MATLAB: Round off problem in matlab

MATLAB

I have 4 variables each have 4 values (1×4) in it and I want to store all in value in 1 variable (4×4) but values are change when i do it.

MAV = [0.0924317411754942 0.0313003411430704 0.00365072354007141 0.00332056432198707]

waveLen = [172.697769103870 68.7603283596702 69.2536410243476 63.2141284111186]

zcdOut = [560 719 6155 6260]

slope_count = [478 212 4949 4830]

HG0_extracted_feature = [waveLen; MAV; zcdOut; slope_count]

it gives

173 69 69 63

0 0 0 0

560 719 6155 6260

478 212 4949 4830

note MAV values become zero, it round off waveLen, etc values…

how i correct it?

Thank you

Best Answer

Based on that display I suspect one or both of the variables zcdOut or slope_count are stored in an integer class. When you combine data stored in an integer class with data stored in a non-integer class, "all elements of the resulting matrix are given the data type of the left-most integer".

Consider converting the variable stored in an integer class to double before combining them into one array, so that the resulting matrix is also a double.

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MATLAB: Problem in applying loop

What is the 'y' you have used in your main script loop? Its not defined anywhere. Use this line in main script:

waveLen(i) = find_waveform_length(x(:,i))

Since you have not shared your MAT file, I tried with a random 30000x4 matrix and it works.

Few other recommendations:

1) Put an 'end' for your function

2) Don't use variable 'i' or any other inbuilt Matlab variable as index in your 'for' loop.

MATLAB: Switch case on an array

Yes, it's a lot clearer. So, the transition to a char array is a complete red herring. It's totally unnecessary.

Using sequentially named variables is a bad idea. It will force you to write complicated code. If you have 10 different numbers that you want to count would you add another 7 cases to your switch? How about 100 numbers, 10000 numbers. Will you write all these cases?

The answer is to use array indexing, T(1) the count of your first, T(2) the count of your 2nd number, etc. Then you can use a loop. For example,one (inneficient) way to then implement your code would be:

%input
x = [12 24 64 24 12 12]
lookup = [12, 24, 64];  %values to compute the histogram of
%code
count = zeros(size(lookup))
for i = 1:numel(x)
    [found, where] = ismember(x(i), lookup);
    if found
        count(where) = count(where) + 1;
    end
end

But the whole thing can be achieve without loops, if or case. You want to compute an histogram, use the histcounts function:

x = [12 24 64 12 12];
values = unique(x)';   %get unique values of
count = histcounts(x, [values; Inf])';  %inf is right edge of last bin
table(values, count)  %for pretty display

returns

ans =
  3×2 table
    values    count
    ______    _____
      12        3  
      24        1  
      64        1

Best Answer

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