I am converting a program from MATLAB 2012 to 2016. I've been getting some strange errors, which I believe some of are due to a lack of precision in MATLAB functions.
For instance, I have a timeseries oldTs as such:
Time Data
-----------------------------
1.00000000000000001 1.277032377439511
1.00000000000000002 1.277032378456123
1.00000000000000003 1.277032380112478
I have another timeseries newTs with similar data, but many more rows. oldTs may have half a million rows, whereas newTs could have a million. I want to interpolate the data from the old timeseries with the new timeseries, for example:
interpolatedTs = interp(oldTs.time, oldTs.data, newTs.time)
This is giving me an error: x values must be distinct
The thing is, my x values are distinct. I think that MATLAB may be truncating some of the data, and therefore believing that some of the data is not unique. I found that other MATLAB functions do this:
test = [1.00000000000000001, 1.00000000000000002, 1.0000000000000000003]
unique(test)
ans =
1
test2 = [10000000000000000001, 10000000000000000002, 10000000000000000003]
unique(test2)
ans =
1.000000000000000e+19
MATLAB thinks that this vector only has one unique value in it instead of three! This is a huge issue for me, as I need to maintain the highest level of accuracy and precision with my data, and I cannot sacrifice any of that precision. Speed/Storage is not a factor.
Do certain MATLAB functions, by default, truncate data at a certain nth decimal? Has this changed from MATLAB 2012 to MATLAB 2016? Is there a way to force MATLAB to use a certain precision for a program? Why does MATLAB do this to begin with?
Any light shed on this topic is much appreciated. Thanks.
No, this has not changed since 2012, nor since the very first version of MATLAB. MATLAB uses, and has always used, double precision floating point values by default (8 bytes). The first value larger than 1 that can be represented is 1 + eps(1), with eps(1) = 2.2204e-16. Basically you have less than 16 decimal digits to play with. Your value 1.00000000000000001 is identical to 1 in double precision floating point representation.
Note that this is not something specific to MATLAB, it is a standard that your hardware conforms to. MATLAB simply uses your hardware's capabilities.
Use the variable precision arithmetic from the Symbolic Math Toolbox to work with higher precision numbers:
data = [vpa(1) + 0.00000000000000001
vpa(1) + 0.00000000000000002
vpa(1) + 0.00000000000000003]
data =
1.00000000000000001
1.00000000000000002
1.00000000000000003
Note that vpa(1.00000000000000001) will not work, as the number is first interpreted as a double-precision float value, and only after converted to VPA, but the damage has already been done at that point.
Note also that arithmetic with VPA is a lot slower, and some operations might not be possible at all.
Related
Why do I get different results when using the same code running in different version of MATLAB (2016 vs 2021) for sum(b.*x1) where b is single and x1 is double. How to avoid such error between MATLAB version?
MATLAB v.2021:
sum(b.*x1)
ans =
single
-0.0013286
MATLAB 2016
sum(b.*x1)
ans =
single
-0.0013283
In R2017b, they changed the behavior of sum for single-precision floats, and in R2020b they made the same changes for other data types too.
The change speeds up the computation, and improves accuracy by reducing the rounding errors. Simply put, previously the algorithm would just run through the array in sequence, adding up the values. The new behavior computes the sum over smaller portions of the array, and then adds up those results. This is more precise because the running total can become a very large number, and adding smaller numbers to it causes more rounding in those smaller numbers. The speed improvement comes from loop unrolling: the loop now steps over, say, 8 values at the time, and in the loop body, 8 running totals are computed (they don’t specify the number they use, the 8 here is an example).
Thus, your newer result is a better approximation to the sum of your array than the old one.
For more details (a better explanation of the new algorithm and the reason for the change), see this blog post.
Regarding how to avoid the difference: you could implement your own sum function, and use that instead of the builtin one. I would suggest writing it as a MEX-file for efficiency. However, do make sure you match the newer behavior of the builtin sum, as that is the better approximation.
Here is an example of the problem. Let's create an array with N+1 elements, where the first one has a value of N and the rest have a value of 1.
N = 1e8;
a = ones(N+1,1,'single');
a(1) = N;
The sum over this array is expected to be 2*N. If we set N large enough w.r.t. the data type, I see this in R2017a (before the change):
>> sum(a)
ans =
single
150331648
And I see this in R2018b (after the change for single-precision sum):
>> sum(a)
ans =
single
199998976
Both implementations make rounding errors here, but one is obviously much, much closer to the expected result (2e8, or 200000000).
I am currently trying to learn MATLAB independently and had a question about a command that used randn().
nddata = fix(8*randn(10,5,3))
I understand what the fix() function does, and the multi dimension array that is created by randn. However, I am not sure what 8 is doing here, it is not multiplying the outcome of the random numbers and it is not part of the limit. So I just want to know the purpose of the 8 here.
