I have tried different repeating functions which tell me the frequency of the numbers, but I want to know the order of the repeating numbers.
For example I have an array which has numbers
a=[ 1,1,1,1,1,2,2,2,2,2,1,1,1,1,4,4,4,5,5,5,7,7,2,2,2,2]
I want my function to give me the answer, i.e. 1,2,1,4,5,7,2.
You can use the diff, you're interested in the first element of a and then every subsequent index where the difference between elements is non-zero
>> a = [ 1,1,1,1,1,2,2,2,2,2,1,1,1,1,4,4,4,5,5,5,7,7,2,2,2,2];
>> b = a( [true, diff(a)~=0] );
b =
1 2 1 4 5 7 2
Related
This question already has answers here:
Generate a matrix containing all combinations of elements taken from n vectors
(4 answers)
Closed 5 years ago.
I want to create all pattern combination which possible occur
For example, I have three ball and I want to pick 3 times. All possible is 27 events but I want to create all possible event in array like this
[1 1 1; 1 1 2; 1 1 3; 1 2 1 ;....]
Dose anyone can help me to write m-file in matlab program, please?
This can be done very easily with base conversion.
If the number of balls does not exceed 10
M = 3; % Number of balls. Must not exceed 10
N = 3; % Number of draws
result = dec2base(0:M^N-1, M)-'0'+1;
Note that dec2base ouputs chars, not numbers, and hence the -'0' part. Characters in arithmetic operations behave like their corresponding ASCII codes. So subtracting '0' transforms characters '0', '1', ..., '9' into the corresponding numbers.
With this approach M cannot exceed 10 because then dec2bin would output '0', '1', ..., '9', 'A', 'B' ... and the character arithmetic would not give the correct result for 'A', 'B', ... But this can be easily solved as follows.
For number of balls up to 36
M = 12; % Number of balls. Must not exceed 36
N = 2; % Number of draws
result = dec2base(0:M^N-1, M)-'0'+1; % same as before
result = result - ('A'-'9'-1)*(result>10); % correction
The new line simply corrects the results for 'A', 'B', ... by subtracting 'A'-'9'-1, to compensate the fact that '9' and 'A' do not have consecutive ASCII codes.
With this approach M cannot exceed 36 because of dec2base restrictions.
You can create the result like this for the specific case:
firstCol = [ones(9,1);2*ones(9,1);3*ones(9,1)];
secondCol = repeat([ones(3,1);2*ones(3,1);3*ones(3,1)],1,3);
thirdCol = repeat([1;2;3],1,9);
result = [firstCol secondCol thirdCol];
First, repeat 9 times 1,2, and 3 for first column. then repeat each of them 3 times for the second column and choose the third column once for each item. Indeed, this generate the all possible choices for each location.
How? If you suppose the first element is 1, you have 3 choice for the second place, and 3 choice for the third place. Hence, you have 9 possible option when the first place is 1. Also, fix the second place, and analyze this. You can generalize this for 2 and 3. The above code, try to generate possibilities based on this explanation.
In the above, ones generate a matrix which all elements are 1 with the specified size and repeat function repeats the specified matrix in the specified size and dimension. You can check the documentation to know more about them.
Hence, You can generalize it for n like the following:
n = 10;
result = zeros(3^n,3);
for idx = 1:n
result(:,idx) = repeat([ones(3^(n-idx),1);2*ones(3^(n-idx),1);3*ones(3^(n-idx),1)],1,3^(idx-1));
end
I need to find all possible combinations of numbers 1:8 such that sum of all elements is equal to 8
The combinations need to be arranged in an ascending order.
Eg
1 7
2 2 4
1 3 5
1 2 2 3
1 1 1 1 1 1 1 1
A number can repeat itself. But a combination must not..
i.e 1 2 2 3 and 2 1 2 3
I need the the solution in ascending order So there will be only one possibility of every combination
I tried a few codes online suggested on Find vector elements that sum up to specific number in MATLAB
VEC = [1:8];
NUM = 8;
n = length(VEC);
finans = zeros(2^n-1,NUM);
for i = 1:(2^n - 1)
ndx = dec2bin(i,n) == '1';
if sum(VEC(ndx)) == NUM
l = length(VEC(ndx));
VEC(ndx)
end
end
but they dont include the possibilities where the numbers repeat.
I found a better approach through recursion and it's more elegant (I like elegant) and faster than my previous attempt (0.00399705213 seconds on my computer).
EDIT: You will need my custom function stretchmat.m that stretches a vector to fit the size of another matrix. Kinda like repmat but stretching the first parameter (see help for details). Very useful!
script.m
% Define funciton to prepend a cell x with a variable i
cellprepend = #(x,i) {[i x]};
% Execute and time function
tic;
a = allcomb(cellprepend,1,8); % Solution in a
toc;
allcomb.m
function a = allcomb( cellprepend, m, n )
% Add entire block as a combination
a{1} = n;
% Exit recursion if block size 1
if n == 1
return;
end
% Recurse cutting blocks at different segments
for i = m:n/2
b = allcomb(cellprepend,i,n-i);
a = [a cellfun( cellprepend, b, num2cell( stretchmat( i, b ) ) )];
end
end
So the idea is simple, for solutions that add to 8 is exhaustive. If you look for only valid answers, you can do a depth first search by breaking up the problem into 2 blocks. This can be written recursively as I did above and is kinda similar to Merge Sort. The allcomb call takes the block size (n) and finds all the ways of breaking it up into smaller pieces.
We want non-zero pieces so we loop it from 1:n-1. It then prepends the first block to all the combinations of the second block. By only doing all comb on one of the blocks, we can ensure that all solutions are unique.
