I am trying to take advantage of vectorization in MATLAB for this, but I might have to resort to for loops. I really don't want to do that! Time to learn algorithms.
Given this (11-by-3) array:
x = [...
4.9000 -0.1000 -5.1000
4.6000 -0.4000 -5.4000
3.0000 -2.0000 -7.0000
2.9000 -2.1000 -7.1000
2.9000 -2.1000 -7.1000
2.9000 -2.1000 -7.1000
2.8000 -2.2000 -7.2000
2.7000 -2.3000 -7.3000
2.7000 -2.3000 -7.3000
2.2000 -2.8000 -7.8000
1.8000 -3.2000 -8.2000
];
I want to find all of the 3^11 = 177147 possible sums of 11 elements in the array, where each of the 11 elements comes from a different row. I want to then store the sums that exceed a threshold value of 16.0, along with the 11 elements that make up each of those sums, in a (12-by-?) array.
Any ideas to get me started? Thanks for the help.
Here's how to do it in a vectorized way:
TR = 16;
sets = num2cell(single(x),2);
c = cell(1, numel(sets));
[c{:}] = ndgrid( sets{:} );
cartProd = cell2mat( cellfun(#(v)v(:), c, 'UniformOutput',false) );
validRows = cartProd(sum(cartProd,2) > TR,:); % output is [353x11]
Notice how I use single to save space and make the computation slightly faster.
The above solution is an adaptation of this answer.
Upon further contemplation, I think I've come up with a way that should be both faster and more memory efficient. We do this by indexing x, and then doing the previous process on the indices. Why is this better, you might ask? This is because we can store the indices as uint8, which consumes considerably less memory than double or even single. Also we get to keep the full double precision of x. Thus:
function validRows = q42933114(x,thresh)
%% Input handling
if nargin < 2
thresh = 16;
end
if nargin < 1
x = [...
4.9000 -0.1000 -5.1000
4.6000 -0.4000 -5.4000
3.0000 -2.0000 -7.0000
2.9000 -2.1000 -7.1000
2.9000 -2.1000 -7.1000
2.9000 -2.1000 -7.1000
2.8000 -2.2000 -7.2000
2.7000 -2.3000 -7.3000
2.7000 -2.3000 -7.3000
2.2000 -2.8000 -7.8000
1.8000 -3.2000 -8.2000
];
end
I = reshape(uint8(1:numel(x)),size(x));
sets = num2cell(I,2);
c = cell(1, numel(sets));
[c{:}] = ndgrid( sets{:} );
cartProd = cell2mat( cellfun(#(v)v(:), c, 'UniformOutput',false) );
validRows = x(cartProd(sum(x(cartProd),2) > thresh,:));
Memory consumption comparison:
Method 1 (old):
>> whos
Name Size Bytes Class Attributes
c 1x11 7795700 cell
cartProd 177147x11 7794468 single
sets 11x1 1364 cell
validRows 353x11 15532 single
Method 2 (new):
>> whos
Name Size Bytes Class Attributes
c 1x11 1949849 cell
cartProd 177147x11 1948617 uint8
sets 11x1 1265 cell
validRows 353x11 31064 double
We see that the memory consumption is indeed smaller (in about 4 times), as expected.
Runtime comparison:
Method 1 -- 0.0110
Method 2 -- 0.0186
Here we see that the 2nd method is actually a bit slower. When profiling this, we see that the cause is x(...) which is relatively expensive.
I did it this way. There is obviously room for improvement in the variable names.
Notice there are 353 matching rows, which agrees with the answer from #Dev-iL.
p = 11;
[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11] = ...
ndgrid(x(1,:),x(2,:),x(3,:),x(4,:),x(5,:),x(6,:),x(7,:),x(8,:),x(9,:),x(10,:),x(11,:));
a = a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+a11;
y = spalloc(p+1,3^p,(p+1)*3^p);
for i = 1:3^p
if a(i) >= 16.1
y(:,i) = [a1(i),a2(i),a3(i),a4(i),a5(i),a6(i),a7(i),a8(i),a9(i),a10(i),a11(i),a(i)];
end
end
nnz(y(p+1,:)); % 353 rows matching the criteria
I don't think you'll have better luck than using a for loop. There could be a Matlab function for generating all 3^11 combinations, and use that as a sort of index, but you'd have a lot of memory consumption that way.
