I am trying to write some code which feeds a function with an unknown number of parameters. The idea is to feed the function the minimum, middle and maximum value within the possible range of values.
For example:
If the function takes 3 parameters
Parameter 1 can accept a range of 0 - 10
Parameter 2 can accept a range of 20 - 40
Parameter 3 can accept a range of 6 - 66
myfunction(para1, para2, para3)
myfunction(min,min,min)
myfunction(min,min,mid)
myfunction(min,min,max)
myfunction(min,mid,min)
myfunction(min,mid,mid)
myfunction(min,mid,max)
etc...
so using our example above:
The first time it loops I need to run
myfunction(0, 20, 0)
next time it loops it needs to run
myfunction(0, 20, 36)
next time it loops it needs to run
myfunction(0, 20, 66)
etc...
For all possible combinations (in this case all 27).
However, if the number of parameters changes to accept 4, it needs to be able to accommodate that and so on. I've looked into doing it as a loop or recursively, but I thought as a loop would be easier to understand, but either would be really helpful.
I don't want to have to do this manually so any help will be much appreciated.
For an arbitrary number of parameters (not fixed to 3), you can use the following code. It makes use of comma-separated lists, which is a powerful tool to handle a variable number of parameters.
Let n be the number of parameters. Then the number of combinations is N = 3^n.
params = {linspace(0,10,3), linspace(20,40,3), linspace(6,66,3)};
%// cell array with arbitrary number of elements. Each cell contains a 3-vector
%// which defines min, mid, max of one parameter.
n = numel(params); %// number of parameters
N = 3^n; %// number of combinations
paramCombs = cell(1,n); %// initialization for output of ndgrid
[paramCombs{end:-1:1}] = ndgrid(params{end:-1:1}); %// generate all combinations
%// in desired order. Gives n matrices, each containing values of one parameter
paramCombs = cellfun(#(c) c(:), paramCombs, 'uni', 0); %// linearize matrices
%// into n column vectors, each with N rows.
paramCombs = [paramCombs{:}]; %// concat column vectors into N x n matrix
paramCombs = mat2cell(paramCombs,ones(N,1),ones(n,1)); %// convert to
%// N x n cell array. Each row contains a combination of parameter values
result = arrayfun(#(n) myFun(paramCombs{n,:}), 1:N, 'uni', 0); %// call myFun
%// with each combination of parameter values
The result variable is a 1 x N cell array, where each cell contains the result of calling myFun with a combination of the n parameters.
Example 1: myFun's output simply replicates the input (as in #thewaywewalk's answer). params is defined as given above, so there are 3 parameters:
>> result{1}
ans =
0 20 6
>> result{2}
ans =
0 20 36
>> result{3}
ans =
0 20 66
>> result{4}
ans =
0 30 6
>> result{5}
ans =
0 30 36
etc.
Example 2: case with 2 parameters: params = {linspace(0,2,3), linspace(0,10,3)}. Again, myFun simply replicates the input:
>> result{1}
ans =
0 0
>> result{2}
ans =
0 5
>> result{3}
ans =
0 10
>> result{4}
ans =
1 0
>> result{5}
ans =
1 5
etc.
The method can be further generalized to an arbitrary (and possibly different) number of values for each parameter, just replacing the line N = 3^n; by
N = prod(cellfun(#numel, params)); %// number of combinations
Example 3: there are 2 parameters; the first with 3 values, and the second with 2: params = {[1 2 3], [10 20]};:
>> result{1}
ans =
1 10
>> result{2}
ans =
1 20
>> result{3}
ans =
2 10
>> result{4}
ans =
2 20
>> result{5}
ans =
3 10
>> result{6}
ans =
3 20
Think about something like this:
function result = permfunction()
% I assume you every parameter-range is defined by 3 values: min, mid, max
% you can define every triple as follows:
para1 = linspace(0,10,3);
para2 = linspace(20,40,3);
para3 = linspace(6,66,3);
% all possible parameters
parameters = [para1(:),para2(:),para3(:)];
% possible combinations of parameters-indices
a = perms(1:3);
% transformed into a cell array with linear indices, to achieve this +3 and +6
% are added to the 2nd and 3rd row.
