I have a 20*120 matrix. For each column in the matrix I need to find the maximum value between all the values, and then sum the remaining values. Then I need to divide the maximum value by the summation of the remaining values. I tried the following code but the result was not correct. What is the problem?
s = 1:z %z=120
for i = 1:x %x=20
maximss = max(Pres_W); %maximum value
InterFss = (sum(Pres_W))-maximss; %remaining values
SIRk(:,s) = (maximss(:,s))./(InterFss(:,s));
end
Instead of answering "what's wrong", I'll first provide a solution explaining how this should be done:
Say we have an example matrix m as follows:
m =
8 5 9 14 10 7 5
10 8 12 11 9 9 12
10 3 7 7 8 4 6
13 11 6 15 13 11 9
Find the maximum value of each column:
col_max = max(m, [], 1)
col_max =
13 11 12 15 13 11 12
Sum all elements in each column, and substract the maximum values:
col_sum = sum(m, 1) - col_max
col_sum =
28 16 22 32 27 20 20
Divide the maximum value by the sum of the other elements:
col_max ./ col_sum
ans =
0.46429 0.68750 0.54545 0.46875 0.48148 0.55000 0.60000
Or, as a one-liner:
max(m,[],1)./(sum(m,1)-max(m,[],1))
ans =
0.46429 0.68750 0.54545 0.46875 0.48148 0.55000 0.60000
By the way: Your code does exactly what you're explaining, it returns the maximum value divided by all values except the maximum value.
Notes regarding best practice:
Vectorize things like this, no need for loops.
max(m, [], 1) is the same as max(m) for 2D-arrays. However, if your matrix for some reason only have one row, it will return the maximum value of the row, thus a single number.
sum(m,1) is the same as sum(m) for 2D-arrays. However, if your matrix for some reason only have one row, it will return the sum of the row, thus a single number.
Related
I have a vector of data for 21 years with daily data and want to create a rolling window of 365 days such as the next period stars one month (30 days) after the previous one. In the question, n_interval defines the difference between the first data point of the next window and the last observation of the previous series.
Let's assume my daily data start from Jan. 1 2000, then the first column would be Jan. 1, 2000 -Jan.1, 2001 and the second column starts from Feb. 1, 2000. and ends on Feb. 1, 2001. and ... the last column will cover Jan. 1, 2017 to Jan. 1, 2018. for example if:
vec = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17]
for a given variable n_interval = 3, with window_size=5, the output matrix should look like:
mat = [[1 4 7 10 13],
[2 5 8 11 14],
[3 6 9 12 15],
[4 7 10 13 16],
[5 8 11 14 17]]
Given your example vector
vec = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17];
we can create an indexing scheme by as follows:
First, we need to determine how many rows there will be in the mat. Assuming we want every element of vec to be expressed in mat at least once then we need to make sure that last index in the last row is greater than or equal to the size of vec. It's fairly easy to see that the index of the last column in mat is described by
last_index = n_interval*(n_rows-1) + n_columns
We want to ensure that last_index >= numel(vec). Substituting in the above expression into the inequality and solving for n_rows gives
n_rows >= (numel(vec) - n_columns)/n_interval + 1
We assign n_rows to be the ceil of this bound so that it is the smallest integer which satisfies the inequality. Now that we know the number of rows we generate the list of starting indices for each row
start_index = 1:n_interval:(n_interval*(n_rows-1)+1);
In the index matrix we want each column to be 1 plus the previous column. In other words we want to offset the column according to the array index_offset = 0:(n_interval-1).
