How do I index outputs from a loop?

1 次查看（过去 30 天）
Nick Lavanture
Nick Lavanture 2022-5-11
回答： Siraj 2023-9-18
Below is my code, but I would like to plot d and dd_dN vs each iteration number
So I believe I need to index and save each outputted d and dd_dN for each iteration, a total of (1+240+1800)*25 = 51025 values for both d and dd_dN
Any help is greatly appreciated, thank you!
clc,clear
format long g
c = 1.1e-10;
m = 3.3;
p500 = [ -0.3142 2.3172 -7.0249 11.2867];
p2000 = [ -1.2577 9.2728 -28.1056 45.1501];
p3150 = [ -1.9807 14.6042 -44.2653 71.1096];
d = 0.015;
for i=1:25
dd_dN = c*((0.706*(p3150(1)+3.150)+0.537*(2*d/pi)*p3150(2)+0.448*(d^2./2)*p3150(3)+0.393*(4*d^3/(3*pi))*p3150(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
for j=1:240
dd_dN = c*((0.706*(p2000(1)+2.000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
end
for j=1:1800
dd_dN = c*((0.706*(p500(1)+.500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
end
end
fprintf('The final crack depth is %18.16f inches',d)
  3 个评论
显示 1更早的评论
Jeffrey Clark
Jeffrey Clark 2022-5-12
% d is changed for every calculation of dd_dN which uses d
% want to create two output vectors for dd_dN and d itteration values
d_vals = zeros(1,(1+240+1800)*25);
dd_dN_vals = d_vals;
val_index = 1;
c = 1.1e-10;
m = 3.3;
p500 = [ -0.3142 2.3172 -7.0249 11.2867];
p2000 = [ -1.2577 9.2728 -28.1056 45.1501];
p3150 = [ -1.9807 14.6042 -44.2653 71.1096];
d = 0.015;
for i=1:25
dd_dN = c*((0.706*(p3150(1)+3.150)+0.537*(2*d/pi)*p3150(2)+0.448*(d^2./2)*p3150(3)+0.393*(4*d^3/(3*pi))*p3150(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
dd_dN_val(val_index) = dd_dN;
d_val(val_index) = d;
val_index = val_index+1;
for j=1:240
dd_dN = c*((0.706*(p2000(1)+2.000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
dd_dN_val(val_index) = dd_dN;
d_val(val_index) = d;
val_index = val_index+1;
end
for j=1:1800
dd_dN = c*((0.706*(p500(1)+.500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
dd_dN_val(val_index) = dd_dN;
d_val(val_index) = d;
val_index = val_index+1;
end
end
% Note that at the end val_index is 1 larger than the size of the _val arrays
fprintf('The final crack depth is %18.16f inches',d)
KSSV
KSSV 2022-5-12
dd_dN_val(val_index) = dd_dN;
d_val(val_index) = d;
The above lines should be:
dd_dN_vals(val_index) = dd_dN;
d_vals(val_index) = d;
Which you have initialized into array right?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Siraj
Siraj 2023-9-18
Ran in:
Hi!
It is my understanding that you would like to plot the variables "d" and "dd_dN" against the iteration number. To plot the variables "d" and "dd_dN" against the iteration number, it is necessary to first store all the values of "d" and "dd_dN" during the iterations.
As the total number of iterations is known, it makes it easier to preallocate arrays to store the values of "d" and "dd_dN" for each iteration. After the loop, you can utilize the "plot" function to plot the values of "d" and "dd_dN" against the iteration number. I have incorporated some additional lines of code into your existing code. The added lines are as follows:
d_vals = zeros(1,(1+240+1800)*25);
dd_dN_vals = zeros(1,(1+240+1800)*25);
val_index = 1;
Here, two arrays, “d_vals” and “dd_dN_vals”, are initialized with zeros. The size of these arrays is determined by the total number of iterations to be performed, which is “(1+240+1800)*25”. The variable “val_index” is set to 1, which will be used as an index to store values in the arrays.
After calculating the value of “dd_dN”, it is stored in the “dd_dN_vals” array at the current “val_index” position. Similarly, the current value of “d” is stored in the “d_vals” array. Then, “val_index” is incremented by 1 to move to the next position in the arrays.
These lines of code are repeated within the nested loops, ensuring that for each iteration, the corresponding values of “dd_dN” and “d” are stored in the respective arrays.
Here is the complete code with the additional lines for your reference and better understanding.
d_vals = zeros(1, (1 + 240 + 1800) * 25); % Preallocate array to store values of "d" for each iteration
dd_dN_vals = zeros(1, (1 + 240 + 1800) * 25); % Preallocate array to store values of "dd_dN" for each iteration
val_index = 1; % Initialize index for storing values in arrays
c = 1.1e-10;
m = 3.3;
p500 = [-0.3142 2.3172 -7.0249 11.2867];
p2000 = [-1.2577 9.2728 -28.1056 45.1501];
p3150 = [-1.9807 14.6042 -44.2653 71.1096];
d = 0.015;
for i = 1:25
dd_dN = c * ((0.706 * (p3150(1) + 3.150) + 0.537 * (2 * d / pi) * p3150(2) + 0.448 * (d^2 / 2) * p3150(3) + 0.393 * (4 * d^3 / (3 * pi)) * p3150(4)) * (sqrt(pi * d)))^m;
d = d + dd_dN;
dd_dN_vals(val_index) = dd_dN; % Store value of "dd_dN" at current index
d_vals(val_index) = d; % Store value of "d" at current index
val_index = val_index + 1; % Increment index
for j = 1:240
dd_dN = c * ((0.706 * (p2000(1) + 2.000) + 0.537 * (2 * d / pi) * p2000(2) + 0.448 * (d^2 / 2) * p2000(3) + 0.393 * (4 * d^3 / (3 * pi)) * p2000(4)) * (sqrt(pi * d)))^m;
d = d + dd_dN;
dd_dN_vals(val_index) = dd_dN; % Store value of "dd_dN" at current index
d_vals(val_index) = d; % Store value of "d" at current index
val_index = val_index + 1; % Increment index
end
for j = 1:1800
dd_dN = c * ((0.706 * (p500(1) + 0.500) + 0.537 * (2 * d / pi) * p500(2) + 0.448 * (d^2 / 2) * p500(3) + 0.393 * (4 * d^3 / (3 * pi)) * p500(4)) * (sqrt(pi * d)))^m;
d = d + dd_dN;
dd_dN_vals(val_index) = dd_dN; % Store value of "dd_dN" at current index
d_vals(val_index) = d; % Store value of "d" at current index
val_index = val_index + 1; % Increment index
end
end
fprintf('The final crack depth is %18.16f inches', d)
The final crack depth is 0.0150000007332667 inches
figure, plot(d_vals)
title("d vs index")
figure, plot(dd_dN_vals)
title("dd\_dN vs index")
After the loops, the plot function is used to create two separate figures. The first figure displays the values of “d” against the index(iteration), while the second figure shows the values of “dd_dN” against the index(iteration).
Hope this helps.

类别

在 Help Center 和 File Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

标签

产品


版本

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by