Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?
The way I intend to use it is to check whether an element index in one matrix is equal to the values stored in another array (where the stored values are the indices of the elements which meet a certain criteria).
So, if the indices of the elements which meet the criteria are stored in the matrix below:
criteriacheck = [3 5 6 8 20];
Going through the main array (called array) and checking if the index matches:
for i = 1:numel(array)
if i == 'Any value stored in criteriacheck'
%# "Do this"
end
end
Does anyone have an idea of how I might go about this?
The excellent answer previously given by #woodchips applies here as well:
Many ways to do this. ismember is the first that comes to mind, since it is a set membership action you wish to take. Thus
X = primes(20);
ismember([15 17],X)
ans =
0 1
Since 15 is not prime, but 17 is, ismember has done its job well here.
Of course, find (or any) will also work. But these are not vectorized in the sense that ismember was. We can test to see if 15 is in the set represented by X, but to test both of those numbers will take a loop, or successive tests.
~isempty(find(X == 15))
~isempty(find(X == 17))
or,
any(X == 15)
any(X == 17)
Finally, I would point out that tests for exact values are dangerous if the numbers may be true floats. Tests against integer values as I have shown are easy. But tests against floating point numbers should usually employ a tolerance.
tol = 10*eps;
any(abs(X - 3.1415926535897932384) <= tol)
you could use the find command
if (~isempty(find(criteriacheck == i)))
% do something
end
Note: Although this answer doesn't address the question in the title, it does address a more fundamental issue with how you are designing your for loop (the solution of which negates having to do what you are asking in the title). ;)
Based on the for loop you've written, your array criteriacheck appears to be a set of indices into array, and for each of these indexed elements you want to do some computation. If this is so, here's an alternative way for you to design your for loop:
for i = criteriacheck
%# Do something with array(i)
end
This will loop over all the values in criteriacheck, setting i to each subsequent value (i.e. 3, 5, 6, 8, and 20 in your example). This is more compact and efficient than looping over each element of array and checking if the index is in criteriacheck.
NOTE: As Jonas points out, you want to make sure criteriacheck is a row vector for the for loop to function properly. You can form any matrix into a row vector by following it with the (:)' syntax, which reshapes it into a column vector and then transposes it into a row vector:
for i = criteriacheck(:)'
...
The original question "Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?" can be solved without any loop.
Just use the setdiff function.
I think the INTERSECT function is what you are looking for.
C = intersect(A,B) returns the values common to both A and B. The
values of C are in sorted order.
http://www.mathworks.de/de/help/matlab/ref/intersect.html
The question if i == 'Any value stored in criteriacheck can also be answered this way if you consider i a trivial matrix. However, you are proably better off with any(i==criteriacheck)
Related
I have a 2D array PointAndTangent of dimension 8500 x 5. The data is row-wise with 8500 data rows and 5 data values for each row. I need to extract the row index of an element in 4th column when this condition is met, for any s:
abs(PointAndTangent[:,3] - s) <= 0.005
I just need the row index of the first match for the above condition. I tried using the following:
index = np.all([[abs(s - PointAndTangent[:, 3])<= 0.005], [abs(s - PointAndTangent[:, 3]) <= 0.005]], axis=0)
i = int(np.where(np.squeeze(index))[0])
which doesn't work. I get the follwing error:
i = int(np.where(np.squeeze(index))[0])
TypeError: only size-1 arrays can be converted to Python scalars
I am not so proficient with NumPy in Python. Any suggestions would be great. I am trying to avoid using for loop as this is small part of a huge simulation that I am trying.
Thanks!
Possible Solution
I used the following
idx = (np.abs(PointAndTangent[:,3] - s)).argmin()
It seems to work. It returns the row index of the nearest value to s in the 4th column.
You were almost there. np.where is one of the most abused functions in numpy. Half the time, you really want np.nonzero, and the other half, you want to use the boolean mask directly. In your case, you want np.flatnonzero or np.argmax:
mask = abs(PointAndTangent[:,3] - s) <= 0.005
mask is a 1D array with ones where the condition is met, and zeros elsewhere. You can get the indices of all the ones with flatnonzero and select the first one:
index = np.flatnonzero(mask)[0]
Alternatively, you can select the first one directly with argmax:
index = np.argmax(mask)
The solutions behave differently in the case when there are no rows meeting your condition. Three former does indexing, so will raise an error. The latter will return zero, which can also be a real result.
