对for循环进行矢量化
是否可以矢量化以下函数（对for循环进行矢量化,r,parallel-processing,data.table,vectorization,R,Parallel Processing,Data.table,Vectorization,是否可以矢量化以下函数（f） x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20) > f(0.5, x) [1] 16.56635
fx <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
xpfx <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
data.tablex <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
# My mock data
x <- data.frame(x=rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20))
# The function to optimise for
f <- function(p,x){
# Generate columns before filling
x$multiplier <- NA
x$cumulative <- NA
for(i in 1:nrow(x)){
# Going through each row systematically
if(i==1){
# If first row do a slightly different set of commands
x[i,'multiplier'] <- 1 * p
x[i,'cumulative'] <- (x[i,'multiplier'] * x[i,'x']) + 1
} else {
# For the rest of the rows carry out these commands
x[i,'multiplier'] <- x[i-1,'cumulative'] * p
x[i,'cumulative'] <- (x[i,'multiplier'] * x[i,'x']) + x[i-1,'cumulative']
}
}
# output the final row's output for the cumulative column
as.numeric(x[nrow(x),'cumulative'])
}
# Checking the function works by putting in a test value of p = 0.5
f(0.5,x)
# Now optimise the function between the interval of p between 0 and 1
optim.p <- optimise(f=f, interval=c(0,1),x, maximum=TRUE)
# Viewing the output of optim.p
optim.p
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
#我的模拟数据
x（编辑-忘记了我写的文章的第一部分，现在就放进去）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
通过检查函数f
的实际功能，可以简化您的问题。因为我很懒，所以我要写<代码> x[i，'乘子'/c>作为MI，<代码> x [ i，'累积' ] < /> >为彝，< <代码> x[i，'x' ] < /> >席。
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
让我们看看f
中的等式。我们首先来看案例i>1
：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
mi=yi-1*p
依＝mi*席＋Y-1～<
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
将上述m_i替换为：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
Y=（Yi-1*P）*席+Y-1//…
Yi＝Y-1*（P*席+ 1）< /P>
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
这样就无需计算乘法器
列
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
现在再仔细看一下你的i==1
案例，我们发现如果我们把y0放在1，那么下面的结果适用于所有i=1，nrow（x）
：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yi=yi-1（pxi+1）----（1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
查看函数f
，您要计算的是yn：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=yn-1（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
如果我们使用（1）替换上述公式中的yn-1，会发生什么情况
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=yn-2（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
现在我们用yn-2的公式替换上述内容：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=yn-3（pxn-2+1）（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
你明白了，对吧？我们一直替换到y1：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=y0（px1+1）（px2+1）…（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
但请记住，y0只是1。因此，要计算f（x，p）
的值，我们只需：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
f（x，p）=（px1+1）（px2+1）…（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
其中n
是nrow（x）
。也就是说，为每个i
计算p*x[i，'x']+1
，并将它们全部相乘
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
要将R中的数字向量相乘，可以使用prod
。因此，如果x
只是一个向量：
f_version2 <- function(p, x) {
return(prod(p * x + 1))
}
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
f_version2（编辑-忘记了我写的文章的第一部分，现在就放进去）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
通过检查函数f
的实际功能，可以简化您的问题。因为我很懒，所以我要写<代码> x[i，'乘子'/c>作为MI，<代码> x [ i，'累积' ] < /> >为彝，< <代码> x[i，'x' ] < /> >席。
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
让我们看看f
中的等式。我们首先来看案例i>1
：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
mi=yi-1*p
依＝mi*席＋Y-1～<
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
将上述m_i替换为：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
Y=（Yi-1*P）*席+Y-1//…
Yi＝Y-1*（P*席+ 1）< /P>
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
这样就无需计算乘法器
列
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
现在再仔细看一下你的i==1
案例，我们发现如果我们把y0放在1，那么下面的结果适用于所有i=1，nrow（x）
：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yi=yi-1（pxi+1）----（1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
查看函数f
，您要计算的是yn：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=yn-1（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
如果我们使用（1）替换上述公式中的yn-1，会发生什么情况
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=yn-2（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
现在我们用yn-2的公式替换上述内容：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=yn-3（pxn-2+1）（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
你明白了，对吧？我们一直替换到y1：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
yn=y0（px1+1）（px2+1）…（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
但请记住，y0只是1。因此，要计算f（x，p）
的值，我们只需：
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
f（x，p）=（px1+1）（px2+1）…（pxn-1+1）（pxn+1）
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
其中n
是nrow（x）
。也就是说，为每个i
计算p*x[i，'x']+1
，并将它们全部相乘
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
要将R中的数字向量相乘，可以使用prod
。因此，如果x
只是一个向量：
f_version2 <- function(p, x) {
return(prod(p * x + 1))
}
x <- rep(c(rep(c(0.2,-0.2),4),0.2,0.2,-0.2,0.2),20)
> f(0.5, x)
[1] 16.56635
> f_version2(0.5, x)
[1] 16.56635
f_version2+6.02*10^23仅最后一句话的评分+6.02*10^23仅最后一句话的评分！