How can I run these codes in a for loop?

Hi! I understand that the variables should not be named dynamically. So, I would like to have an idea on how can I run these kind of code in a for loop without repeating the variable inthe future. A good idea will help me to think differently and more efficiently. Thank you in advance!!

idx_0 = hour(result.HourSeries1)==00;
m0 = mean(Climatology1.H1(idx_0,:))

idx_1 = hour(result.HourSeries1)==01;
m1 = mean(Climatology1.H1(idx_1,:))

idx_2 = hour(result.HourSeries1)==02;
m2 = mean(Climatology1.H1(idx_2,:))

idx_3 = hour(result.HourSeries1)==03;
m3 = mean(Climatology1.H1(idx_3,:))

idx_4 = hour(result.HourSeries1)==04;
m4 = mean(Climatology1.H1(idx_4,:))

I tried to do this way but it shows "Array indices must be positive integers or logical values."

x = [0:23];
for i =1:24;
    idx{i-1} = hour(result.HourSeries1)==i-1; 
    m{i-1} = mean(Climatology1.H1(idx{i-1},:))
end

Answer 1

[Voss]

If the means are all scalars

m = zeros(5,1);
for ii = 1:5
    idx = hour(result.HourSeries1) == ii-1;
    m(ii) = mean(Climatology1.H1(idx,:));
end

Otherwise

m = zeros(5,size(Climatology1.H1,2));
for ii = 1:5
    idx = hour(result.HourSeries1) == ii-1;
    m(ii,:) = mean(Climatology1.H1(idx,:), 1);
end

[Ashfaq Ahmed]

Hi! Yes! It worked. Thank you so much. But can you please also tell me what would be the case if the ==00 or ==01 were actually a string, e.g., =='00' or =='01'. How could I change the code in that case?

[Voss]

The built-in hour function returns a double, so I assume the hour you're using is your own function or variable.

idx = strcmp(hour(result.HourSeries1),sprintf('%02d',ii-1))

[Campion Loong]

Adding to @Voss's answer, this is basically subscripting with "hour" in result.HourSeries1. Depending on whether that is a datetime or text, I would prefer holding the "Climatology" data in a timetable or table so your data labels (i.e. time, text) go with your data, like this (if you use datetime)

% Making up random multi-column data for H1
ResultTimes = datetime(2023,3,13) + hours(1:1000)';
Climatology = timetable(ResultTimes, rand(1000,5),'VariableNames',"H1")

Climatology = 1000×1 timetable
        ResultTimes                                      H1                              
    ____________________    _____________________________________________________________

    13-Mar-2023 01:00:00      0.84146      0.79722      0.82259      0.66337      0.65927
    13-Mar-2023 02:00:00      0.67373      0.89206      0.58789     0.019026      0.66615
    13-Mar-2023 03:00:00      0.88871       0.4429     0.046643      0.40312      0.23974
    13-Mar-2023 04:00:00       0.3896      0.68458      0.30343      0.40802       0.7763
    13-Mar-2023 05:00:00        0.439      0.53479      0.88126      0.19547      0.12425
    13-Mar-2023 06:00:00      0.62536      0.44538      0.67953      0.84661     0.028268
    13-Mar-2023 07:00:00      0.77342      0.29515       0.9852      0.16639      0.63207
    13-Mar-2023 08:00:00      0.76506      0.46657      0.59155      0.37077      0.52225
    13-Mar-2023 09:00:00       0.6151      0.41233      0.91669      0.58814      0.19279
    13-Mar-2023 10:00:00      0.81799      0.22129      0.95078     0.087836      0.56349
    13-Mar-2023 11:00:00     0.046164      0.73673      0.84577      0.67314      0.39125
    13-Mar-2023 12:00:00      0.66193     0.093104     0.021965      0.13976      0.70648
    13-Mar-2023 13:00:00     0.045651      0.78472       0.6179      0.78254      0.74244
    13-Mar-2023 14:00:00    0.0089332      0.28782      0.48349      0.50385      0.23653
    13-Mar-2023 15:00:00      0.22407      0.03468      0.75578      0.57565      0.24565
    13-Mar-2023 16:00:00       0.2304     0.056612      0.17365      0.93395      0.48146

% Pull out data based on time-of-day (e.g. 3AM)
Climatology( hour(Climatology.ResultTimes) == 3, :).H1

ans = 42×5

0.8887 0.4429 0.0466 0.4031 0.2397 0.1195 0.3022 0.0428 0.5548 0.8576 0.5340 0.6981 0.6842 0.1122 0.8263 0.0722 0.5172 0.1391 0.0341 0.3207 0.4562 0.6380 0.0116 0.8380 0.2409 0.9217 0.0622 0.4751 0.9748 0.9114 0.8217 0.4114 0.9609 0.5626 0.0504 0.8715 0.5308 0.1174 0.4100 0.5577 0.3256 0.1523 0.9762 0.7915 0.0644 0.7501 0.0962 0.4406 0.8895 0.5859

% Now applying @Voss's answer for non-scalar means
m = zeros(24,size(Climatology.H1,2));
ResultTimes_HourOfDay = hour(Climatology.ResultTimes);
for ii = 1:24 % note HoD ranges from 0-23
    m(ii,:) = mean( Climatology(ResultTimes_HourOfDay == ii-1,:).H1 );
end

m

m = 24×5

0.4870 0.6044 0.4125 0.4095 0.4529 0.5072 0.5466 0.5409 0.5480 0.5396 0.5507 0.5194 0.5247 0.4759 0.5053 0.5690 0.5028 0.4895 0.4708 0.5433 0.5019 0.5100 0.4935 0.4068 0.5119 0.5362 0.4813 0.4724 0.4415 0.5209 0.5907 0.4943 0.5441 0.4915 0.5329 0.5106 0.4688 0.5590 0.5524 0.5145 0.4397 0.5091 0.5410 0.4844 0.5222 0.5857 0.4629 0.4559 0.4624 0.4481

Answer 2

[Torsten]

for i = 0:4
    idx(i+1) = hour(result.HourSeries1)==strcat('0',string(i));
    m(i+1) = mean(Climatology1.H1(idx(i+1),:));
end

[Ashfaq Ahmed]

Excellent. But what happens when there are more than 9 count?

strcat('0',string(i))

won't be working then. Because the strings are ...07,08,09,10,11,12,..

[Torsten]

if i < 10
    str = strcat('0',string(i));
else
    str = string(i);
end

And to compare strings, you shouldn't use ==, but strcmp.

[Voss]

str = sprintf("%02d",i);

[Ashfaq Ahmed]

Thank you both. You guys are awesome.