The code would be hard to read, as well.
However, recent versions of Matlab don't behave that badly wrt for-loops, because the they JIT the code. Before it was just interpreted, or JIT-ing was used for specific purposes. You wouldn't want to reimplement the matrix routines in Matlab because of this, but for simple code like this one it should perform well.
Related
I have a matrix:
S = [ -1.0400 4.9100 4.1000 -3.5450 -0.6600 -0.9300 4.3950 -1.0650 2.9850 -4.9800 0.2100;
-0.5200 -4.3150 -3.0950 0.5700 4.4700 1.1500 3.1350 0.6450 0.3750 -4.9150 -2.1150;
5.0000 5.0000 5.0000 5.0000 5.0000 5.0000 5.0000 5.0000 5.0000 5.0000 5.0000 ];
I want to convert the columns to unit vectors, so I use a for loop
for i=1:size(S,2)
S(:,i) = S(:,i) / norm( S(:,i) );
end
Is there a way to do this more efficiently in MATLAB?
TLDR
If you have MATLAB 2016b or newer, and no compatibility concerns, I would use
S = S ./ sqrt(sum(S.^2,1));
Edit: See benchmark at the bottom for performance benchmark of alternatives.
Context
We can just manually calculate the norm and divide column-wise.
By definition, norm(x) = sqrt( sum( x(:).^2 ) ). I've used (:) here to show that norm is calculated over the whole matrix. What's useful for us is that sum works column-wise by default, so the column-wise norm is defined like so:
nrm = sqrt( sum( x.^2 ) );
Note that if there's a possibility of your matrix S only having 1 row, you should ecplicitly enforce column-wise summation using nrm = sqrt(sum(x.^2,1)).
Now we have several options for division:
Implicit expansion (MATLAB R2016b or newer)
S = S ./ nrm;
Implicit expansion using bsxfun (all MATLAB versions)
S = bsxfun( #mrdivide, S, nrm );
Manual expansion using repmat (all MATLAB versions)
S = S ./ repmat(nrm, size(S,1), 1);
If you have MATLAB R2017b or newer, and again no compatability concerns, you can use vecnorm, which can be used in place of the manual norm calculation
S = S ./ vecnorm(S, 2, 1);
Benchmark:
Since you asked for performance, here is a simple benchmark for testing the speed of these different methods. Specifically the original loop in your question versus implicit expansion with either vecnorm or the manual calculation.
Results (run using R2017b)
size(S): 1e3*1e2 1e5*1e3 1e3*1e6
Looping: 0.0005 1.0186 12.7788
Implicit manual: 0.0001 1.1236 10.4031
Implicit vecnorm: 0.0002 0.5774 6.8058
Conclusions
For relatively small arrays, all of the methods are very fast and I would opt for code clarity over performance.
If you only want to use MATLAB versions which support it, vecnorm is approximately twice as quick as other methods for large matrices.
For matrices of the order 1e5*1e3, looping is comparable to implicit expansion.
Code
function benchie()
S = rand( 1e3, 1e2 )*5;
f1 = #() loopingNorm(S);
f2 = #() implicitManual(S);
f3 = #() implicitVecnorm(S);
fprintf( 'Looping: %.4f\nImplicit manual: %.4f\nImplicit vecnorm: %.4f\n', ...
