Normalising multiple columns of a matrix to a fraction of its maximum value

3 次查看（过去 30 天）
Nikodem Podlaszewski
编辑： Voss 2024-5-23
Data: Measurements of a bridge's natural frequencies / eigenfrequencies (master's degree, civil engineering)
1) Data set example (paint):
Notes:
a) f_1, f_2, f_3, f_n => natural frequencies [Hz],
b) a_i, b_i, c_i, n_i => displacement of i-th node / measuring point [mm],
c) there are always 21 measuring points => there are always 21 rows
d) measurements (data sets) are different => n is sometimes 5, sometimes 10, sometimes 15 but I need to analyze all data sets
2) Problem to solve (paint):
Notes:
a) a_i/a_max, b_i/b_max, n_i/n_max => fractions of maximum displacement [mm/mm=no unit]
Comment:
I need to write a code that would remake a matrix consisting of displacement values [mm] so it divides all values by the value's column's maximum value [mm/mm=no unit]. So in each column there should be exactly one number equal to 1,00. I know how to divide a whole matrix by a maximum value (of a matrix or just a row), but I have no idea how to divide each column by a different number. Complicated, I know. I need this for further calculations (comparing experimental and finite element method modal shapes).
Thanks in advance! I hope you liked my Paint talent :)

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采纳的回答

Voss
Voss 2024-5-20
Ran in:
To divide each column by its maximum value:
data_normalized = data./max(data,[],1);
Example:
data = rand(21,5); % random data
data_normalized = data./max(data,[],1);
disp(data)
0.3794 0.8516 0.6192 0.9974 0.0485 0.5473 0.9512 0.3346 0.7293 0.3087 0.7468 0.8790 0.5080 0.3312 0.6452 0.1507 0.2931 0.9213 0.8985 0.8317 0.9915 0.1097 0.6171 0.5542 0.1940 0.4685 0.6239 0.4579 0.9900 0.1844 0.0630 0.8840 0.3104 0.5871 0.2056 0.1789 0.5089 0.1588 0.2956 0.6751 0.2515 0.0815 0.3480 0.3586 0.7300 0.4045 0.6555 0.6625 0.3284 0.6248 0.1053 0.6238 0.3821 0.0726 0.0529 0.9091 0.3172 0.9784 0.4816 0.8802 0.1635 0.8599 0.1088 0.6291 0.4188 0.0465 0.9600 0.4690 0.0528 0.3343 0.5934 0.2478 0.5415 0.7363 0.8638 0.9241 0.0415 0.9797 0.0765 0.9500 0.1802 0.7070 0.6599 0.7147 0.4607 0.9711 0.4433 0.7790 0.4921 0.0153 0.8528 0.8070 0.3096 0.7092 0.8844 0.2960 0.1858 0.3051 0.1203 0.8197 0.9482 0.7046 0.2076 0.3336 0.2242
disp(data_normalized)
0.3827 0.8870 0.6321 1.0000 0.0511 0.5520 0.9908 0.3415 0.7313 0.3249 0.7533 0.9156 0.5185 0.3321 0.6792 0.1520 0.3053 0.9404 0.9009 0.8756 1.0000 0.1143 0.6299 0.5556 0.2042 0.4725 0.6499 0.4674 0.9926 0.1941 0.0635 0.9208 0.3169 0.5887 0.2164 0.1804 0.5301 0.1621 0.2963 0.7106 0.2537 0.0849 0.3552 0.3595 0.7685 0.4079 0.6828 0.6763 0.3293 0.6577 0.1062 0.6497 0.3900 0.0728 0.0556 0.9170 0.3304 0.9987 0.4828 0.9265 0.1649 0.8957 0.1110 0.6307 0.4409 0.0469 1.0000 0.4787 0.0530 0.3519 0.5985 0.2581 0.5527 0.7382 0.9093 0.9321 0.0432 1.0000 0.0767 1.0000 0.1817 0.7364 0.6736 0.7166 0.4850 0.9794 0.4618 0.7951 0.4934 0.0161 0.8601 0.8406 0.3160 0.7111 0.9310 0.2985 0.1936 0.3114 0.1206 0.8629 0.9563 0.7340 0.2119 0.3344 0.2360
  2 个评论
Nikodem Podlaszewski
Nikodem Podlaszewski 2024-5-23
编辑：Nikodem Podlaszewski 2024-5-23
Hello, thanks for help. It works but I need to modify this function:
How can I modify this function to divide each column by its maximum or abs(minimum value), depending on:
if column's max value is greater than abs(min value), divide by maximum value,
if column's abs(min value) is greater than max value, divide my abs(min value)
I need to do it because I want each column to have values from -1 to 1
Thanks in advance
Voss
Voss 2024-5-23
编辑：Voss 2024-5-23

To divide by max of max value and abs(min value), as you describe:

data_normalized = data./max(abs(data),[],1);

But that won't give you values from -1 to 1. If you want that, it would be:

mi = min(data,[],1);
ma = max(data,[],1);
normalized_data = (data-mi)./(ma-mi)*2-1;

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更多回答（1 个）

Catalytic
Catalytic 2024-5-21
Data=normalize(Data,1,'norm',inf)
  1 个评论
Nikodem Podlaszewski
Nikodem Podlaszewski 2024-5-23
Hello, thanks for help. I need to modify this function:
How can I modify this function to divide each column by its maximum or abs(minimum value), depending on:
if column's max value is greater than abs(min value), divide by maximum value,
if column's abs(min value) is greater than max value, divide my abs(min value)
I need to do it because I want each column to have values from -1 to 1
Thanks in advance

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