Python论坛的帖子：

Mon Jan 9 11:23:58 HKT 2006

大小写问题?Image->image?
按理说windows不区分大小写啊,怪.你的lib目录也有些怪,一会c:,一会d:的
>------------------------------
>
>Message: 5
>Date: Mon, 9 Jan 2006 05:53:41 +0800
>From: " ?? " <ibrick at 163.com>
>Subject: [python-chinese] ?idle?????DOS???????????
>To: "python-chinese" <python-chinese at lists.python.cn>
>Message-ID: <43C18D66.09B6DB.25282>
>Content-Type: text/plain;	charset="gb2312"
>
>python-chinese，您好！
>
>	哪位牛人来指点指点，程序目的是为了输出一组编码的图像，
>在IDLE中运行完成，用python.exe g.py就不行了，太怪了=:<
>
>=================================================================
>错误信息：
>raceback (most recent call last):
>  File "g.py", line 5, in ?
>    import Image
>  File "C:\Python24\Lib\site-packages\PIL\Image.py", line 68, in ?
>    import os, string, sys
>  File "D:\Python24\Lib\string.py", line 83, in ?
>    import re as _re
>  File "D:\Python24\lib\re.py", line 20, in ?
>    
>AttributeError: 'module' object has no attribute 'search'
>
>*******************************************************************
>
>
>程序内容：
>#--------------------------------------------------------------
># -*- coding: cp936 -*-
>
>from Tkinter import *
>from tkFileDialog import *
>import Image
>import ImageFont
>import ImageDraw
>
>outpath = askdirectory ()
>font = 'D:/WINDOWS/Fonts/ARIAL.TTF'     # 字体文件路径
>list = [(32,2), (14,4), (26,8)]
>strls = []
>all = 0
>accls = {}
>
>for i in list:
>	all += 9*i[0]*i[1]
>	for l in range(1, i[0] + 1):
>		for g in range(1,10):
>			n = g*100+l
>			for b in range(i[1]):
>				strls.append(str(n))
>				
>if all != len(strls):
>	print "len = %d, all = %d" % (len(strls),all)
>	print "Error"
>	sys.exit(1)
>
>for c in strls:
>	if accls.has_key(c):
>		accls[c] += 1
>	else:
>		accls[c] = 1
>
>allstr = accls.keys()
>allstr.sort()
>
>endls = []
>for ch in allstr:
>	endls.append((ch,accls[ch]))
>			 
>
>for text in endls:
>	im = Image.new("RGB",(260,88),(66,17,17))
>	draw = ImageDraw.Draw(im)
>	draw.setfont(ImageFont.truetype(font,95))
># Arial = ImageFont.truetype(font,95)
>
>	for i in [0,1,2]:
>		draw.text((48+i*55,-11),text[0][i],(255,211,0))
>		fname = outpath + '/' + u'哑膜PP背胶-0.06×0.02-%d张-%s.tif' % (text[1],text[0])
>		im.save(fname)
># --------------------------- end ------------------------------------
>
>错误信息：
>raceback (most recent call last):
>  File "g.py", line 5, in ?
>    import Image
>  File "C:\Python24\Lib\site-packages\PIL\Image.py", line 68, in ?
>    import os, string, sys
>  File "D:\Python24\Lib\string.py", line 83, in ?
>    import re as _re
>  File "D:\Python24\lib\re.py", line 20, in ?
>    
>AttributeError: 'module' object has no attribute 'search'
>
>
>
>　　　　　　　　致
>礼！
> 				
>
>　　　　　　　　冷眼
>　　　　　　　　ibrick at 163.com
>　　　　　　　　　　2006-01-09
>
>------------------------------
>
>Message: 6
>Date: Mon, 9 Jan 2006 08:14:08 +0800
>From: Qu Jianning <redfox.qu at gmail.com>
>Subject: [python-chinese] Re: python-chinese Digest, Vol 25, Issue 38
>To: python-chinese at lists.python.cn
>Message-ID: <62b1fc650601081614j5a1c6a3ak at mail.gmail.com>
>Content-Type: text/plain; charset="gb2312"
>
>我觉得还是魏忠的好一点，起码代码清楚，重要的是符合题意
>
>在06-1-8，python-chinese-request at lists.python.cn <
>python-chinese-request at lists.python.cn> 写道：
>>
>> Send python-chinese mailing list submissions to
>>        python-chinese at lists.python.cn
>>
>> To subscribe or unsubscribe via the World Wide Web, visit
>>        http://python.cn/mailman/listinfo/python-chinese
>> or, via email, send a message with subject or body 'help' to
>>        python-chinese-request at lists.python.cn
>>
>> You can reach the person managing the list at
>>        python-chinese-owner at lists.python.cn
>>
>> When replying, please edit your Subject line so it is more specific
>> than "Re: Contents of python-chinese digest..."
