Мне просто интересно, есть ли общий способ связать файл rmarkdown в файл Word docx, но иметь его таким образом, чтобы я также мог сохранить соответствующие цифры и строки кода. Я хочу иметь возможность легко цитировать и ссылаться на конкретную строку кода, ссылаясь на ее номерную строку, когда я объясняю код.
Например, я хотел бы иметь возможность сказать: «В строке кода 223, xy и z случаются "
Вот два примера фрагментов кода:
## Monetary value of harm from accident
D = 5
## Maximum and minimum effort costs
maxw = 5
minw = 1
## Ex-ante average distribution cost of effort
wbar = (maxw+minw)/2
#True a priori negligece standard based on ex-ante average of distribution of cost of effort
estar = 1 - wbar/(2*D)
## Number (100) of simulations run in the loop
iteration = (1)
## Series of empty vectors used later for aggregating values from each simulation. In order of appearance, we have vector of driver 1
age1 = c()
age2 = c()
age3 = c()
age4 = c()
aga = c()
agb = c()
agna = c()
## Series of empty vectors used later for aggregating drawn costs by each driver in each simulation.
w1 = c()
w2 = c()
w3 = c()
w4 = c()
Этот фрагмент кода состоит из пошаговой схемы последовательности воспроизведения для iteration количество симуляций.
#The simulation is encapsulated by a for-loop that repeats the simulation for the number of iterations that is set above.
for (i in iteration) {
# Before the simulation begins, I create a series of empty vectors that serve as placeholders for the changes in the legal threshold, the decisions of each driver, and the number of periods that have passed since a driver has been in an accident. These are all encapsulated by a "reset" call object, which resets all the vectors back to their empty forms for the next iteration/run-through of the simulation.
reset = {
t = 1
a = c() #lower bound of belief in estar
b = c() #upper bound of belief in estar
e1 = c() #empty vector of driver one's decisions
e2 = c()
e3 = c()
e4 = c()
e = c() #two empty vectors that will serve as placeholders for effort levels of drivers that get in an accident
eaccident = c()
na = c() #number of periods that take place before an accident takes place.
na1 = c() #number of periods that take place before Driver One has an accident.
na2 = c()
na3 = c()
na4 = c()
# Each driver draws a constant marginal cost of effort from a uniform four-point distribution. This constant marginal cost will be fixed for the entirety of the simulation.
w1[i] = sample(c(1,2,4,5),size = 1, replace = TRUE, prob = c(0.25,0.25,0.25,0.25))
w2[i] = sample(c(1,2,4,5),size = 1, replace = TRUE, prob = c(0.25,0.25,0.25,0.25))
w3[i] = sample(c(1,2,4,5),size = 1, replace = TRUE, prob = c(0.25,0.25,0.25,0.25))
w4[i] = sample(c(1,2,4,5),size = 1, replace = TRUE, prob = c(0.25,0.25,0.25,0.25))
}
# Below is a while-loop imbedded within the for-loop above
while (TRUE){
if(t==1){a[t]=0.40;b[t]=0.90}
if(t==1){na1[t]=0;na2[t]=0;na3[t]=0;na4[t]=0}
e[t] = 0
##Drivers problem: pick the e that will minimize your cost --- piecewise optimization. First object is if (e<=a), second is if (a<e<b), and third is when (e>=b).
e11 = (1-(w1[i]/(2*D)))
e12 = ((2+b[t])-sqrt((b[t]^2)-2*b[t]+1+3*(w1[i]/D)*(b[t]-a[t])))/3
e13 = b[t]
c11 = (w1[i]*e11)+(((e11-1)^2)*D)
c12 = (w1[i]*e12)+(((e12-1)^2)*((b[t]-e12)/(b[t]-a[t]))*D)
c13 = w1[i]*e13
## Same for Driver Two
e21 = (1-(w2[i]/(2*D)))
e22 = ((2+b[t])-sqrt((b[t]^2)-2*b[t]+1+3*(w2[i]/D)*(b[t]-a[t])))/3
e23 = b[t]
c21 = (w2[i]*e21)+(((e21-1)^2)*D)
c22 = (w2[i]*e22)+(((e22-1)^2)*((b[t]-e22)/(b[t]-a[t]))*D)
c23 = w2[i]*e23
## Same for Driver Three
e31 = (1-(w3[i]/(2*D)))
e32 = ((2+b[t])-sqrt((b[t]^2)-2*b[t]+1+3*(w3[i]/D)*(b[t]-a[t])))/3
e33 = b[t]
c31 = (w3[i]*e31)+(((e31-1)^2)*D)
c32 = (w3[i]*e32)+(((e32-1)^2)*((b[t]-e32)/(b[t]-a[t]))*D)
c33 = w3[i]*e33
## Same for Driver Four
e41 = (1-(w4[i]/(2*D)))
e42 = ((2+b[t])-sqrt((b[t]^2)-2*b[t]+1+3*(w4[i]/D)*(b[t]-a[t])))/3
e43 = b[t]
c41 = (w4[i]*e41)+(((e41-1)^2)*D)
c42 = (w4[i]*e42)+(((e42-1)^2)*((b[t]-e42)/(b[t]-a[t]))*D)
c43 = w4[i]*e43
## These lines of code
c1star = min(c11,c12,c13)
c2star = min(c21,c22,c23)
c3star = min(c31,c32,c33)
c4star = min(c41,c42,c43)
e1[t] = if(c1star==c11){e11} else if(c1star==c12 & (e12 < e13)){e12} else if(c1star==c13 || (e12 >= e13)){e13}
e2[t] = if(c2star==c21){e21} else if(c2star==c22 & (e22 < e23)){e22} else if(c2star==c23 || (e22 >= e23)){e23}
e3[t] = if(c3star==c31){e31} else if(c3star==c32 & (e32 < e33)){e32} else if(c3star==c33 || (e32 >= e33)){e33}
e4[t] = if(c4star==c41){e41} else if(c4star==c42 & (e42 < e43)){e42} else if(c4star==c43 || (e42 >= e43)){e43}
proba1 = function(e1){(e1[t]-1)^2} #probability of accident only given driver1's effort level.