As for the sorting, I'm not quite sure what you mean by ascending. From what I see, you appear to be sorting from the last number in ascending order. Can you confirm? Any sort can be appended to the end of script.m.
EDIT 2/3 Notes
For the permutatively unique case, the code can be found here
Thanks to #Simon for helping me QA the code multiple times
EDIT: Look at my second more efficient answer!
The Naive approach! Where the cartprod.m function can be found here.
% Create all permutations
p(1:8) = {0:8};
M = fliplr( cartprod( p{:} ) );
% Check sums
r = sum( M, 2 ) == 8;
M = M(sum( M, 2 ) == 8,:); % Solution here
There are definitely more efficient solutions than this but if you just need a quick and dirty solution for small permutations, this will work. Please note that this made Matlab take 3.5 GB of RAM to temporarily store the permutations.
First save all combinations with repetitions in a cell array. In order to do that, just use nmultichoosek.
v = 1 : 8;
combs = cell(length(v),0);
for i = v
combs{i} = nmultichoosek(v,i);
end
In this way, each element of combs contains a matrix where each row is a combination. For instance, the i-th row of combs{4} is a combination of four numbers.
Now you need to check the sum. In order to do that to all the combinations, use cellfun
sums = cellfun(#(x)sum(x,2),combs,'UniformOutput',false);
sums contains the vectors with the sum of all combinations. For
instance, sums{4} has the sum of the number in combination combs{4}.
The next step is check for the fixed sum.
fixed_sum = 10;
indices = cellfun(#(x)x==fixed_sum,sums,'UniformOutput',false);
indices contains arrays of logical values, telling if the combination satisfies the fixed sum. For instance, indices{4}(1) tells you if the first combination with 4 numbers sums to fixed_sum.
Finally, retrieve all valid combinations in a new cell array, sorting them at the same time.
valid_combs = cell(length(v),0);
for i = v
idx = indices{i};
c = combs{i};
valid_combs{i} = sortrows(c(idx,:));
end
valid_combs is a cell similar to combs, but with only combinations that sum up to your desired value, and sorted by the number of numbers used: valid_combs{1} has all valid combinations with 1 number, valid_combs{2} with 2 numbers, and so on. Also, thanks to sortrows, combinations with the same amount of numbers are also sorted. For instance, if fixed_sum = 10 then valid_combs{8} is
1 1 1 1 1 1 1 3
1 1 1 1 1 1 2 2
This code is quite efficient, on my very old laptop I am able to run it in 0.016947 seconds.
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
I have a vector that could look like this:
v = [1 1 2 2 2 3 3 3 3 2 2 1 1 1];
that is, the number of equal elements can vary, but they always increase and decrease stepwise by 1.
What I want is an easy way to be left with a new vector looking like this:
v2 = [ 1 2 3 2 1];
holding all the different elements (in the right order as they appear in v), but only one of each. Preferably without looping, since generally my vectors are about 10 000 elements long, and already inside a loop that's taking for ever to run.
Thank you so much for any answers!
You can use diff for this. All you're really asking for is: Delete any element that's equal to the one in front of it.
diff return the difference between all adjacent elements in a vector. If there is no difference, it will return 0. v(ind~=0) will give you all elements that have a value different than zero. The 1 in the beginning is to make sure the first element is counted. As diff returns the difference between elements, numel(diff(v)) = numel(v)-1.
v = [1 1 2 2 2 3 3 3 3 2 2 1 1 1];
ind = [1 diff(v)];
v(ind~=0)
ans =
1 2 3 2 1
This can of course be done in a single line if you want:
v([1, diff(v)]~=0)
You could try using diff which, for a vector X, returns [X(2)-X(1) X(3)-X(2) ... X(n)-X(n-1)] (type help diff for details on this function). Since the elements in your vector always increase or decrease by 1, then
diff(v)
will be a vector (of size one less than v) with zeros and ones where a one indicates a step up or down. We can ignore all the zeros as they imply repeated numbers. We can convert this to a logical array as
logical(diff(v))
so that we can index into v and access its elements as
v(logical(diff(v)))
which returns
1 2 3 2
This is almost what you want, just without the final number which can be added as
[v(logical(diff(v))) v(end)]
Try the above and see what happens!
If i have a vector (array) in matlab, is it possible to remove the values from that vector based on some restriction (e.g.. all non negative numbers).
Can you please advise me on the best approach to do that.
Yes, you can either use logical indexing to keep the values which meet a criterion or use the find function to get the indexes which hold values that meet a criterion.
logical indexing
the find function
An example of logical indexing where we want to remove all the values from a vector which are not greater than three:
>> x=[1,2,3,4,5,6]
x =
1 2 3 4 5 6
>> x=x(x>3)
x =
4 5 6
You can also ask for multiple criteria as you would expect. In the following example, we want to keep every value which is greater than three, but not five.
>> x=[1,2,3,4,5,6]
x =
1 2 3 4 5 6
>> x=x(x>3 & x~=5)
x =
4 6
Finally, the find function can come in handy when you need the indexes of values which meet a criterion.
>> x=[1,1,2,2,5,5]
x =
1 1 2 2 5 5
>> ind=find(x>3)
ind =
5 6
Logical indexing and find can also be applied to matrices with more than one row/column.
Thanks #Alan for helping me improve the answer.
You may want to look into logical indexing, as it neatly handles your problem.
To use the example you gave, if you have a vector a of numbers, and you want to remove all negative numbers you could do the following:
b = a(a >= 0);
which would create a vector b containing only the positive elements of a, or you could try:
a(a < 0) = [];
would set any elements in the vector a to []