>>
>>
>> Today's Topics:
>>
>>   1. Re: ?????????? (yzhh)
>>   2. Re: ?????????? (???)
>>   3. Re: Re: ?????????? (??)
>>   4. Re: Re: ?????????? (??)
>>   5. Re: Re: ?????????? (???)
>>
>>
>> ----------------------------------------------------------------------
>>
>> Message: 1
>> Date: Sun, 08 Jan 2006 18:04:27 +0800
>> From: yzhh <yezonghui at gmail.com>
>> Subject: [python-chinese] Re: ??????????
>> To: python-chinese at lists.python.cn
>> Message-ID: 1 at sea.gmane.org>
>> Content-Type: text/plain; charset=gb2312
>>
>> 魏忠 wrote:
>>
>> > 据说这道题来自google的面试：
>> > 已知：
>> > f(1)=1
>> > f(10)=2
>> > f(11)=4
>> > f(n)表示从1到n的全部整数中1这个数字出现的次数
>> > 写一个尽可能短的程序，计算出下一次f(n)=n时，n等于多少
>> > 分析清楚之后，用python实现很容易。忙累了的高手也可以休息一下试试哦！
>> >
>> > 这里给出答案：199981
>> > D:\py>python -m profile f11.py
>> > 答案是:199981
>> >          199985 function calls in 1.582 CPU seconds
>> >
>> > 开飞机的舒克
>> > http://www.lvye.org/shuke
>> > msn:weizhong at netease.com
>>
>> 这是我的代码，比较快，请指教
>>
>> def fac(n):
>>    """Comput factorial of n"""
>>    r = 1
>>    i = 1
>>    while (i<=n):
>>        r *= i
>>        i += 1
>>    return r
>>
>> def C(n,m):
>>    """Comput combinatory: select m from n"""
>>    return fac(n)/(fac(m)*fac(n-m))
>>
>> def f(n):
>>    if n == 0:
>>        return 0
>>
>>    d = 0
>>    m = 1
>>    while m <= n :
>>        m *= 10
>>        d += 1
>>    m /= 10
>>    d -= 1  #now m == 10**d <= n < 10**(d+1)
>>
>>    count = 0
>>    for i in range(1,d+1):
>>        #count += No. of '1's where there is i occurrence of '1' in a
>> number
>>        count += i * C(d,i) * (9 **(d-i))
>>    r = (n / m) * count
>>    if (n / m ) == 1:
>>        r += n - m + 1
>>    else:
>>        r += m
>>    r += f(n%m)
>>    return r
>>
>> #import pdb
>> #pdb.set_trace()
>>
>> n2 = 2
>> while n2 > f(n2):
>>    n2 += n2 - f(n2)
>> #    print "f(" + str(n2) + ") = " + str(f(n2))
>> n1 = n2 / 2
>> n = n2
>> while (n != f(n)):
>>    n = (n1 + n2) / 2
>>    if n > f(n):
>>        n1 = n
>>    else:
>>        n2 = n
>> #    print "f(" + str(n) + ") = " + str(f(n))
>>
>> print n
>> --
>>   regards,
>> yzhh
>>
>>
>>
>> ------------------------------
>>
>> Message: 2
>> Date: Sun, 8 Jan 2006 11:05:02 +0000
>> From: ??? <leexiaofeng at gmail.com>
>> Subject: Re: [python-chinese] ??????????
