C++博客-清风竹林-随笔分类-python

C++博客-清风竹林-随笔分类-pythonhttp://www.cppblog.com/xmli/category/10513.htmlぷ雪飘绛梅映残红 <br>    ぷ花舞霜飞映苍松<br>      ----- Do more,suffer lesszh-cnMon, 11 May 2009 10:30:17 GMTMon, 11 May 2009 10:30:17 GMT60Python Challenge lv5: peak hellhttp://www.cppblog.com/xmli/archive/2009/05/11/82588.html李现民李现民Mon, 11 May 2009 08:54:00 GMThttp://www.cppblog.com/xmli/archive/2009/05/11/82588.htmlhttp://www.cppblog.com/xmli/comments/82588.htmlhttp://www.cppblog.com/xmli/archive/2009/05/11/82588.html#Feedback0http://www.cppblog.com/xmli/comments/commentRss/82588.htmlhttp://www.cppblog.com/xmli/services/trackbacks/82588.htmlhttp://www.pythonchallenge.com/pc/def/peak.html
这是反序列化的一道题pickle，页面源码中的banner.p的字样，为序列化文件地址。但解析的结果并不是最终的答案，花了半天时间终于弄明白结果是一个用‘#’字组成的图形，呜呼：程序输出：

import pickle

import urllib.request

if __name__ == '__main__':

url = 'http://www.pythonchallenge.com/pc/def/banner.p'

request= urllib.request.Request(url)

# my pc must use proxy to connect

request.set_proxy('172.16.0.252:80', 'http')

try:

response= urllib.request.urlopen(request)

banner= pickle.load(response)

response.close()

for line in banner:

print(''.join(map(lambda x: x[0]* x[1], line)))

except Exception as ex:

print(ex)

程序输出：


              #####                                                                      #####
               ####                                                                       ####
               ####                                                                       ####
               ####                                                                       ####
               ####                                                                       ####
               ####                                                                       ####
               ####                                                                       ####
               ####                                                                       ####
      ###      ####   ###         ###       #####   ###    #####   ###          ###       ####
   ###   ##    #### #######     ## ###      #### #######   #### #######     ### ###     ####
###     ### #####    ####   ###   ####    #####    #### #####    ####   ###     ###   ####
###           ####     ####   ###    ###    ####     #### ####     #### ###      #### ####
###           ####     ####          ###    ####     #### ####     #### ###       ### ####
####           ####     ####     ##   ###    ####     #### ####     #### ####       ### ####
####           ####     ####   ##########    ####     #### ####     #### ############## ####
####           ####     #### ###    ####    ####     #### ####     #### ####            ####
####           ####     #### ####     ###    ####     #### ####     #### ####            ####
###           ####     #### ####     ###    ####     #### ####     #### ###            ####
###      ## ####     #### ###    ####    ####     #### ####     ####   ###      ##   ####
   ###    ##   ####     ####   ###########   ####     #### ####     ####    ###    ##    ####
      ###     ######    #####    ##    #### ######    ###########    #####      ###      ######

得到下一关地址： channel

李现民 2009-05-11 16:54 发表评论

]]>Python Challenge lv4: follow the chainhttp://www.cppblog.com/xmli/archive/2009/05/11/82578.html李现民李现民Mon, 11 May 2009 08:05:00 GMThttp://www.cppblog.com/xmli/archive/2009/05/11/82578.htmlhttp://www.cppblog.com/xmli/comments/82578.htmlhttp://www.cppblog.com/xmli/archive/2009/05/11/82578.html#Feedback0http://www.cppblog.com/xmli/comments/commentRss/82578.htmlhttp://www.cppblog.com/xmli/services/trackbacks/82578.htmlhttp://www.pythonchallenge.com/pc/def/linkedlist.php

说实话，好不容易通过google搞清楚题目的要求：通过不断的从服务器取得一个web page，然后从源码中找出下一个链接的地址。需要注意的是：虽然页面的源码很简单，但并不是其中所有的数字都是有效的，需要使用正则表达式找出正确的pattern形式才可以，对本题而言r'nothing is (\d+)'是一个可用的pattern，使用''.join([x for x in text if x.isdigit()] 将所有的数字都粘连起来了，结果跟踪到4000多还没结束，才知道上当了。。。

import re

import urllib.request

if __name__ == '__main__':

url = 'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing='

index = '17675'

counter = 1

pattern = re.compile(r'nothing is (\d+)')

while True:

try:

request= urllib.request.Request(url+index)

# my pc must use proxy to connect

request.set_proxy('172.16.0.252:80', 'http')

response= urllib.request.urlopen(request)

content=str(response.read().decode())

response.close()

print(counter, content)

result = pattern.search(content)

if not result:

break

index = result.group(1)

counter += 1

except Exception as ex:

print(ex)

break

程序输出：
1 and the next nothing is 8511
2 and the next nothing is 89456
3 and the next nothing is 43502
4 and the next nothing is 45605
5 and the next nothing is 12970
6 and the next nothing is 91060
7 and the next nothing is 27719
8 and the next nothing is 65667
9 peak.html

得到下一个题目的地址peak.html (注：我的index初始值是17675，题目中最早给出的可不是这个值，我是从地址列表的后一部分选了一个数字而已，因此不要担心)