idx = mat2cell( [a(:,1) , a(:,2)+3 , a(:,3)+6] , ones(length(a),1) );
% all parameter combinations
combinations = cellfun(#(x) parameters(x),idx,'uni',0');
% apply to your function myfunction (custom)
result = cellfun(#myfunction, combinations,'uni',0' );
end
function y = myfunction( parametertriple )
%just pass the input to the output
y = parametertriple;
end
you finally get a cell array with the results of myfunction for all your combinations of parameters. In this case I just passed the parameters to the output:
>> celldisp(ans)
ans{1} =
10 30 6
ans{2} =
10 20 36
ans{3} =
5 40 6
ans{4} =
5 20 66
ans{5} =
0 30 66
ans{6} =
0 40 36
This is how I would solve it:
%dummy function
testfun=#(a,b,c)fprintf('call function for %d,%d,%d\n',a,b,c);
%functon name
testfunname='testfun';
%intended inputs
input={[0,5,10],[20,30,40],[6,36,66]};
%generate all inputs
eval(['[' sprintf('a{%d} ',1:numel(input)) ']=meshgrid(input{:});']);
%call function
eval(['arrayfun(' testfunname sprintf(',a{%d} ',1:numel(input)) ');']);
Using eval is a dirty solution, but I don't find an alternative which allows variable input size.
Related
This question is motivated by very specific combinatorial optimization problem, where search space is defined as a space of permuted subsets of vector unsorted set of discrete values with multiplicities.
I am looking for effective (fast enough, vectorized or any other more clever solution) function which is able to find indices of subsets in the following manner:
t = [1 1 3 2 2 2 3 ]
is unsorted vector of all possible values, including its multiplicities.
item = [2 3 1; 2 1 2; 3 1 1; 1 3 3]
is a list of permuted subsets of vector t.
I need to find list of corresponding indices of subsets item which corresponds to the vector t. So, for above mentioned example we have:
item =
2 3 1
2 1 2
3 1 1
1 3 3
t =
1 1 3 2 2 2 3
ind = item2ind(item,t)
ind =
4 3 1
4 1 5
3 1 2
1 3 7
So, for item = [2 3 1] we get ind = [4 3 1], which means, that:
first value "2" at item corresponds to the first value "2" at t on position "4",
second value "3" at item corresponds to the first value "3" at t on position "3" and
third value "1" at item corresponds to the first value "1" at t on position "1".
In a case item =[ 2 1 2] we get ind = [4 1 5], which means, that:
first value "2" at item corresponds to the first value "2" at t on position "4",
second value "1" at item corresponds to the first value "1" at t on position "1", and
third value "2" at item corresponds to the second(!!!) value "1" at t on position "5".
For
item = [1 1 1]
does not exist any solution, because vector t contains only two "1".
My current version of function "item2ind" is very trivial serial code, which is possible simple parallelized by changing of "for" to "parfor" loop:
function ind = item2ind(item,t)
[nlp,N] = size(item);
ind = zeros(nlp,N);
for i = 1:nlp
auxitem = item(i,:);
auxt = t;
for j = 1:N
I = find(auxitem(j) == auxt,1,'first');
if ~isempty(I)
auxt(I) = 0;
ind(i,j) = I;
else
error('Incompatible content of item and t.');
end
end
end
end
But I need something definitely more clever ... and faster:)
Test case for larger input data:
t = 1:10; % 10 unique values at vector t
t = repmat(t,1,5); % unsorted vector t with multiplicity of all unique values 5
nlp = 100000; % number of item rows
[~,p] = sort(rand(nlp,length(t)),2); % 100000 random permutations
item = t(p); % transform permutations to items
item = item(:,1:30); % transform item to shorter subset
tic;ind = item2ind(item,t);toc % runing and timing of the original function
tic;ind_ = item2ind_new(item,t);toc % runing and timing of the new function
isequal(ind,ind_) % comparison of solutions
To achieve vectorizing the code, I have assumed that the error case won't be present. It should be discarded first, with a simple procedure I will present below.
