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How to check if a number is a power of 2
...his problem:
bool IsPowerOfTwo(ulong x)
{
return (x & (x - 1)) == 0;
}
Note, this function will report true for 0, which is not a power of 2. If you want to exclude that, here's how:
bool IsPowerOfTwo(ulong x)
{
return (x != 0) && ((x & (x - 1)) == 0);
}
Explanation
Fi...
Bin size in Matplotlib (Histogram)
... boundaries. They can be unequally distributed, too:
plt.hist(data, bins=[0, 10, 20, 30, 40, 50, 100])
If you just want them equally distributed, you can simply use range:
plt.hist(data, bins=range(min(data), max(data) + binwidth, binwidth))
Added to original answer
The above line works for...
Ruby on Rails and Rake problems: uninitialized constant Rake::DSL
...
A tweet from DHH earlier. Rake .9.0 breaks Rails and several other things, you need to:
gem "rake", "0.8.7"
in your Gemfile.
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How to create the most compact mapping n → isprime(n) up to a limit N?
...ent every odd number with one bit e.g. for the given range of numbers (1, 10], starts at 3: 1110
31 Answers
...
CSS: bolding some text without changing its container's size
...h a little compromise, I used text-shadow instead.
li:hover {text-shadow:0px 0px 1px black;}
Here's a working example:
body {
font-family: segoe ui;
}
ul li {
display: inline-block;
border-left: 1px solid silver;
padding: 5px
}
.textshadow :hover {
text-shadow: 0px 0p...
UITableView - scroll to the top
...e top. But I cannot guarantee that the first object is going to be section 0, row 0. May be that my table view will start from section number 5.
...
HSL to RGB color conversion
...
308
Garry Tan posted a Javascript solution on his blog (which he attributes to a now defunct mjijac...
Permutations in JavaScript?
...
108
If you notice, the code actually splits the chars into an array prior to do any permutation, so...
Determine the data types of a data frame's columns
...
220
Your best bet to start is to use ?str(). To explore some examples, let's make some data:
set...
Sorting arrays in NumPy by column
... copy:
In [1]: import numpy as np
In [2]: a = np.array([[1,2,3],[4,5,6],[0,0,1]])
In [3]: np.sort(a.view('i8,i8,i8'), order=['f1'], axis=0).view(np.int)
Out[3]:
array([[0, 0, 1],
[1, 2, 3],
[4, 5, 6]])
To sort it in-place:
In [6]: a.view('i8,i8,i8').sort(order=['f1'], axis=0) #&...