I am trying to find the sum of the following matrix in matlab [1 1 1 1; 1 2 1 2; 4 5 3 2; 1 3 2 4; 10 11 1 1; 90 9 2 1]
I am trying to do so using a nested for statement yet i keep getting errors. please help
Must use nested for
My code:
A = [1 1 1 1; 1 2 1 2; 4 5 3 2; 1 3 2 4; 10 11 1 1; 90 9 2 1];
for j=1:4,
for i=1:6,
sum = A(j,:)+A(j+1,:)+A(j+2,:)
end
end
You will need to change your code from this:
A = [1 1 1 1; 1 2 1 2; 4 5 3 2; 1 3 2 4; 10 11 1 1; 90 9 2 1];
for j=1:4,
for i=j:6,
sum = A(j,:)+A(j+1,:)+A(j+2,:);
end
end
to this:
A = [1 1 1 1; 1 2 1 2; 4 5 3 2; 1 3 2 4; 10 11 1 1; 90 9 2 1];
sum = 0;
for j=1:4,
for i=1:6,
sum = sum + A(j,i);
end
end
Note various modifications:
Initialize sum=0. If you're using this in the interpreter, you'll be starting off with the previous result, guranteeing you don't get the right result.
Cumulate the values. If you assign to sum at each iteration, you'll throw away the result of other iterations.
There is no point in writing the outer loop if you're going to hardcode j+1, j+2, etc. in the inner loop.
Fix the inner loop so that it starts iterating at 1.
Suppress output in the inner loop by using a semicolon to get a clean result.
I will not post the corrected code, I'll instead add comments to the code you posted:
A = [1 1 1 1; 1 2 1 2; 4 5 3 2; 1 3 2 4; 10 11 1 1; 90 9 2 1];
% you are missing sum initialization here - you should first set sum to zero
for j=1:4, % there is no comma needed at the end
for i=j:6, % you want to iterate all the rows, from 1 to 6
sum = A(j,:)+A(j+1,:)+A(j+2,:) % you should be adding to the sum - i.e sum is sum + current field A(j, i)
end
end
Why don't you just use sum()?
Related
Given a vector A that contains a sequence of numbers.
The objective is to find all series (longer than a given number "threshold") that contain the same value. The result should be the position of both first and last values of that series.
Example: given a vector A where:
A = [1 1 1 2 1 3 3 3 1 1 1 1 1 4 3 2 2 2 2 2 2 2 3 4];
and a threshold B = 5;
The results would be:
[9 13] % a series contain only the number 1 with length equal to 5
[16 22] % a series contain only the number 2 with length equal to 7
A=[1 1 1 2 1 3 3 3 1 1 1 1 1 4 3 2 2 2 2 2 2 2 3 4];
B = 5;
[l c]= size(A); % to know the size of 'A'
K=1; % to define the length of the series
W=1; % a value used to save the positions of the wanted series.
For i=1:c-1
If A(i)==A(i+1)
K=k+1;
Else
If k>= B % to test of the actual series is equal or longer than the given threshold
S(w,1)=i;
S(w,2)= S(w,1)-k+1; % saving the first position and the last position of the series in 'S'
w=w+1;
end
k=1;
end
S % the final result which is a table contain all wanted series.
the result is as follow:
S 13 9 % 13: the last position of the wanted series and 9 is the first position
16 22
This one work soo good. But still... it is soo slow when its come to a big table.
A faster, vectorized option is to modify the approach from this solution for finding islands of zeroes:
A = [1 1 1 2 1 3 3 3 1 1 1 1 1 4 3 2 2 2 2 2 2 2 3 4]; % Sample data
B = 5; % Threshold
tsig = (diff(A) ~= 0);
dsig = diff([1 tsig 1]);
startIndex = find(dsig < 0);
endIndex = find(dsig > 0)-1;
duration = endIndex-startIndex+1;
stringIndex = (duration >= (B-1));
result = [startIndex(stringIndex); endIndex(stringIndex)+1].';
And the results:
result =
9 13
16 22
I have two matrices, one of which (Let's say matrix H) is 4x2 and the other one (matrix N) is 100x2.
I want to make a combination for every pair of N, containing every pair of H.