Using bsxfun we generate the index matrix by computing the sums of all pairs between the start_index and index_offset arrays
index = bsxfun(#plus, index_offset, start_index');
The final thing we need to worry about is going out of bounds. To handle this we apply the mod function to wrap the out of bounds indicies:
index_wrapped = mod(index-1, numel(vec))+1;
Then we simply sample the vector according to index_wrapped
mat = vec(index_wrapped);
The complete code is
n_interval = 3;
n_columns = 5;
vec = 1:17;
n_rows = ceil((numel(vec)-n_columns)/n_interval + 1);
start_index = 1:n_interval:(n_interval*(n_rows-1)+1);
index_offset = 0:(n_columns-1);
index = bsxfun(#plus, index_offset, start_index');
index_wrapped = mod(index-1, numel(vec))+1;
mat = vec(index_wrapped);
I have data of integers in x = 500 X 612 matrix. I need a new variable xx in a 500 X 612 matrix but I need to apply cumsum along each row (500) across 12 column steps and applying cumsum like this 51 times --> 500 X (12 X 51) matrix. Then I need a for loop to produce 51 plots of the 500 rows and 12 columns of the cumsum time series. thank you!
I will rephrase what the question is asking to benefit those who are reading.
The OP wishes to segment a matrix into chunks by splitting up the matrix into a bunch of columns. A cumsum is applied to each row individually for each column and are then concatenated together to build a final matrix. As such, given this source matrix:
x =
1 2 3 4 5 6 7 8 9 10 11 12
13 14 15 16 17 18 19 20 21 22 23 24
Supposing that we wish to split up the matrix by columns 3, 6 and 9 and 12, we will have four chunks to work with. We do a cumsum on each of these blocks individually and piece the final result together. So the result would like the following:
xx =
1 3 6 4 9 15 7 15 24 10 21 33
13 27 42 16 33 51 19 39 60 22 45 69
First, you need to determine how many columns you want to break up the matrix into. In your case, we wish to segment the matrix into 4 chunks: Columns 1 - 3, columns 4 - 6, columns 7 - 9, and columns 10 - 12. As such, I'm going to reshape this matrix so that each column is an individual row from a chunk in this matrix. We then apply cumsum over this reshaped matrix and we then reshape it back to what you had originally.
Therefore, do this:
num_chunks = 4; %// Columns 3, 6, 9, 12
divide_point = size(x,2) / num_chunks; %// Determine how many elements are in a row for a cumsum
x_reshape = reshape(x.', divide_point, []); %// Get reshaped matrix
xy = cumsum(x_reshape); %// cumsum over all columns individually
xx = reshape(xy, size(x,2), size(x,1)).'; %// Reconstruct matrix
In the third line of code, x_reshape = reshape(x.', divide_point, []); may seem a bit daunting, but it's actually not that bad. I had to transpose the matrix first because you want to take each row of a chunk and place them into individual columns so we can perform a cumsum on each column. When you reshape something in MATLAB, it collects values column-wise and reshapes the input into an output of a specified size. Therefore, to collect the rows, we need to collect row-wise and so we must transpose this matrix. Next, divide_point tells you how many elements we have for a single row in one chunk. As such, we want to construct a matrix that is of size divide_point x N where divide_point tells you how many elements we have in a row of a chunk and N is the total number of rows over all chunks. Because I don't want to calculate how many there are (am rather lazy actually....), the [] syntax is to automatically infer this number so that we can get a reshaped matrix that respects the total number of elements in the original input. We then perform cumsum on each of these columns, and then we need to reshape this back into the original shape of the input. With this, we use reshape again on the cumsum result, but in order to get it back into the row-order that you want, we have to determine the transpose as reshape takes values in column-major order, then re-transpose that result.
We get:
xx =
1 3 6 4 9 15 7 15 24 10 21 33
13 27 42 16 33 51 19 39 60 22 45 69
In general, the total number of elements to sum over for a row needs to be evenly divisible by the total number of columns that your matrix contains. For example, given the above, if you were to try to segment this matrix into 5 chunks, you would certainly get an error as the number of rows to cumsum over is not symmetric.
As another example, let's say we wanted to break up the matrix into 6 chunks. Therefore, by setting num_chunks = 6, we get:
xx =
1 3 3 7 5 11 7 15 9 19 11 23
13 27 15 31 17 35 19 39 21 43 23 47
You can see that cumsum restarts at every second column, as we desired 6 chunks and to get 6 chunks with a matrix of 12 columns, a chunk is created at every second column.
