快速处理模拟中的规则 [英] Fast handling of rules in a simulation
问题描述
如果离散事件模拟中只有几个规则，那么这并不重要，但是如果有很多规则并且它们可以互相干扰，则您可能希望跟踪它们的哪个"和哪里"使用.
If you only have a few rules in a discrete event simulation this is not critical but if you have a lot of them and they can interfere with each other and you may want to track the "which" and "where" they are used.
- 有人知道如何像原始函数一样快速地获取下面的代码吗?
- 是否有比
eval(parse(...)更好的选择?
- Does anybody know how to get the code below as fast as the original function?
- Are there better options than
eval(parse(...)?
这是一个简单的示例，显示我将速度降低了100 .假设您运行了一个模拟，其中一个(许多规则中)是:选择时间少于5的状态:
Here is an simple example which shows that I loose a factor 100 in speed. Assume you run a simulation and one (of many rules) is: Select the states with time less 5:
> a <- rnorm(100, 50, 10)
> print(summary(microbenchmark::microbenchmark(a[a < 5], times = 1000L, unit = "us")))
expr min lq mean median uq max neval
a[a < 5] 0.76 1.14 1.266745 1.141 1.52 11.404 1000
myfun <- function(a0) {
return(eval(parse(text = myrule)))
}
> myrule <- "a < a0" # The rule could be read from a file.
print(summary(microbenchmark::microbenchmark(a[myfun(5)], times = 1000L, unit = "us")))
expr min lq mean median uq max neval
a[myfun(5)] 137.61 140.271 145.6047 141.411 142.932 343.644 1000
注意:我认为我不需要额外的 rete包，可以有效地进行簿记.但是，如果还有其他意见，请告诉我...
Note: I don't think that I need an extra rete package which can do the book keeping efficiently. But if there are other opinions, let me know...
推荐答案
让我们对此进行配置:
Rprof()
for (i in 1:1e4) a[myfun(5)]
Rprof(NULL)
summaryRprof()
#$by.self
# self.time self.pct total.time total.pct
#"parse" 0.36 69.23 0.48 92.31
#"structure" 0.04 7.69 0.06 11.54
#"myfun" 0.02 3.85 0.52 100.00
#"eval" 0.02 3.85 0.50 96.15
#"stopifnot" 0.02 3.85 0.06 11.54
#"%in%" 0.02 3.85 0.02 3.85
#"anyNA" 0.02 3.85 0.02 3.85
#"sys.parent" 0.02 3.85 0.02 3.85
#
#$by.total
# total.time total.pct self.time self.pct
#"myfun" 0.52 100.00 0.02 3.85
#"eval" 0.50 96.15 0.02 3.85
#"parse" 0.48 92.31 0.36 69.23
#"srcfilecopy" 0.12 23.08 0.00 0.00
#"structure" 0.06 11.54 0.04 7.69
#"stopifnot" 0.06 11.54 0.02 3.85
#".POSIXct" 0.06 11.54 0.00 0.00
#"Sys.time" 0.06 11.54 0.00 0.00
#"%in%" 0.02 3.85 0.02 3.85
#"anyNA" 0.02 3.85 0.02 3.85
#"sys.parent" 0.02 3.85 0.02 3.85
#"match.call" 0.02 3.85 0.00 0.00
#"sys.function" 0.02 3.85 0.00 0.00
大部分时间都用在parse中.我们可以使用基准进行确认:
Most of the time is spent in parse. We can confirm this with a benchmark:
microbenchmark(a[myfun(5)], times = 1000L, unit = "us")
#Unit: microseconds
# expr min lq mean median uq max neval
# a[myfun(5)] 67.347 69.141 72.12806 69.909 70.933 160.303 1000
a0 <- 5
microbenchmark(parse(text = myrule), times = 1000L, unit = "us")
#Unit: microseconds
# expr min lq mean median uq max neval
# parse(text = myrule) 62.483 64.275 64.99432 64.787 65.299 132.903 1000
如果从文件中以文本形式读取规则非常困难，那么我认为没有办法加快速度.当然，您不应该重复解析相同的规则，但现在我假设您是这样.
If reading the rules as text from a file is a hard requirement, I don't think there is a way to speed this up. Of course, you should not parse the same rule repeatedly, but I assume you now that.
根据评论进行编辑以提供更多说明:
您应该将规则存储为带引号的表达式(例如，如果需要将其作为文件使用，请使用saveRDS列表):
You should store your rules as quoted expressions (e.g., in a list using saveRDS if you need them as a file):
myrule1 <- quote(a < a0)
myfun1 <- function(rule, a, a0) {eval(rule)}
microbenchmark(a[myfun1(myrule1, a, 30)], times = 1000L, unit = "us")
#Unit: microseconds
# expr min lq mean median uq max neval
# a[myfun1(myrule1, a, 30)] 1.792 2.049 2.286815 2.304 2.305 30.217 1000
为方便起见，您可以将表达式列表设为S3对象，并为其创建漂亮的print方法，以便获得更好的概览.
For convenience, you could then make that list of expressions an S3 object and create a nice print method for it in order to get a better overview.
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