spring.cloud.gcp.pubsub.subscriber.executor-threads
spring.cloud.gcp.pubsub.subscriber.parallel-pull-count
I understand that the concurrency = executor-threads * parallel-pull-count. What I'm asking is how different combination of them will be better than the other.
Suppose I need to have concurrency of 6, each message can be processed in around 5 seconds. Which combination will be best?
parallel-pull-count = 1 and executor-threads = 6 OR
parallel-pull-count = 6 and executor-threads = 1 OR
parallel-pull-count = 2 and executor-threads = 3 OR
parallel-pull-count = 3 and executor-threads = 2
Related
I am trying to create two vectors
first:
[ 10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9 10 ]
second:
[ 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1 ]
I had almost no problems with the first one:
i = 1:10;
for t = 1: 2*length(i)-1
y1(t) = abs(length(x)-t)+1;
end
But there are some problems with the second...
Does anyone have any idea how I can create it using the same for loop?Thanks in advance
If you want to do it with a loop:
N = 10;
for t = 1:2*N-1
y2(t) = -abs(t-N)+N;
end
But its probably easier to use the following, first make an array 1:N, and then concatenate the array N-1:-1:1:
y2 = [1:N, N-1:-1:1]
I have a data set of size 11490x1. the data is recorded every 0.25 second(i.e. 4hz). So, 1 second accounts for 4 data points. The goal here is to further create sub sets every 3 seconds, meaning that I want to look at data every 3 seconds and analyze it. for example: if I had data such as [1 2 3 4 5 6 8 2 4 2 4 3 2 4 2 5 2 5 24 2 5 1 5 1], I want to have a sub set [1 2 3 4 5 6 8 2 4 2 4 3 ] and so on...
Any help would be appreciate.
It really depends on how you plan to "analyse" your data. The simplest way is to use a loop:
n = 4*3;
breaks = 0:n:numel(data)
for i = 1:numel(breaks)-1
sub = data(breaks(i)+1:breaks(i+1));
%// do analysis
%// OR sub{i} = data(breaks(i)+1:breaks(i+1));
end
A vectorized approach might use reshape(data,[],12) after padding data so that mod(numel(data),12)==0
A third way might be to break your matrix up into a cell array using mat2cell or in a for loop like above but instead of sub=... rather use sub{i}=...
I have a matrix
[1 2
3 6
7 1
2 1]
and would like to remove mirror imaged pairs..i.e. output would be either:
[1 2
3 6
7 1]
or
[3 6
7 1
2 1]
Is there a simple way to do this? I can imagine a complicated for loop, something like (or a version which wouldn't delete the original pair..only the duplicates):
for i=1:y
var1=(i,1);
var2=(i,2);
for i=1:y
if array(i,1)==var1 && array(i,2)==var2 | array(i,1)==var2 && array(i,2)==var1
array(i,1:2)=[];
end
end
end
thanks
How's this for simplicity -
A(~any(tril(squeeze(all(bsxfun(#eq,A,permute(fliplr(A),[3 2 1])),2))),2),:)
Playing code-golf? Well, here we go -
A(~any(tril(pdist2(A,fliplr(A))==0),2),:)
If dealing with two column matrices only, here's a simpler version of bsxfun -
M = bsxfun(#eq,A(:,1).',A(:,2)); %//'
out = A(~any(tril(M & M.'),2),:)
Sample run -
A =
1 2
3 6
7 1
6 5
6 3
2 1
3 4
>> A(~any(tril(squeeze(all(bsxfun(#eq,A,permute(fliplr(A),[3 2 1])),2))),2),:)
ans =
1 2
3 6
7 1
6 5
3 4
>> A(~any(tril(pdist2(A,fliplr(A))==0),2),:)
ans =
1 2
3 6
7 1
6 5
3 4
Here a not so fancy, but hopefully understandable and easy way.
% Example matrix
m = [1 2; 3 6 ; 7 1; 2 1; 0 3 ; 3 0];
Comparing m with its flipped version, the function ismember returns mirror_idx, a 1D-vector with each row containing the index of the mirror-row, or 0 if there's none.
