Source: my own notes taken during Standford’s Machine Learning course in Coursera, by Andrew Ng. All credits to him and this organization. The course is freely available for anybody to be taken at their own pace. The images are made by myself using LaTeX (formulas) and R (graphics). Hypothesis function Logistic regression is used when … Read more
It looks like Java actually has a log10 function: Math.log10(x) Otherwise, just use math: Math.log(x) / Math.log(10) http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html
A very simple solution is to use a decent table-driven approximation. You don’t actually need a lot of data if you reduce your inputs correctly. exp(a)==exp(a/2)*exp(a/2), which means you really only need to calculate exp(x) for 1 < x < 2. Over that range, a runga-kutta approximation would give reasonable results with ~16 entries IIRC. … Read more
In case you don’t want to return a BigInt, then the following might work for you too: function log10(bigint) { if (bigint < 0) return NaN; const s = bigint.toString(10); return s.length + Math.log10(“0.” + s.substring(0, 15)) } function log(bigint) { return log10(bigint) * Math.log(10); } function natlog(bigint) { if (bigint < 0) return NaN; … Read more
This really depends on the domain of what values you want to compute a logarithm of. For IEEE doubles, many processors can take logarithms in a single assembly instruction; x86 has the FYL2X and FYL2XP1 instructions, for example. Although typically instructions like these will only take the logarithm in some fixed base, they can be … Read more
logsumexp works by evaluating the right-hand side of the equation log(∑ exp[a]) = max(a) + log(∑ exp[a – max(a)]) I.e., it pulls out the max before starting to sum, to prevent overflow in exp. The same can be applied before doing vector dot products: log(exp[a] ⋅ exp[b]) = log(∑ exp[a] × exp[b]) = log(∑ exp[a … Read more
let matplotlib take the log for you: fig = plt.figure() ax = plt.gca() ax.scatter(data[‘o_value’] ,data[‘time_diff_day’] , c=”blue”, alpha=0.05, edgecolors=”none”) ax.set_yscale(‘log’) ax.set_xscale(‘log’) If you are using all the same size and color markers, it is faster to use plot fig = plt.figure() ax = plt.gca() ax.plot(data[‘o_value’] ,data[‘time_diff_day’], ‘o’, c=”blue”, alpha=0.05, markeredgecolor=”none”) ax.set_yscale(‘log’) ax.set_xscale(‘log’)
Your code is doing a log of a number that is less than or equal to zero. That’s mathematically undefined, so Python’s log function raises an exception. Here’s an example: >>> from math import log >>> log(-1) Traceback (most recent call last): File “<pyshell#59>”, line 1, in <module> log(-1) ValueError: math domain error Without knowing … Read more
I think it does not matter what is the base of the log as the relative complexity is the same irrespective of the base used. So you can think of it as O(log2X) = O(log10X) Also to mention that logarithms are related by some constant. So log₁₀(x) = log₂(x) / log₂(10) So most of the … Read more
Use this identity: logb(n) = loge(n) / loge(b) Where log can be a logarithm function in any base, n is the number and b is the base. For example, in Java this will find the base-2 logarithm of 256: Math.log(256) / Math.log(2) => 8.0 Math.log() uses base e, by the way. And there’s also Math.log10(), … Read more