I do have a three dimensional array A(m,n,t), (8 x 60 x 8), filled with positive numbers.
What I am trying to do is to set every element of this array to zero if the third index (t) is smaller than the first (m).
So for examle A(5,42,3) should be set to 0 as m=5 > t=3. However A(5,13,7) should not be changed, hence m=5 <= t=7.
If one would look at squeeze(A(:,val,:)) where val is any number from 1 to 60, then what I am trying to do is to set the lower triangular matrix to zero.
My approach was a loop over the first index doing
for ii=2:8
A(ii,:,1:ii-1)=0;
end
However this uses a loop and I am almost certain there should be a (smart) way without it.
So how does one do this without the use of a loop?
try this:
[xg,yg,zg]=ndgrid(1:size(arr,1),1:size(arr,2),1:size(arr,3));
arr(xg>zg)=0;
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.
I am working with N-dimensional array and have a problem with the array indexing. I have a task to find an (N-1)-dimensional array in the middle N-dimensional array.
Let me explain in detail with 3D array. A is a 3-dimensional array that has split into groups. In each group, there are b - number of 2-dimensional arrays in the group. I have simulated it as:
b=5;
A=rand(2,2,20);
groups = reshape(A, size(A,1), size(A,2),b, []);
groups is 4-dimensional array, the 4-th dimension is a number of groups ( here it 4).
To find a middle in each group I have added the following loop:
for ii=1:size(groups,4) % Loop over all groups/slices
middle(:,:,ii) = groups(:,:,(w-1)/2+1,ii); % 1 2 3 4 5 : the middle is 3
end
middle is 3-dimensional array that collects middle array in each group.
As you see in my example I have used b=5( odd number). My problem is with even number b.
I have tried to implement it as ( rewrite the loop above);
l=rem(w,2);
for ii=1:size(groups,4) % Loop over all groups/slices
if l==1
middle(:,:,ii) = groups(:,:,(w-1)/2+1,ii);
else
middle(:,:,ii) = groups(:,:,(w-1)/2,ii);
end
end
But it doesn't work. Matlab gives me an error in the line l=rem(w,2);
Could you suggest to me how I can fix it? Is there another way to implement it?
You should use floor of ceil to round the index to whichever element you want:
middle_index = floor((w-1)/2+1);
Here, the middle of 4 is 2, using ceil you’d pick index 3.
Next, you can extract the arrays in a single indexing operation:
middle = groups(:,:,middle_index,:);
Finally, use squeeze or reshape to get rid of the 3rd index:
middle = squeeze(middle);
I have a numeric array sized 1000x1 which have values 0 and 1, called conditionArray.
I have a cell array called netNames with the same size (1000x1) and its cells contain string values (which are name of some circuit nets).
I want to extract net names which from netNames which their pairwise condition bit is 1 in conditionArray.
E.g. if conditionArray(100) is equal to extract its net name from netNames{100}.
Output of this process can be stored in an string array or cell array.
Are there any ways to do this operation with pairwise operations or I should use a for statement for this?
You shall check out cellfun in Matlab anytime you want to manipulate each element inside a cellarray without using a for loop.
As I understand, you have:
N = 1000;
% an array with 0s and 1s (this generates random 0s and 1s):
conditionArray = randi([0,1],N);
% a cell array with strings (this generates random 5-character strings):
netNames = cell(N);
netNames = cellfun(#(c)char(randi([double('a'),double('z')],1,5)), netNames, 'UniformOutput',false);
To extract the elements from netNames where conditionArray is 1, you can do:
netNames(conditionArray==1)
This uses logical indexing into the cell array.
I have the following cell matrix, which will be used as a confusion matrix:
confusion=cell(25,25);
Then, I have two other cell arrays, on which each line contains predicted labels (array output) and another cell matrix containing the real labels (array groundtruth).
whos output
Name Size Bytes Class Attributes
output 702250x1 80943902 cell
whos groundtruth
Name Size Bytes Class Attributes
groundtruth 702250x1 84270000 cell
Then, I created the following script to create the confusion matrix
function confusion=write_confusion_matrix(predict, groundtruth)
confusion=cell(25,25);
for i=1:size(predict,1)
confusion{groundtruth{i},predict{i}}=confusion{groundtruth{i}, predict{i}}+1;
end
end
But when I run it in matlab I have the following error:
Index exceeds matrix dimensions.
Error in write_confusion_matrix (line 4)
confusion{groundtruth{i},predict{i}}=confusion{groundtruth{i}, predict{i}}+1;
I was curious to print output's and groundtruth's values to see what was happening
output{1}
ans =
2
groundtruth{1}
ans =
1
So, nothing seems to be wrong with the values, so what is wrong here? is the confusion matrix's indexing right in the code?
The error occurs in a for loop. Checking the first iteration of the loop is not sufficient in this case. Index exceeds matrix dimensions means there exists an i in the range of 1:size(output,1) for which either groundtruth{i} or output{i} is greater than 25.
You can find out which one has at least one element bigger than the range:
% 0 means no, there is none above 25. 1 means yes, there exists at least one:
hasoutlier = any(cellfun(#(x) x > 25, groundtruth)) % similar for 'output'
Or you can count them:
outliercount = sum(cellfun(#(x) x > 25, groundtruth))
Maybe you also want to find these elements:
outlierindex = find(cellfun(#(x) x > 25, groundtruth))
By the way, I am wondering why are you working with cell arrays in this case? Why not numeric arrays?
I have a question in algorithm design about arrays, which should be implement in C language.
Suppose that we have an array which has n elements. For simplicity n is power of '2' like 1, 2, 4, 8, 16 , etc. I want to separate this to 2 parts with (n/2) elements. Condition of separating is lowest absolute difference between sum of all elements in two arrays for example if I have this array (9,2,5,3,6,1,4,7) it will be separate to these arrays (9,5,1,3) and (6,7,4,2) . summation of first array's elements is 18 and the summation of second array's elements is 19 and the difference is 1 and these two arrays are the answer but two arrays like (9,5,4,2) and (7,6,3,1) isn't the answer because the difference of element summation is 4 and we have found 1 . so 4 isn't the minimum difference. How to solve this?
Thank you.
This is the Partition Problem, which is unfortunately NP-Hard.
However, since your numbers are integers, if they are relatively low, there is a pseudo polynomial O(W*n^2) solution using Dynamic Programming (where W is sum of all elements).
The idea is to create the DP matrix of size (W/2+1)*(n+1)*(n/2+1), based on the following recursive formula:
D(0,i,0) = true
D(0,i,k) = false k != 0
D(x,i,k) = false x < 0
D(x,0,k) = false x > 0
D(x,i,0) = false x > 0
D(x,i,k) = D(x,i-1,k) OR D(x-arr[i], i-1,k-1)
The above gives a 3d matrix, where each entry D(x,i,k) says if there is a subset containing exactly k elements, that sums to x, and uses the first i elements as candidates.
Once you have this matrix, you just need to find the highest x (that is smaller than SUM/2) such that D(x,n,n/2) = true
Later, you can get the relevant subset by going back on the table and "retracing" your choices at each step. This thread deals with how it is done on a very similar problem.
For small sets, there is also the alternative of a naive brute force solution, which basically splits the array to all possible halves ((2n)!/(n!*n!) of those), and picks the best one out of them.