[~, mirror_idx] = ismember(m,fliplr(m),'rows');
Go through the indices of the mirror-rows. If you find one "mirrored" row (mirror_idx > 0), set its counter-part to "not mirrored".
for ii = 1:length(mirror_idx)
if (mirror_idx(ii) > 0)
mirror_idx(mirror_idx(ii)) = 0;
end
end
Take only the rows that are marked as not having a mirror.
m_new = m(~mirror_idx,:);
Greetings
I have two vectors a and b as an example:
a = [1 2 3 4];
b = [5 6 7 8];
I want to create strings from the indices of a and b:
c1 = a(1):b(1) = [1 2 3 4 5];
c2 = a(2):b(2) = [2 3 4 5 6];
c3 = a(3):b(3) = [3 4 5 6 7];
c4 = a(4):b(4) = [4 5 6 7 8];
Then I want to concatenate the obtained strings:
C = cat(2, c1, c2, c3, c4) = [1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8];
I would like a general solution to help me automatize this algorithm.
Solution
This should do the trick without using a loop:
>> a = [1 3 4 5];
>> b = [5 6 7 8];
>> resultstr = num2str(cell2mat(arrayfun(#(x,y) x:y,a,b,'UniformOutput',false)))
resultstr =
1 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8
Performance
I tried to do a quick comparison between this method, Luis Mendo's method and the for loop (see e.g. A. Visser's answer). I created two arrays with pseudo-random numbers from 1 to 50 and 501 to 1000 and timed the calculation for array sizes from 1 to 300, disregarding the string conversion.
Luis Mendo's anwer is the clear winner, in terms of time complexity arrayfun seems to be on par with bsxfun. The for loop fares much worse, except for very small array sizes, but I'm not sure the timing can be trusted there.
The code can be found here. I'd be very happy to get some feedback, I'm a bit unsure about those measurements.
This can be done without loops (be they for, arrayfun or cellfun) using bsxfun's masking capability. This works also if the "strings" have different lengths, as in the following example.
a = [1 2 3];
b = [3 5 6];
m = (0:max(b-a)).'; %'
C = bsxfun(#plus, a, m);
mask = bsxfun(#le, m, b-a);
C = C(mask).';
The result in this example is:
C =
1 2 3 2 3 4 5 3 4 5 6
Try something like:
A = [1:5; 5:8];, C = [];
for i = A, C = [C i(1):i(2)]; end
Cstr = num2str(C);
I would do this:
a = [1 3 4 5]; b = [5 6 7 8];
c =[];
for i =1:length(a)
c = [c [a(i):b(i)]];
end
num2str(c)
First of all, don't assign variables as C1, C2, C3 etc, use cells for that.
I'd use a for-loop for this, though there probably is a better alternative.
a = [1 3 4 5]; b = [5 6 7 8];
C = [];
for ii = 1:length(a)
tmp = a(ii):b(ii);
C = [C tmp];
end
This way tmp stores your separate arrays, then the following line adds it to the pre-existing array C.
Just for fun:
a = [1 2 3 4]; b = [5 6 7 8];
A=fliplr(gallery('circul',fliplr([a,b])));
B=A(1:numel(a)+1,1:numel(a));
C=num2str(B(:).')
This question already has answers here:
A similar function to R's rep in Matlab [duplicate]
(4 answers)
Closed 7 years ago.
A = [1 4 5 2 1 2]
How could I concisely replicate each element n times whilst maintaining the overall order e.g. if n = 3, the desired result would be:
[1 1 1 4 4 4 5 5 5 2 2 2 1 1 1 2 2 2]
For Matlab R2015a or higher use repelem
n = 3;
u = repelem(A,n)
For older versions use bsxfun
n = 3;
u = bsxfun(#mtimes ,A(:).',ones(n,1))
u = u(:)
You could do the following:
reshape(repmat(a',[3 1]),[],1)