probn1 = 1 - proba1(e1) #probability of not getting in an accident given driver1's effort level
proba2 = function(e2){(e2[t]-1)^2}
probn2= 1 - proba2(e2)
proba3 = function(e3){(e3[t]-1)^2}
probn3= 1 - proba3(e3)
proba4 = function(e4){(e4[t]-1)^2}
probn4= 1 - proba4(e4)
while(e[t]!="e1[t]" & e[t]!="e2[t]" & e[t] != "e3[t]" & e[t] != "e4[t]"){e[t] = sample(c("e1[t]","e2[t]","e3[t]","e4[t]",0,0,0,0),size = 1, replace = TRUE, prob = c(proba1(e1),proba2(e2), proba3(e3),proba4(e4), probn1, probn2, probn3, probn4))
if(e[t]== "0"){
na[t] = na[t]+1
na1[t] = na1[t]+1
na2[t] = na2[t]+1
na3[t] = na3[t]+1
na4[t] = na4[t]+1}}
show(e[t])
eaccident[t]={if(e[t]=="e1[t]"){"Driver 1"}
else if(e[t]=="e2[t]"){"Driver 2"}
else if(e[t]=="e3[t]"){"Driver 3"}
else if(e[t]=="e4[t]"){"Driver 4"}}
{if(e[t]=="e1[t]"){
na1[t+1] = 0
na2[t+1] = na2[t]+1
na3[t+1] = na3[t]+1
na4[t+1] = na4[t]+1}
else if(e[t]=="e2[t]"){
na2[t+1] = 0
na1[t+1] = na1[t]+1
na3[t+1] = na3[t]+1
na4[t+1] = na4[t]+1}
else if(e[t]=="e3[t]"){
na3[t+1] = 0
na2[t+1] = na2[t]+1
na1[t+1] = na1[t]+1
na4[t+1] = na4[t]+1}
else if(e[t]=="e4[t]"){
na4[t+1] = 0
na2[t+1] = na2[t]+1
na1[t+1] = na1[t]+1
na3[t+1] = na3[t]+1}}
if (e[t]=="e1[t]"){e[t] = e1[t];e = as.numeric(e)} else if (e[t]=="e2[t]"){e[t] = e2[t];e = as.numeric(e)} else if (e[t]=="e3[t]"){e[t] = e3[t];e = as.numeric(e)} else if (e[t]=="e4[t]"){e[t] = e4[t];e = as.numeric(e)}
probg = function(e,a,b){(b[t]-e[t])/(b[t]-a[t])} ### probability of being found guilty given e in the ambiguous region
probi = function(e,a,b){1-probg(e,a,b)} ### probability of being found innocent given e in the ambiguous region
rulings = if(e[t]>=b[t]){ruling = print("Not Negligent")
} else if(e[t]<=a[t]) {ruling = print("Negligent")
} else if(e[t]>a[t] & e[t]<b[t]) {ruling = if(e[t]<estar){"guilty"} else if(e[t]>=estar){"not guilty"}; show(ruling)}
if(ruling == "guilty") {a[t+1] = e[t]; b[t+1] = b[t]} else if (ruling == "not guilty") {b[t+1] = (e[t]); a[t+1]=a[t]} else {a[t+1]=a[t]; b[t+1] = b[t]; print("thresholds unchanged")}
e = round(e,digits= 5)
e1 = round(e1,digits = 5)
e2 = round(e2, digits = 5)
e3 = round(e3, digits = 5)
e4 = round(e4, digits = 5)
a = round(a, digits = 5)
b = round(b, digits = 5)
threshold = 0.00009
lower_limit = e - threshold
upper_limit = e + threshold
if((a[t+1]==a[t]) && (b[t+1] == b[t]) && (all(sapply(c(e1[t], e2[t], e3[t],e4[t]), function(c) any(c >= lower_limit & c <= upper_limit)))) && ((e1[t] <= a[t]) | (e1[t] >= b[t])) && ((e2[t] <= a[t]) | (e2[t] >= b[t])) && ((e3[t] <= a[t]) | (e3[t] >= b[t])) && ((e4[t] <= a[t]) | (e4[t] >= b[t]))){break} else {print("not equilibrium")}
t= t+1
}
print("Simulation End")
age1[[i]] = e1
age2[[i]] = e2
age3[[i]] = e3
age4[[i]] = e4
aga[[i]] = a
agb[[i]] = b
}
Спасибо