timeit(f1), timeit(f2), timeit(f3) );
end
function S = loopingNorm(S)
for ii = 1:size(S,2)
S(:,ii) = S(:,ii) / norm( S(:,ii) );
end
end
function S = implicitManual(S)
S = S ./ sqrt(sum(S.^2,1));
end
function S = implicitVecnorm(S)
S = S ./ vecnorm( S, 2, 1 );
end
I need to exclude some error data from matrix. I know what data is correct and i am trying to interpolate values between so I can get decent diagrams with not so big errors. I must use that form of matrix and I must preserve its shape. I must only substitute some data that is marked as errors. I will show you my work so far:
M=[0.1000
0.6000
0.7000
0.8000
0.9000
0.9500
1.0000
1.0500
1.1000
1.1500
1.2000
1.2500
1.3000
1.5000
1.7500
2.0000
2.2500
2.5000
3.0000];
CZ1=[ 9.4290
9.5000
9.3250
9.2700
9.2950
9.4350
9.6840
10.0690
10.1840
10.2220
10.2160
9.6160
9.6890
9.4880
9.5000
9.5340
9.3370
9.0990
8.5950];
N1=11;
Nn=13;
Mx1=M(N1);
Mx2=M(Nn);
Mx=[Mx1 Mx2]';
CN1=CZ1(N1);
CN2=CZ1(Nn);
CNy=[C1 C2]';
y1=interp1q(Mx,CNy,M(N1:Nn));
CNf=CZ1;
NEWRangeC=y1;
Cfa=changem(CZ1,[NEWRangeC], [CNf(N1:Nn)]);
figure
plot(M,Cf,'-*b',M,Cfa,'r')
So far as you can see I used points 11 and 13 and i excluded point 12 interpolating that point from 11 to 13. This is working but i want to make a modification.
My question is: How can I select values that are errors and remove them but interpolate space between their neighbors. I want to use a M matrix values as my reference (not points as my example).
Assuming you know which elements are incorrect, you can use Matlab's interp1 function to interpolate them (this will only work if the M matrix is actually a vector`:
error_indices = [11 13];
all_indices = 1:length(M)
% Get the indices where we have valid data
all_correct_indices = setdiff(all_indices, error_indices)
% the first two arguments are the available data.
% the third arguments is what indices you are looking for
M_new = interp1(all_correct_indices, M(all_correct_indices), all_indices)
The above interpolates values at all_indices -- including the missing elements. Where you already have valid data (all_correct_indices), Matlab will return that data. In other places, it will interpolate using the two nearest neighbors.
Try help interp1 for more information on how this function works.
Update - an example
x = 1:10; % all indices
y = x*10;
e = 3:7; % the unknown indices
s = setdiff(x, e); % the known indices
y_est = interp1(s, y(s), x)
ans =
10 20 30 40 50 60 70 80 90 100
And we see that interp1 had interpolated all values from 30 to 70 linearly using the available data (specifically the adjacent points 20 and 80).
Well, you can start out by finding the elements that are errors, with the find command (this will return the indices). This should also work for matrices.
You can then grab the elements around each of the indices, and interpolate between, as you did.
Given two vectors containing numerical values, say for example
a=1.:0.1:2.;
b=a+0.1;
I would like to select only the differing values. For this Matlab provides the function setdiff. In the above example it is obvious that setdiff(a,b) should return 1. and setdiff(b,a) gives 2.1. However, due to computational precision (see the questions here or here) the result differs. I get
>> setdiff(a,b)
ans =
1.0000 1.2000 1.4000 1.7000 1.9000
Matlab provides a function which returns a lower limit to this precision error, eps. This allows us to estimate a tolerance like tol = 100*eps;
My question now, is there an intelligent and efficient way to select only those values whose difference is below tol? Or said differently: How do I write my own version of setdiff, returning both values and indexes, which includes a tolerance limit?
I don't like the way it is answered in this question, since matlab already provides part of the required functionality.
Introduction and custom function
In a general case with floating point precision issues, one would be advised to use a tolerance value for comparisons against suspected zero values and that tolerance must be a very small value. A little robust method would use a tolerance that uses eps in it. Now, since MATLAB basically performs subtractions with setdiff, you can use eps directly here by comparing for lesser than or equal to it to find zeros.
This forms the basis of a modified setdiff for floating point numbers shown here -
function [C,IA] = setdiff_fp(A,B)
%//SETDIFF_FP Set difference for floating point numbers.
%// C = SETDIFF_FP(A,B) for vectors A and B, returns the values in A that
%// are not in B with no repetitions. C will be sorted.
%//
%// [C,IA] = SETDIFF_FP(A,B) also returns an index vector IA such that
%// C = A(IA). If there are repeated values in A that are not in B, then
%// the index of the first occurrence of each repeated value is returned.
