I have made this scripts that calculates the frequency of a given dataset, but matlab is not precise enough, is it possible to make matlab read in more accurat numbers and not cut off the numbers? I want it to use 8 digits (0.12345678) instead of 4 (0.1234) that is does now
fid = fopen('forceCoeffs.dat','rt');
A = textscan(fid, '%f%f%f%f%f%f', 'HeaderLines',9,'Collect', 9);
A = A{1};
fclose(fid);
t = A(:,1);
Fs = 1/(A(1,1));
x = A(:,2)
x = detrend(x,0);
xdft = fft(x);
freq = 0:Fs/length(x):Fs/2;
xdft = xdft(1:length(x)/2+1);
plot(freq,abs(xdft));
[~,I] = max(abs(xdft));
fprintf('Maximum occurs at %d Hz.\n',freq(I));
File: https://drive.google.com/file/d/0B9CEsYCSSZUSb1JmcHRkbFdWYUU/view?usp=sharing
Thank you for including the forceCoeffs.dat file as it allowed me to run your code. Here is an explanation of what you are seeing.
First I want to point out that MATLAB is not rounding anything. You can check the data type of A to ensure you have enough precision.
>> class(A)
ans =
double
And since you are reading in the file using %f for each column, MATLAB will use all the bits provided by the double type. Ok, now take a look at the contents of your file. The first column has only 2 decimals of precision at most.
0.05 -7.013874e-09 1.410717e+02 -6.688450e-02 -3.344226e-02 -3.344224e-02
...
349.95 -1.189524e-03 1.381022e+00 -2.523909e-01 -1.273850e-01 -1.250059e-01
350 -1.423947e-03 1.380908e+00 -2.471767e-01 -1.250123e-01 -1.221644e-01
Since no more is needed MATLAB only prints four decimal places when you look at the variable in the variable explorer. Try looking at one of the other columns to see what I am talking about. I commented out the A = A{1} part of your code and looked at the second column. When clicking on the number you see the full precision.
You can use a long type to display 16 digits
To get more than 4 digits precision, you can use
format long
However, to get exactly 8 digits, you need to round it. If your number is a then let use:
format long
round(1e8*a)*1e-8
Related
I have MATLAB code:
clear;
clc;
syms x;
f=log(x)*sin(x^2);
a=vpa(subs(f,x,2),100)
fprintf('a=%.100f\n',a);
doublea=double(a);
fprintf('a=%.100f\n',doublea);
and the result is
a =
-0.5245755158634217064842071630254785076113576311088295152384038229263081153172372089356742060202648499
a=-0.5245755158634216600000000000000000000000000000000000000000000000000000000000000000000000000000000000
doublea =
-0.5246
a=-0.5245755158634216600000000000000000000000000000000000000000000000000000000000000000000000000000000000
>>
Why if I use fprintf the decimal precision is only up to 16 digits, even though I use 100 digits precision?
Also why if I convert a to double then a is only up to 16 digits?
Can it cause an error in the calculation if I want use more than 16 digits precision? How to fix it?
Have you tried using num2str or vpa coupled with disp?
a1 = vpa(a,100);
disp(a1);
Or
disp(['a = ' num2str(a,100)])
Edit: Hadi's link in the comments has a great explanation.
I am working with binary information, read from file. This is a sequence of numbers, in 40 bit format, where first 8bit should be neglected in my case, and another 32 bits are "shuffled" 32bit single precision IEEE 754 format. This "shuffle" is pretty straightforward: the correct IEEE 754 binary32 I get, when I take following order of bits: 24-32, 17-24, 9-16
All this is simulated with code below.
Question: How can I improve the code below to make it faster, get rid of "for" loop and use efficient MATLAB matrix operations?
a = (1:5*8*1000000)'; % represent indices of bits binary information, read from file
tic
a_reshaped = reshape(a, 8, [])';
toc %Elapsed time is 0.176375 seconds.
n_elem = size(a_reshaped,1)/5;
result = zeros(n_elem,8*4);
for i = 1:n_elem
result(i,:) = [a_reshaped(5*i,:) a_reshaped(5*i-1,:) a_reshaped(5*i-2,:) a_reshaped(5*i-3,:)];
end
toc %Elapsed time is 4.243868 seconds.
Try this:
ind = size(a_reshaped,1):-1:1;
ind(end:-5:1) = []; %remove the indices for rows you don't need
a_reduced = a_reshaped(ind,:);
result = flipud(reshape(a_reduced',8*4,[])');
the "for" loop can be replaced within following one line code, which is much faster:
result(1:n_elem,:) = [a_reshaped(5*(1:n_elem),:) a_reshaped(5*(1:n_elem)-1,:) a_reshaped(5*(1:n_elem)-2,:) a_reshaped(5*(1:n_elem)-3,:)]; %Elapsed time is 0.766840 seconds.
