How can I write y0 code in one line and still I can get this output.

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SAHIL SAHOO
SAHIL SAHOO 2023-5-2
评论： VBBV 2023-5-3
y0= [ (1e-2)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
(1e-2)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
(1e-2)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1)); ]
================================================
output
y0 = 0.0016
0.0079
0.6824
0.0017
0.0060
1.2281
0.0069
0.0075
-1.1515

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回答（1 个）

VBBV
VBBV 2023-5-2
Ran in:
y0= [ (1e-2)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
(1e-2)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
(1e-2)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1)); ]
y0 = 9×1
0.0082 0.0023 -0.4870 0.0071 0.0082 1.1632 0.0008 0.0039 1.7553
y0 = [1e-2*rand(6,1);repmat((-3.14)*rand(1)+(3.14)*rand(1),3,1)]
y0 = 9×1
0.0018 0.0012 0.0068 0.0078 0.0044 0.0079 -1.7388 -1.7388 -1.7388
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SAHIL SAHOO
SAHIL SAHOO 2023-5-3
Here I'm having same problem that you have, the first two no.
0.0031
0.0010
0.8000
0.0083
0.0089
1.8618
0.0020
0.0010
1.6583
are less than 0.0031 but the 3rd no. is between -3.14 to 3.14,
therefore I want to modify my code in such manner that I can satisfy my given condition.
VBBV
VBBV 2023-5-3
Ran in:
how is it less than 0.0031 ? take a look at the values, they are just randome values multiplied by 1e-2. if you want the first two values to higher than 0.0031, then you can use randi instead of rand
y0 = [ 0.0096
0.0081
0.0060
0.0044
0.0098
0.0030
-0.3596
-0.0365
-0.3486]
y0 = 9×1
0.0096 0.0081 0.0060 0.0044 0.0098 0.0030 -0.3596 -0.0365 -0.3486
y0 = [1e-2*randi([0 10],6,1);[(-3.14)*rand(1)+(3.14)*rand(1)].*rand(3,1)]
y0 = 9×1
0.0700 0.1000 0.0500 0.0600 0.0600 0.0700 -0.2265 -0.4150 -0.3119

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