Both can be written as a one-liner by replacing mask with the expression that was assigned to it.
There is an integer array, for eg.
{3,1,2,7,5,6}
One can move forward through the array either each element at a time or can jump a few elements based on the value at that index. For e.g., one can go from 3 to 1 or 3 to 7, then one can go from 1 to 2 or 1 to 2(no jumping possible here), then one can go 2 to 7 or 2 to 5, then one can go 7 to 5 only coz index of 7 is 3 and adding 7 to 3 = 10 and there is no tenth element.
I have to only count the number of possible paths to reach the end of the array from start index.
I could only do it recursively and naively which runs in exponential time.
Somebody plz help.
My recommendation: use dynamic programming.
If this key word is sufficient and you want the challenge to find a possible solution on your own, dont read any further!
Here a possible DP-algorithm on the example input {3,1,2,7,5,6}. It will be your job to adjust on the general problem.
create array sol length 6 with just zeros in it. the array will hold the number of ways.
sol[5] = 1;
for (i = 4; i>=0;i--) {
sol[i] = sol[i+1];
if (i+input[i] < 6 && input[i] != 1)
sol[i] += sol[i+input[i]];
}
return sol[0];
runtime O(n)
As for the directed graph solution hinted in the comments :
Each cell in the array represents a node. Make an directed edge from each node to the node accessable. Basically you can then count more easily the number of ways by just looking at the outdegrees on the nodes (since there is no directed cycle) however it is a lot of boiler plate to actual program it.
Adjusting the recursive solution
another solution would be to pruning. This is basically equivalent to the DP-algorithm. The exponentiel time comes from the fact, that you calculate values several times. Eg function is recfunc(index). The initial call recFunc(0) calls recFunc(1) and recFunc(3) and so on. However recFunc(3) is bound to be called somewhen again, which leads to a repeated recursive calculation. To prune this you add a Map to hold all already calculated values. If you make a call recFunc(x) you lookup in the map if x was already calculated. If yes, return the stored value. If not, calculate, store and return it. This way you get a O(n) too.
Say I have a vector, for example, x <- 1:10, then x[0] returns a zero-length vector of the same class as x, here integer(0).
I was wondering if there is a reason behind that choice, as opposed to throwing an error, or returning NA as x[11] would? Also, if you can think of a situation where having x[0] return integer(0) is useful, thank you for including it in your answer.
As seen in ?"["
NA and zero values are allowed: rows of an index matrix containing a
zero are ignored, whereas rows containing an NA produce an NA in the
result.
So an index of 0 just gets ignored. We can see this in the following
x <- 1:10
x[c(1, 3, 0, 5, 0)]
#[1] 1 3 5
So if the only index we give it is 0 then the appropriate response is to return an empty vector.
My crack at it as I am not a programmer and certainly do not contribute to R source. I think it may be because you need some sort of place holder to state that something occurred here but nothing was returned. This becomes more apparent with things like tables and split. For instance when you make a table of values and say there are zero of that cell you need to hold that that cell made from a string in a vector has no values. it would not be a appropriate to have x[0]==0 as it's not the numeric value of zero but the absence of any value.
So in the following splits we need a place holder and integer(0) holds the place of no values returned which is not the same as 0. Notice for the second one it returns numeric(0) which is still a place holder stating it was numeric place holder.
with(mtcars, split(as.integer(gear), list(cyl, am, carb)))
with(mtcars, split(gear, list(cyl, am, carb)))
So in a way my x[FALSE] retort is true in that it holds the place of the non existent zero spot in the vector.
All right this balonga I just spewed is true until someone disputes it and tears it down.
PS page 19 of this guide (LINK) state that integer() and integer(0) are empty integer.
Related SO post: How to catch integer(0)?
Since the array indices are 1-based, index 0 has no meaning. The value is ignored as a vector index.
I'm trying to check if my arrays are returning nonsense by accessing out of bounds elements, in fortran. And I want to check these values are less than one, and if they are, change them to one.
This is the piece of my code causing issues:
lastNeighLabel=(/clusterLabel(jj-1,kk,ll), clusterLabel(jj,kk-1,ll), clusterLabel(jj,kk,ll-1)/)
LastNeighLabel contains the cluster label (between 1 and n, where n isthe total number of unique seperate clusters found) for the last neighbour in the x,y,z direction respectively.
