使用方程式计算变量，然后使用生成的值生成新的值
Calculate variables using equations then use the generated values to generate new one
很难解释，但我会一步一步来
假设我有 2 辆车，一前一后，我知道前车的速度，我想计算两辆车之间的距离，我们可以使用多个方程式计算距离，我也知道后车的初速和两车之间的距离
Following_Car_Speed = 13.68490 m/s
Distance = 17.024 m
Lead_Car_Speed = c(13.784896, 13.745834, 13.880556, 13.893577, 13.893577, 13.923959,
13.945661, 13.919619, 13.897917, 14.002257, 14.002257, 13.980556,
13.980556, 14.067536, 14.063195, 14.080556, 14.123959, 14.163021,
14.236806, 14.167188)
Delta_Speed = Lead_Car_Speed[1]-Following_Car_Speed = 13.784896-13.68490 = 0.1
Gap <- 1.554 + Following_Car_Speed*0.878- (Following_Car_Speed*Delta_Speed)/(2*sqrt(0.8418*0.8150))=
1.554+ 13.68490*0.878- (13.68490*0.1)/(2*sqrt(0.8418*0.8150) = 12.74325
Acceleration <- 0.8418*(1-(Following_Car_Speed/29.2)^3.52-(Gap/Distance)^2)=0.3116923
现在我已经计算了加速度，所以我要计算后面的车的新速度。
Following_Car_Speed <- Following_Car_Speed + Acceleration*0.1
所以现在我要计算前车和后车之间的新速度差值
Delta_Speed <- Lead_Car_Speed[2]-Following_Car_Speed
Distance<- Distance+(Delta_Speed[2]+Delta_Speed[1])/2*0.1
然后继续使用相同的方程式，直到我们结束以下汽车的所有值。
使用 For 循环很容易做到这一点，但我想获得更有效的方法，我尝试使用 dplyr，但很难，我没有得到答案。
所以请帮助我。
myfun <- function(list, lcs,lcs2){
ds <- lcs - list[[1]]
Distance <- list[[1]]*D_Time - (list[[1]] * ds) / (2*sqrt(M_Acc*Com_Acc))
if (Distance < 0|is.na(Distance)) {Distance <- 0}
gap <- Gap_J + Distance
acc <- M_Acc * (1 - (list[[1]] / D_Speed)^Beta - (gap / list[[2]])^2)
fcs_new <- list[[1]] + acc * 0.1
ds_new <- lcs2- fcs_new
di_new <- list[[2]]+(ds_new+ds)/2*0.1
return(list(Speed = fcs_new,Distance = di_new))
}
Generated_Data <- data %>%group_by(Driver,FileName)%>%
mutate(Speed_Distance_Calibrated = accumulate2( .init = list(Filling_Speed[1],
Filling_Range[1]),.x = Lead_Veh_Speed_F,.y = Lead_Veh_Speed_F2, myfun)[-1])%>%ungroup()
我已经添加了 lead_car_speed 的领先优势，我也想使用新的距离和新的速度，所以我将它放入列表并放入 .initla
您的操作取决于迭代或某种递归来完成（由于可能 NA、Inf 等）- 看起来像 for 或 while需要。
我的替代版本是将函数分解为更简单的单元，以便对更大的示例进行缩放和单元测试，或者用另一个函数调用环绕。
我假设您的 var Distance 由于某种原因已修复并且未提供预期的输出，因此我无法完全验证函数。
Following_Car_Speed <- 13.68490
Lead_Car_Speed <- c(13.784896, 13.745834, 13.880556, 13.893577, 13.893577,
13.923959, 13.945661, 13.919619, 13.897917, 14.002257,
14.002257, 13.980556, 13.980556, 14.067536, 14.063195,
14.080556, 14.123959, 14.163021, 14.236806, 14.167188)
Distance <- 17.024
#init
i <- length(Lead_Car_Speed)
Delta_Speed <- c(0.1, vector(mode = "numeric", length = i-1))
Gap <- c(12.74325, vector(mode = "numeric", length = i-1))
Acceleration <- c(0.3116923, vector(mode = "numeric", length = i-1 ))
Following_Car_Speed <- c(Following_Car_Speed, vector(mode="numeric", length = i-1 ))
cal.following.car.speed <- function(Following_Car_Speed, Acceleration){
#calculate and return current following-car-speed: the i-th following-car-speed
current.follow.car.speed <- Following_Car_Speed + Acceleration*0.1
return(current.follow.car.speed)
}
cal.delta.speed <- function(Lead_Car_Speed, following.car.speed){
#calculate and return current delta speed: the i-th delta speed
current.delta.speed <- Lead_Car_Speed - following.car.speed
return(current.delta.speed)
}
cal.gap <- function(Following_Car_Speed, Delta_Speed){
#calculate and return current gap: the i-th gap
current.gap <- 1.554 + Following_Car_Speed*0.878 - (Following_Car_Speed*Delta_Speed)/(2*sqrt(0.8418*0.8150))
return(current.gap)
}
cal.acceleration <- function(Following_Car_Speed, Gap, Distance){
#calculate and return current acceleration: the i-th acceleration