Method First, let's compute the indexes of all elements in t:
t = t(:);
mct = max(accumarray(t,1));
G = accumarray(t,1:length(t),[],#(x) {sort(x)});
G = cellfun(#(x) padarray(x.',[0 mct-length(x)],0,'post'), G, 'UniformOutput', false);
G = vertcat(G{:});
Explanation: after putting input in column vector shape, we compute the max number of occurences of each possible value in t using accumarray. Now, we form array of all indexes of all numbers. It forms a cell array as there may be not the same number of occurences for each value. In order to form a matrix, we pad each array independently to the max length (naming mct). Then we can transform the cell array into a matrix. At this step, we have:
G =
1 11 21 31 41
2 12 22 32 42
3 13 23 33 43
4 14 24 34 44
5 15 25 35 45
6 16 26 36 46
7 17 27 37 47
8 18 28 38 48
9 19 29 39 49
10 20 30 40 50
Now, we process item. For that, let's figure out how to create the cumulative sum of occurences of values inside a vector. For example, if I have:
A = [1 1 3 2 2 2 3];
then I want to get:
B = [1 2 1 1 2 3 2];
Thanks to implicit expansion, we can have it in one line:
B = diag(cumsum(A==A'));
As easy as this. The syntax A==A' expands into a matrix where each element is A(i)==A(j). Making the cumulative sum in only one dimension and taking the diagonal gives us the good result: each column in the cumulative sum of occurences over one value.
To use this trick with item which 2-D, we should use a 3D array. Let's call m=size(item,1) and n=size(item,2). So:
C = cumsum(reshape(item,m,1,n)==item,3);
is a (big) 3D matrix of all cumulatives occurences. Last thing is to select the columns that are on the diagonal along dimension 2 and 3:
ia = C(sub2ind(size(C),repelem((1:m).',1,n),repelem(1:n,m,1),repelem(1:n,m,1)));
Now, with all these matrices, indexing is easy:
ind = G(sub2ind(size(G),item,ia));
Finally, let's recap the code of the function:
function ind = item2ind_new(item,t)
t = t(:);
[m,n] = size(item);
mct = max(accumarray(t,1));
G = accumarray(t,1:length(t),[],#(x) {sort(x)});
G = cellfun(#(x) padarray(x.',[0 mct-length(x)],0,'post'), G, 'UniformOutput', false);
G = vertcat(G{:});
C = cumsum(reshape(item,m,1,n)==item,3);
ia = C(sub2ind(size(C),repelem((1:m).',1,n),repelem(1:n,m,1),repelem(1:n,m,1)));
ind = G(sub2ind(size(G),item,ia));
Results Running the provided script on an old 4-core, I get:
Elapsed time is 4.317914 seconds.
Elapsed time is 0.556803 seconds.
ans =
logical
1
Speed up is substential (more than 8x), along with memory consumption (with matrix C). I guess some improvements can be done with this part to save more memory.
EDIT For generating ia, this procedure can cost a lost of memory. A way to save memory is to use a for-loop to generate directly this array:
ia = zeros(size(item));
for i=unique(t(:)).'
ia = ia+cumsum(item==i, 2).*(item==i);
end
In all cases, when you have ia, it's easy to test if there is an error in item compared to t:
any(ind(:)==0)
A simple solution to get items in error (as a mask) is then
min(ind,[],2)==0
Overview
An n×m matrix A and an n×1 vector Date are the inputs of the function S = sumdate(A,Date).
The function returns an n×m vector S such that all rows in S correspond to the sum of the rows of A from the same date.
For example, if
A = [1 2 7 3 7 3 4 1 9
6 4 3 0 -1 2 8 7 5]';
Date = [161012 161223 161223 170222 160801 170222 161012 161012 161012]';
Then I would expect the returned matrix S is
S = [15 9 9 6 7 6 15 15 15;
26 7 7 2 -1 2 26 26 26]';
Because the elements Date(2) and Date(3) are the same, we have
S(2,1) and S(3,1) are both equal to the sum of A(2,1) and A(3,1)
S(2,2) and S(3,2) are both equal to the sum of A(2,2) and A(3,2).
