I have two arrays of data that I'm trying to amalgamate. One contains actual latencies from an experiment in the first column (e.g. 0.345, 0.455... never more than 3 decimal places), along with other data from that experiment. The other contains what is effectively a 'look up' list of latencies ranging from 0.001 to 0.500 in 0.001 increments, along with other pieces of data. Both data sets are X-by-Y doubles.
What I'm trying to do is something like...
for i = 1:length(actual_latency)
row = find(predicted_data(:,1) == actual_latency(i))
full_set(i,1:4) = [actual_latency(i) other_info(i) predicted_info(row,2) ...
predicted_info(row,3)];
end
...in order to find the relevant row in predicted_data where the look up latency corresponds to the actual latency. I then use this to created an amalgamated data set, full_set.
I figured this would be really simple, but the find function keeps failing by throwing up an empty matrix when looking for an actual latency that I know is in predicted_data(:,1) (as I've double-checked during debugging).
Moreover, if I replace find with a for loop to do the same job, I get a similar error. It doesn't appear to be systematic - using different participant data sets throws it up in different places.
Furthermore, during debugging mode, if I use find to try and find a hard-coded value of actual_latency, it doesn't always work. Sometimes yes, sometimes no.
I'm really scratching my head over this, so if anyone has any ideas about what might be going on, I'd be really grateful.
You are likely running into a problem with floating point comparisons when you do the following:
predicted_data(:,1) == actual_latency(i)
Even though your numbers appear to only have three decimal places of precision, they may still differ by very small amounts that are not being displayed, thus giving you an empty matrix since FIND can't get an exact match.
One feature of floating point numbers is that certain numbers can't be exactly represented, since they aren't an integer power of 2. This occurs with the numbers 0.1 and 0.001. If you repeatedly add or multiply one of these numbers you can see some unexpected behavior. Amro pointed out one example in his comment: 0.3 is not exactly equal to 3*0.1. This can also be illustrated by creating your look-up list of latencies in two different ways. You can use the normal colon syntax:
vec1 = 0.001:0.001:0.5;
Or you can use LINSPACE:
vec2 = linspace(0.001,0.5,500);
You'd think these two vectors would be equal to one another, but think again!:
>> isequal(vec1,vec2)
ans =
0 %# FALSE!
This is because the two methods create the vectors by performing successive additions or multiplications of 0.001 in different ways, giving ever so slightly different values for some entries in the vector. You can take a look at this technical solution for more details.
When comparing floating point numbers, you should therefore do your comparisons using some tolerance. For example, this finds the indices of entries in the look-up list that are within 0.0001 of your actual latency:
tolerance = 0.0001;
for i = 1:length(actual_latency)
row = find(abs(predicted_data(:,1) - actual_latency(i)) < tolerance);
...
The topic of floating point comparison is also covered in this related question.
You may try to do the following:
row = find(abs(predicted_data(:,1) - actual_latency(i))) < eps)
EPS is accuracy of floating-point operation.
Have you tried using a tolerance rather than == ?
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 working on a code in Least Square Non Negative solution recovery context on Matlab, and I need (with no more details because it's not that important for this question) to know the number of non zero elements in my matrices and arrays.
The function NNZ on matlab does exactly what I want, but it happens that I need more information about what Matlab thinks of a "zero element", it could be 0 itself, or the numerical zero like 1e-16 or less.
Does anybody has this information about the NNZ function, cause I couldn't get the original script
Thanks.
PS : I am not an expert on Matlab, so accept my apologies if it's a really simple task.
I tried "open nnz", on Matlab but I only get a small script of commented code lines...
Since nnz counts everything that isn't an exact zero (i.e. 1e-100 is non-zero), you just have to apply a relational operator to your data first to find how many values exceed some tolerance around zero. For a matrix A:
n = nnz(abs(A) > 1e-16);
Also, this discussion of floating-point comparison might be of interest to you.
You can add in a tolerance by doing something like:
nnz(abs(myarray)>tol);
This will create a binary array that is 1 when abs(myarray)>tol and 0 otherwise and then count the number of non-zero entries.