current.acceleration <- 0.8418*(1-(Following_Car_Speed/29.2)^3.52-(Gap/Distance)^2)
return(current.acceleration)
}
#main
counter <- 1
while(counter != i){
if(counter == 1) {counter = counter + 1} #skip 1st ith as we have init
else{
Gap[counter] <- cal.gap(Following_Car_Speed[counter-1], Delta_Speed[counter-1])
Acceleration[counter] <- cal.acceleration(Following_Car_Speed[counter-1], Gap[counter-1], Distance)
Following_Car_Speed[counter] <- cal.following.car.speed(Following_Car_Speed[counter-1], Acceleration[counter])
Delta_Speed[counter] <- cal.delta.speed(Lead_Car_Speed[counter], Following_Car_Speed[counter])
counter = counter + 1
}
}
cbind(Delta_Speed, Gap, Acceleration, Following_Car_Speed)
这是使用 purrr 包中的 accumulate 的简单方法，它是 tidyverse 的一部分。
首先我定义了一个函数 myfun 来更新 following_car_speed (fcs)。
myfun <- function(fcs, lcs, di){
ds <- lcs - fcs
gap <- 1.554 + fcs*0.878 - fcs * ds / (2*sqrt(0.8418*0.8150))
acc <- 0.8418 * (1 - (fcs / 29.2)^3.52 - (gap / di)^2)
fcs_new <- fcs + acc * 0.1
return(fcs_new)
}
library(tidyverse)
tibble(lead_car_speed = c(13.784896, 13.745834, 13.880556, 13.893577, 13.893577, 13.923959,
13.945661, 13.919619, 13.897917, 14.002257, 14.002257, 13.980556,
13.980556, 14.067536, 14.063195, 14.080556, 14.123959, 14.163021,
14.236806, 14.167188)) %>%
mutate(following_car_speed = accumulate(lead_car_speed, myfun, .init = 13.68490, di = 17.024)[-1])^
# A tibble: 20 x 2
lead_car_speed following_car_speed
<dbl> <dbl>
1 13.8 13.7
2 13.7 13.7
3 13.9 13.8
4 13.9 13.8
5 13.9 13.8
6 13.9 13.9
7 13.9 13.9
8 13.9 13.9
9 13.9 13.9
10 14.0 14.0
11 14.0 14.0
12 14.0 14.0
13 14.0 14.0
14 14.1 14.1
15 14.1 14.1
16 14.1 14.1
17 14.1 14.1
18 14.2 14.1
19 14.2 14.2
20 14.2 14.2
如果距离也发生变化，您可以使用 accumulate2 而不是 accumulate。
很难解释，但我会一步一步来
假设我有 2 辆车，一前一后，我知道前车的速度，我想计算两辆车之间的距离，我们可以使用多个方程式计算距离，我也知道后车的初速和两车之间的距离
Following_Car_Speed = 13.68490 m/s
Distance = 17.024 m
Lead_Car_Speed = c(13.784896, 13.745834, 13.880556, 13.893577, 13.893577, 13.923959,
13.945661, 13.919619, 13.897917, 14.002257, 14.002257, 13.980556,
13.980556, 14.067536, 14.063195, 14.080556, 14.123959, 14.163021,
14.236806, 14.167188)
Delta_Speed = Lead_Car_Speed[1]-Following_Car_Speed = 13.784896-13.68490 = 0.1
Gap <- 1.554 + Following_Car_Speed*0.878- (Following_Car_Speed*Delta_Speed)/(2*sqrt(0.8418*0.8150))=
1.554+ 13.68490*0.878- (13.68490*0.1)/(2*sqrt(0.8418*0.8150) = 12.74325
Acceleration <- 0.8418*(1-(Following_Car_Speed/29.2)^3.52-(Gap/Distance)^2)=0.3116923
现在我已经计算了加速度，所以我要计算后面的车的新速度。
Following_Car_Speed <- Following_Car_Speed + Acceleration*0.1
所以现在我要计算前车和后车之间的新速度差值
Delta_Speed <- Lead_Car_Speed[2]-Following_Car_Speed
Distance<- Distance+(Delta_Speed[2]+Delta_Speed[1])/2*0.1
然后继续使用相同的方程式，直到我们结束以下汽车的所有值。
使用 For 循环很容易做到这一点，但我想获得更有效的方法，我尝试使用 dplyr，但很难，我没有得到答案。
所以请帮助我。
myfun <- function(list, lcs,lcs2){
ds <- lcs - list[[1]]
Distance <- list[[1]]*D_Time - (list[[1]] * ds) / (2*sqrt(M_Acc*Com_Acc))
if (Distance < 0|is.na(Distance)) {Distance <- 0}
gap <- Gap_J + Distance
acc <- M_Acc * (1 - (list[[1]] / D_Speed)^Beta - (gap / list[[2]])^2)
fcs_new <- list[[1]] + acc * 0.1
ds_new <- lcs2- fcs_new
di_new <- list[[2]]+(ds_new+ds)/2*0.1
return(list(Speed = fcs_new,Distance = di_new))
}
Generated_Data <- data %>%group_by(Driver,FileName)%>%
mutate(Speed_Distance_Calibrated = accumulate2( .init = list(Filling_Speed[1],
Filling_Range[1]),.x = Lead_Veh_Speed_F,.y = Lead_Veh_Speed_F2, myfun)[-1])%>%ungroup()
