I am new in matlab and I have a file contains 657 columns and 97 rows and I want to normalize these data set between 0 and 1.
The same way you can do in any code or calculation, being A you matrix, divide by the maximum of A:
As noted by #AnderBiguri, If you data is not starting from zero, you need to do some math:
A_normalized= (A-min(A(:)))/(max(A(:))-min(A(:)))
If is starting from zero:
A_normalized=A/max(A(:))
Note that A(:) get all the numbers, no need to the the max in each column.
If this is not what you want, give it some comments.
Related
given the 10x10 matrix m:(10 10)#100?1
Find the average of every row and column of the matrix
Find the average of the matrix
Take the first element from the first row, the second from the second row etc
find the diagonal elements of a matrix
First I'm going to change the input so you can better see the solution. The 100?1 will give a list of 100 0s which I don't think is what you want. Maybe you wanted either 100 random (0;1) or 100 random values between 0-1. I went for the latter.
q)show m:(10 10)#100?1.
0.4655548 0.8455166 0.7281041 0.7403385 0.5199511 0.199172 0.9548708 0.498..
0.86544 0.3112134 0.3520122 0.4485896 0.6742543 0.2357538 0.7589261 0.318..
0.7053699 0.8153197 0.5051956 0.7546554 0.08613905 0.7824787 0.2080171 0.282..
So, now the questions.
Find the average of every row and column of the matrix.
q)meanRows:avg each m
q)meanCols:avg each flip m
UPDATE From Comment. You can get the average of a matrix column without using each or flip but will return null if any element is null. Another note is that if a column is length 10, 5 of which are null. Then the avg will only consider the average of the 5 non-null values.
So, if you believe there are nulls you may want to get rid of them and then get the mean values:
q)m:^[0;m] //Replace null with 0s if necessary
q)meanCols:avg m //Get avg without flipping or using each
Find the average of the matrix
q)avg avg each m
I think that^ is the quickest way to get the overall mean because it doesn't require razing or flipping.
Take the first element from the first row, the second from the second row etc
q)getVector:{[mtx]mtx'[c;c:til count mtx]}
q)getVector m
0.4655548 0.3112134 0.5051956 0.6333324 0.7258795 0.8671843 0.7556175 0.17954..
Let me know if you have any further questions.
I need to save the indices of 4 maximum numbers
for example, I need to get the indices of rows 10,9,7,5
5.0259327
4.7127487
4.8435524
4.8538644
5.1048996
6.2441973
5.9413803
6.2912638
5.1117512
5.8309519
5.7419509
6.9663477
5.9958319
6.9519781
6.5802736
6.7327558
7.6765878
I have used
[mA,nA]=max(distA)
where mA is the row and nA is the column
in getting one maximum number but I cannot figure out how to choose another maximum number without duplication. I cannot sort because I need the indices.
You can use the gsort function:
[S,ind]=gsort(distA,"g","d");
The index of the 4 largest elements is the given by
ind(1:4)
You can use this little trick.
[output_val, output_index] = max(input_mat(input_mat < max(input_mat)))
This will give you the value and index of the second largest element. And then similarly, you can do it for 4 numbers.
I have a matrix 1x5000 with numbers. Now I am interested in getting values from the matrix in different positions, more precisely in six different places of the matrix. The places should be based on the length, these are the numbers I want to get out:
Number in 1/6 of the matrix length
Number in 2/6 of the matrix length
Number in 3/6 of the matrix length
Number in 4/6 of the matrix length
Number in 5/6 of the matrix length
Number in 6/6 of the matrix length
These values could be printed out in another matrix, so assume the matrix is 1x5000, 3/6 would give the number in the middle of the matrix. I am new in Matlab and therefore the help is much appreciated!
Since your question is unclear I can try to give you an example.
First of all you can use numel function to get matrix's size.
It's easy to get necessary element in Matlab: you can address directly to any element if you know its number (index). So:
x(100) returns 100th element.
Now you got size and know what to do. Last moment - what to do if numel(x) / 6 return non integer?
