将data.table聚合到原始值的间隔行

Question

我有一些data.table数量列如下:

n = 1e5
set.seed(1)

dt <- data.table(id = 1:n, amount = pmax(0,rnorm(n, mean = 5e3, sd = 1e4)))

并给出了一个断点矢量:

breaks <- as.vector( c(0, t(sapply(c(1, 2.5, 5, 7.5), function(x) x * 10^(1:4))) ) )

对于这些中断定义的每个间隔,我想使用data.table语法:

1. 得到amount包含的数量
2. 获得amount等于或大于左边界的计数(基本上n * (1-cdf(amount))

对于1,这主要起作用,但不返回空间隔的行:

dt[, .N, keyby = breaks[findInterval(amount,breaks)] ] #would prefer to get 0 for empty intvl

对于2,我试过:

dt[, sum(amount >= thresh[.GRP]), keyby = breaks[findInterval(amount,breaks)]  ]

但它不起作用,因为sum仅限于集团内部,而不是超出.因此提出了一个解决方法,它也返回空间隔:

dt[, cbind(breaks, sapply(breaks, function(x) sum(amount >= x)))] # desired result

那么,data.table修复我的方法是什么？并获得两者的空间隔？

Answer 1

我会考虑这样做:

mybreaks = c(-Inf, breaks, Inf)
dt[, g := cut(amount, mybreaks)]
dt[.(g = levels(g)), .N, on="g", by=.EACHI]


                  g     N
 1:        (-Inf,0] 30976
 2:          (0,10]    23
 3:         (10,25]    62
 4:         (25,50]    73
 5:         (50,75]    85
 6:        (75,100]    88
 7:       (100,250]   503
 8:       (250,500]   859
 9:       (500,750]   916
10:     (750,1e+03]   912
11: (1e+03,2.5e+03]  5593
12: (2.5e+03,5e+03]  9884
13: (5e+03,7.5e+03]  9767
14: (7.5e+03,1e+04]  9474
15: (1e+04,2.5e+04] 28434
16: (2.5e+04,5e+04]  2351
17: (5e+04,7.5e+04]     0
18:  (7.5e+04, Inf]     0

cumsum如果你想要CDF,你可以使用.