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Check if a value is within a range of numbers
...e. You don't need "multiple if" statements to do it, either:
if (x >= 0.001 && x <= 0.009) {
// something
}
You could write yourself a "between()" function:
function between(x, min, max) {
return x >= min && x <= max;
}
// ...
if (between(x, 0.001, 0.009)) {
//...
Python: print a generator expression?
...ator expression is a "naked" for expression. Like so:
x*x for x in range(10)
Now, you can't stick that on a line by itself, you'll get a syntax error. But you can put parenthesis around it.
>>> (x*x for x in range(10))
<generator object <genexpr> at 0xb7485464>
This is som...
Returning first x items from array
...ray_slice returns a slice of an array
$sliced_array = array_slice($array, 0, 5)
is the code you want in your case to return the first five elements
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System.IO.Packaging
I have my project set to .NET Framework 4.0. When I add System.IO.Packaging , it says that it doesn't exist. It also doesn't show up when I try to add it as a reference to the project.
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awk without printing newline
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answered Jan 7 '10 at 16:56
CodeRainCodeRain
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Standard deviation of a list
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Since Python 3.4 / PEP450 there is a statistics module in the standard library, which has a method stdev for calculating the standard deviation of iterables like yours:
>>> A_rank = [0.8, 0.4, 1.2, 3.7, 2.6, 5.8]
>>> import statis...
Using [UIColor colorWithRed:green:blue:alpha:] doesn't work with UITableView seperatorColor?
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You need to divide by 255.0
Because I hardly ever use values between 1.0 and 0.0, I created a very simple UIColor category that does the messy looking division by itself: (from http://github.com/Jon889/JPGeneral)
//.h file
@interface UIColor (JPExtr...
Why am I seeing “TypeError: string indices must be integers”?
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answered May 20 '11 at 21:16
TamásTamás
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Find the most frequent number in a numpy vector
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edited Aug 5 at 18:08
B. Willems
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answered Jun 6 '11 at 13:01
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Is there a NumPy function to return the first index of something in an array?
...mensions and it contained your item at two locations then
array[itemindex[0][0]][itemindex[1][0]]
would be equal to your item and so would
array[itemindex[0][1]][itemindex[1][1]]
numpy.where
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