How to change angles to 0 to 180

Question

TTA 2023-5-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1975269-how-to-change-angles-to-0-to-180

评论： TTA 2023-5-30

采纳的回答： Sam Chak

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I have an array of angles ranging from [-180, 180]

please I want to change to [0, 180], how can I do this?

I have tried this code below but it's giving me [90, 180]

Angles180 = @(a) rem(180+a, 360)-90;
Result = Angles180([-90, 0, 90])

please how can I do this?

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Answer 1

Sam Chak 2023-5-29

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Hi @TTA

Are you looking for the conversion like this?

Angles180 = @(a) a/2 + 90;
Result1   = Angles180([-180, 0, 180])
Result1 = 1×3
     0    90   180
Result2   = Angles180([-90, 0, 90])
Result2 = 1×3
    45    90   135

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TTA 2023-5-29

编辑：TTA 2023-5-29

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Honestly, the answer I accepted didn't do what I expected either. I am looking for a btter answer to accept.

What I wanted to do is I want to put all the following angles in the first quadrant ([0,90] so the ones that are already within 0 t 90 does not change and the ones that are with negative can taken as absolute.

30.9334909952858
3.52761847675680
4.59390438411553
0.673955187737078
-1.64646093428910
-5.88284857785972
-7.63488379965276
-13.2174529305943
-18.5463116650289
-22.0385436652521
-22.2420277734567
-17.4941779611427
-8.56583948331003
-0.252059075009702
5.98003127980421
10.1438497196693
3.20791262170253
-7.02580499273734
-16.6977290939165
-20.5936340375392
-21.3959894853361
-17.6311801870524
-9.70194857048854
11.7676931872348
43.0540344675305
72.0416817854922
89.0448914429762
95.9922408810587
100.875499683777
103.330610504292
104.413850582525
103.364277947976
100.617226400410
96.4283763466032
91.8382501208323
85.8745260293066
81.0699629969629
75.3550207618278
69.3012945930419
5.02597534734592
3.01291530518864
1.66779490750663
-0.0732445094082720
1.06958022532632
3.10405918566257
6.9150301883856

DGM 2023-5-29

编辑：DGM 2023-5-29

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So then should the answer be

[0 0 0 30 60 90]

or

[90 60 30 30 60 90]

or

[60 30 60 30 60 30]

or something else?

TTA 2023-5-30

Thanks for your effort

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Answer 2

Image Analyst 2023-5-29

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How about just adding 180 to all angles less than 0, so for example -135 becomes +45 degrees.

mask = angles < 180;
angles(mask) = angles(mask) + 180; % Only add 180 to negative angles.

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DGM 2023-5-29

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angles = mod(angles,180);

TTA 2023-5-30

编辑：TTA 2023-5-30

@DGM you are right.

for example, To find the equivalent angle within the range of 0 to 90 degrees for -20 degrees, you can use the modulo operation and add multiples of 90 until the angle falls within the desired range.

Here's how you can calculate the equivalent angle:

Take -20 modulo 360: -20 % 360 = 340. This step ensures the angle is within the range of 0 to 360 degrees.
Since 340 is greater than 90, subtract multiples of 90 to bring it within the range of 0 to 90 degrees.

340 - 270 = 70

Thanks

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Answer 3

Image Analyst 2023-5-29

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You say "What I wanted to do is I want to put all the following angles in the first quadrant ([0,90] so the ones that are already within 0 t 90 does not change and the ones that are with negative can taken as absolute."

Well, what about this:

angles = abs(angles);

???