Thanks
randn generated a standard normally distributed matrix of random numbers (standard in this context is defined as mean = 0 and standard deviation = 1). The 8 factor simply stretches this distribution along the x-axis; a scalar multiplication for each value in the 3D matrix. The fix function then rounds each value to the nearest integer towards 0, i.e. -3.9 becomes -3.0. This effectively reduces the standard deviation of the data.
To see this for yourself, split the expression up and create temporary variables for each operation, and step through it with the debugger.
I have to import a large integer (1000 digits) into matlab to run some calculations on it. However, when I import it I seem to loose accuracy due to the fact that matlab uses the scientific notation.
Is there any way that I can get the actual integer?
Here's the actual data I have to import:
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Such a large integer cannot be represented in IEEE floating point standard. Check out this answer for the largest double that can be represented without losing precision (its 1.7977e+308). That can be obtained by typing realmax in MATLAB.
You can use vpi (available here, as mentioned in comment) or you can use the MATLAB in-built vpa.
This is how you use vpa
R=vpa('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450');
You can check the following:
vpa('R+1000-R')
The answer of the above is 1000 as expected. Do not forget to put your expression in quotes. Otherwise, you are passing inifinity to vpa instead of the 1000 digit number.
If you want to use vpi, its a beautiful toolbox, go ahead, download it. Go into its root directory and run the following command:
a=vpi('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')
Well, the advantage with vpi is as follows:
The output of vpi:
a=vpi(<<Your 1000 digit number in quotes>>); %output prints 1000 digits on screen.
The output of vpa:
R=vpa(<<Your 1000 digit number in quotes>>);
this prints:
R =
7.3167176531330624919225119674427e999
Also, with vpi, you can do something like this:
a=vpi('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')
b=a+1
b-a %output of this yields 1.
I somehow cannot do the operation of b-a in vpa and obtain the answer 1.
I have a question about adding the number 1 to very small numbers. Right now, I am trying to plot a circular arc in the complex plane centered around the real number 1. My code looks like:
arc = 1 + rho .* exp(1i.*theta);
The value rho is a very small number, and theta runs from 0 to pi, so whenever 1 is added to the real part of arc, MATLAB seems to just round it to 1, so when I type in plot(real(arc),imag(arc)), all I see is a spike instead of a semicircle around 1. Does anyone know how to remedy this so that MATLAB will not round 1 + real(arc) to 1, and instead conserve the precision?
Thanks
rho=1e-6; theta=0:pi/100:pi; arc=1+rho*exp(1i.*theta); plot(arc); figure(); plot(arc-1);
Shows, that the problem is in plot, not in loss of precision. After rho<1e-13 there will be expected trouble with precision.
The two other possible misconceptions:
- doubles have finite precision. 16 decimal digits or 1+2^-52 is the limit with doubles.
- format short vs. format long -- matlab shows by default only 6 or 7 digits
It also happens to be that 6-7 digits is the limit of a 32-bit float, which could explain also that perhaps the plot function in Octave 3.4.3 is also implemented with floats.
Left: 1+1e-6*exp, Right: (1+1e-6*exp)-1
There is a builtin solution for exactly this probem:
exp1m()
log1p()
explicitly:
log(arc)=log1p(rho*exp(1i*theta))
to get what you need.
Of course you need to work in log space to represent this precision, but this is the typical way this is done.
In double precision floating point representations, the smallest number strictly greater than 1 that can be represented is 1 + 2^-52.
This is a limitation imposed by the way non-integer numbers are represented on most machines that can be avoided in software, but not easily. See this question about approaches for MATLAB.
In MATLAB, is there any way at all to make it do all calculations using only 4 digits after decimal ?
If I write fprintf('%85.83f\n',single(1.1566)), its not exactly 1.1566. For my specific problem I cannot afford to accumulate those extra digits.
I want all internal calculations to be done on this precision only.
By default MATLAB conducts all operations with double precision. If you want your operations to use a lower precision, you can use variable precision arithmetic, which is part of the Symbolic Math Toolbox. In your case, first set the precision to 4 significant digits using the digits function, and then declare and manipulate variables using the vpa function:
old_precision = digits; % Save the old precision
digits(4);
vpa(1/3 + 1/2)
digits(old_precision); % Set the precision back
will output
>> vpa(1/3+1/2)
ans =
0.8333
Another example:
>> digits(3);
>> a = 5.4;
>> vpa(a)+1/3
ans =
5.73
The MathWorks offer a Fixed-Point Toolbox which might meet OP's needs.
However, if OP's statement I cannot afford to accumulate those extra digits is code for I want faster execution and don't mind the loss of precision I fear that OP may be out of luck. On modern digital computers with either 32- or 64-bit (or both) floating-point computations built into the hardware, computations at a non-hardware-precision will have to be done in software and are likely to be noticeably slower.