When jj or kk or ll are 1, they try and access the 0th element in the array, and as FORTRAN counts from 1 in arrays, it tries to destroy the universe. I'm currently in a tangled mess of about 8 if/elseif statements trying to code for every eventuality. But I was hoping there was a way of operating on each element. So basically I'd like to say where((/jj-1,kk-1,ll-1/).lt.1) do clusterLabel(jj-1,kk,ll)=0 etc depending on which element is causing the problem.
But I can't think of a way to do that because where will only manipulate the variables passed to it, not a different array at the same index. Or am I wrong?
Will gladly edit if this doesn't make sense.
It is not obligatory that Fortran accesses arrays starting from one. Any starting value is allowed. If it more convenient to you to have a zero indexed array, declare the array as:
real, dimension (0:N-1, 0:M-1) :: array
Or
real, dimension (0:N, 0:M) :: array
and have the 0 indices be extra to catch special cases.
This might be another solution to your problem, since zero index values would be legal.
Another possible way to approach this, is to create an extended cluster label array (with index bounds starting at 0), which is equal to the cluster label array with a layer of zeroes tacked on the outside. You can then let your loop run safely over all values of jj, kk, and ll. It depends on the size of the array if this is a feasible solution.
integer :: extended_cluster_label(0:size(cluster_label,1), &
0:size(cluster_label,2), &
0:size(cluster_label,3) &
)
extended_cluster_label(0,:,:) = 0
extended_cluster_label(:,0,:) = 0
extended_cluster_label(:,:,0) = 0
extended_cluster_label(1:, 1:, 1:) = cluster_label
Maybe you could use a function?
real function f(A,i,j,k)
real :: A(:,:,:)
integer :: i,j,k
if (i==0.or.j==0.or.k==0) then
f=0
else
f=A(i,j,k)
endif
end function f
and then use f(clusterLabel,jj-1,kk,ll) etc.
I have made some c code for a program, which does some psycho-acoustics on sound data.
There is a piece of code which runs very slowly.
I think it would be best to use a look up table. How would one go about implementing it?
Any pointers or help would be appreciated! :)
Your values are not equidistant so it is not that easy. But its still possible: take your greatest common divisor of all your condition-values (thats here 50) and then make your table
byteout = lut[difference/50 + 12];
And in the lookup table you can just use your values in the posted order, where you duplicate the entries in case your stepping is 100.
Btw it just see, there is a mistake, all your negative cases are catched by your first <=0 (my example assumes that you want to omit the first case).
Firstly, take a look at where you want that first check against 0, as it makes all of your negative checks pointless.
Secondly, I would probably construct a lookup table as an array of 1300 elements, offset by 500 (your lowest negative value). Each element would be the result you want when you look up that number. If you are looking for something less than -500, don't check the array.
So it would look something like this:
table[0] = 0b0110; // -500 through -599
table[1] = 0b0110;
...
table[100] = 0b0101; // -400 through -499
table[101] = 0b0101;
...
Lookup would be:
if (value <= -600) {
return 0b0111;
}
else {
return table[value + 600];
}
It's a small enough number of values that the size of the array is not prohibitive. Initialize with a loop at the beginning of your program.
Binary search for the win.
Store all possible values in an array, and be sure to sort them.
Start in the middle and see if difference is less than that value. If so, move to the middle of what's left of your cursor and try again. If not, move to the right. Keep going until you find the value you want, and then use that.
Your array could be of structs that have the minimum value and the corresponding byteout value.
EDIT: To clear up possible misunderstandings, with "every possible value" I don't mean every number between -1400 and 1400, just values you check against in your original code.
Let's look at the first bit:
if (difference <= 0)
byteout = 0b0000;
else if (difference <= -600)
byteout = 0b0111;
Let's say you have a value of -601.
Is it <= 0? Yes, so byteout = 0b0000;
you never get to the -600. So effectively, ALL negative values are 0b0000. This may or may not be by design, but if it is you can get rid of all other negative values.
Otherwise, I would consider reducing this to a formula (with as few branches as possible) or going with #Ebomike's solution of a precomputed lookup table and binary search.