Basically, if my
H = [2 2; 2 4; 4 2; 4 4]
N = [1 1; 1 2; 1 3; ...;
10 8; 10 9; 10 10]
I would like to have a final matrix
M = [1 2 2 1; 1 2 4 1; 1 4 2 1; 1 4 4 1; 1 2 2 2; 1 2 4 2; ...; 10 4 4 10]
of a size 100x4 (because every pair of N will be multiplied |H|=4 times.)
So all the pairs of H matrix will be between all pairs of my N matrix.
I hope I am clear enough.
Use the follwing syntax:
%calculates the Cartesian multipication of 1:size(h,1) and 1:size(N,1)
sets = {1:size(H,1), 1:size(N,1)};
[hInds, nInds] = ndgrid(sets{:});
%generates the output matrix
outRes = [N( nInds(:),1),H( hInds(:),1),H( hInds(:),2),N( nInds(:),2)];
Partial results (displaying just the first rows of the output):
outRes =
1 2 2 1
1 2 4 1
1 4 2 1
1 4 4 1
1 2 2 2
1 2 4 2
1 4 2 2
1 4 4 2
1 2 2 3
1 2 4 3
1 4 2 3
1 4 4 3
...
Notice that if N is 4x2 and N is 10x2, the final matrix size will be 40x4 and not 100x4 as you mentioned.
Try this:
H= [2 2; 2 4; 4 2; 4 4];
N= fix(100*(rand(10,2))) % Replace this with your N matrix
iter=0;
for i=1:10
for j=1:4
iter=iter+1;
A(iter,:)=[N(i,1), H(j,1:2), N(i,2)];
end
end
A
I have a list of numbers, [1:9], that I need to divide three groups. Each group must contain at least one number. I need to enumerate all of the combinations (i.e. order does not matter). Ideally, the output is a x by 3 array. Any ideas of how to do this in matlab?
Is this what you want:
x = 1:9;
n = length(x);
T=3;
out = {};
%// Loop over all possible solutions
for k=1:T^n
s = dec2base(k, T, n);
out{k}{T} = [];
for p=1:n
grpIndex = str2num(s(p))+1;
out{k}{grpIndex} = [out{k}{grpIndex} x(p)];
end
end
%// Print result. size of out is the number of ways to divide the input. out{k} contains 3 arrays with the values of x
out
Maybe this is what you want. I'm assuming that the division in groups is "monotonous", that is, first come the elements of the first group, then those of the second etc.
n = 9; %// how many numbers
k = 3; %// how many groups
b = nchoosek(1:n-1,k-1).'; %'// "breaking" points
c = diff([ zeros(1,size(b,2)); b; n*ones(1,size(b,2)) ]); %// result
Each column of c gives the sizes of the k groups:
c =
Columns 1 through 23
1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 4 5
1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 1
7 6 5 4 3 2 1 6 5 4 3 2 1 5 4 3 2 1 4 3 2 1 3
Columns 24 through 28
5 5 6 6 7
2 3 1 2 1
2 1 2 1 1
This produces what I was looking for. The function nchoosekr_rec() is shown below as well.
for x=1:7
numgroups(x)=x;
end
c=nchoosekr_rec(numgroups,modules);
i=1;
d=zeros(1,modules);
for x=1:length(c(:,1))
c(x,modules+1)=sum(c(x,1:modules));
if c(x,modules+1)==length(opt_mods)
d(i,:)=c(x,1:modules);
i=i+1;
end
end
numgroups=[];
for x=1:length(opt_mods)
numgroups(x)=x;
end
count=0;
for x=1:length(d(:,1))
combos=combnk(numgroups,d(x,1));
for y=1:length(combos(:,1))
for z=1:nchoosek(9-d(x,1),d(x,2))
new_mods{count+z,1}=combos(y,:);
numgroups_temp{count+z,1}=setdiff(numgroups,new_mods{count+z,1});
end
count=count+nchoosek(9-d(x,1),d(x,2));
end
end
count=0;
for x=1:length(d(:,1))
for y=1:nchoosek(9,d(x,1))
combos=combnk(numgroups_temp{count+1},d(x,2));
for z=1:length(combos(:,1))
new_mods{count+z,2}=combos(z,:);
new_mods{count+z,3}=setdiff(numgroups_temp{count+z,1},new_mods{count+z,2});
end
count=count+length(combos(:,1));
end
end
function y = nchoosekr_rec(v, n)
if n == 1
y = v;
else
v = v(:);
y = [];
m = length(v);
if m == 1
y = zeros(1, n);
y(:) = v;
else
for i = 1 : m
y_recr = nchoosekr_rec(v(i:end), n-1);
s_repl = zeros(size(y_recr, 1), 1);
s_repl(:) = v(i);
y = [ y ; s_repl, y_recr ];
end
end
end
My input is the following:
X = [1 1; 1 2; 1 3; 1 4; 2 5; 1 6; 2 7; 1 8];
X =
1 1
1 2
1 3
1 4
2 5
1 6
2 7
1 8
I am looking to output a new matrix based on the value of the first column. If the value is equal to 1 -- the output will remain the same, when the value is equal to 2 then I would like to output two of the values contained in the second row. Like this:
Y =
1
2
3
4
5
5
6
7
7
8
Where 5 is output two times because the value in the first column is 2 and the same for 7
Here it is (vectorized):
C = cumsum(X(:,1))
A(C) = X(:,2)
D = hankel(A)
D(D==0) = inf
Y = min(D)
Edit:
Had a small bug, now it works.