I would like to partition a number into an almost equal number of values in each partition. The only criteria is that each partition must be in between 60 to 80.
For example, if I have a value = 300, this means that 75 * 4 = 300.
I would like to know a method to get this 4 and 75 in the above example. In some cases, all partitions don't need to be of equal value, but they should be in between 60 and 80. Any constraints can be used (addition, subtraction, etc..). However, the outputs must not be floating point.
Also it's not that the total must be exactly 300 as in this case, but they can be up to a maximum of +40 of the total, and so for the case of 300, the numbers can sum up to 340 if required.
Assuming only addition, you can formulate this problem into a linear programming problem. You would choose an objective function that would maximize the sum of all of the factors chosen to generate that number for you. Therefore, your objective function would be:
(source: codecogs.com)
.
In this case, n would be the number of factors you are using to try and decompose your number into. Each x_i is a particular factor in the overall sum of the value you want to decompose. I'm also going to assume that none of the factors can be floating point, and can only be integer. As such, you need to use a special case of linear programming called integer programming where the constraints and the actual solution to your problem are all in integers. In general, the integer programming problem is formulated thusly:
You are actually trying to minimize this objective function, such that you produce a parameter vector of x that are subject to all of these constraints. In our case, x would be a vector of numbers where each element forms part of the sum to the value you are trying to decompose (300 in your case).
You have inequalities, equalities and also boundaries of x that each parameter in your solution must respect. You also need to make sure that each parameter of x is an integer. As such, MATLAB has a function called intlinprog that will perform this for you. However, this function assumes that you are minimizing the objective function, and so if you want to maximize, simply minimize on the negative. f is a vector of weights to be applied to each value in your parameter vector, and with our objective function, you just need to set all of these to -1.
Therefore, to formulate your problem in an integer programming framework, you are actually doing:
(source: codecogs.com)
V would be the value you are trying to decompose (so 300 in your example).
The standard way to call intlinprog is in the following way:
x = intlinprog(f,intcon,A,b,Aeq,beq,lb,ub);
f is the vector that weights each parameter of the solution you want to solve, intcon denotes which of your parameters need to be integer. In this case, you want all of them to be integer so you would have to supply an increasing vector from 1 to n, where n is the number of factors you want to decompose the number V into (same as before). A and b are matrices and vectors that define your inequality constraints. Because you want equality, you'd set this to empty ([]). Aeq and beq are the same as A and b, but for equality. Because you only have one constraint here, you would simply create a matrix of 1 row, where each value is set to 1. beq would be a single value which denotes the number you are trying to factorize. lb and ub are the lower and upper bounds for each value in the parameter set that you are bounding with, so this would be 60 and 80 respectively, and you'd have to specify a vector to ensure that each value of the parameters are bounded between these two ranges.
Now, because you don't know how many factors will evenly decompose your value, you'll have to loop over a given set of factors (like between 1 to 10, or 1 to 20, etc.), place your results in a cell array, then you have to manually examine yourself whether or not an integer decomposition was successful.
num_factors = 20; %// Number of factors to try and decompose your value
V = 300;
results = cell(1, num_factors);
%// Try to solve the problem for a number of different factors
for n = 1 : num_factors
x = intlinprog(-ones(n,1),1:n,[],[],ones(1,n),V,60*ones(n,1),80*ones(n,1));
results{n} = x;
end
You can then go through results and see which value of n was successful in decomposing your number into that said number of factors.
One small problem here is that we also don't know how many factors we should check up to. That unfortunately I don't have an answer to, and so you'll have to play with this value until you get good results. This is also an unconstrained parameter, and I'll talk about this more later in this post.
