# How to rearrange time array in a matrix of years and days?

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A LL on 31 May 2022
Commented: A LL on 1 Jun 2022
I have an array of values (let say x) representing daily data for 43 years.
Is there a straightforward way and without using a loop to rearrange x as a matrix with years as rows and days as columns?
%Array of values along time
x = rand(365*43,1);
I want:
x_matrix = [[Year 1]; [Year 2]; [Year 3]; ... ; [Year 43]]
where [Year n] = [Year n day 1, Year n day 2, ..., Year n day 365]
Thank you

Voss on 31 May 2022
% some x
x = (1:365*43).';
% reshape x to have years as rows and days as columns
x_matrix = reshape(x,365,[]).'
x_matrix = 43×365
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 1096 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 1123 1124 1125 1461 1462 1463 1464 1465 1466 1467 1468 1469 1470 1471 1472 1473 1474 1475 1476 1477 1478 1479 1480 1481 1482 1483 1484 1485 1486 1487 1488 1489 1490 1826 1827 1828 1829 1830 1831 1832 1833 1834 1835 1836 1837 1838 1839 1840 1841 1842 1843 1844 1845 1846 1847 1848 1849 1850 1851 1852 1853 1854 1855 2191 2192 2193 2194 2195 2196 2197 2198 2199 2200 2201 2202 2203 2204 2205 2206 2207 2208 2209 2210 2211 2212 2213 2214 2215 2216 2217 2218 2219 2220 2556 2557 2558 2559 2560 2561 2562 2563 2564 2565 2566 2567 2568 2569 2570 2571 2572 2573 2574 2575 2576 2577 2578 2579 2580 2581 2582 2583 2584 2585 2921 2922 2923 2924 2925 2926 2927 2928 2929 2930 2931 2932 2933 2934 2935 2936 2937 2938 2939 2940 2941 2942 2943 2944 2945 2946 2947 2948 2949 2950 3286 3287 3288 3289 3290 3291 3292 3293 3294 3295 3296 3297 3298 3299 3300 3301 3302 3303 3304 3305 3306 3307 3308 3309 3310 3311 3312 3313 3314 3315
Voss on 1 Jun 2022
Edited: Voss on 1 Jun 2022
Here's one way you might could deal with that:
x = (1:365*43+11).';
% If my counting is correct (check it) these are the indices of the
% Feb 29ths. Jan 1, 1979, is index 1, so Feb 29, 1980, is index 365+31+29,
% with each Feb 29th after that coming some multiple of 365*3+366 days
% later, in this particular case:
idx = 365+31+29+(0:10)*(365*3+366)
idx = 1×11
425 1886 3347 4808 6269 7730 9191 10652 12113 13574 15035
x(idx) = []; % remove the leap days
% reshape x to have years as rows and days as columns
x_matrix = reshape(x,365,[]).'
x_matrix = 43×365
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 1123 1124 1125 1126 1462 1463 1464 1465 1466 1467 1468 1469 1470 1471 1472 1473 1474 1475 1476 1477 1478 1479 1480 1481 1482 1483 1484 1485 1486 1487 1488 1489 1490 1491 1827 1828 1829 1830 1831 1832 1833 1834 1835 1836 1837 1838 1839 1840 1841 1842 1843 1844 1845 1846 1847 1848 1849 1850 1851 1852 1853 1854 1855 1856 2193 2194 2195 2196 2197 2198 2199 2200 2201 2202 2203 2204 2205 2206 2207 2208 2209 2210 2211 2212 2213 2214 2215 2216 2217 2218 2219 2220 2221 2222 2558 2559 2560 2561 2562 2563 2564 2565 2566 2567 2568 2569 2570 2571 2572 2573 2574 2575 2576 2577 2578 2579 2580 2581 2582 2583 2584 2585 2586 2587 2923 2924 2925 2926 2927 2928 2929 2930 2931 2932 2933 2934 2935 2936 2937 2938 2939 2940 2941 2942 2943 2944 2945 2946 2947 2948 2949 2950 2951 2952 3288 3289 3290 3291 3292 3293 3294 3295 3296 3297 3298 3299 3300 3301 3302 3303 3304 3305 3306 3307 3308 3309 3310 3311 3312 3313 3314 3315 3316 3317
However, I'm obliged to point out that you can write more robust code for dealing with these types of date/time considerations, by using datetime arrays.
A LL on 1 Jun 2022
I will have a look at the datetime arrays. But this should work even if it is not the most robust.
Thanks again!

Jon on 31 May 2022
X = reshape(X,365,43)'

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