% untested code:
Y = []; % would be better to pre-allocate
for ii = 1:size(X,1)
Y = [Y; X(ii,2)*ones(X(ii,1),1)];
end
I have a matrix which has the following form:
M =
[1 4 56 1;
1 3 5 1;
1 3 6 4;
2 3 5 0;
2 0 0 0;
3 1 2 3;
3 3 3 3]
I want to split this matrix based on the number given in the first column. So I want to split the matrix into this:
A =
[1 4 56 1;
1 3 5 1;
1 3 6 4]
B =
[2 3 5 0;
2 0 0 0]
C =
[3 1 2 3;
3 3 3 3]
I tried this by making the following loop, but this gave me the desired matrices with rows of zeros:
for i = 1:length(M)
if (M(i,1) == 1)
A(i,:) = M(i,:);
elseif (M(i,1) == 2)
B(i,:) = M(i,:);
elseif (M(i,1) == 3)
C(i,:) = M(i,:);
end
end
The result for matrix C is then for example:
C =
[0 0 0 0;
0 0 0 0;
0 0 0 0;
2 3 5 0;
2 0 0 0]
How should I solve this issue?
Additional information:
The actual data has a date in the first column in the form yyyymmdd. The data set spans several years and I want to split this dataset in matrices for each year and after that for each month.
You can use arrayfun to solve this task:
M = [
1 4 56 1;
1 3 5 1;
1 3 6 4;
2 3 5 0;
2 0 0 0;
3 1 2 3;
3 3 3 3]
A = arrayfun(#(x) M(M(:,1) == x, :), unique(M(:,1)), 'uniformoutput', false)
The result A is a cell array and its contents can be accessed as follows:
>> a{1}
ans =
1 4 56 1
1 3 5 1
1 3 6 4
>> a{2}
ans =
2 3 5 0
2 0 0 0
>> a{3}
ans =
3 1 2 3
3 3 3 3
To split the data based on an yyyymmdd format in the first column, you can use the following:
yearly = arrayfun(#(x) M(floor(M(:,1)/10000) == x, :), unique(floor(M(:,1)/10000)), 'uniformoutput', false)
monthly = arrayfun(#(x) M(floor(M(:,1)/100) == x, :), unique(floor(M(:,1)/100)), 'uniformoutput', false)
If you don't know how many outputs you'll have, it is most convenient to put the data into a cell array rather than into separate arrays. The command to do this is MAT2CELL. Note that this assumes your data is sorted. If it isn't use sortrows before running the code.
%# count the repetitions
counts = hist(M(:,1),unique(M(:,1));
%# split the array
yearly = mat2cell(M,counts,size(M,2))
%# if you'd like to split each cell further, but still keep
%# the data also grouped by year, you can do the following
%# assuming the month information is in column 2
yearByMonth = cellfun(#(x)...
mat2cell(x,hist(x(:,2),unique(x(:,2)),size(x,2)),...
yearly,'uniformOutput',false);
You'd then access the data for year 3, month 4 as yearByMonth{3}{4}
EDIT
If the first column of your data is yyyymmdd, I suggest splitting it into three columns yyyy,mm,dd, like below, to facilitate grouping afterward:
ymd = 20120918;
yymmdd = floor(ymd./[10000 100 1])
yymmdd(2:3) = yymmdd(2:3)-100*yymmdd(1:2)