Because the elements Date(1), Date(7), Date(8) and Date(9) are the same, we have
S(1,1), S(7,1), S(8,1), S(9,1) equal the sum of A(1,1), A(7,1), A(8,1), A(9,1)
S(1,2), S(7,2), S(8,2), S(9,2) equal the sum of A(1,2), A(7,2), A(8,2), A(9,2)
The same for S([4,6],1) and S([4,6],2)
As the element Date(5) does not repeat, so S(5,1) = A(5,1) = 7 and S(5,2) = A(5,2) = -1.
The code I have written so far
Here is my try on the code for this task.
function S = sumdate(A,Date)
S = A; %Pre-assign S as a matrix in the same size of A.
Dlist = unique(Date); %Sort out a non-repeating list from Date
for J = 1 : length(Dlist)
loc = (Date == Dlist(J)); %Compute a logical indexing vector for locating the J-th element in Dlist
S(loc,:) = repmat(sum(S(loc,:)),sum(loc),1); %Replace the located rows of S by the sum of them
end
end
I tested it on my computer using A and Date with these attributes:
size(A) = [33055 400];
size(Date) = [33055 1];
length(unique(Date)) = 2645;
It took my PC about 1.25 seconds to perform the task.
This task is performed hundreds of thousands of times in my project, therefore my code is too time-consuming. I think the performance will be boosted up if I can eliminate the for-loop above.
I have found some built-in functions which do special types of sums like accumarray or cumsum, but I still do not have any ideas on how to eliminate the for-loop.
I would appreciate your help.
You can do this with accumarray, but you'll need to generate a set of row and column subscripts into A to do it. Here's how:
[~, ~, index] = unique(Date); % Get indices of unique dates
subs = [repmat(index, size(A, 2), 1) ... % repmat to create row subscript
repelem((1:size(A, 2)).', size(A, 1))]; % repelem to create column subscript
S = accumarray(subs, A(:)); % Reshape A into column vector for accumarray
S = S(index, :); % Use index to expand S to original size of A
S =
15 26
9 7
9 7
6 2
7 -1
6 2
15 26
15 26
15 26
Note #1: This will use more memory than your for loop solution (subs will have twice the number of element as A), but may give you a significant speed-up.
Note #2: If you are using a version of MATLAB older than R2015a, you won't have repelem. Instead you can replace that line using kron (or one of the other solutions here):
kron((1:size(A, 2)).', ones(size(A, 1), 1))
I want to select a random subset of a vector, much like datasample(data,k), but I want them in order.
I have an ODE which has [t,y] as output and it's the y that I want a subset of. I cannot just do a sort because y is not linear and so I somehow have to sort it with respect to t.
Any ideas how I can to this?
If I understand correctly, you want to sample the elements maintaining their original order. You can do it this way:
randomly sample the indices rather than the values;
sort the sampled indices;
use them to access the selected values;
that is:
result = data(sort(randsample(numel(data), k)));
The above uses the randsample function from the Statistics Toolbox. Alternatively, in recent Matlab versions you can use the two-input form of randperm:
result = data(sort(randperm(numel(data), k)));
For example, given
data = [61 52 43 34 25 16];
k = 4;
a possible result is
result =
61 43 34 25
This can be solved using a combination of randperm and intersect:
function q40673112
% Create a vector:
v = round(sin(0:0.6:6),3); disp(['v = ' mat2str(v)]);
% Set the size of sample we want:
N = 5;
% Create the random indices:
inds = intersect(1:numel(v), randperm(numel(v),N)); disp(['inds = ' mat2str(inds)]);
% Sample from the vector:
v_samp = v(inds); disp(['v_samp = ' mat2str(v_samp)]);
Example output:
% 1 2 3 4 5 6 7 8 9 10 11
v = [0 0.565 0.932 0.974 0.675 0.141 -0.443 -0.872 -0.996 -0.773 -0.279]
inds = [4 6 9 10 11]
v_samp = [0.974 0.141 -0.996 -0.773 -0.279]
I'm trying to elegantly split a vector. For example,
vec = [1 2 3 4 5 6 7 8 9 10]
