Matlab 创建一个迭代，从数组中选取n个元素进行计算，并在下一个迭代中增加n个元素

我有一个matlab脚本，可以计算冷藏运输拖车一天的每小时热负荷。使用的环境温度是特定年份每小时温度数据的数组，因此为8760x1个元素。我想创建一个for循环或某种迭代，从环境温度数组中每隔24小时提取一次数据，因此是1:24、25:48，等等；执行计算并将其存储在矩阵中。因此，最终它将产生一个24x365

clc; clear; close all;
%%
load('Heerlen_TMY');  
T_out = T2m(1:24);
T_in = zeros(length(T_out),1);
% T_in(:,1) = -18;
T_in(:,1) = 3;
% T_out(:,1) = 33;
RH = RH(1:24);

dT = (T_out - T_in);
%% Parameters
l_i = 13.36; w_i = 2.50; h_i = 2.50; % Trailer internal parameters
l_e = 13.54; w_e = 2.60; h_e = 2.75; % Trailer external parameters

m = zeros(length(T_out),1);    %Initializing cargo mass
m_c = 20.0*1e3;         % Total cargo mass[kg]
m_c = [m_c 0.87*m_c 0.75*m_c 0.62*m_c 0.5*m_c 0.37*m_c 0.25*m_c 0.12*m_c];
Delivery = [6 8 10 12 14 16 18 20]; % Delivery time during 24-hr period
On_road = [7 9 11 13 15 17 19 21];  % Driving times in a 24-hr period
m(Delivery)= m_c;   % For product load during delivery
m(On_road) = m_c;   % For calculating product load on road

%% Transmission load
K = 0.6; %0.4; %Overall heat transfer coefficient
S_i = ((l_i*w_i)+(l_i*h_i)+(w_i*h_i))*2; % Inner surface area;
S_e = ((l_e*w_e)+(l_e*h_e)+(w_e*h_e))*2; % Outer surface area;
S = sqrt(S_i*S_e);
deltaT = zeros(length(T_out),1);
deltaT(5) = dT(5);
deltaT(Delivery)= dT(Delivery);   % For product load during delivery
deltaT(On_road) = dT(On_road);
Q_tr = K*S*deltaT.*1e-3;
Q_tr = abs(Q_tr)*1.23;

figure(1)
plot(Q_tr)
%% Precooling
V_a = l_i*w_i*h_i;    % Volume payload [m^3]
rho_a = 1.27;         % Density of air at 0 degrees [kg/m^3]
c_a = 1.005;          % specific heat of air [kJ/kg/K]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Assuming initial temperature inside trailer = ambient temperature
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
PreC = zeros(length(T_out),1); % Initializing precooling load
P = 3600;   % Precooling duration
PreC(5) = P;
% Q_pre = (c_pt*V_pt*rho_pt + V_a*rho_a*c_a)*dT.*PreC./(3600*24);
Q_pre = (V_a*rho_a*c_a)*dT.*PreC./(3600);
figure(2); plot(Q_pre)
%% Product load
T1 = zeros(length(T_out),1); 
T1(:,1) = 5;        % Initial loading temperature of product
dT_pro = T1 - T_in;
h_r = 14.67;        % heat of respiration [mW/kg]
c_l = 3.81;         % Specific heat of dry drugs [kJ/kgK]
Q_pull = m.*c_l.*dT_pro; % loaded products to be cooled to setpoint temp
Q_resp = (h_r*m)./(1e6); % Load caused by chilled perishable products
Q_pro = Q_resp + (abs(Q_pull)./(3600*(length(Delivery)+length(On_road)+1)));

%% Infiltration load
W_d = 2.50;       % Door width
H_d = 2.69;       % Door height
D = [300 180];   % Loading and delivery duration
Theta_o = zeros(24,1);
Theta_o(Delivery) = D(2);
Theta_o(Delivery(1)) = D(1);
Theta_d = [1 0.25 0.5];
E = 0;              % Effectiveness of doorway protective device
D_t = Theta_o/(3600* Theta_d(1));
%%%----------------------------------------------------
% Calculate sensible heat ratio Rs
%%%----------------------------------------------------
h = zeros(length(T_out),2);
y = h;
for k=1:numel(RH)
    y(k,1) = Psychrometricsnew('phi',RH(k),'Tdb',T_out(k));
    y(k,2) = Psychrometricsnew('phi',90,'Tdb',T_in(k));
end

%% Infiltration load - Method 1
D_f = zeros(length(T_in),1);
for n = 1:length(T_in)
    if dT(n) >= 11
        D_f1 = 0.5;
    else
        D_f1 = 0.8;
    end
D_f(n)= D_f1;
%     disp(D_f1)
end
%%%----------------------------------------------------
% Estimate sensible heat load of infiltration Qs/Ad
%%%----------------------------------------------------
Q_s = 4.5; 
R_s = c_a*dT./((y(:,1) - y(:,2)).*1e-3);
q = 0.577*W_d*H_d^(1.5)*Q_s*(1./R_s);
% 
Q_inf = q.*D_t.*D_f.*(1-E);

%% Method 2
rho_o = 1.205; rho_i = 1.275;
g = 9.81;
Enth = ((y(:,1) - y(:,2))./1e3);
Fm = (2/(1+(rho_i/rho_o)^(1/3)))^(1.5);
Q_inf_ = 0.221*W_d*H_d*rho_i*sqrt(1-(rho_o/rho_i))*...
    sqrt(g*H_d)*Fm*(1-E);
Q_inf_1 = Q_inf_.*Enth.*D_f.*D_t;

%% Total load
% Q_pro = 0; %Q_inf = 0;
% Safety factor of 20% to account for 
Q_tot = (Q_tr + Q_pre*1e-3 + Q_pro + Q_inf)*1.2;
Q_tot_1 = (Q_tr + Q_pre*1e-3 + Q_pro + Q_inf_1)*1.2;
% Q_tot = Q_tot*0.9;
figure()
plot(Q_tot)
hold on
plot(Q_tot_1)

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正如mimocha在评论中所建议的，您只需要将每小时温度的8760x1矩阵重塑为24x365矩阵，其中每列包含一天的数据，每行包含该天内每小时的数据

这是由

A = reshape(input, [24, 465]);

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其中，输入是8760x1矩阵，A是尺寸为24x365的重塑阵列。

您的意思是要将一些8760x1阵列转换为24x365阵列吗？因为这可以通过命令

重塑（input_array，[24365]）

完成。这不是我最初的想法，但这肯定比使用for循环更快。我已经实施了它。非常感谢。是的，成功了。谢谢