A boolean vector has been created based on some rule and we need to identify the indices where the rule holds for 2 observations in a row. The following code does that
indices:0101001101b
runs:{0 x\x}"f"$;
where 2=runs indices
Could you please help me understand how the scan operator is used in the definition of the runs function? Appreciate your help.
It's using this special shorthand commonly used in calculating exponential moving averages: https://code.kx.com/q/ref/accumulators/#alternative-syntax
So {0 x\x} is equivalent to:
q){z+x*y}\[0;indices;indices]
0 1 0 1 0 0 1 2 0 1
What this is doing is essentially using the booleans as an on/off switch (via the boolean multiplication) for the rolling sum. It adds (z+) until it hits a negative boolean in which case the rolling sum resets back to zero.
In english: nextValue + [currentValue (starting at 0) * nextValue]
When nextValue is 1, 1 gets added. When nextValue is 0 the result is zero (resetting the rolling sum).
Something like this can achieve the same thing, though no less easy to read at a glance (and using two scans instead of one):
q){s-maxs not[x]*s:sums x}indices
0 1 0 1 0 0 1 2 0 1i
Terry has answered your question about how runs works.
Comparing adjacent items is common. You might prefer to use the prior keyword. Certainly easier to see what it is doing.
q)where (and) prior indices
,7
I want to create a vector without the number 1 .
x=-10:1:10;
To avoid this:
for(n=0:21)
if(x(n)==1)
x(n)=[];
end
end
What can I do ?
I would use setdiff
>> setdiff(-5:5,1)
ans =
-5 -4 -3 -2 -1 0 2 3 4 5
Instead of manually generating a vector from -10 to 10 and removing the entry that has the value of 1, you can always use colon / : and not include 1 in the vector instead. Something like:
x = [-10:0 2:10];
Because it's such a small vector, you probably won't gain much by doing it this way in comparison to fully generating the vector and removing one entry as per David's suggestion. I do agree with David though. Learn logical indexing! It's one of the backbones for making any MATLAB code fast.
You can try setting it manually to " ".
eg x(10)=[];
I'm trying to create a random "path" on a coordinate system on Matlab. I am doing this by creating a for loop where for each iteration it fills in a new value on a matrix that has initial values of zeros.
For example, I have 5 points so I have an initial matrix a=[0 0 0 0 0; 0 0 0 0 0] (row1 = x values, row2 = y values).
The path can move right/left or up/down (no diagonals). In my for loop, I call randi(4) and say something like "if randi(4)=1, then move 1 point to the left (x-1). if randi(4)=2, then move to the right (x+1), etc."
The problem is that you cannot visit a specific point more than once. For example, the path can start at (0,0), then go to (0,1), then (1,1), then (1,0), and then it CANNOT go back to (0,0).. in my current code I don't have this restriction so I was hoping I could get some suggestions..
Since in this example the matrix would look something like a=[0 0 1 1 0; 0 1 1 0 0].
I was thinking of maybe subtracting each new coordinate (here (0,0)) from each column on the matrix a and if any of the columns give me values of zero for both rows (since it's the same coordinate subtracted from itself), then go back one step and let randi(4) run again.. but
How could I tell it to "go back one step" (or two or three)?
How do you compare one column against each column of the already established matrix?
This was just an idea.. are there any functions in Matlab that would let me do this? or maybe compare if two columns are the same within a matrix?
To your questions.
to go back - I suppose this means just throwing away the rightmost columns in your matrix.
to find if it is present you could use ismember
unfortunately it only takes rows so you will need to transpose. Snippet:
a = [1:10; repmat(1:2,1,5)]'
test = ismember(a,[3,2],'rows')
any(test) % not found
test = ismember(a,[3,1],'rows')
any(test) % found
Of course your idea would also work.
I can answer this:
How do you compare one column against each column of the already
established matrix?
Use two different matrices. Compare them using the setdiff() function: http://www.mathworks.com/help/matlab/ref/setdiff.html
I'm a newbie to Matlab and just stumped how to do a simple task that can be easily performed in excel. I'm simply trying to get the percent change between cells in a matrix. I would like to create a for loop for this task. The data is setup in the following format:
DAY1 DAY2 DAY3...DAY 100
SUBJECT RESULTS
I could only perform getting the percent change between two data points. How would I conduct it if across multiple days and multiple subjects? And please provide explanation
Thanks a bunch
FOR EXAMPLE, FOR DAY 1 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5), DAY 2 SUBJECT1(RESULT=2), SUBJECT2(RESULT=8), SUBJECT3(RESULT=10), DAY 3 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5).
I WANT THE PERCENT CHANGE SO OUTPUT WILL BE DAY 2 SUBJECT1(RESULT=100%), SUBJECT2(RESULT=100%), SUBJECT3(RESULT=100%). DAY3 SUBJECT1(RESULT=50%), SUBJECT2(RESULT=50%), SUBJECT3(RESULT=50%)
updated:
Hi thanks for responding guys. sorry for the confusion. zebediah49 is pretty close to what I'm looking for. My data is for example a 10 x 10 double. I merely wanted to get the percentage change from column to column. For example, if I want the percentage change from rows 1 through 10 on all columns (from columns 2:10). I would like the code to function for any matrix dimension (e.g., 1000 x 1000 double) zebediah49 could you explain the code you posted? thanks
updated2:
zebediah49,
(data(1:end,100)- data(1:end,99))./data(1:end,99)
output=[data(:,2:end)-data(:,1:end-1)]./data(:,1:end-1)*100;
Observing the code above, How would I go about modifying it so that column 100 is used as the index against all of the other columns(1-99)? If I change the code to the following:
(data(1:end,100)- data(1:end,:))./data(1:end,:)
matlab is unable because of exceeding matrix dimensions. How would I go about implementing that?