>> To: python-chinese at lists.python.cn
>> Message-ID: <87af76540601080305o28798527i at mail.gmail.com>
>> Content-Type: text/plain; charset=GB2312
>>
>> 在 06-1-7，魏忠<weizhong2004 at gmail.com> 写道：
>> > 据说这道题来自google的面试：
>> > 已知：
>> > f(1)=1
>> > f(10)=2
>> > f(11)=4
>> > f(n)表示从1到n的全部整数中1这个数字出现的次数
>> > 写一个尽可能短的程序，计算出下一次f(n)=n时，n等于多少
>> > 分析清楚之后，用python实现很容易。忙累了的高手也可以休息一下试试哦！
>> >
>> > 这里给出答案：199981
>> > D:\py>python -m profile f11.py
>> > 答案是:199981
>> >          199985 function calls in 1.582 CPU seconds
>> >
>> 一个非常蹩脚的实现，验证了你的答案，请指教
>> #!/usr/bin/env python
>> def howmany1( n ):
>>        count = 0
>>        while(n != 0 ):
>>                r = n % 10
>>                n = n / 10
>>                if( r != 1 ):
>>                        continue
>>                count += 1
>>
>>        return count
>>
>> def f11(n):
>>        count = 0
>>        while(n>0):
>>                count += howmany1(n)
>>                n -= 1
>>        return count
>>
>> print f11(199981)
>>
>> ------------------------------
>>
>> Message: 3
>> Date: Sun, 8 Jan 2006 19:07:43 +0800
>> From: ?? <weizhong2004 at gmail.com>
>> Subject: Re: [python-chinese] Re: ??????????
>> To: python-chinese at lists.python.cn
>> Message-ID: <acd2ebf20601080307h29b513c5p at mail.gmail.com>
>> Content-Type: text/plain; charset="gb2312"
>>
>> 老兄的代码果真飞快!
>> 俺的慢代码是这样的:
>>
>> result=[1]
>> temp=1
>> x=2
>> while True:
>>   jg=str(x).count('1')+temp
>>   if jg==x:print x;break;
>>   x+=1
>>   temp=jg
>>
>>
>> 在06-1-8，yzhh <yezonghui at gmail.com> 写道：
>> >
>> > 魏忠 wrote:
>> >
>> > > 据说这道题来自google的面试：
>> > > 已知：
>> > > f(1)=1
>> > > f(10)=2
>> > > f(11)=4
>> > > f(n)表示从1到n的全部整数中1这个数字出现的次数
>> > > 写一个尽可能短的程序，计算出下一次f(n)=n时，n等于多少
>> > > 分析清楚之后，用python实现很容易。忙累了的高手也可以休息一下试试哦！
>> > >
>> > > 这里给出答案：199981
>> > > D:\py>python -m profile f11.py
>> > > 答案是:199981
>> > >          199985 function calls in 1.582 CPU seconds
>> > >
>> > > 开飞机的舒克
>> > > http://www.lvye.org/shuke
>> > > msn:weizhong at netease.com
>> >
>> > 这是我的代码，比较快，请指教
>> >
>> > def fac(n):
>> >     """Comput factorial of n"""
>> >     r = 1
>> >     i = 1
>> >     while (i<=n):
>> >         r *= i
>> >         i += 1
>> >     return r
>> >
>> > def C(n,m):
>> >     """Comput combinatory: select m from n"""
>> >     return fac(n)/(fac(m)*fac(n-m))
>> >
>> > def f(n):
>> >     if n == 0:
>> >         return 0
>> >
>> >     d = 0
>> >     m = 1
>> >     while m <= n :
>> >         m *= 10
>> >         d += 1
>> >     m /= 10
>> >     d -= 1  #now m == 10**d <= n < 10**(d+1)
>> >
>> >     count = 0
>> >     for i in range(1,d+1):
>> >         #count += No. of '1's where there is i occurrence of '1' in a
>> > number
>> >         count += i * C(d,i) * (9 **(d-i))
>> >     r = (n / m) * count
>> >     if (n / m ) == 1:
>> >         r += n - m + 1
>> >     else:
>> >         r += m
>> >     r += f(n%m)
>> >     return r
>> >
>> > #import pdb
>> > #pdb.set_trace()
>> >
>> > n2 = 2
>> > while n2 > f(n2):
>> >     n2 += n2 - f(n2)
>> > #    print "f(" + str(n2) + ") = " + str(f(n2))
>> > n1 = n2 / 2
>> > n = n2
>> > while (n != f(n)):
>> >     n = (n1 + n2) / 2
>> >     if n > f(n):
>> >         n1 = n
>> >     else:
>> >         n2 = n
>> > #    print "f(" + str(n) + ") = " + str(f(n))
>> >
>> > print n
>> > --
>> >    regards,
>> > yzhh
>> >
>> > _______________________________________________
>> > python-chinese
>> > Post: send python-chinese at lists.python.cn
>> > Subscribe: send subscribe to python-chinese-request at lists.python.cn
>> > Unsubscribe: send unsubscribe to  python-chinese-request at lists.python.cn
>> > Detail Info: http://python.cn/mailman/listinfo/python-chinese
>> >
>>
>>
>>
>> --
>>
>> 开飞机的舒克
>> http://www.lvye.org/shuke
>> msn:weizhong at netease.com
>> -------------- next part --------------
>> An HTML attachment was scrubbed...
