The simplest and best solution is to use torch.sum(). To sum all elements of a tensor: torch.sum(x) # gives back a scalar To sum over all rows (i.e. for each column): torch.sum(x, dim=0) # size = [ncol] To sum over all columns (i.e. for each row): torch.sum(x, dim=1) # size = [nrow] It should be … Read more
select sum(A),sum(B),sum(C) from mytable where id in (1,2,3);
You can use AVG like so: SELECT AVG(P1_Score)
I suggest something like: find . -regex ‘.*\.bak’ -print0 | du –files0-from=- -ch | tail -1 Some notes: The -print0 option for find and –files0-from for du are there to avoid issues with whitespace in file names The regular expression is matched against the whole path, e.g. ./dir1/subdir2/file.bak, not just file.bak, so if you modify … Read more
You could use the Array.reduce method: const myData = [ [‘2013-01-22’, 0], [‘2013-01-29’, 0], [‘2013-02-05’, 0], [‘2013-02-12’, 0], [‘2013-02-19’, 0], [‘2013-02-26’, 0], [‘2013-03-05’, 0], [‘2013-03-12’, 0], [‘2013-03-19’, 0], [‘2013-03-26’, 0], [‘2013-04-02’, 21], [‘2013-04-09’, 2] ]; const sum = myData .map( v => v[1] ) .reduce( (sum, current) => sum + current, 0 ); console.log(sum); See … Read more
The expression (sum(row) for row in M) creates what’s called a generator. This generator will evaluate the expression (sum(row)) once for each row in M. However, the generator doesn’t do anything yet, we’ve just set it up. The statement next(G) actually runs the generator on M. So, if you run next(G) once, you’ll get the … Read more
Two ways: Use indexes: double sum = 0; for(int i = 0; i < m.size(); i++) sum += m.get(i); return sum; Use the “for each” style: double sum = 0; for(Double d : m) sum += d; return sum;
Short Answer: Just generate N random numbers, compute their sum, divide each one by the sum and multiply by M. Longer Answer: The above solution does not yield a uniform distribution which might be an issue depending on what these random numbers are used for. Another method proposed by Matti Virkkunen: Generate N-1 random numbers … Read more
You can use filter function >>> sum(filter(None, [1,2,3,None])) 6 Updated from comments Typically filter usage is filter(func, iterable), but passing None as first argument is a special case, described in Python docs. Quoting: If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.
array.map(&:cash).inject(0, &:+) or array.inject(0){|sum,e| sum + e.cash }