I have a vector of values which represent an index of a row to be removed in some matrix M (an image). There's only one row value per column in this vector (i.e. if the image is 128 x 500, my vector contains 500 values).
I'm pretty new to MATLAB so I'm unsure if there's a more efficient way of removing a single pixel (row,col value) from a matrix so I've come here to ask that.
I was thinking of making a new matrix with one less row, looping through each column up until I find the row whose value I wish to remove, and "shift" the column up by one and then move onto the next column to do the same.
Is there a better way?
Thanks
Yes, there is a solution which avoids loops and is thus faster to write and to execute. It makes use of linear indexing, and exploits the fact that you can remove a matrix entry by assigning it an empty value ([]):
% Example data matrix:
M = [1 5 9 13 17
2 6 10 14 18
3 7 11 15 19
4 8 12 16 20];
% Example vector of rows to be removed for each column:
vector = [2 3 4 1 3];
[r c] = size(M);
ind = sub2ind([r c],vector,1:c);
M(ind) = [];
M = reshape(M,r-1,c);
This gives the result:
>> M =
1 5 9 14 17
3 6 10 15 18
4 8 11 16 20
I have a matrix of 2d lets assume the values of the matrix
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
17 24 1 8 15
11 18 25 2 9
This matrix is going to be divided into three different matrices randomly let say
b =
17 24 1 8 15
23 5 7 14 16
c =
4 6 13 20 22
11 18 25 2 9
d =
10 12 19 21 3
17 24 1 8 15
How can i know the index of the vectors in matrix d for example in the original matrix a,note that the values of the matrix can be duplicated.
for example if i want to know the index of {10 12 19 21 3} in matrix a?
or the index of {17 24 1 8 15} in matrix a,but for this one should return only on index value?
I would appreciate it so much if you can help me with this. Thank you in advance
You can use ismember with the 'rows' option. For example:
tf = ismember(a, c, 'rows')
Should produce:
tf =
0
0
1
0
0
1
To get the indices of the rows, you can apply find on the result of ismember (note that it's redundant if you're planning to use this vector for matrix indexing). Here find(tf) return the vector [3; 6].
If you want to know the number of the row in matrix a that matches a single vector, you either use the method explained and apply find, or use the second output parameter of ismember. For example:
[tf, loc] = ismember(a, [10 12 19 21 3], 'rows')
returns loc = 4 for your example. Note that here a is the second parameter, so that the output variable loc would hold a meaningful result.
Handling floating-point numbers
If your data contains floating point numbers, The ismember approach is going to fail because floating-point comparisons are inaccurate. Here's a shorter variant of Amro's solution:
x = reshape(c', size(c, 2), 1, []);
tf = any(all(abs(bsxfun(#minus, a', x)) < eps), 3)';
Essentially this is a one-liner, but I've split it into two commands for clarity:
x is the target rows to be searched, concatenated along the third dimension.
bsxfun subtracts each row in turn from all rows of a, and the magnitude of the result is compared to some small threshold value (e.g eps). If all elements in a row fall below it, mark this row as "1".
It depends on how you build those divided matrices. For example:
a = magic(5);
d = a([2 1 2 3],:);
then the matching rows are obviously: 2 1 2 3
EDIT:
Let me expand on the idea of using ismember shown by #EitanT to handle floating-point comparisons:
tf = any(cell2mat(arrayfun(#(i) all(abs(bsxfun(#minus, a, d(i,:)))<1e-9,2), ...
1:size(d,1), 'UniformOutput',false)), 2)
not pretty but works :) This would be necessary for comparisons such as: 0.1*3 == 0.3
(basically it compares each row of d against all rows of a using an absolute difference)
I have a vector called time, which contains time values. I'd like obtain a vector of indexes of time in which the value is between threshold x and threshold y.
This is undoubltedly trivial to do, but I'm struggling with Matlab syntax a bit, here. Any help would be greatly appreciated.
Blz
time=5:20
idx = find(time > 10 & time < 15) % indices
values=time(time(:)>10 & time(:)<15) % values
which give
time =
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
idx =
7 8 9 10
values =
11 12 13 14