According to another vector of 0's and 1's of the same length where the 1's indicate where the vector should be split - or rather cut:
cut = [0 0 0 1 0 0 0 0 1 0]
Giving us a cell output similar to the following:
[1 2 3] [5 6 7 8] [10]
Solution code
You can use cumsum & accumarray for an efficient solution -
%// Create ID/labels for use with accumarray later on
id = cumsum(cut)+1
%// Mask to get valid values from cut and vec corresponding to ones in cut
mask = cut==0
%// Finally get the output with accumarray using masked IDs and vec values
out = accumarray(id(mask).',vec(mask).',[],#(x) {x})
Benchmarking
Here are some performance numbers when using a large input on the three most popular approaches listed to solve this problem -
N = 100000; %// Input Datasize
vec = randi(100,1,N); %// Random inputs
cut = randi(2,1,N)-1;
disp('-------------------- With CUMSUM + ACCUMARRAY')
tic
id = cumsum(cut)+1;
mask = cut==0;
out = accumarray(id(mask).',vec(mask).',[],#(x) {x});
toc
disp('-------------------- With FIND + ARRAYFUN')
tic
N = numel(vec);
ind = find(cut);
ind_before = [ind-1 N]; ind_before(ind_before < 1) = 1;
ind_after = [1 ind+1]; ind_after(ind_after > N) = N;
out = arrayfun(#(x,y) vec(x:y), ind_after, ind_before, 'uni', 0);
toc
disp('-------------------- With CUMSUM + ARRAYFUN')
tic
cutsum = cumsum(cut);
cutsum(cut == 1) = NaN; %Don't include the cut indices themselves
sumvals = unique(cutsum); % Find the values to use in indexing vec for the output
sumvals(isnan(sumvals)) = []; %Remove NaN values from sumvals
output = arrayfun(#(val) vec(cutsum == val), sumvals, 'UniformOutput', 0);
toc
Runtimes
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.068102 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.117953 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 12.560973 seconds.
Special case scenario: In cases where you might have runs of 1's, you need to modify few things as listed next -
%// Mask to get valid values from cut and vec corresponding to ones in cut
mask = cut==0
%// Setup IDs differently this time. The idea is to have successive IDs.
id = cumsum(cut)+1
[~,~,id] = unique(id(mask))
%// Finally get the output with accumarray using masked IDs and vec values
out = accumarray(id(:),vec(mask).',[],#(x) {x})
Sample run with such a case -
>> vec
vec =
1 2 3 4 5 6 7 8 9 10
>> cut
cut =
1 0 0 1 1 0 0 0 1 0
>> celldisp(out)
out{1} =
2
3
out{2} =
6
7
8
out{3} =
10
For this problem, a handy function is cumsum, which can create a cumulative sum of the cut array. The code that produces an output cell array is as follows:
vec = [1 2 3 4 5 6 7 8 9 10];
cut = [0 0 0 1 0 0 0 0 1 0];
cutsum = cumsum(cut);
cutsum(cut == 1) = NaN; %Don't include the cut indices themselves
sumvals = unique(cutsum); % Find the values to use in indexing vec for the output
sumvals(isnan(sumvals)) = []; %Remove NaN values from sumvals
output = {};
for i=1:numel(sumvals)
output{i} = vec(cutsum == sumvals(i)); %#ok<SAGROW>
end
As another answer shows, you can use arrayfun to create a cell array with the results. To apply that here, you'd replace the for loop (and the initialization of output) with the following line:
output = arrayfun(#(val) vec(cutsum == val), sumvals, 'UniformOutput', 0);
That's nice because it doesn't end up growing the output cell array.
The key feature of this routine is the variable cutsum, which ends up looking like this:
cutsum =
0 0 0 NaN 1 1 1 1 NaN 2
Then all we need to do is use it to create indices to pull the data out of the original vec array. We loop from zero to max and pull matching values. Notice that this routine handles some situations that may arise. For instance, it handles 1 values at the very beginning and very end of the cut array, and it gracefully handles repeated ones in the cut array without creating empty arrays in the output. This is because of the use of unique to create the set of values to search for in cutsum, and the fact that we throw out the NaN values in the sumvals array.
