Python Challenge-Level 1

2019独角兽企业重金招聘Python工程师标准>>>

http://www.pythonchallenge.com/pc/def/map.html

http://wiki.pythonchallenge.com/index.php?title=Level1:Main_Page

#!/usr/bin/env python
#-*- coding: utf-8 -*-
#Level1_Sol1.py
import string
text = "g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."
def Sol1():
    #First, load the text and create a translation table.
    #string.ascii_lowercase: The lowercase letters 'abcdefghijklmnopqrstuvwxy'. This value is not locale-dependent and will not change.
    table = string.maketrans(string.ascii_lowercase, string.ascii_lowercase[2:]+string.ascii_lowercase[:2])
    print "table is ", repr(table), "\nlen(table) is ", len(table), "\ntype(table) is ", type(table)
    print "------------------------"
    #Now, we apply the translation table on the string.
    newstr1 = string.translate(text, table)
    print "newstr1 is ", newstr1
    print "------------------------"
    #Alternatively, just use the translate on the "text" variable.
    newstr2 = text.translate(table)
    print "text is ", text
    print ">>>>>>>>>>>>>>>>>>>>>>>>"
    print "newstr2 is ", newstr2
def Sol2():
    '''
        Solved without translate
    '''
    tmp_str = ""
    for x in text:
        if ord(x) >= ord('a') and ord(x) <= ord('z'):
            tmp_str += chr((ord(x) + 2 - ord('a'))%26 + ord('a'))
        else:
            tmp_str += x
    print "tmp_str is ", tmp_str
def Sol3():
    '''
        Similar to Sol2() but with some of 2.5's nested ternary operators
    '''
    for x in text:
        print chr(ord(x) if ord(x)+2 < ord('a') else ord(x) + 2 if ord(x)+2 < ord('z') else ord(x) - 24),
def Sol4():
    '''
        With a dictionary
    '''
    """
    zip([iterable, ...])zip()是python的一个内建函数，它接受一系列可迭代的对象作为参数，可把两个或多个序列中的相应项合并在一起，将对象中对应的元素打包成一个个tuple元组，然后返回由这些tuples组成的list（列表）。若输入参数的长度不等，则返回list的长度和参数中长度最短的对象相同。利用*号操作符，可以将list unzip（解压）
    >>> a = [1,2,3]
    >>> b = [4,5,6]
    >>> c = [7,8,9,0]
    >>> zipped = zip(a,b)
    >>> zipped
    [(1, 4), (2, 5), (3, 6)]
    >>> zipped = zip(a,b,c)
    >>> zipped
    [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
    >>> zip(a,c)
    [(1, 7), (2, 8), (3, 9)]
    >>> zip(*zipped)
    [(1, 2, 3), (4, 5, 6), (7, 8, 9)]
    
    dict() -> new empty dictionary
    dict(mapping) -> new dictionary initialized from a mapping object's(key, value) pairs
    dict(iterable) -> new dictionary initialized as if via:
        d = {}
        for k, v in iterable:
            d[k] = v
    dict(**kwargs) -> new dictionary initialized with the name=value pairs in the keyword argument list.  For example:  dict(one=1, two=2)
    """
    cypher = dict(zip(string.lowercase, string.lowercase[2:] + string.lowercase[:2]))
    #dict.get(key, default=None)对字典dict中的键key，返回它对应的值value,如果字典中不存在此键，则返回default的值。注意，参数default的默认值为None.
    print "cypher is ", cypher
    print "".join(cypher.get(c,c) for c in text)
        
if __name__ == '__main__':
    #Sol1()
    #Sol2()
    #Sol3()
    Sol4()

Do by myself:

str = "g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."
lists = str.split()
list1 = []
for str1 in lists:
    str2 = ''
    for ch in str1:
        if ch == '.' or ch == '\'' or ch == '(' or ch == ')':
            str2 += ch
        else:
            str2 += chr(ord(ch)+2)
    str2 = str2.replace('{','a')
    str2 = str2.replace('|','b')
    list1.append(str2)
print(" ".join(list1))

转载于:https://my.oschina.net/u/1178546/blog/165808