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How do I calculate a point on a circle’s circumference?
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The parametric equation for a circle is
x = cx + r * cos(a)
y = cy + r * sin(a)
Where r is the radius, cx,cy the origin, and a the angle.
That's pretty easy to adapt into any language with basic trig functions. Note that most languages will use radians for the a...
Difference between Inheritance and Composition
...aving an instance of another class C as a field of your class, instead of extending C. A good example where composition would've been a lot better than inheritance is java.util.Stack, which currently extends java.util.Vector. This is now considered a blunder. A stack "is-NOT-a" vector; you should no...
Hidden Features of JavaScript? [closed]
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Side-by-side plots with ggplot2
...de-by-side (or n plots on a grid)
The function grid.arrange() in the gridExtra package will combine multiple plots; this is how you put two side by side.
require(gridExtra)
plot1 <- qplot(1)
plot2 <- qplot(1)
grid.arrange(plot1, plot2, ncol=2)
This is useful when the two plots are not bas...
do { … } while (0) — what is it good for? [duplicate]
I've been seeing that expression for over 10 years now. I've been trying to think what it's good for. Since I see it mostly in #defines, I assume it's good for inner scope variable declaration and for using breaks (instead of gotos.)
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How to convert JSON data into a Python object
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# Parse JSON into an object with attributes corresponding to dict keys.
x = json.loads(data, object_hook=lambda d: SimpleNamespace(**d))
print(x.name, x.hometown.name, x.hometown.id)
OLD ANSWER (Python2)
In Python2, you can do it in one line, using namedtuple and object_hook (but it's very slow ...
How can I count the occurrences of a list item?
...you want to count all items, or even just multiple items, use Counter, as explained in the other answers.
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Compare two DataFrames and output their differences side-by-side
I am trying to highlight exactly what changed between two dataframes.
14 Answers
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Rounding up to next power of 2
I want to write a function that returns the nearest next power of 2 number. For example if my input is 789, the output should be 1024. Is there any way of achieving this without using any loops but just using some bitwise operators?
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Pythonic way to check if a list is sorted or not
...haw is looking for. Here is the one liner:
all(l[i] <= l[i+1] for i in xrange(len(l)-1))
For Python 3:
all(l[i] <= l[i+1] for i in range(len(l)-1))
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