You could use -1 instead of NaN as the signal flag for the cut locations to not use, but I like NaN for readability. The -1 value would probably be more efficient, as all you'd have to do is truncate the first element from the sumvals array. It's just my preference to use NaN as a signal flag.
The output of this is a cell array with the results:
output{1} =
1 2 3
output{2} =
5 6 7 8
output{3} =
10
There are some odd conditions we need to handle. Consider the situation:
vec = [1 2 3 4 5 6 7 8 9 10 11 12 13 14];
cut = [1 0 0 1 1 0 0 0 0 1 0 0 0 1];
There are repeated 1's in there, as well as a 1 at the beginning and end. This routine properly handles all this without any empty sets:
output{1} =
2 3
output{2} =
6 7 8 9
output{3} =
11 12 13
You can do this with a combination of find and arrayfun:
vec = [1 2 3 4 5 6 7 8 9 10];
N = numel(vec);
cut = [0 0 0 1 0 0 0 0 1 0];
ind = find(cut);
ind_before = [ind-1 N]; ind_before(ind_before < 1) = 1;
ind_after = [1 ind+1]; ind_after(ind_after > N) = N;
out = arrayfun(#(x,y) vec(x:y), ind_after, ind_before, 'uni', 0);
We thus get:
>> celldisp(out)
out{1} =
1 2 3
out{2} =
5 6 7 8
out{3} =
10
So how does this work? Well, the first line defines your input vector, the second line finds how many elements are in this vector and the third line denotes your cut vector which defines where we need to cut in our vector. Next, we use find to determine the locations that are non-zero in cut which correspond to the split points in the vector. If you notice, the split points determine where we need to stop collecting elements and begin collecting elements.
However, we need to account for the beginning of the vector as well as the end. ind_after tells us the locations of where we need to start collecting values and ind_before tells us the locations of where we need to stop collecting values. To calculate these starting and ending positions, you simply take the result of find and add and subtract 1 respectively.
Each corresponding position in ind_after and ind_before tell us where we need to start and stop collecting values together. In order to accommodate for the beginning of the vector, ind_after needs to have the index of 1 inserted at the beginning because index 1 is where we should start collecting values at the beginning. Similarly, N needs to be inserted at the end of ind_before because this is where we need to stop collecting values at the end of the array.
Now for ind_after and ind_before, there is a degenerate case where the cut point may be at the end or beginning of the vector. If this is the case, then subtracting or adding by 1 will generate a start and stopping position that's out of bounds. We check for this in the 4th and 5th line of code and simply set these to 1 or N depending on whether we're at the beginning or end of the array.
The last line of code uses arrayfun and iterates through each pair of ind_after and ind_before to slice into our vector. Each result is placed into a cell array, and our output follows.
We can check for the degenerate case by placing a 1 at the beginning and end of cut and some values in between:
vec = [1 2 3 4 5 6 7 8 9 10];
cut = [1 0 0 1 0 0 0 1 0 1];
Using this example and the above code, we get:
>> celldisp(out)
out{1} =
1
out{2} =
2 3
out{3} =
5 6 7
out{4} =
9
out{5} =
10
Yet another way, but this time without any loops or accumulating at all...
lengths = diff(find([1 cut 1])) - 1; % assuming a row vector
lengths = lengths(lengths > 0);
data = vec(~cut);
result = mat2cell(data, 1, lengths); % also assuming a row vector
The diff(find(...)) construct gives us the distance from each marker to the next - we append boundary markers with [1 cut 1] to catch any runs of zeros which touch the ends. Each length is inclusive of its marker, though, so we subtract 1 to account for that, and remove any which just cover consecutive markers, so that we won't get any undesired empty cells in the output.
For the data, we mask out any elements corresponding to markers, so we just have the valid parts we want to partition up. Finally, with the data ready to split and the lengths into which to split it, that's precisely what mat2cell is for.
Also, using #Divakar's benchmark code;
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.272810 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.436276 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 17.112259 seconds.
