将所有逻辑规则与数据帧匹配（需要超快功能）
我有一个函数可以检查数据帧中是否存在逻辑序列
fu <- function(dat , rule , res.only=T){
debug.vec <- rep("no",nrow(dat)) # control of rule triggers
rule.id <- 1 # rule number in vector
for(i in 1:nrow(dat)){
# check if the rule "rule[rule.id]" has worked on this "i" index in dat[i,]
current_rule <- with(data = dat[i,] , expr = eval(parse(text = rule[rule.id])) )
if(current_rule){ # if the rule is triggered
debug.vec[i] <- rule[rule.id]
if( rule.id==length(rule) ) break # stop if there are no more rules
rule.id <- rule.id+1 # go to the next rule
}}
if(!res.only) return( cbind(dat,debug.vec) )
return( sum(debug.vec!="no")==length(rule) )
}
例如我有一些数据
set.seed(123)
dat <- as.data.frame(matrix(data = sample(10,30,replace = T),ncol = 3))
colnames(dat) <- paste0("x" ,1:ncol(dat))
..
dat
x1 x2 x3
1 3 5 9
2 3 3 3
3 10 9 4
4 2 9 1
5 6 9 7
6 5 3 5
7 4 8 10
8 6 10 7
9 9 7 9
10 10 10 9
还有一个带有规则的向量
rule <- c("x1>5 & x2>2" , "x1>x2" , "x3!=4" )
该函数检查数据帧中是否存在这样的逻辑序列并给出逻辑答案
> fu(dat = dat, rule = rule, res.only = T)
[1] TRUE
或者您可以更改标志res.only = F并查看序列在debug.vec列中的位置
> fu(dat = dat, rule = rule, res.only = F)
x1 x2 x3 debug.vec
1 3 5 9 no
2 3 3 3 no
3 10 9 4 x1>5 & x2>2
4 2 9 1 no
5 6 9 7 no
6 5 3 5 x1>x2
7 4 8 10 x3!=4
8 6 10 7 no
9 9 7 9 no
10 10 10 9 no
我需要这个函数的最快版本，也许使用Rccp 包或类似的东西..
UPD=======================
该Waldi功能与我的功能不同，出现问题
UPD_2_====================================
# Is this correct?
是的，这是正确的，如果触发了规则 [k] 然后搜索规则 [k+1] 从新的一行数据开始
原谅我的问题不够准确，这是我的错
我的函数返回FALSE是因为最后一条规则"x3!=4"不起作用，应该是
dat <- structure(list(x1 = c(2L, 5L, 1L, 3L, 9L, 2L, 6L, 3L, 3L, 9L),
x2 = c(2L, 1L, 6L, 10L, 8L, 10L, 10L, 4L, 6L, 4L),
x3 = c(4L, 9L, 8L, 7L, 10L, 1L, 2L, 8L, 3L, 10L)),
class = "data.frame", row.names = c(NA, -10L))
dat
rule <- c("x1>5 & x2>2" , "x1>x2" , "x3!=4" )
my_fu(dat = dat, rule = rule, res.only = F)
只有两条规则有效
> my_fu(dat = dat, rule = rule, res.only = F)
x1 x2 x3 debug.vec
1 2 2 4 no
2 5 1 9 no
3 1 6 8 no
4 3 10 7 no
5 9 8 10 x1>5 & x2>2
6 2 10 1 no
7 6 10 2 no
8 3 4 8 no
9 3 6 3 no
10 9 4 10 x1>x2
它应该是
> my_fu(dat = dat, rule = rule, res.only = T)
[1] FALSE
回答
更新
根据您的更新，我编写了一个新fu函数，即TIC_fu()
TIC_fu <- function(dat, rule, res.only = TRUE) {
m <- with(dat, lapply(rule, function(r) eval(str2expression(r))))
idx <- na.omit(
Reduce(
function(x, y) {
k <- which(y)
ifelse(all(k <= x), NA, min(k[k > x]))
}, m,
init = 0, accumulate = TRUE
)
)[-1]
if (!res.only) {
fidx <- head(idx, length(rule))
debug.vec <- replace(rep("no", nrow(dat)), fidx, rule[seq_along(fidx)])
return(cbind(dat, debug.vec))
}
length(idx) >= length(rule)
}
你会看到
> TIC_fu(dat, rule, FALSE)
x1 x2 x3 debug.vec
1 2 2 4 no
2 5 1 9 no
3 1 6 8 no
4 3 10 7 no
5 9 8 10 x1>5 & x2>2
6 2 10 1 no
7 6 10 2 no
8 3 4 8 no
9 3 6 3 no
10 9 4 10 x1>x2
> TIC_fu(dat,rule)
[1] FALSE
用于基准测试
> microbenchmark(
+ TIC_fu(dat, rule, FALSE),
+ fu(dat, rule, FALSE),
+ unit = "relative"
+ )
Unit: relative
expr min lq mean median uq max
TIC_fu(dat, rule, FALSE) 1.000000 1.000000 1.000000 1.000000 1.0000 1.000000
fu(dat, rule, FALSE) 4.639093 4.555523 3.383911 4.450056 4.3993 1.007532
neval
100
100
上一个答案
这里有一些类似于@Waldi所做的选项，但唯一的区别是parse,str2lang和str2expression
microbenchmark::microbenchmark(
any(with(dat, rowSums(sapply(rule, function(rule) eval(parse(text = rule))))==length(rule))),
any(with(dat, rowSums(sapply(rule, function(rule) eval(str2lang(rule))))==length(rule))),
any(with(dat, rowSums(sapply(rule, function(rule) eval(str2expression(rule))))==length(rule))),
any(with(dat, eval(str2expression(paste0(rule,collapse = " & ")))))
)
你会看到
Unit: microseconds
expr
any(with(dat, rowSums(sapply(rule, function(rule) eval(parse(text = rule)))) == length(rule)))
any(with(dat, rowSums(sapply(rule, function(rule) eval(str2lang(rule)))) == length(rule)))
any(with(dat, rowSums(sapply(rule, function(rule) eval(str2expression(rule)))) == length(rule)))
any(with(dat, eval(str2expression(paste0(rule, collapse = " & ")))))
min lq mean median uq max neval
94.0 98.6 131.431 107.35 121.90 632.7 100
37.5 39.2 48.887 44.05 48.50 174.1 100
36.8 39.6 51.627 46.20 48.45 241.4 100
12.7 15.8 19.786 17.00 19.75 97.9 100