I want to calculate the percentage of accuracy. I have the code below. But it give unexpected output like this "The accuracy is 2.843137e+01x37".
While expected result is "The accuracy is 28.43%"
y %Amount of correct data
j %Amount of all data
a = 'The accuracy is %dx%d.';
percent = '%.0f%%';
format short
acc = 100 * double(y/j);
sprintf (a,acc)
How to fix it?
Any help would be so much appreciated.
Thank you.
You almost have what you expected, just put it together the right way.
The correct format specifier for 28.43% is %.2f%%. This gives you two digits after the decimal point and adds the %-sign at the end. You have that defined in the variable percent, except that .0 should be .2 for two digits as you have written in the expected result. If you look closely, you'll realize that percent is never used.
Let's come to the conclusion. Change the format specifier to the following:
a = 'The accuracy is %.2f%%';
That's all you need to do. The line defining percent can be omitted as well as format short unless you need this for something later on.
Something important regarding the cast to double: What you currently have just casts the result. If necessary, do the cast individually to y and/or j before the division. Probably you don't need any casting in your case.
The whole code with an assumption for y and j is:
y = 28.43137; %// Amount of correct data
j = 100; %// Amount of all data
a = 'The accuracy is %.2f%%';
acc = 100 * (y/j); %// no cast
% acc = 100 * (double(y)/double(j)); %// with cast
sprintf(a,acc);
Output:
ans =
The accuracy is 28.43%
Try,
a = 'The accuracy is %f.';
acc = 100 * double(y/j);
sprintf (a,acc)
Let's say I have a random variable a=1.2400, and I want to print it with four significant figures, i.e., 1.240. How would I go about that?
fprintf('%0.4g',a) % drops rightmost zero
fprintf('%0.3f',a) % give too many sig figs if a >= 10
Using '%g' drops the important zeros, and with '%f' I can only specify the number of digits after the decimal, which results in too many significant figures if, say, a=10.04. I'm not too familiar with formatting ,but there has to be a simple method. I haven't found it in my searches.
If the values to be printed are all less than 10000, you can do the following. (Sorry, only tested in octave.)
octave:62> a = 1.24
a = 1.2400
octave:63> sprintf('%.*f\n', 3-floor(log10(abs(a))), a)
ans = 1.240
octave:64> a = 234.56
a = 234.56
octave:65> sprintf('%.*f\n', 3-floor(log10(abs(a))), a)
ans = 234.6
For more about the expression floor(log10(abs(a))), see How can I get the exponent of each number in a np.array?
If you don't mind exponential notation, another alternative is to use '%.3e' to always get the same number of signficant digits:
octave:70> a = 1.24
a = 1.2400
octave:71> sprintf('%.3e\n', a)
ans = 1.240e+00
octave:72> a = 234.56
a = 234.56
octave:73> sprintf('%.3e\n', a)
ans = 2.346e+02
I decided to build on the answer by Warren, and I wrote a function that should work for both small and large numbers alike. Perhaps someone will improve on this, but I am pleased with it.
function str=sigfigstr(a,sigfigs)
numdecimal = floor(log10(abs(a)));
if sigfigs - numdecimal < 0
str=sprintf('%.0f',round(a,sigfigs,'significant'));
else
str=strip(sprintf('%.*f\n', sigfigs-floor(log10(abs(a))), a));
end
Here are a few examples if it in action in Matlab
>> sigfigstr(.000012431634,3)
ans = '0.0000124'
>> sigfigstr(26666,3)
ans = '26700'
Here is a very specific example
>> S = num2str(12345,'%6.0e')
S =
1e+04
and that's just great since I want only my first digit and an exponential notation. However I also want to add leading zeros to the exponent in order to fill the width, but I cannot quite find the way to get the following...
1e+004
Meanwhile it's very straighforward to pad the significant digits with leading zeros
>> S = num2str(12345,'%06.0e')
S =
01e+04
So is there an appropriate formatting for what I want? Or a trick to accomplish it quickly?
The exponent is always a zero-padded two-digit value. To add, say, two zeros you can use
regexprep(num2str(12345, '%6.0e'), '\+', '\+00')
and achieve
ans =
1e+0004
Edit: To cover negative exponents you may use
regexprep(num2str(0.12345, '%6.0e'), '(\+|\-)', '$100')
to achieve
ans =
1e-0001
And, to cover three-digit exponents
regexprep(num2str(1e-100, '%6.0e'), '(\+|\-)(\d{2,3})$', {'$10$2', '$10$2'})
ans =
1e-0100
regexprep(num2str(1e-10, '%6.0e'), '(\+|\-)(\d{2,3})$', {'$10$2', '$10$2'})
ans =
1e-0010
Well, I think you have to edit, what you say you want is wat you get :D
however, if I understood correctly what you are looking for, this function will help you
function printX(x, digits)
format = sprintf('\t%%.%de', digits - 1);
strcat(inputname(1), ' = ', sprintf(format, x))
end