As has been pointed out already, you’re not looking for the derivative. You’re really looking for a “significant change” detection algorithm for a time series. You’ll certainly want a smoothing filter (and the moving average filter is fine — see Bjorn’s answer for this part). But in addition to the smoothing filter, you will also … Read more
Drepper’s What Every Programmer Should Know About Memory [pdf] is a good reference to one aspect of low-level optimisation.
Big Letter: chr(ord(‘@’)+number) 1 -> A 2 -> B … Small Letter: chr(ord(‘`’)+number) 1 -> a 2 -> b …
All numbers in JavaScript are 64-bit floating point numbers. Ref: http://www.hunlock.com/blogs/The_Complete_Javascript_Number_Reference http://www.crockford.com/javascript/survey.html
Something like this: int quick_pow10(int n) { static int pow10[10] = { 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000 }; return pow10[n]; } Obviously, can do the same thing for long long. This should be several times faster than any competing method. However, it is quite limited if you have lots of … Read more
Try this: ORDER BY CAST(thecolumn AS int)
While the sum of squares algorithm works fine most of the time, it can cause big trouble if you are dealing with very large numbers. You basically may end up with a negative variance… Plus, don’t never, ever, ever, compute a^2 as pow(a,2), a * a is almost certainly faster. By far the best way … Read more
The fractions module can do that >>> from fractions import Fraction >>> Fraction(98, 42) Fraction(7, 3) There’s a recipe over here for a numpy gcd. Which you could then use to divide your fraction >>> def numpy_gcd(a, b): … a, b = np.broadcast_arrays(a, b) … a = a.copy() … b = b.copy() … pos = … Read more
In python 2.6 it is portable if the CPU supports it The float() function will now turn the string nan into an IEEE 754 Not A Number value, and +inf and -inf into positive or negative infinity. This works on any platform with IEEE 754 semantics.
For a given IEEE-754 floating point number X, if 2^E <= abs(X) < 2^(E+1) then the distance from X to the next largest representable floating point number (epsilon) is: epsilon = 2^(E-52) % For a 64-bit float (double precision) epsilon = 2^(E-23) % For a 32-bit float (single precision) epsilon = 2^(E-10) % For a … Read more