| odd_numbers = [x in range(0,100) if x % 2 == 1] | |
| even_numver = [x in range(0,100) if x % 2 == 0] |
Author: Ben Brady Licensed: MIT
Allows selected elements to be changed (shown/hidden or callback) based on a set of field validation rules.
This is useful for large forms where you may want to hide/disable various fieldsets which are relevant only if a particular input element is selected/checked.
| """ | |
| HG Fogbugz Mercurial Extension | |
| Prevents committing bad BugIDs | |
| Add to .hgrc or mercurial.ini | |
| [extensions] | |
| fogbugz = C:\Users\User\hg_extensions\hgfogbugz.py | |
| """ |
| class Counter(object): | |
| """ This class counts stuff... """ | |
| def __init__(self, start = 0, end = 10, increment = 1): | |
| self.start = start | |
| self.end = end | |
| self.increment = increment | |
| def __iter__(self): | |
| # this method is called when a Counter object is used as an iterator (e.g. in a for loop). | |
| i = self.start |
| ''' | |
| See http://projecteuler.net/index.php?section=problems&id=17 | |
| If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. | |
| If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used? | |
| NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage. | |
| ''' | |
| def num_letters(number): | |
| digit_len = {1:3, 2:3, 3:5, 4:4, 5:4, 6:3, 7:5, 8:5, 9:4} |
| from collections import defaultdict | |
| import re | |
| class Location(): | |
| name = None | |
| msg_arrive = 'Hi {player}, Welcome to {location_name}' | |
| msg_depart = 'You leave {location_name}' | |
| msg_look = 'You see {location_name}' |