UPDATE 3
zebediah49,
Worked perfectly!!! Originally I created a new variable for the index and repmat the index to match the matrices which was not a good idea. It took forever to replicate when dealing with large numbers.
Thanks for you contribution once again.
Thanks Chris for your contribution too!!! I was looking more on how to address and manipulate arrays within a matrix.
It's matlab; you don't actually want a loop.
output=input(2:end,:)./input(1:end-1,:)*100;
will probably do roughly what you want. Since you didn't give anything about your matlab structure, you may have to change index order, etc. in order to make it work.
If it's not obvious, that line defines output as a matrix consisting of the input matrix, divided by the input matrix shifted right by one element. The ./ operator is important, because it means that you will divide each element by its corresponding one, as opposed to doing matrix division.
EDIT: further explanation was requested:
I assumed you wanted % change of the form 1->1->2->3->1 to be 100%, 200%, 150%, 33%.
The other form can be obtained by subtracting 100%.
input(2:end,:) will grab a sub-matrix, where the first row is cut off. (I put the time along the first dimension... if you want it the other way it would be input(:,2:end).
Matlab is 1-indexed, and lets you use the special value end to refer to the las element.
Thus, end-1 is the second-last.
The point here is that element (i) of this matrix is element (i+1) of the original.
input(1:end-1,:), like the above, will also grab a sub-matrix, except that that it's missing the last column.
I then divide element (i) by element (i+1). Because of how I picked out the sub-matrices, they now line up.
As a semi-graphical demonstration, using my above numbers:
input: [1 1 2 3 1]
input(2,end): [1 2 3 1]
input(1,end-1): [1 1 2 3]
When I do the division, it's first/first, second/second, etc.
input(2:end,:)./input(1:end-1,:):
[1 2 3 1 ]
./ [1 1 2 3 ]
---------------------
== [1.0 2.0 1.5 0.3]
The extra index set to (:) means that it will do that procedure across all of the other dimension.
EDIT2: Revised question: How do I exclude a row, and keep it as an index.
You say you tried something to the effect of (data(1:end,100)- data(1:end,:))./data(1:end,:). Matlab will not like this, because the element-by-element operators need them to be the same size. If you wanted it to only work on the 100th column, setting the second index to be 100 instead of : would do that.
I would, instead, suggest setting the first to be the index, and the rest to be data.
Thus, the data is processed by cutting off the first:
output=[data(2:end,2:end)-data(2:end,1:end-1)]./data(2:end,1:end-1)*100;
OR, (if you neglect the start, matlab assumes 1; neglect the end and it assumes end, making (:) shorthand for (1:end).
output=[data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100;
However, you will probably still want the indices back, in which case you will need to append that subarray back:
output=[data(1,1:end-1) data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100];
This is probably not how you should be doing it though-- keep data in one matrix, and time or whatever else in a separate array. That makes it much easier to do stuff like this to data, without having to worry about excluding time. It's especially nice when graphing.
Oh, and one more thing:
(data(:,2:end)-data(:,1:end-1))./data(:,1:end-1)*100;
is identically equivalent to
data(:,2:end)./data(:,1:end-1)*100-100;
Assuming zebediah49 guessed right in the comment above and you want
1 4 5
2 8 10
1 4 5
to turn into
1 1 1
-.5 -.5 -.5
then try this:
data = [1,4,5; 2,8,10; 1,4,5];
changes_absolute = diff(data);
changes_absolute./data(1:end-1,:)
ans =
1.0000 1.0000 1.0000
-0.5000 -0.5000 -0.5000
You don't need the intermediate variable, you can directly write diff(data)./data(1:end,:). I just thought the above might be easier to read. Getting from that result to percentage numbers is left as an exercise to the reader. :-)
Oh, and if you really want 50%, not -50%, just use abs around the final line.
I have a line of code in matlab for which i am selecting a subset of a matrix:
A(3:5,1:3);
Now i want to adapt this line, to only select rows for which all three values are larger than zero:
(A(3:5,1:3) > 0);
But apparently i am not doing this right. How do i select part of the matrix, and also make sure that only the rows (for which all three values are) larger than zero are selected?
EDIT: To clarify: lets say that i have a matrix of coordinates called A, that looks like this:
Matrix A [5,3]
3 4 0
0 1 0
0 3 1
0 0 0
4 8 7
Now i want to select only part [3:5,1:3], and of that part i only want to select row 3 and 5. How do i do that?
The expression:
A(find(sum(A(3:5,:),2)~=0),:)
will return only the rows of A(3:5,:) which have a row-sum not equal to zero.
If you had posted syntactically correct Matlab it would have been easier for me to cut and paste your test data into my Matlab session.
I'm modelling this answer off of A(find( A > 0 ))
distances = pdist(find( pdist(medoidContainer(i,1:3)) > 0 ));
This will give you a vector of values in the distances variable. The reason the pdist(medoidContainer(i,1:3) > 0) does not work is because it first, finds the indices specified by i,1:3 in medoidContainer. Then it finds the indices in medoidContainer(i,1:3) that are greater than 0. However, since medoidContainer(i,1:3) and pdist now likely have different dimensions, the comparison does not give the right indexes.