%// Get 2D matrix of absolute difference between each element of A against
%// each element of B
abs_diff_mat = abs(bsxfun(#minus,A,B.')); %//'
%// Compare each element against eps to "negate" the floating point
%// precision issues. Thus, we have a binary array of true comparisons.
abs_diff_mat_epscmp = abs_diff_mat<=eps;
%// Find indices of A that are exclusive to it
A_ind = ~any(abs_diff_mat_epscmp,1);
%// Get unique(to account for no repetitions and being sorted) exclusive
%// A elements for the final output alongwith the indices
[C,IA] = intersect(A,unique(A(A_ind)));
return;
Example runs
Case1 (With integers)
This will verify that setdiff_fp works with integer arrays just the way setdiff does.
A = [2 5];
B = [9 8 8 1 2 1 1 5];
[C_setdiff,IA_setdiff] = setdiff(B,A)
[C_setdiff_fp,IA_setdiff_fp] = setdiff_fp(B,A)
Output
A =
2 5
B =
9 8 8 1 2 1 1 5
C_setdiff =
1 8 9
IA_setdiff =
4
2
1
C_setdiff_fp =
1 8 9
IA_setdiff_fp =
4
2
1
Case2 (With floating point numbers)
This is to show that setdiff_fp produces the correct results, while setdiff doesn't. Additionally, this will also test out the output indices.
A=1.:0.1:1.5
B=[A+0.1 5.5 5.5 2.6]
[C_setdiff,IA_setdiff] = setdiff(B,A)
[C_setdiff_fp,IA_setdiff_fp] = setdiff_fp(B,A)
Output
A =
1.0000 1.1000 1.2000 1.3000 1.4000 1.5000
B =
1.1000 1.2000 1.3000 1.4000 1.5000 1.6000 5.5000 5.5000 2.6000
C_setdiff =
1.2000 1.4000 1.6000 2.6000 5.5000
IA_setdiff =
2
4
6
9
7
C_setdiff_fp =
1.6000 2.6000 5.5000
IA_setdiff_fp =
6
9
7
For Tolerance of 1 epsilon This should work:
a=1.0:0.1:2.0;
b=a+0.1;
b=[b b-eps b+eps];
c=setdiff(a,b)
The idea is to expand b to include also its closest values.
This question already has answers here:
Return Unique Element with a Tolerance
(7 answers)
Closed 9 years ago.
I use unique function to find a full rank matrix (unique rows of matrix), now I need to unique
function that returns unique rows with a tolerance. Is there anyone who know how can I do this?
I am gratefule to you for your help.
I think you can do it with sortrows (which is what unique uses internally):
A = [1 2 3; 1 2.1 3; 1.001 2.001, 3.001; -4 5 6] % example data
tol = 0.01; % tolerance
[~, ii] = sortrows(A);
ii_unique = ii(logical([1; any(diff(A(ii,:))>tol,2)]));
A_unique_tol = A(sort(ii_unique),:) % result
The result in this example is:
A =
1.0000 2.0000 3.0000
1.0000 2.1000 3.0000
1.0010 2.0010 3.0010
-4.0000 5.0000 6.0000
A_unique_tol =
1.0000 2.0000 3.0000
1.0000 2.1000 3.0000
-4.0000 5.0000 6.0000
As usual, for old Matlab versions that don't support ~ as an output argument, change [~, ii] = sortrows(A); by [vv, ii] = sortrows(A);
This is hard to define well, assume you have a tolerance of 1.
Then what would be the outcome of [1; 2; 3; 4]?
When you have multiple columns a definition could become even more challenging.
However, if you are mostly worried about rounding issues, you can solve most of it by one of these two approaches:
Round all numbers (considering your tolerance), and then use unique
Start with the top row as your unique set, use ismemberf to determine whether each new row is unique and if so, add it to your unique set.
The first approach has the weakness that 0.499999999 and 0.500000000 may not be seen as duplicates. Whilst the second approach has the weakness that the order of your input matters.
I want to delete several specific values from a matrix (if they exist). It is highly probable that there are multiple copies of the values in the matrix.
For example, consider an N-by-2 matrix intersections. If the pairs of values [a b] and [c d] exist as rows in that matrix, I want to delete them.
