scalar matrix operations / vector arithemtics

2022 12 14
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更新时间：2022 12 15
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Ran in:
Hello!
I have a problem when I want to calculate percentage deviations from values. Ie. I have the following vecotrs:
a = [-3.67843750000000;
-0.149062500000000;
-0.0284375000000000;
1.52562500000000;
0.217812500000000;
0.00562500000000000;
-0.00625000000000000;
-0.00875000000000000;
-0.00750000000000000;
-0.00937500000000000;
-0.00937500000000000;
-0.00937500000000000]
a = 12×1
-3.6784 -0.1491 -0.0284 1.5256 0.2178 0.0056 -0.0063 -0.0088 -0.0075 -0.0094
b = [-5;
-0.0116649326387481;
-2.72141306933060e-05;
4.98833506736125;
0.255995664691759;
0.0123648464061628;
0.000597234436888179;
2.88470200832711e-05;
1.39333989516825e-06;
6.72997092199941e-08;
3.25064320400358e-09;
1.57009314931713e-10]
b = 12×1
-5.0000 -0.0117 -0.0000 4.9883 0.2560 0.0124 0.0006 0.0000 0.0000 0.0000
When I want to see on how much percentage each element in vector a differs from the same index element in vector b, i thought I can do this with a for loop:
for i=1:1:length(a)
deviation(i,1) = (a(i,1) - b(i,1))*100/b(i,1);
end
deviation
deviation = 12×1
1.0e+09 * -0.0000 0.0000 0.0001 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0005 -0.0139
but as you can see the result is wrong, or I have formated it wrong.
I then thought ok, maybe i can do it with scalar multiplication and division:
deviation_scalar = (a - b)*100./b
deviation_scalar = 12×1
1.0e+09 * -0.0000 0.0000 0.0001 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0005 -0.0139
but this is also not returning the correct result. I.e. for the first entry the calculation would be:
(-3.6784 - -5)*100/-5
ans = -26.4320
which gives me now the correct result. But i don't understand what I do wrong with the vector arithmetic. Does anyone have a clue?

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Stephen23
Stephen23 2022-12-14
编辑：Stephen23 2022-12-15
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Ran in:
"but as you can see the result is wrong.."
No, I do not see that. I see the default format which has four digits after the decimal point and a common multiplier.
"...or I have formated it wrong."
Well, the default format SHORT uses four digits after the decimal point and a common multiplier:
Question: what is -26 to the nearest 1e9 (the common multiplier), when displayed with just four fractional digits?
Answer: -0.0000 * 1e9
And that is exactly what MATLAB displays. If you want, you can change the FORMAT():
a = [-3.6784375;-0.14906250;-0.0284375;1.525625;0.2178125;0.005625;-0.00625;-0.00875;-0.0075;-0.009375;-0.009375;-0.009375]
a = 12×1
-3.6784 -0.1491 -0.0284 1.5256 0.2178 0.0056 -0.0063 -0.0088 -0.0075 -0.0094
b = [-5;-0.0116649326387481;-2.72141306933060e-05;4.98833506736125;0.255995664691759;0.0123648464061628;0.000597234436888179;2.88470200832711e-05;1.39333989516825e-06;6.72997092199941e-08;3.25064320400358e-09;1.57009314931713e-10]
b = 12×1
-5.0000 -0.0117 -0.0000 4.9883 0.2560 0.0124 0.0006 0.0000 0.0000 0.0000
deviation_scalar = (a - b)*100./b
deviation_scalar = 12×1
1.0e+09 * -0.0000 0.0000 0.0001 -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0.0005 -0.0139
format long G
deviation_scalar % exactly the same numbers, different precision and multiplier
deviation_scalar = 12×1
1.0e+00 * -26.43125 1177.8685022565 104395.345893944 -69.4161482859845 -14.9155512995639 -54.5081288094575 -1146.49022460341 -30432.4224642333 -538374.976982149 -13930323.6349259
format long
deviation_scalar % exactly the same numbers, different precision and multiplier
deviation_scalar = 12×1
1.0e+09 * -0.000000026431250 0.000001177868502 0.000104395345894 -0.000000069416148 -0.000000014915551 -0.000000054508129 -0.000001146490225 -0.000030432422464 -0.000538374976982 -0.013930323634926
format short E
deviation_scalar % exactly the same numbers, different precision and multipliers
deviation_scalar = 12×1
1.0e+00 * -2.6431e+01 1.1779e+03 1.0440e+05 -6.9416e+01 -1.4916e+01 -5.4508e+01 -1.1465e+03 -3.0432e+04 -5.3837e+05 -1.3930e+07
These are all the same numbers, just displayed to different precisions and exponents.

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Benjamin Brammer
Benjamin Brammer 2022-12-15
Dear Stephen23,
thanks for your answer. But to be true, I don't understand how -0.0000 x 1.0 e+9 is the same as -26. And why do I get different result when use slightly different solution ways.
With this logic a lot of different numbers would get me the same -0.0000 x 1.0e+9 result and this would be catastrophic.
How do I know when matlab will make the "correct" format assumption as with (-3.6784 - -5)*100/-5 and the "not correct" format assunmption? I believe there is a logic to this but this will get me in the responsibility to check for the correct format.
Btw, why is my 20 year old calculator showing me a "correct", so with a meaningful format, result? But one of the powerful mathematical softwares not?
Off course, one might argue it's me who has to know what he's been doing, but if I do experiments where I don't know what results are to expected, I definetely don't want to question every mathematical calculation MATLAB does for me.
Benjamin Brammer
Benjamin Brammer 2022-12-15
Ok, I think I misunderstood, that this affects only the displaying format of numbers. So MATLAB - internally - does allways have the correct calculation result. So its up to me to decide what displaying format I like and force matlab to use it.
Benjamin Brammer
Benjamin Brammer 2022-12-15
Hey Stephen23,
thanks again for your efforts on explaining matters to me and your examples. I think I get the point. I have found that format shortG gives me my "correct" expectation of values.
What I found not intuitive is that MATLAB used a common multiplier to display elements in a vector, which is - in my opinion - not suitable if you have values that differ in several magnitudes. So I would have found it intuitive if MATLAB would have used the shortG formating to show me values in a - for me - reasonable way.
Steven Lord
Steven Lord 2022-12-15
If you want to change the default display format for floating-point numbers you can do that in the Preferences. See the "Format Floating-Point Numbers" section on this documentation page for instructions.
Benjamin Brammer
Benjamin Brammer 2022-12-15
Thanks for the hint and help!

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