mittenchops’s gists

mittenchops / inkscapeSVG2PNGafolder.sh

Created August 7, 2013 22:01

For each SVG in a directory, makes a faithful PNG copy.

for i in *.svg; do inkscape -f "$i" -e "$i.png"; done

mittenchops / addweeks.py

Created August 5, 2013 18:54

Add weeks to time in MMM DDth YYYY format.

	from dateutil import parser
	from datetime import timedelta

	date = "Mar 11th 2013"

	def addweeks(date,week=0):
	x = parser.parse(date)
	new = x + timedelta(weeks=week)
	return(new.strftime("%b %d %Y"))

mittenchops / nonemptykeypct.py

Created July 30, 2013 20:45

Count percentage of keys in a weirdly nested json document. Very useful for working with mongodb. Works like variety.js but at the item-level.

	from __future__ import division
	from functools import partial

	def nonemptykeypct(x):
	"""
	Returns the percentage of keys that have non-null values in an object (as decimal).

	Usage:
	>>> x = {'a': 'meow', 'b': {'c': 'asd'}, 'd': [{'e': 'stuff', 'f': 1}, {'e': 'more stuff', 'f': 2}], 'g': ''}
	>>> nonemptykeypct(x)

mittenchops / listNkeys.py

Created July 30, 2013 20:41

This lets you get all the keys in a weirdly nested json document. Very useful for working with mongodb. Works like variety.js but at the item-level.

	def listNkeys(obj, prefix=''):
	"""
	This lets you list ALL the keys of large nested dictionaries a la mongodb documents.
	from: http://stackoverflow.com/questions/17952981/return-a-list-of-all-variable-names-in-a-python-nested-dict-json-document-in-dot

	Usage:
	>>> x = {
	'a': 'meow',
	'b': {
	'c': 'asd'

mittenchops / unique.py

Created July 23, 2013 18:42

Return a list of only unique elements.

	def unique(x):
	return(list(set(x)))

mittenchops / tagger.py

Last active December 20, 2015 03:19

Regexy thing for matching tags out of strings. You define a thing, ARTS, which is just a list of what you want to match, then follow syntax below to return either a dict of counts or a list of things with nonzero counts. I'm actually pretty pleased with the performance of this that it keeps working fairly quickly even as the string size approach…

	import re

	ARTS = ['dance','photography','art therapy']

	def string_found(string1, string2):
	if re.search(r"\b" + re.escape(string1) + r"\b", string2):
	return 1
	return 0

	def string_count(string1, string2):

mittenchops / dontrepeatpubkey.sh

Created July 22, 2013 18:24

How to not have to repeat your pubkey in unix. This weakens your security, so only use it on throwaway keys. You're generally better off just typing your password again.

	$ eval "$(ssh-agent)"
	$ ssh-add ~/.ssh/id_rsa

mittenchops / findbytheword.sh

Created July 17, 2013 20:09

Find all files that contain a particular word, recursively in the current directory. This is the version that doesn't use xargs, which I think would look more like this: http://www.cyberciti.biz/faq/howto-recursively-search-all-files-for-words/

find . -type f -exec grep -l "wordofinterest" {} +

mittenchops / multiprocess.py

Last active December 19, 2015 18:39

How to use multiprocessing

	from multiprocessing import Pool
	from functools import partial
	x = [1,2,3]
	y = 10

	# NOTE, f CANNOT be a lambda function: http://stackoverflow.com/questions/17657571/python-multiprocessing-map-function-error
	def f(x,y):
	return(x**2+y)

	# ordinary map works:

mittenchops / flattendict.py

Created July 12, 2013 14:03

flatten a dictionary of lists

	# http://feldboris.alwaysdata.net/blog/python-trick-how-to-flatten-dictionaries-values-composed-of-iterables.html

	from itertools import chain
	def flatten_dict_values(dictionary):
	return chain(*dictionary.values())

adam mittenchops