However, intlinprog was only released in recent versions of MATLAB. If you want to do the same thing without it, you can use linprog, which is the floating point version of integer programming... actually, it's just the core linear programming framework itself. You would call linprog this way:
x = linprog(f,A,b,Aeq,beq,lb,ub);
All of the variables are the same, except that intcon is not used here... which makes sense as linprog may generate floating point numbers as part of its solution. Due to the fact that linprog can generate floating point solutions, what you can do is if you want to ensure that for a given value of n, you could loop over your results, take the floor of the result and subtract with the final result, and sum over the result. If you get a value of 0, this means that you had a completely integer result. Therefore, you'd have to do something like:
num_factors = 20; %// Number of factors to try and decompose your value
V = 300;
results = cell(1, num_factors);
%// Try to solve the problem for a number of different factors
for n = 1 : num_factors
x = linprog(-ones(n,1),[],[],ones(1,n),V,60*ones(n,1),80*ones(n,1));
results{n} = x;
end
%// Loop through and determine which decompositions were successful integer ones
out = cellfun(#(x) sum(abs(floor(x) - x)), results);
%// Determine which values of n were successful in the integer composition.
final_factors = find(~out);
final_factors will contain which number of factors you specified that was successful in an integer decomposition. Now, if final_factors is empty, this means that it wasn't successful in finding anything that would be able to decompose the value into integer factors. Noting your problem description, you said you can allow for tolerances, so perhaps scan through results and determine which overall sum best matches the value, then choose whatever number of factors that gave you that result as the final answer.
Now, noting from my comments, you'll see that this problem is very unconstrained. You don't know how many factors are required to get an integer decomposition of your value, which is why we had to semi-brute-force it. In fact, this is a more general case of the subset sum problem. This problem is NP-complete. Basically, what this means is that it is not known whether there is a polynomial-time algorithm that can be used to solve this kind of problem and that the only way to get a valid solution is to brute-force each possible solution and check if it works with the specified problem. Usually, brute-forcing solutions requires exponential time, which is very intractable for large problems. Another interesting fact is that modern cryptography algorithms use NP-Complete intractability as part of their ciphertext and encrypting. Basically, they're banking on the fact that the only way for you to determine the right key that was used to encrypt your plain text is to check all possible keys, which is an intractable problem... especially if you use 128-bit encryption! This means you would have to check 2^128 possibilities, and assuming a moderately fast computer, the worst-case time to find the right key will take more than the current age of the universe. Check out this cool Wikipedia post for more details in intractability with regards to key breaking in cryptography.
In fact, NP-complete problems are very popular and there have been many attempts to determine whether there is or there isn't a polynomial-time algorithm to solve such problems. An interesting property is that if you can find a polynomial-time algorithm that will solve one problem, you will have found an algorithm to solve them all.
The Clay Mathematics Institute has what are known as Millennium Problems where if you solve any problem listed on their website, you get a million dollars.
Also, that's for each problem, so one problem solved == 1 million dollars!
(source: quickmeme.com)
The NP problem is amongst one of the seven problems up for solving. If I recall correctly, only one problem has been solved so far, and these problems were first released to the public in the year 2000 (hence millennium...). So... it has been about 14 years and only one problem has been solved. Don't let that discourage you though! If you want to invest some time and try to solve one of the problems, please do!
Hopefully this will be enough to get you started. Good luck!
Here's some simple code that shows what I've been seeing:
A = randn(1,5e6)+1i*randn(1,5e6);
B = randn(1,5e6)+1i*randn(1,5e6);
sum(A.*conj(B)) - A*B'
sum(A.*conj(B)) - mtimes(A,B')
A*B' - mtimes(A,B')
Now, the three methods shown on the bottom are supposed to do the same thing, so the answers should be zero, right? Wrong! The differences are small, though not small enough that I would consider them negligible. In addition, the error increases as the length of A and B increases.
Does anyone know what the actual difference between these methods is? I understand that there are probably shortcuts written into the code, but I would like to quantify that if possible. Does Matlab post the differences anywhere? I've looked around, but have not found anything.
It probably has something to do with the order in which the operations are performed. For example,
sum(A.*conj(B)) - fliplr(A)*fliplr(B)'
gives a result different than
sum(A.*conj(B)) - A*B'
Or, more strikingly,
A*B' - fliplr(A)*fliplr(B)'
gives a nonzero result, of the same order as your tests.
So my bet is that depending on the method (sum or *) Matlab internally does the operations in a different order, and that may well account for the different roundoff errors you observed.
Given the size of each number, the rounding error in simple operations on this number is of the order 10^-14
You have 5*10^6 numbers, hence if you are really unlucky the rounding error can become anything upto 5*10^-8.