You can use rounding functions: ceil, floor or round.
index = ceil(numel(x)/6) %if you want NEXT element always
result = x(index)
Next step: there are a lot of ways to divide data. For example now you have just 6 numbers (1/6, 2/6 and so on) but what if there are 1000 of them? You can't do it manually. So you can use for loop, or you can use matrix of indexes or perfect comment Stewie Griffin.
My example:
divider = [6 5 4 3 2 1] % lets take 1/6 1/5 1/4 1/3 1/2 and 1/1
ind = ceil( numel(x)./divider)
res = x(ind)
The colon notation in MATLAB provides an easy way to extract a range of elements from v:
v(3:7) %Extract the third through the seventh elements
You could either manually input range or use a function to convert fractions into suitable ranges
My goal is to create a random, 20 by 5 array of integers, sort them by increasing order from top to bottom and from left to right, and then calculate the mean in each of the resulting 20 rows. This gives me a 1 by 20 array of the means. I then have to find the column whose mean is closest to 0. Here is my code so far:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
MeanArray= mean(transpose(NewArray(:,:)))
X=min(abs(x-0))
How can I store the column number whose mean is closest to 0 into a variable? I'm only about a month into coding so this probably seems like a very simple problem. Thanks
You're almost there. All you need is a find:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
% MeanArray= mean(transpose(NewArray(:,:))) %// gives means per row, not column
ColNum = find(abs(mean(NewArray,1))==min(abs(mean(NewArray,1)))); %// gives you the column number of the minimum
MeanColumn = RandomArray(:,ColNum);
find will give you the index of the entry where abs(mean(NewArray)), i.e. the absolute values of the mean per column equals the minimum of that same array, thus the index where the mean of the column is closest to 0.
Note that you don't need your MeanArray, as it transposes (which can be done by NewArray.', and then gives the mean per column, i.e. your old rows. I chucked everything in the find statement.
As suggested in the comment by Matthias W. it's faster to use the second output of min directly instead of a find:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
% MeanArray= mean(transpose(NewArray(:,:))) %// gives means per row, not column
[~,ColNum] = min(abs(mean(NewArray,1)));
MeanColumn = RandomArray(:,ColNum);
Given a positive integer n, I want to generate all possible n bit combinations in matlab.
For ex : If n=3, then answer should be
000
001
010
011
100
101
110
111
How do I do it ?
I want to actually store them in matrix. I tried
for n=1:2^4
r(n)=dec2bin(n,5);
end;
but that gave error "In an assignment A(:) = B, the number of elements in A and B must be the same.
Just loop over all integers in [0,2^n), and print the number as binary. If you always want to have n digits (e.g. insert leading zeros), this would look like:
for ii=0:2^n-1,
fprintf('%0*s\n', n, dec2bin(ii));
end
Edit: there are a number of ways to put the results in a matrix. The easiest is to use
x = dec2bin(0:2^n-1);
which will produce an n-by-2^n matrix of type char. Each row is one of the bit strings.
If you really want to store strings in each row, you can do this:
x = cell(1, 2^n);
for ii=0:2^n-1,
x{ii} = dec2bin(ii);
end
However, if you're looking for efficient processing, you should remember that integers are already stored in memory in binary! So, the vector:
x = 0 : 2^n-1;
Contains the binary patterns in the most memory efficient and CPU efficient way possible. The only trade-off is that you will not be able to represent patterns with more than 32 of 64 bits using this compact representation.
This is a one-line answer to the question which gives you a double array of all 2^n bit combinations:
bitCombs = dec2bin(0:2^n-1) - '0'
So many ways to do this permutation. If you are looking to implement with an array counter: set an array of counters going from 0 to 1 for each of the three positions (2^0,2^1,2^2). Let the starting number be 000 (stored in an array). Use the counter and increment its 1st place (2^0). The number will be 001. Reset the counter at position (2^0) and increase counter at 2^1 and go on a loop till you complete all the counters.