## For loop and storing results values from an iteration algorithm

H, i am a beginner in R and I have problems dealing with for loops. What I am trying to do is store 25 values of tau, b0 and b1. In order to find b1 we have to find the lowest value of Out function in the code ( b1 should be close to the value 1 ) and I want use this b1 in the for loop to find b0 and b1 and should be stored in a vector of length 25 ( I should have a list of b0 ,b1 tau). Please help and thanks in advance! I have added what I intend to do

I want

b1[1], b1[2],..., b1[25] t[1],t[2],...,t[25] b0[1],b0[2],...,b0[25]

by solving the problem in the picture

set.seed(22) #Inverse Transformation on CDF Simburr.f1 <- function(n, tau) { u=runif(n) x<- runif(n) lambda = exp(1+x) y= (1/(u^(1/lambda))-1)^(1/tau) y } y33 = Simburr.f1(25,0.5) #------------------------------------ # We fix Tau = 0.5 est3 = function(m) { x=runif(25) Out = rep(0,m) k= seq(0.05,2,l=m) for ( i in 1:m){ Out[i]= log(mean(x)) -log(mean(exp(k[i]*x)*log(1+y33^0.5)*x) ) + log(mean(log(1+y33^0.5)*exp(k[i]*x))) } Out } p=est3(200) k[which.min(abs(p-0))] # = B1 #-------------------------------------- x =runif(25) tau[1]= 0.5 b0[1] = 1 b1[1] = 1 # Iterative process for ( i in 2:25){ b0[i]= -log(mean(log(1+y33^(tau[i-1]))*exp(b1[i-1]*x))) tau[i] =1/( mean((exp(b0[i-1]+b1[i-1]*x)+1)*(y33^tau[i-1])*log(y33)/(1+y33^tau[i-1])) -mean(log(y33))) }

In terms of your original question, you're nearly there. But you can't declare variables by specifically defining their first element with `[1]`

. Rather you need to either declare an empty array then write to it:

b0 = NULL b0[1] = 1

Or you can simply declare `b0`

by writing `1`

to it:

b0 = 1

In both cases, this will produce a vector of length 1. The code you're using will extend the length of the vector as it runs.

**Note:** your code still won't run, as you've referenced the variable `k`

which only exists in the scope of a function, you'll need to correct that too. But I'm not sure what you're wanting to acheive with `k[which.min(abs(p-0))] # = B1`

so I can't help with that bit.

**Magic Functions to Obtain Results from for Loops in R,** For example, let's calculate squared values for 1 to 3. Moreover, you may want to store the result as a data.frame with the iteration numbers. After the iteration is completed entirely, the value of count is incremented by 1 . 5th iteration: count is 5. The test condition count<=num is not satisfied as (5<=4) is false.

I don't quite get what the question is.
But, I clearly understand that your code isn't working because you can't put the value of `tau[1]`

, `b0[1]`

, and `b1[1]`

before explicitly declare the variable as list or matrix.
You need to add this code before you put the value into the first index:

tau <- matrix(c(0),ncol=1,nrow=25) b0 <- matrix(c(0),ncol=1,nrow=25) b1 <- matrix(c(0),ncol=1,nrow=25) tau[1]= 0.5 b0[1] = 1 b1[1] = 1

and here might be what you're looking for:

for ( i in 2:25){ p <- est3(200) b1[i] = p[which(abs(p-0)==min(abs(p-0)))] b0[i]= -log(mean(log(1+y33^(tau[i-1]))*exp(b1[i-1]*x))) tau[i] =1/( mean((exp(b0[i-1]+b1[i-1]*x)+1)*(y33^tau[i-1])*log(y33)/(1+y33^tau[i-1])) -mean(log(y33))) }

b1 values are from the value of est3(200) that closest to zero

The complete code:

