如何解决如何在语法 gramEvol 包中生成数据帧
我需要在 gramEvol 包中制作数据框
set.seed(123)
dt <- as.data.frame(matrix(ncol = 4,nrow = 0))
colnames(dt) <- c("pat","price","mindist","StopNext")
# make dataframe with random rows
for(i in 1:sample(5:15,1)){
pat <- sample(c("O","H","L"),1)
price <- sample(-5:5,size = 1)
mindist <- sample(seq(0,0.5,by = 0.1),1)
StopNext <- sample(c("stop","next"),1,prob = c(0.3,0.7))
res <- data.frame(pat,price,mindist,StopNext)
dt <- rbind.data.frame(dt,res)
}
喜欢这个
dt
pat price mindist StopNext
1 H -3 0.1 next
2 H 5 0.4 stop
3 H 0 0.0 next
4 L -1 0.2 stop
5 O 3 0.0 next
6 L 2 0.1 next
7 H 3 0.5 next
但我不明白如何在 gramEvol
这是我得到的
library("gramEvol")
ruleDef <- list( result = grule( data.frame(pat,StopNext) ),pat = grule("O",price = grule( -5,-4,-3,-2,-1,2,3,4,5),mindist = grule( 0.01,0.02,0.03,0.04,0.05,0.1,0.2,0.3,0.4,0.5 ),StopNext = grule("stop","next")
)
grammarDef <- CreateGrammar(ruleDef)
GrammarRandomExpression(grammarDef,numExpr = 3)
[[1]]
expression(data.frame("H","next"))
[[2]]
expression(data.frame("O","next"))
[[3]]
expression(data.frame("O","stop"))
但我想获得三个日期框架
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