Ben Felder pykong

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Tech priest at the Adeptus Mechanicus, Biochemist, Pythonista, MLOps Guru at day, AI M.Sc. Student at night and Builder of Joyful Tools.

39 followers · 30 following

Adeptus Mechanicus
Germany
19:10 (UTC +02:00)
https://resume.github.io/?pykong

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pykong / sync_with_main_repo.sh

Created November 2, 2017 18:32

	# list remotes
	git remote -v

	# add remote
	git remote add upstream https://github.com/SublimeLinter/SublimeLinter3.git

	# sync with main repo
	git pull upstream master

pykong / ssh_key.sh

Last active December 13, 2018 08:03

https://serverfault.com/questions/241588/how-to-automate-ssh-login-with-password https://stackoverflow.com/questions/112396/how-do-i-remove-the-passphrase-for-the-ssh-key-without-having-to-create-a-new-ke

	Don't use a password. Generate a passphraseless SSH key and push it to your VM.

	If you already have an SSH key, you can skip this step… Just hit Enter for the key and both passphrases:

	$ ssh-keygen -t rsa -b 2048
	Generating public/private rsa key pair.
	Enter file in which to save the key (/home/username/.ssh/id_rsa):
	Enter passphrase (empty for no passphrase):
	Enter same passphrase again:
	Your identification has been saved in /home/username/.ssh/id_rsa.

pykong / repl_last_char.py

Last active November 12, 2017 09:09

Also, note the elegant usage of an index to slice string. https://stackoverflow.com/questions/14496006/finding-last-occurrence-of-substring-in-string-replacing-that

	old_string = "this is going to have a full stop. some written sstuff!"
	k = old_string.rfind(".")
	new_string = old_string[:k] + ". - " + old_string[k+1:]

pykong / dict_pop.py

Last active November 12, 2017 15:44

Note that if the second argument, i.e. None is not given, KeyError is raised if the key is not in the dictionary. Providing the second argument prevents the conditional exception. This is unlike the mydict.get() method, which defaults to None. https://stackoverflow.com/questions/15411107/delete-a-dictionary-item-if-the-key-exists

mydict.pop("key", None)

pykong / infinite_loop.py

Last active November 13, 2017 19:40

https://stackoverflow.com/questions/23416381/circular-list-iterator-in-python

	from itertools import cycle

	lst = ['a', 'b', 'c']

	pool = cycle(lst)

	for item in pool:
	print(item)

pykong / build_nested_dict.py

Last active November 13, 2017 22:07

Use this logic instead of a sequence of key checks.

	l0 = {}

	l1 = l0.setdefault("key_l1", {})
	l2 = l1.setdefault("key_l2", [])
	l2.append("3")

	{'key_l1': {'key_l2': ['3']}}

pykong / compare_tuples.py

Created November 18, 2017 14:50

	# Write this:
	if (mark.a, mark.b) == (region.a, region.b):
	...

	# much cleaner than:
	if mark.a == region.a and mark.b == region.b:
	...

pykong / get_active_view.py

Created November 18, 2017 19:41

Often used.

	def get_active_view(self):
	"""Return the active view in the currently active window."""
	return sublime.active_window().active_view()

pykong / get_index.py

Created November 19, 2017 23:44

https://stackoverflow.com/questions/176918/finding-the-index-of-an-item-given-a-list-containing-it-in-python

	>>> ["foo", "bar", "baz"].index("bar")
	1

pykong / get_neighbours.py

Last active December 5, 2017 18:23

Gets both numbers adjacent to input number.

	import bisect


	def get_neighbours(num, interval):
	interval = set(interval)
	interval.discard(num)
	interval = list(interval)
	interval.sort()

	if num < interval[0] or interval[-1] < num:

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