Let's say I want to delete rows like [-2.0 0.5] and [7 7] in the following matrix:
intersections =
-4.0000 0.5000
-2.0000 0.5000
2.0000 3.0000
4.0000 0.5000
-2.0000 0.5000
So that after deletion I get:
intersections =
-4.0000 0.5000
2.0000 3.0000
4.0000 0.5000
What's the most efficient/elegant way to do this?
Try this one-liner (where A is your intersection matrix and B is the value to remove):
A = [-4.0 0.5;
-2.0 0.5;
2.0 3.0;
4.0 0.5;
-2.0 0.5];
B = [-2.0 0.5];
A = A(~all(A == repmat(B,size(A,1),1),2),:);
Then just repeat the last line for each new B you want to remove.
EDIT:
...and here's another option:
A = A((A(:,1) ~= B(1)) | (A(:,2) ~= B(2)),:);
WARNING: The answers here are best used for cases where small floating point errors are not expected (i.e. with integer values). As noted in this follow-up question, using the "==" and "~=" operators can cause unwanted results. In such cases, the above options should be modified to use relational operators instead of equality operators. For example, the second option I added would be changed to:
tolerance = 0.001; % Or whatever limit you want to set
A = A((abs(A(:,1)-B(1)) > tolerance) | (abs(A(:,2)-B(2)) > tolerance),:);
Just a quick head's up! =)
SOME RUDIMENTARY TIMING:
In case anyone was really interested in efficiency, I just did some simple timing for three different ways to get the subindex for the matrix (the two options I've listed above and Fanfan's STRMATCH option):
>> % Timing for option #1 indexing:
>> tic; for i=1:10000, index = ~all(A == repmat(B,size(A,1),1),2); end; toc;
Elapsed time is 0.262648 seconds.
>> % Timing for option #2 indexing:
>> tic; for i=1:10000, index = (A(:,1) ~= B(1)) | (A(:,2) ~= B(2)); end; toc;
Elapsed time is 0.100858 seconds.
>> % Timing for STRMATCH indexing:
>> tic; for i=1:10000, index = strmatch(B,A); end; toc;
Elapsed time is 0.192306 seconds.
As you can see, the STRMATCH option is faster than my first suggestion, but my second suggestion is the fastest of all three. Note however that my options and Fanfan's do slightly different things: my options return logical indices of the rows to keep, and Fanfan's returns linear indices of the rows to remove. That's why the STRMATCH option uses the form:
A(index,:) = [];
while mine use the form:
A = A(index,:);
However, my indices can be negated to use the first form (indexing rows to remove):
A(all(A == repmat(B,size(A,1),1),2),:) = []; % For option #1
A((A(:,1) == B(1)) & (A(:,2) == B(2)),:) = []; % For option #2
The simple solution here is to look to set membership functions, i.e., setdiff, union, and ismember.
A = [-4 0.5;
-2 0.5;
2 3;
4 0.5;
-2 0.5];
B = [-2 .5;7 7];
See what ismember does with the two arrays. Use the 'rows' option.
ismember(A,B,'rows')
ans =
0
1
0
0
1
Since we wish to delete rows of A that are also in B, just do this:
A(ismember(A,B,'rows'),:) = []
A =
-4 0.5
2 3
4 0.5
Beware that set membership functions look for an EXACT match. Integers or multiples of 1/2 such as are in A satisfy that requirement. They are exactly represented in floating point arithmetic in MATLAB.
Had these numbers been real floating point numbers, I'd have been more careful. There I'd have used a tolerance on the difference. In that case, I might have computed the interpoint distance matrix between the two sets of numbers, removing a row of A only if it fell within some given distance of one of the rows of B.
You can also abuse the strmatch function to suit your needs: the following code removes all occurences of a given row b in a matrix A
A(strmatch(b, A),:) = [];
If you need to delete more than one row, such as all rows from matrix B, iterate over them:
for b = B'
A(strmatch(b, A),:) = [];
end
Not sure when this function was introduced (using 2012b) but you can just do:
setdiff(A, B, 'rows')
ans =
-4.0000 0.5000
2.0000 3.0000
4.0000 0.5000
Based on:
A = [-4.0 0.5;
-2.0 0.5;
2.0 3.0;
4.0 0.5;
-2.0 0.5];
B = [-2.0 0.5];