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What's the difference between “groups” and “captures” in .NET regular expressions?
...istory tracker. When the regex makes his match, it goes through the string from left to right (ignoring backtracking for a moment) and when it encounters a matching capturing parentheses, it will store that in $x (x being any digit), let's say $1.
Normal regex engines, when the capturing parenthes...
Check if at least two out of three booleans are true
...e a&&b || b&&c || a&&c version wins then.
Results from Tofubeer with the latest code running OS X:
a&&b || b&&c || a&&c : 1358 ms
a ? b||c : b&&c : 1187 ms
a&b | b&c | c&a : 410 ms
a + b + c >= 2 : 602 ms
...
What's the strangest corner case you've seen in C# or .NET? [closed]
...t if you simply round up, you will end up with potentially huge difference from the sum of the non-rounded numbers. Very bad if you are doing financial calculations!
– Tsvetomir Tsonev
Oct 12 '08 at 10:05
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How does zip(*[iter(s)]*n) work in Python?
...re you're passing the same iterator 3 times to zip(), and it pulls an item from the iterator each time.
x = iter([1,2,3,4,5,6,7,8,9])
print zip(x, x, x)
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How to ISO 8601 format a Date with Timezone Offset in JavaScript?
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The sign indicates the offset of the local time from GMT
– Steven Moseley
Jul 2 '13 at 0:50
1
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Pairs from single list
...
My favorite way to do it:
from itertools import izip
def pairwise(t):
it = iter(t)
return izip(it,it)
# for "pairs" of any length
def chunkwise(t, size=2):
it = iter(t)
return izip(*[it]*size)
When you want to pair all elements you...
Mac OSX Lion DNS lookup order [closed]
...cal webserver running on 127.0.0.1:80 and your browser receives a response from the webserver (error or otherwise), no AAAA query is issued, as it seems to be satisfied that a TCP connection was at least possible.
On a related note, if you make heavy use of the hosts file (for adblocking, local w...
Python: fastest way to create a list of n lists
...ly way which is marginally faster than
d = [[] for x in xrange(n)]
is
from itertools import repeat
d = [[] for i in repeat(None, n)]
It does not have to create a new int object in every iteration and is about 15 % faster on my machine.
Edit: Using NumPy, you can avoid the Python loop using
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Erasing elements from a vector
I want to clear a element from a vector using the erase method. But the problem here is that the element is not guaranteed to occur only once in the vector. It may be present multiple times and I need to clear all of them. My code is something like this:
...
Rebasing a branch including all its children
...ased on top of master. Also isn't C^ not the same as B? so we are rebasing from B(excluding?) to each branch containing C on top of ... B. Wouldn't the result be exactly the same as before?
– Marenz
Jan 27 '14 at 15:57
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