Your observed error is of size 10^10, which is well in the expected range.
Note that the difference is not caused by the complex transpose, but by the product of the sum vs the matrix product.
A = randn(1,5e6)+1i*randn(1,5e6);
B = randn(1,5e6)+1i*randn(1,5e6);
B1 = conj(B);
B2 = B';
isequal(B1(:),B2(:)) % This returns true
A*transpose(conj(B)) - A*B' % Hence this returns zero
sum(A.*transpose(B')) - A*B' % But this returns something like 1e-10
A similar effect occurs for non complex A and B:
N=1e6;
A = 1:N;
B=1:N;
(N * (N + 1) * (2*N + 1))/6 % This will give exactly the right answer
A*B'
fliplr(A)*fliplr(B)'
Note that the two lowest answers only vary a few hundred from eachother, whilst they are actually over 2000 from the correct answer. If this is a problem consider using the symbolic toolbox. That allows you to calculate with arbitrary precision.
I have made an array of doubles and when I want to use the find command to search for the indices of specific values in the array, this yields an empty matrix which is not what I want. I assume the problem lies in the precision of the values and/or decimal places that are not shown in the readout of the array.
command:
peaks=find(y1==0.8236)
array readout:
y1 =
Columns 1 through 11
0.2000 0.5280 0.8224 0.4820 0.8239 0.4787 0.8235 0.4796 0.8236 0.4794 0.8236
Columns 12 through 20
0.4794 0.8236 0.4794 0.8236 0.4794 0.8236 0.4794 0.8236 0.4794
output:
peaks =
Empty matrix: 1-by-0
I tried using the command
format short
but I guess this only truncates the displayed values and not the actual values in the array.
How can I used the find command to give an array of indices?
By default, each element of a numerical matrix in Matlab is stored using floating point double precision. As you surmise in the question format short and format long merely alter the displayed format, rather than the actual format of the numbers.
So if y1 is created using something like y1 = rand(100, 1), and you want to find particular elements in y1 using find, you'll need to know the exact value of the element you're looking for to floating point double precision - which depending on your application is probably non-trivial. Certainly, peaks=find(y1==0.8236) will return the empty matrix if y1 only contains values like 0.823622378...
So, how to get around this problem? It depends on your application. One approach is to truncate all the values in y1 to a given precision that you want to work in. Funnily enough, a SO matlab question on exactly this topic attracted two good answers about 12 hours ago, see here for more.
If you do decide to go down this route, I would recommend something like this:
a = 1e-4 %# Define level of precision
y1Round = round((1/a) * y1); %# Round to appropriate precision, and leave y1 in integer form
Index = find(y1Round == SomeValue); %# Perform the find operation
Note that I use the find command with y1Round in integer form. This is because integers are stored exactly when using floating point double, so you won't need to worry about floating point precision.
An alternative approach to this problem would be to use find with some tolerance for error, for example:
Index = find(abs(y1 - SomeValue) < Tolerance);
Which path you choose is up to you. However, before adopting either of these approaches, I would have a good hard look at your application and see if it can be reformulated in some way such that you don't need to search for specific "real" numbers from among a set of "reals". That would be the most ideal outcome.
EDIT: The code advocated in the other two answers to this question is neater than my second approach - so I've altered it accordingly.
Testing for equality with floating-point numbers is almost always a bad idea. What you probably want to do is test to see which numbers are close enough to the target value:
peaks = find( abs( y - .8236 ) < .0001 );
The problem is indeed with the precision. The array that you see displayed is not the actual array, as the actual array has more digits for each of the numbers. Changing the format just changes the way in which the array is displayed, so it doesn't solve the problem.
You have two options, either modify the array or modify what you are looking for. It is probably better to modify what you are looking for, since then you are not changing the actual values.
So instead of looking for equality, you can look for proximity (so the difference between the number you are searching for and the number in the array is at most some small epsilon):
peaks = find( abs(y1-0.8236) < epsilon )
In general, when you are dealing with floats, always try to avoid exact comparisons and use some error thresholds, since the representation of these numbers is limited so they are often stored with small inaccuracies.