I have 4x4 matrix A
[1 2 3 4;
2 2 2 3;
5 5 5 5;
4 4 4 4]
I know how to locate all values less than 4. A<4. But I'm not sure how to write an 'if' statement for; three or more values, all which are less than 4, contained in the same row. For instance; see above A(1,:) and A(2,:) satisfies my conditions.
You can basically do A<4 to know which ones are smaller. If you want to know which rows contain N values smaller than 4 then you can do
rows=find(sum(A<4,2)>=3)
This basically does:
find smaller than 4
Count how many of them in each row (sum(_,2))
find if they are 3 or more
give the row index of those find()
Related
I'd like to take a single array lets say 3x5 as follows
1 3 5
1 3 5
2 8 6
4 5 7
4 5 8
and write a code that will create a new array that adds the third column together with the previous number if the numbers in the first and second columns equal the numbers in the row below it.
since the first two values in row 1 and 2, then add the third elements in row 1 and 2 together
so the output from the array above should look like this
1 3 10
2 8 6
4 5 15
The function accumarray(subs,val) accumulate elements of vector val using the subscripts subs. So we can use this function to sum the elements in the third column having the same value in the first and second column. We can use unique(...,'rows') to determine which pairs of value are unique.
%Example data
A = [1 3 5,
1 3 5,
2 3 6,
4 5 7,
4 5 7]
%Get the unique pair of value based on the first two column (uni) and the associated index.
[uni,~,sub] = unique(A(:,1:2),'rows');
%Create the result using accumarray.
R = [uni, accumarray(sub,A(:,3))]
If the orders matters the script would be a little bit more complex:
%Get the unique pair of value based on the first two column (uni) and the associated index.
[uni,~,sub] = unique(A(:,1:2),'rows');
%Locate the consecutive similar row with this small tricks
dsub = diff([0;sub])~=0;
%Create the adjusted index
subo = cumsum(dsub);
%Create the new array
R = [uni(sub(dsub),:), accumarray(subo,A(:,3))]
Or you can get an identical result with a for loop:
R = A(1,:)
for ii = 2:length(A)
if all(A(ii-1,1:2)==A(ii,1:2))
R(end,3) = R(end,3)+A(ii,3)
else
R(end+1,:) = A(ii,:)
end
end
Benchmark:
With an array A of size 100000x3 on the mathworks live editor:
The for loop take about 5.5s (no pre-allocation, so it's pretty slow)
The vectorized method take about 0.012s
Given a 2-column matrix, for instance. The input is:
[ 1,2;
3,4;
5,5]
The expected output is:
[1,2;
3,4;]
Does anyone know how do accomplish this? Many thanks for your time and attention.
You could use logical indexing:
A = [1 2;3 4;5 5];
match = A(:,1) == A(:,2); // 1 where row has the same elements in both columns
A(match,:) = []; // make the match columns empty
You would need to make this more generic for another case, but for two columns and your example this will work.
Your question suggests your matrix may have an arbitrary number of columns. In that case you may want to delete a row if it has (a) any two elements equal, or (b) all elements equal.
One possible approach is:
Apply sort along each row;
Use diff to compute differences between consecutive elements;
Generate a logical index with all to (a) keep rows for which all such differences are non-zero, or with any to (b) keep rows for which any such difference is non-zero:
So:
X = [1 2 3;
3 4 3;
5 5 5];
Y = X(all(diff(sort(X,2),[],2),2),:);
Z = X(any(diff(sort(X,2),[],2),2),:);
gives
Y =
1 2 3
Z =
1 2 3
3 4 3
Given A = [3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 8]
Output B = [3 4 5 6 8]
Is there a Matlab function or command to get this result? I am new to Matlab. Just now I am doing it going through for each element and keeping a counter for it. I have very big array so this is taking too much time.
Use a combination of unique and histc:
uA = unique(A); %// find unique values
B = uA(histc(A, uA)>=2); %// select those that appear at least twice
The above code gives the values that appear at least twice. If you want values that appear exactly twice, replace >= by ==.
I'm working in R and I'd like to call a selection of values from a data frame by their column and row indices. However doing this yields a matrix rather than an array. I shall demonstrate:
Given the data.frame:
a = data.frame( a = array(c(1,2,3,4,5,6,7,8,9), c(3,3)) )
(for those of you who don't want to plug it in, it looks like this)
a.1 a.2 a.3
1 1 4 7
2 2 5 8
3 3 6 9
And lets say I have two arrays pointing to the values I'd like to grab
grab_row = c(3,1,2)
grab_col = c(1,2,1)
Now I'd expect this to be the code I want...
a[ grab_row, grab_col ]
To get these results...
[1] 3 4 2
But that comes out as a 3x3 matrix, which makes enough sense in and of itself
a.1 a.2 a.1.1
3 3 6 3
1 1 4 1
2 2 5 2
Alright, I also see my answer is in the diagonal of the 3x3 matrix... but I'd really rather stick to an array as the output.
Any thoughts? Danka.
Passing the row and column indices in as a two-column matrix (here constructed using cbind()) will get you the elements you were expecting:
a[cbind(grab_row, grab_col)]
[1] 3 4 2
This form of indexing is documented in ?"[":
Matrices and array:
[...snip...]
A third form of indexing is via a numeric matrix with the one
column for each dimension: each row of the index matrix then
selects a single element of the array, and the result is a vector.
Try this:
> mapply(function(i,j)a[i,j], grab_row, grab_col)
[1] 3 4 2
Works for both dataframes and matrices.
I want to group my elements using the repeated segments in the array. The breaking is basically depend on where the repeated segments are, in my real data contains ~10000 elements and I want to know if there is a easier way to do that.
Here is a short example to clarify what I want:
Let's say I have an array,
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
What I want is to break A into [1 5 3],[6 9], and [9 5 2];
What is the easiest to code this using matlab??
Thanks.
For a vectorized solution, you can find out the places where either forward or backward differences to the neighbor are zero, and then use bwlabel (from the Image Processing Toolbox) and accumarray to gather the data.
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
d = diff(A)==0;
%# combine forward and backward difference
%# and invert to identify non-repeating elments
goodIdx = ~([d,false]|[false,d]);
%# create list of group labels using bwlabel
groupIdx = bwlabel(goodIdx);
%# distribute the data into cell arrays
%# note that the first to inputs should be n-by-1
B = accumarray(groupIdx(goodIdx)',A(goodIdx)',[],#(x){x})
EDIT
Replace the last two lines of code with the following if you want the repeating elements to appear in the cell array as well
groupIdx = cumsum([1,abs(diff(goodIdx))]);
B = accumarray(groupIdx',A',[],#(x){x})
EDIT2
If you want to be able to split consecutive groups of identical numbers as well, you need to calculate groupIdx as follows:
groupIdx = cumsum([1,abs(diff(goodIdx))|~d.*~goodIdx(2:end)])
Here is a solution that works if I understand the question correctly. It can probably be optimised further.
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
% //First get logical array of non consecutive numbers
x = [1 (diff(A)~=0)];
for nn=1:numel(A)
if ~x(nn)
if x(nn-1)
x(nn-1)=0;
end
end
end
% //Make a cell array using the logical array
y = 1+[0 cumsum(diff(find(x))~=1)];
x(x~=0) = y;
for kk = unique(y)
B{kk} = A(x==kk);
end
B{:}