set.seed(22) #Inverse Transformation on CDF Simburr.f1 <- function(n, tau) { u=runif(n) x<- runif(n) lambda = exp(1+x) y= (1/(u^(1/lambda))-1)^(1/tau) y } y33 = Simburr.f1(25,0.5) #------------------------------------ # We fix Tau = 0.5 est3 = function(m) { x=runif(25) Out = rep(0,m) k= seq(0.05,2,l=m) for ( i in 1:m){ Out[i]= log(mean(x)) -log(mean(exp(k[i]*x)*log(1+y33^0.5)*x) ) + log(mean(log(1+y33^0.5)*exp(k[i]*x))) } Out } #-------------------------------------- x =runif(25) tau <- matrix(c(0),ncol=1,nrow=25) b0 <- matrix(c(0),ncol=1,nrow=25) b1 <- matrix(c(0),ncol=1,nrow=25) tau[1]= 0.5 b0[1] = 1 b1[1] = 1 # Iterative process for ( i in 2:25){ p=est3(200) b1[i] = p[which(abs(p-0)==min(abs(p-0)))] b0[i]= -log(mean(log(1+y33^(tau[i-1]))*exp(b1[i-1]*x))) tau[i] =1/abs(( mean((exp(b0[i-1]+b1[i-1]*x)+1)*(y33^tau[i-1])*log(y33)/(1+y33^tau[i-1])) -mean(log(y33)))) }

**For loop and storing results values from an iteration algorithm,** H, i am a beginner in R and I have problems dealing with for loops. What I am trying to do is store 25 values of tau, b0 and b1. In order to find b1 Storing results from a for-loop in a vector of Learn more about for loop, iteration, storing values, vector of zeros, preallocating vectors, portfolio optimization, rebalancing

Before you declare your initial `tau`

, `b0`

and `b1`

values you need to declare the variables by adding the following line;
`tau <- b0 <- b1 <- vector("numeric", 25L)`

.

**EDIT**

set.seed(22) #Inverse Transformation on CDF Simburr.f1 <- function(n, tau) { u=runif(n) x<- runif(n) lambda = exp(1+x) y= (1/(u^(1/lambda))-1)^(1/tau) y } y33 = Simburr.f1(25,0.5) #------------------------------------ # We fix Tau = 0.5 est3 = function(m) { x=runif(25) Out = rep(0,m) k= seq(0.05,2,l=m) for ( i in 1:m){ Out[i]= log(mean(x)) -log(mean(exp(k[i]*x)*log(1+y33^0.5)*x) ) + log(mean(log(1+y33^0.5)*exp(k[i]*x))) } Out } p=est3(200) k <- seq(0.05,2, length.out = 200) k[which.min(abs(p-0))] # = B1 #-------------------------------------- x =runif(25) tau <- b0 <- b1 <- vector("numeric", 25) tau[1]= 0.5 b0[1] = 1 b1[1] = 1 # Iterative process for ( i in 2:25){ b0[i]= -log(mean(log(1+y33^(tau[i-1]))*exp(b1[i-1]*x))) tau[i] =1/( mean((exp(b0[i-1]+b1[i-1]*x)+1)*(y33^tau[i-1])*log(y33)/(1+y33^tau[i-1])) -mean(log(y33))) }

**7. Iteration,** Because iteration is so common, Python provides several language features to Each item in turn is (re-)assigned to the loop variable, and the body of the loop is For example, if the starting value is a power of two, then the value of n will be When the loop exits, count contains the result — the total number of times the put() is the default function to store values in magicalized for loops. It allows to take any number of variables and can display them. It allows to take any number of variables and can display them. 2.

**Save loop data - MATLAB Answers - MATLAB Central,** Learn more about for loop, save MATLAB. I have a for loop, but every iteration overwrites the variable and I have only both should give you similar results. To many may result in a value function moving further from the true one since the policy function is not the optimal policy. We also add an additional step to the value function iteration algorithm between the existing Step 4 and Step 5, that we’ll name Step 4.5, as follows. Step 4.5 Complete the following steps nh times. a) Set V(0)(K) = V(1)(K)

**How can I store a value from each iteration of a for loop, ready for ,** So just i am getting increment numbers for for loop instead of actual values. Example: now nTimeLen = 1, 2, 3 till end earlier loop was 0, 0.01, We often use a loop to work with items of a list when programming. This section introduces a new type of loop, called a for-loop, that is particularly well suited to dealing with items of a list. Anything you can accomplish with a for-loop you can also accomplish with a while-loop, so we are not getting any more power out of for-loops.

**For Loops - Simon Ejdemyr,** Summary For loops can be useful when you want to iterate a process in R — e.g., run a We can convert the values in v1 from centimeters to meters using a for loop: However, the example illustrates the following points about writing a for loop: Begin by creating an object that can store the results of your for loop. You will see updates in your activity feed. You may receive emails, depending on your notification preferences. however after each loop, the previous value is overwritten, is there a way I can save them all? Sign in to answer this question. Basically, the above code will store all the values in each loop in a matrix x.