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Given a very long string, replace chars with numbers and obtain cumulative sum vector

Asked by Paolo Binetti on 17 Dec 2016
Latest activity Commented on by Image Analyst
on 17 Dec 2016
As input I have a vector of a few million char, which can be 'A', 'C', 'G', 'T'. The vector is called sequence.
As output I want a vector with the cumulative sums of the elements of the input vector, after I have converted its chars in numbers. This code works, but takes forever. What might be the problem?
n = numel(sequence);
skew = zeros(1,n+1,'int32');
for i = 1:n
switch sequence(i)
case 'C'
skew(i+1) = skew(i)-1;
case 'G'
skew(i+1) = skew(i)+1;
otherwise
skew(i+1) = skew(i);
end
end
By the way, I am not interested in the cumulative sums vectors as is. The only thing I care is the index of its minimum.

  2 Comments

Cumsum crashing due to memory issues, on ONLY a few million integers? Please show us what you did, as that is virtually impossible. A loop on only a few million terms taking forever? Again, any reasonable loop of that size will NOT take forever. So you need to show what you are doing. A few million is simply not a very long vector.
v = int32(floor(rand(1,1e7)*3)) - 1;
V = cumsum(v);
whos
Name Size Bytes Class Attributes
V 1x10000000 40000000 int32
v 1x10000000 40000000 int32
So 10 million elements. 40 megabytes per vector? I tried to watch the memory required, but it happened too fast to see a blip.
You are right! I made a mistake, the problem was not cumsum, but another one which is not covered by this question. I will edit this question to deal only with the for-loop, and then create a new question on the memory issue.

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3 Answers

Answer by Image Analyst
on 17 Dec 2016
 Accepted Answer

Then you're doing something wrong. Are you preallocating your cdf array? It should be only a fraction of a second. For example:
numPoints = 20000000; % Twenty million elements
m = randi(-1,1, 1, numPoints);
tic;
cdf = zeros(1, numPoints);
cdf(1) = m(1);
for k = 2 : length(m)
cdf(k) = cdf(k-1) + m(k);
end
toc;
The elapsed time on my computer for twenty million elements is 0.21 seconds. How long is it taking on your computer?

  3 Comments

Thank you. I copy-pasted your code, just edited the randi function because I am using Octave on this computer. It takes 4 seconds with numPoints = 200000 ! Of course I did not even try going to 20 millions.
So now I have no idea of what could be wrong, maybe I have the wrong version of Octave. Maybe I should try the same on Matlab.
Well, two thumbs up for MATLAB and two thumbs down for Octave. The Mathworks spent a lot of time about 5 years ago or so to really speed up for loops quite a bit. That probably was not done with open source Octave. But there are still people who hold on to the believe that for loops are horribly slow and should be avoided at all costs. While they may still be slower than vectorized solutions, the belief that they're really slow is now a myth (except for in Octave I guess).
So Octave might be the problem ... I will check this out and try also on Matlab, of which I only have an old version however.

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Answer by Star Strider
on 17 Dec 2016

If your memory is that limited, this may be a work-around:
V = int32(randi([-1 1], 1, 1000)); % Create Data
Vr = reshape(V, 100, []); % Create Matrix (Can Be Assigned As ‘V’, Separate Here To Check Code)
vctmin = Inf; % Initialise ‘vctmin’
endsum = 0; % Initialise ‘endsum’
for k1 = 1:size(Vr,2)
colsum = cumsum(Vr(:,k1))+endsum; % Calculate Column ‘cumsum’
endsum = colsum(end); % End Value
[colmin,idx] = min(colsum);
if (colmin < vctmin) % Test Minima & Replace
vctmin = colmin; % New Vector Minimum
minidx = idx + (k1 - 1)*size(Vr,1); % New Minimum Index
end
end
vctmin
minidx
[testmin,testidx] = min(cumsum(V))
It creates a matrix from your vector. (I gave them different names here so I could check the code, but assigning the reshaped vector to ‘V’ instead of ‘Vr’ would not use any additional memory.) The code then does the cumulative sum on each column of the matrix, stores the minimum and its index, and proceeds through the matrix. The advantage is that only one column of the matrix at a time is in your workspace, so memory should not be an issue.
When I checked the matrix approach with directly calculating the minimum and its index, the results were the same over several test runs.

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Answer by Star Strider
on 17 Dec 2016

With respect to your creating ‘skew’, this may be more efficient:
bases = {'A','C','T','G'}; % Cell Array
sequence = bases(randi(4, 1, 20));
skew = zeros(1, length(sequence)+1,'int32');
Cix = find(ismember(sequence, 'C'));
Gix = find(ismember(sequence, 'G'));
skew(Cix+1) = -1;
skew(Gix+1) = +1;
I don’t know what ‘sequence’ is, so I created it as a cell array here.

  3 Comments

Thanks, string is actually a char array generated like this:
sequence = fileread('source.txt');
Can I adapt your code to work with a char array?
Yes! It works the same way, producing the same (correct) result:
bases = ['A','C','T','G']; % Character Array
sequence = bases(randi(4, 1, 20));
skew = zeros(1, length(sequence)+1,'int32');
Cix = find(ismember(sequence, 'C'));
Gix = find(ismember(sequence, 'G'));
skew(Cix+1) = -1;
skew(Gix+1) = +1;
If you're going to ask another question, then please also attach 'source.txt'.

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