李现民 2009-05-11 16:05 发表评论

]]>Python Challenge lv3: rehttp://www.cppblog.com/xmli/archive/2009/05/11/82577.html李现民李现民Mon, 11 May 2009 07:52:00 GMThttp://www.cppblog.com/xmli/archive/2009/05/11/82577.htmlhttp://www.cppblog.com/xmli/comments/82577.htmlhttp://www.cppblog.com/xmli/archive/2009/05/11/82577.html#Feedback0http://www.cppblog.com/xmli/comments/commentRss/82577.htmlhttp://www.cppblog.com/xmli/services/trackbacks/82577.htmlhttp://www.pythonchallenge.com/pc/def/equality.html

One small letter, surrounded by EXACTLY three big bodyguards on each of its sides. 何为big bodyguards? 我估计是大写字母，即要求找到由三个大写字母包围着的小写字母。事实上，题目要求的确如此。不过我在一开始的时候走了弯路：我把连续的七个字母全部都打印出来了（左边三个大写字母，中间一个小写字母，右边三个大写字母），费了半天才搞明白原来要求的只是中间的小写字母。
另外注意EXACTLY 这个词，即“当且仅当”左右两边均为“三”个大写字母的pattern才算。

代码到也简单：

import re

if __name__ == '__main__':

finpath = 'fin.txt'

with open(finpath) as fin:

# translate text into a single string

text = ''.join([line.rstrip() for line in fin.read()])

pattern = re.compile(r'[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]')

print(''.join(pattern.findall(text)))

程序输出：linkedlist

参考答案

李现民 2009-05-11 15:52 发表评论

]]>Python Challenge lv2: ocrhttp://www.cppblog.com/xmli/archive/2009/05/11/82574.html李现民李现民Mon, 11 May 2009 07:40:00 GMThttp://www.cppblog.com/xmli/archive/2009/05/11/82574.htmlhttp://www.cppblog.com/xmli/comments/82574.htmlhttp://www.cppblog.com/xmli/archive/2009/05/11/82574.html#Feedback0http://www.cppblog.com/xmli/comments/commentRss/82574.htmlhttp://www.cppblog.com/xmli/services/trackbacks/82574.htmlhttp://www.pythonchallenge.com/pc/def/ocr.html
根据提示，题目要求是从html页面源文件的一段文本中找出rare characters。何为rare，暂时不知道，不过不要紧，先把整段文本存放于一个叫fin.txt的文件中，预处理一下：

if __name__ == '__main__':

finpath = 'fin.txt'

with open(finpath) as fin:

# translate text into a single string

text = ''.join([line.rstrip() for line in fin.read()])

d= {}

for c in text:

d[c] = d.get(c, 0) +1

for k, v in d.items():

print(k, v)

输出结果：

! 6079
# 6115
% 6104
$ 6046
& 6043
) 6186
( 6154
+ 6066
* 6034
@ 6157
[ 6108
] 6152
_ 6112
^ 6030
a 1
e 1
i 1
l 1
q 1
u 1
t 1
y 1
{ 6046
} 6105

好了，很显然了， rare characters指的就是个数为1的这几个字母，于是将代码稍微改一下即可打印得到结果：

if __name__ == '__main__':

finpath = 'fin.txt'

with open(finpath) as fin:

# translate text into a single string

text = ''.join([line.rstrip() for line in fin.read()])

d= {}

for c in text:

d[c] = d.get(c, 0) +1

print(''.join([c for c in text if d[c] ==1]))

程序输出： equality

考虑到结果集中未输出的都是非字母，因此可以考虑如下方法求解：

if __name__ == '__main__':

finpath = 'fin.txt'

with open(finpath) as fin:

# translate text into a single string

text = ''.join([line.rstrip() for line in fin.read()])

# only print letters

print(''.join([c for c in text if c.isalpha()]))

# another method

print(''.join(filter(lambda x: x.isalpha(), text)))

参考答案

李现民 2009-05-11 15:40 发表评论

]]>Python Challenge lv1: What about making trans?http://www.cppblog.com/xmli/archive/2009/05/11/82567.html李现民李现民Mon, 11 May 2009 07:05:00 GMThttp://www.cppblog.com/xmli/archive/2009/05/11/82567.htmlhttp://www.cppblog.com/xmli/comments/82567.htmlhttp://www.cppblog.com/xmli/archive/2009/05/11/82567.html#Feedback0http://www.cppblog.com/xmli/comments/commentRss/82567.htmlhttp://www.cppblog.com/xmli/services/trackbacks/82567.htmlhttp://www.pythonchallenge.com/pc/def/map.html
由于仅仅是第二道题，所以题目还是很易懂的，按图片中的指示，将每个字母替换成它在字母表中顺延两位的那个字母就可以了，解答如下：、

def trans(char):
    '''Translate a letter as the one two places after it'''
    if char.isalpha():
        return chr((ord(char)- ord('a') + 2)%26 + ord('a'))
    return char

if __name__ == '__main__':
    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. '''
    print(''.join(map(trans, text)))

打印结果： i hope you didnt translate it by hand. thats what computers are for. doing it in by hand is inefficient and that's why this text is so long. using string.maketrans() is recommended. now apply on the url.

参考答案链接

李现民 2009-05-11 15:05 发表评论

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