by=new_src]
new_src topv topn n
1: cdn.adnxs.com 3 6 9
2: s1.2mdn.net 1 1 1若要处理关系,请向order(-N, ...)添加更多参数。
嵌套不是链接by=,而是另一种选择:
DF[, .SD[, .N, by=action][order(-N), c(.SD[1], .(totn = sum(.N)))], by=new_src]
new_src action N totn
1: cdn.adnxs.com 3 6 2
2: s1.2mdn.net 1 1 1不过,我发现很难理解;而且由于 is optimized,它可能会慢一些。
https://stackoverflow.com/questions/40552051
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