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Regular expression that matches valid IPv6 addresses
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30 Answers
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What is the easiest way to remove the first character from a string?
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I kind of favor using something like:
asdf = "[12,23,987,43"
asdf[0] = ''
p asdf
# >> "12,23,987,43"
I'm always looking for the fastest and most readable way of doing things:
require 'benchmark'
N = 1_000_000
puts RUBY_VERSION
STR = "[12,23,987,43"
Benchmark.bm(7) do |b|
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Validating IPv4 addresses with regexp
...tion, but without much luck. It seemed at one point I had had it with (25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(\.|$)){4} , but it produces some strange results:
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How to find list intersection?
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answered Sep 13 '10 at 1:32
Mark ByersMark Byers
683k155155 gold badges14681468 silver badges13881388 bronze badges
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How to increase the gap between text and underlining in CSS
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370
No, but you could go with something like border-bottom: 1px solid #000 and padding-bottom: 3px.
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how does multiplication differ for NumPy Matrix vs Array classes?
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Neil G
27.5k3030 gold badges127127 silver badges228228 bronze badges
answered Oct 8 '10 at 16:49
Joe KingtonJoe Ki...
How to profile a bash shell script slow startup?
... you'd like to begin a trace in any Bash script):
PS4='+ $(date "+%s.%N")\011 '
exec 3>&2 2>/tmp/bashstart.$$.log
set -x
add
set +x
exec 2>&3 3>&-
at the end of ~/.bashrc (or at the end of the section of any Bash script you'd like tracing to stop). The \011 is an octal ...
Why are these numbers not equal?
...cially true because some values which are simple, finite decimals (such as 0.1 and 0.05) are not represented exactly in the computer and so the results of arithmetic on them may not give a result that is identical to a direct representation of the "known" answer.
This is a well known limitation of ...
Regular expression for matching HH:MM time format
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Your original regular expression has flaws: it wouldn't match 04:00 for example.
This may work better:
^([0-1]?[0-9]|2[0-3]):[0-5][0-9]$
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Apply multiple functions to multiple groupby columns
...f = pd.DataFrame(np.random.rand(4,4), columns=list('abcd'))
df['group'] = [0, 0, 1, 1]
df
a b c d group
0 0.418500 0.030955 0.874869 0.145641 0
1 0.446069 0.901153 0.095052 0.487040 0
2 0.843026 0.936169 0.926090 0.041722 1
3 0.635846 ...