计算通过和失败每种条件的观察值 [英] Count observations that pass and fail each condition
问题描述
这是对上一个问题的后续操作，该问题显示在此处。
This is a follow-up to a previous question shown here.
我有类似以下数据：
set.seed(1)
dt <- data.table(ID=1:10, Status=c(rep("OUT",2),rep("IN",2),"ON",rep("OUT",2),rep("IN",2),"ON"),
t1=round(rnorm(10),1), t2=round(rnorm(10),1), t3=round(rnorm(10),1),
t4=round(rnorm(10),1), t5=round(rnorm(10),1), t6=round(rnorm(10),1),
t7=round(rnorm(10),1),t8=round(rnorm(10),1))
ID Status t1 t2 t3 t4 t5 t6 t7 t8
1: 1 OUT -0.6 1.5 0.9 1.4 -0.2 0.4 2.4 0.5
2: 2 OUT 0.2 0.4 0.8 -0.1 -0.3 -0.6 0.0 -0.7
3: 3 IN -0.8 -0.6 0.1 0.4 0.7 0.3 0.7 0.6
4: 4 IN 1.6 -2.2 -2.0 -0.1 0.6 -1.1 0.0 -0.9
5: 5 ON 0.3 1.1 0.6 -1.4 -0.7 1.4 -0.7 -1.3
6: 6 OUT -0.8 0.0 -0.1 -0.4 -0.7 2.0 0.2 0.3
7: 7 OUT 0.5 0.0 -0.2 -0.4 0.4 -0.4 -1.8 -0.4
8: 8 IN 0.7 0.9 -1.5 -0.1 0.8 -1.0 1.5 0.0
9: 9 IN 0.6 0.8 -0.5 1.1 -0.1 0.6 0.2 0.1
10: 10 ON -0.3 0.6 0.4 0.8 0.9 -0.1 2.2 -0.6
我将约束应用于从csv读取的 dt ：
I apply constraints to dt that are read in from a csv:
dt_constraints <- data.table(columns=c("t1","t3","t7","t8"), operator=c(rep(">=",2),rep("<=",2)),
values=c(-.6,-.5,2.4,.5))
columns operator values
1 t1 >= -0.6
2 t3 >= -0.5
3 t7 <= 2.4
4 t8 <= 0.5
@akrun提供了一个非常有效的解决方案：
@akrun provided a very efficient solution:
dt[eval(parse(text=do.call(paste, c(dt_constraints, collapse= ' & '))))]
我还需要另外两个信息：通过每个观察的数量约束以及未通过每个约束的观察数。输出可能如下所示：
I need two additional bits of information: the number of observations that pass each constraint and the number of observations that fail each constraint. The output could look like this:
columns operator values obs_pass obs_fail
1 t1 >= -0.6 8 2
2 t3 >= -0.5 8 2
3 t7 <= 2.4 10 0
4 t8 <= 0.5 9 1
任何想法都将不胜感激。谢谢。
Any ideas will be greatly appreciated. Thanks.
推荐答案
您可以构造 on = 表达式并执行加入每一行：
You could construct the on= expression and do a join for each row:
dt_constraints[, onexpr := sprintf("%s%sv", columns, operator)]
dt_constraints[, n_pass := dt[.(v = values), on=onexpr, .N, by=.EACHI]$N, by=onexpr]
dt_constraints[, n_fail := nrow(dt) - n_pass ]
by = .EACHI 应该允许您将相同的 onexpr 与多个不同的值一起使用，尽管我猜有更好的选择做到这一点的方法。如果 onexpr 是唯一的，那么...
The by=.EACHI should allow you to use the same onexpr with multiple different values, though I guess there are better ways to do that. If onexpr are unique, there's instead...
dt_constraints[, n_pass := dt[.(v = values), on=onexpr, .N], by=onexpr]
dt_constraints[, n_fail := nrow(dt) - n_pass ]
使用此方法一次过滤所有约束：*
To filter on all constraints at once using this approach:*
n_grp = nrow(dt_constraints)
w = seq_len(nrow(dt))
dt[dt_constraints[, {
w <- w[dt[w][.(v = values), on=onexpr, which=TRUE]]
if (.GRP == n_grp) w
}, by=onexpr]$V1]
ID Status t1 t2 t3 t4 t5 t6 t7 t8
1: 1 OUT -0.6 1.5 0.9 1.4 -0.2 0.4 2.4 0.5
2: 2 OUT 0.2 0.4 0.8 -0.1 -0.3 -0.6 0.0 -0.7
3: 5 ON 0.3 1.1 0.6 -1.4 -0.7 1.4 -0.7 -1.3
4: 7 OUT 0.5 0.0 -0.2 -0.4 0.4 -0.4 -1.8 -0.4
5: 9 IN 0.6 0.8 -0.5 1.1 -0.1 0.6 0.2 0.1
6: 10 ON -0.3 0.6 0.4 0.8 0.9 -0.1 2.2 -0.6
这可以迭代l像一个循环，只检查那些先前具有约束的行（ w ）。
This works iteratively like a loop, only checking those rows (w) for which prior constraints held.
*假定没有重复的 onexpr ，因为那没有道理..您可以先将每个 onexpr 过滤为最大 和> = 的值< = 的最小值。
*assuming no repeated onexpr, since that wouldn't make sense.. you could first filter per onexpr to max values for >= and min for <=.
OP的后续问题：
我将如何更改代码以执行以下操作：（1）给出第一个约束的通过/失败计数； （2）使用来自第一约束的通过子集的数据，给出第二约束的通过/失败计数； （3）使用第二个约束的通过次数子集的数据，给出第三个约束的通过/失败计数，依此类推。
How would I alter the code to perform the following: (1)give pass/fail counts for the first constraint; (2)using data subsetted by passes from the first constraint, give the pass/fail counts for the second constraint; (3)using data subsetted by passes from the second constraint, give the pass/fail counts for the third constraint, etc.
# iteratively count pass/fail
w = seq_len(nrow(dt))
dt_constraints[, {
w0 <- w
w <- w[dt[w][.(v = values), on=onexpr, which=TRUE]]
.(n_pass = length(w), n_fail = length(w0) - length(w))
}, by=onexpr]
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