我已经添加了 lead_car_speed 的领先优势，我也想使用新的距离和新的速度，所以我将它放入列表并放入 .initla
您的操作取决于迭代或某种递归来完成（由于可能 NA、Inf 等）- 看起来像 for 或 while需要。
我的替代版本是将函数分解为更简单的单元，以便对更大的示例进行缩放和单元测试，或者用另一个函数调用环绕。
我假设您的 var Distance 由于某种原因已修复并且未提供预期的输出，因此我无法完全验证函数。
Following_Car_Speed <- 13.68490
Lead_Car_Speed <- c(13.784896, 13.745834, 13.880556, 13.893577, 13.893577,
13.923959, 13.945661, 13.919619, 13.897917, 14.002257,
14.002257, 13.980556, 13.980556, 14.067536, 14.063195,
14.080556, 14.123959, 14.163021, 14.236806, 14.167188)
Distance <- 17.024
#init
i <- length(Lead_Car_Speed)
Delta_Speed <- c(0.1, vector(mode = "numeric", length = i-1))
Gap <- c(12.74325, vector(mode = "numeric", length = i-1))
Acceleration <- c(0.3116923, vector(mode = "numeric", length = i-1 ))
Following_Car_Speed <- c(Following_Car_Speed, vector(mode="numeric", length = i-1 ))
cal.following.car.speed <- function(Following_Car_Speed, Acceleration){
#calculate and return current following-car-speed: the i-th following-car-speed
current.follow.car.speed <- Following_Car_Speed + Acceleration*0.1
return(current.follow.car.speed)
}
cal.delta.speed <- function(Lead_Car_Speed, following.car.speed){
#calculate and return current delta speed: the i-th delta speed
current.delta.speed <- Lead_Car_Speed - following.car.speed
return(current.delta.speed)
}
cal.gap <- function(Following_Car_Speed, Delta_Speed){
#calculate and return current gap: the i-th gap
current.gap <- 1.554 + Following_Car_Speed*0.878 - (Following_Car_Speed*Delta_Speed)/(2*sqrt(0.8418*0.8150))
return(current.gap)
}
cal.acceleration <- function(Following_Car_Speed, Gap, Distance){
#calculate and return current acceleration: the i-th acceleration
current.acceleration <- 0.8418*(1-(Following_Car_Speed/29.2)^3.52-(Gap/Distance)^2)
return(current.acceleration)
}
#main
counter <- 1
while(counter != i){
if(counter == 1) {counter = counter + 1} #skip 1st ith as we have init
else{
Gap[counter] <- cal.gap(Following_Car_Speed[counter-1], Delta_Speed[counter-1])
Acceleration[counter] <- cal.acceleration(Following_Car_Speed[counter-1], Gap[counter-1], Distance)
Following_Car_Speed[counter] <- cal.following.car.speed(Following_Car_Speed[counter-1], Acceleration[counter])
Delta_Speed[counter] <- cal.delta.speed(Lead_Car_Speed[counter], Following_Car_Speed[counter])
counter = counter + 1
}
}
cbind(Delta_Speed, Gap, Acceleration, Following_Car_Speed)
这是使用 purrr 包中的 accumulate 的简单方法，它是 tidyverse 的一部分。
首先我定义了一个函数 myfun 来更新 following_car_speed (fcs)。
myfun <- function(fcs, lcs, di){
ds <- lcs - fcs
gap <- 1.554 + fcs*0.878 - fcs * ds / (2*sqrt(0.8418*0.8150))
acc <- 0.8418 * (1 - (fcs / 29.2)^3.52 - (gap / di)^2)
fcs_new <- fcs + acc * 0.1
return(fcs_new)
}
library(tidyverse)
tibble(lead_car_speed = c(13.784896, 13.745834, 13.880556, 13.893577, 13.893577, 13.923959,
13.945661, 13.919619, 13.897917, 14.002257, 14.002257, 13.980556,
13.980556, 14.067536, 14.063195, 14.080556, 14.123959, 14.163021,
14.236806, 14.167188)) %>%
mutate(following_car_speed = accumulate(lead_car_speed, myfun, .init = 13.68490, di = 17.024)[-1])^
# A tibble: 20 x 2
lead_car_speed following_car_speed
<dbl> <dbl>
1 13.8 13.7
2 13.7 13.7
3 13.9 13.8
4 13.9 13.8
5 13.9 13.8
6 13.9 13.9
7 13.9 13.9
8 13.9 13.9
9 13.9 13.9
10 14.0 14.0
11 14.0 14.0
12 14.0 14.0
13 14.0 14.0
14 14.1 14.1
15 14.1 14.1
16 14.1 14.1
17 14.1 14.1
18 14.2 14.1
19 14.2 14.2
20 14.2 14.2
如果距离也发生变化，您可以使用 accumulate2 而不是 accumulate。