-------------------- With mat2cell
Elapsed time is 0.084207 seconds.
...just sayin' ;)
Here's what you need:
function spl = Splitting(vec,cut)
n=1;
j=1;
for i=1:1:length(b)
if cut(i)==0
spl{n}(j)=vec(i);
j=j+1;
else
n=n+1;
j=1;
end
end
end
Despite how simple my method is, it's in 2nd place for performance:
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.264428 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.407963 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 18.337940 seconds.
-------------------- SIMPLE
Elapsed time is 0.271942 seconds.
Unfortunately there is no 'inverse concatenate' in MATLAB. If you wish to solve a question like this you can try the below code. It will give you what you looking for in the case where you have two split point to produce three vectors at the end. If you want more splits you will need to modify the code after the loop.
The results are in n vector form. To make them into cells, use num2cell on the results.
pos_of_one = 0;
% The loop finds the split points and puts their positions into a vector.
for kk = 1 : length(cut)
if cut(1,kk) == 1
pos_of_one = pos_of_one + 1;
A(1,one_pos) = kk;
end
end
F = vec(1 : A(1,1) - 1);
G = vec(A(1,1) + 1 : A(1,2) - 1);
H = vec(A(1,2) + 1 : end);
I believe most functions in MATLAB should be able to receive matrix input and return the output in the form of matrix.
For example sqrt([1 4 9]) would return [1 2 3].
However, when I tried this recurring factorial function:
function k = fact(z)
if z ~= 0
k = z * fact(z-1);
else
k = 1;
end
end
It works perfectly when a number is input into fact. However, when a matrix is input into fact, it returns the matrix itself, without performing the factorial function.
E.g.
fact(3) returns 6
fact([1 2 3]) returns [1 2 3] instead of [1 2 6].
Any help is appreciated. Thank you very much!
Since MATLAB is not known to be good with recursive functions, how about a vectorized approach? Try this for a vector input -
mat1 = repmat([1:max(z)],[numel(z) 1])
mat1(bsxfun(#gt,1:max(z),z'))=1
output1 = prod(mat1,2)
Sample run -
z =
1 2 7
output1 =
1
2
5040
For the sake of answering your original question, here's the annoying loopy code for a vector or 2D matrix as input -
function k1 = fact1(z1)
k1 = zeros(size(z1));
for ii = 1:size(z1,1)
for jj = 1:size(z1,2)
z = z1(ii,jj);
if z ~= 0
k1(ii,jj) = z .* fact1(z-1);
else
k1(ii,jj) = 1;
end
end
end
return
Sample run -
>> fact1([1 2 7;3 2 1])
ans =
1 2 5040
6 2 1
You can use the gamma function to compute the factorial without recursion:
function k = fact(z)
k = gamma(z+1);
Example:
>> fact([1 2 3 4])
ans =
1 2 6 24
Not sure if all of you know, but there is an actual factorial function defined in MATLAB that can take in arrays / matrices of any size, and computes the factorial element-wise. For example:
k = factorial([1 2 3 4; 5 6 7 8])
k =
1 2 6 24
120 720 5040 40320
Even though this post is looking for a recursive implementation, and Divakar has provided a solution, I'd still like to put my two cents in and suggest an alternative. Also, let's say that we don't have access to factorial, and we want to compute this from first principles. What I would personally do is create a cell array that's the same size as the input matrix, but each element in this cell array would be a linear index array from 1 up to the number defined for each location in the original matrix. You would then apply prod to each cell element to compute the factorial. A precondition is that no number is less than 1, and that all elements are integers. As such:
z1 = ... ; %// Define input matrix here
z1_matr = arrayfun(#(x) 1:x, z1, 'uni', 0);
out = cellfun(#prod, z1_matr);
If z1 = [1 2 3 4; 5 6 7 8];, from my previous example, we get the same output with the above code:
out =
1 2 6 24
120 720 5040 40320
This will obviously be slower as there is an arrayfun then cellfun call immediately after, but I figured I'd add another method for the sake of just adding in another method :) Not sure how constructive this is, but I figured I'd add my own method and join Divakar and Luis Mendo :)