kgriffs

Exploring the cost of defining a noop function inline, vs. calling a free noop function:

In [215]: %timeit test_inline()
1000000 loops, best of 3: 261 ns per loop

In [216]: %timeit test_external()
1000000 loops, best of 3: 215 ns per loop

In [217]: %timeit y.test_inline()

1000000 loops, best of 3: 306 ns per loop

kgriffs

In [56]: %timeit datetime.datetime.utcnow()
1000000 loops, best of 3: 875 ns per loop

In [57]: %timeit time.time()
10000000 loops, best of 3: 110 ns per loop

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[global]
timeout = 10
index-url = https://pypi.python.org/simple/

# This doesn't work for a few packages for some reason, but mostly works and is a little faster
;index-url = https://simple.crate.io/

# Try the standard pypi mirrors if the above times out
use-mirrors = True

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{
    "cmd": ["tox", "-e", "py27,pep8"],
    "working_dir": "${file_path}",
    "env": {
        "PATH": "/usr/local/bin:/usr/local/sbin:/usr/local/share/python:$PATH"
    },
    "file_regex": "^[ ]*File \"(...*?)\", line ([0-9]*)",
    "selector": "source.python"
}

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#!/usr/bin/env bash

/Applications/VLC.app/Contents/MacOS/VLC -I rc "$@"

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class Catalog(object):
    """Represents the mapping between queues and shard drivers."""

    __slots__ = ['_shards']

    _singleton = None

    def __init__(self):
        self._shards = {}

kgriffs

       / .'
 .---. \/
(._.' \()
 ^"""^"

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# Examples demonstrating the rules for combining URLs

In [63]: urlparse.urljoin("https://marconi.example.com/v1", "/queues/foo")
Out[63]: 'https://marconi.example.com/queues/foo'

In [58]: urlparse.urljoin("https://marconi.example.com/v1", "queues/foo")
Out[58]: 'https://marconi.example.com/queues/foo'

In [59]: urlparse.urljoin("https://marconi.example.com/v1/", "queues/foo")
Out[59]: 'https://marconi.example.com/v1/queues/foo'

kgriffs

import six

class FooError(Exception):

    message = u'An unknown exception occurred.'

    def __str__(self):
        if six.PY3:
            return self.message

kgriffs

import time

def check(points):
    indexes = []
    prev = points[0]
    for p, i in enumerate(points[1:]):
        if p == prev:
            indexes.append(i)
        prev = p
    return indexes