Python lexicographically problematic instances and python dictionary instances

Python lexicographically problematic instances and python dictionary instances

This article describes the python lexicographic issue and shares it with you for your reference. The details are as follows:

Problem description:

The order of letters from left to right is the same as that in the alphabet, and each character can appear up to once .. for example, a, B, AB, bc, and xyz are all ascending strings. sort all ascending strings of the letter A with A length not greater than 6 in alphabetical order and encode them as follows:

1	2	..	26	27	28	...
A	B	..	Z	AB	Ac	..

Quickly calculate the encoding of an ascending string in the preceding dictionary.

The implementation code is as follows:

import stringall_letter = string.ascii_lowercasedef gen_dict():  result = {}  list_num_one = [ a_letter for a_letter in all_letter ]  list_num_two = [ i + j for i in all_letter for j in all_letter[all_letter.find(i)+1:]]  list_num_three = [ i + j + k for i in all_letter            for j in all_letter[all_letter.find(i)+1:]           for k in all_letter[all_letter.find(j)+1:]]  list_num_four = [ i + j + k + l for i in all_letter            for j in all_letter[all_letter.find(i)+1:]           for k in all_letter[all_letter.find(j)+1:]           for l in all_letter[all_letter.find(k)+1:]]  list_num_five = [ i + j + k + l + m for i in all_letter            for j in all_letter[all_letter.find(i)+1:]           for k in all_letter[all_letter.find(j)+1:]           for l in all_letter[all_letter.find(k)+1:]           for m in all_letter[all_letter.find(l)+1:]]  list_num_six = [ i + j + k + l + m + n  for i in all_letter      for j in all_letter[all_letter.find(i)+1:]      for k in all_letter[all_letter.find(j)+1:]      for l in all_letter[all_letter.find(k)+1:]      for m in all_letter[all_letter.find(l)+1:]      for n in all_letter[all_letter.find(m)+1:]      ]  for key,value in enumerate(list_num_one + list_num_two + list_num_three + list_num_four + list_num_five + list_num_six):    result.setdefault(key+1,value)  return result  my_dict = gen_dict()value_to_get = 'abcdef'for key,value in my_dict.iteritems():  if value == value_to_get:    print key

Result: 83682

That is, the abcdef encoding in the dictionary.

I hope this article will help you with Python programming.

How to sort Python dictionaries, for example

In Python2.7.x, OrderedDict is added to the collections class. The usage is as follows:
 

In Python2.7.x, OrderedDict is added to the collections class. The usage is as follows:

Pywugw @ pywugw-laptop :~ $/Usr/local/bin/python2.7
Python 2.7b1 (r27b1: 79927, Apr 26 2010, 11:44:19)
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> From collections import OrderedDict
>>> D = {'banana ': 3, 'apple': 4, 'pear': 1, 'Orange ': 2}

# Sort by key
>>> OrderedDict (sorted (d. items (), key = lambda t: t [0])
OrderedDict ([('apple', 4), ('banana ', 3), ('Orange', 2), ('pear ', 1)])

# Sort by value
>>> OrderedDict (sorted (d. items (), key = lambda t: t [1])
OrderedDict ([('pear ', 1), ('Orange', 2), ('banana ', 3), ('apple', 4)])
# Sort by key length
>>> OrderedDict (sorted (d. items (), key = lambda t: len (t [0])
OrderedDict ([('pear ', 1), ('apple', 4), ('Orange', 2), ('banana ', 3)])

Sort dictionaries in python

>>> D
{'A': 1, 'World': 11, 'z': 9, 'Hello': 10}
>>> K = d. keys ()
>>> K. sort ()
>>> K
['A', 'Hello', 'World', 'z']
>>> T = map (lambda key :( key, d [key]), k)
>>> T
[('A', 1), ('hello', 10), ('World', 11), ('Z', 9)]