>> URL:
>> http://lists.exoweb.net/pipermail/python-chinese/attachments/20060108/038989f6/attachment-0001.htm
>>
>> ------------------------------
>>
>> Message: 4
>> Date: Sun, 8 Jan 2006 19:28:42 +0800
>> From: ?? <weizhong2004 at gmail.com>
>> Subject: Re: [python-chinese] Re: ??????????
>> To: python-chinese at lists.python.cn
>> Message-ID: <acd2ebf20601080328i70c8f6bfp at mail.gmail.com>
>> Content-Type: text/plain; charset="gb2312"
>>
>> 老兄的代码速度很快,但俺没有看懂.能再加点注释么?
>> 在俺的机器上,你的代码
>> D:\py>python -m profile f112.py
>> 199981
>>         8557 function calls (7799 primitive calls) in 0.118 CPU seconds
>> 当俺将你代码中的 range函数改为 xrange 之后,性能又提高了将近一倍
>>
>>
>> 在06-1-8，yzhh <yezonghui at gmail.com> 写道：
>> >
>> > 魏忠 wrote:
>> >
>> > > 据说这道题来自google的面试：
>> > > 已知：
>> > > f(1)=1
>> > > f(10)=2
>> > > f(11)=4
>> > > f(n)表示从1到n的全部整数中1这个数字出现的次数
>> > > 写一个尽可能短的程序，计算出下一次f(n)=n时，n等于多少
>> > > 分析清楚之后，用python实现很容易。忙累了的高手也可以休息一下试试哦！
>> > >
>> > > 这里给出答案：199981
>> > > D:\py>python -m profile f11.py
>> > > 答案是:199981
>> > >          199985 function calls in 1.582 CPU seconds
>> > >
>> > > 开飞机的舒克
>> > > http://www.lvye.org/shuke
>> > > msn:weizhong at netease.com
>> >
>> > 这是我的代码，比较快，请指教
>> >
>> > def fac(n):
>> >     """Comput factorial of n"""
>> >     r = 1
>> >     i = 1
>> >     while (i<=n):
>> >         r *= i
>> >         i += 1
>> >     return r
>> >
>> > def C(n,m):
>> >     """Comput combinatory: select m from n"""
>> >     return fac(n)/(fac(m)*fac(n-m))
>> >
>> > def f(n):
>> >     if n == 0:
>> >         return 0
>> >
>> >     d = 0
>> >     m = 1
>> >     while m <= n :
>> >         m *= 10
>> >         d += 1
>> >     m /= 10
>> >     d -= 1  #now m == 10**d <= n < 10**(d+1)
>> >
>> >     count = 0
>> >     for i in range(1,d+1):
>> >         #count += No. of '1's where there is i occurrence of '1' in a
>> > number
>> >         count += i * C(d,i) * (9 **(d-i))
>> >     r = (n / m) * count
>> >     if (n / m ) == 1:
>> >         r += n - m + 1
>> >     else:
>> >         r += m
>> >     r += f(n%m)
>> >     return r
>> >
>> > #import pdb
>> > #pdb.set_trace()
>> >
>> > n2 = 2
>> > while n2 > f(n2):
>> >     n2 += n2 - f(n2)
>> > #    print "f(" + str(n2) + ") = " + str(f(n2))
>> > n1 = n2 / 2
>> > n = n2
>> > while (n != f(n)):
>> >     n = (n1 + n2) / 2
>> >     if n > f(n):
>> >         n1 = n
>> >     else:
>> >         n2 = n
>> > #    print "f(" + str(n) + ") = " + str(f(n))
>> >
>> > print n
>> > --
>> >    regards,
>> > yzhh
>> >
>> > _______________________________________________
>> > python-chinese
>> > Post: send python-chinese at lists.python.cn
>> > Subscribe: send subscribe to python-chinese-request at lists.python.cn
>> > Unsubscribe: send unsubscribe to  python-chinese-request at lists.python.cn
>> > Detail Info: http://python.cn/mailman/listinfo/python-chinese
>> >
>>
>>
>>
>> --
>>
>> 开飞机的舒克
>> http://www.lvye.org/shuke
>> msn:weizhong at netease.com
>> -------------- next part --------------
>> An HTML attachment was scrubbed...
>> URL:
>> http://lists.exoweb.net/pipermail/python-chinese/attachments/20060108/507404fc/attachment-0001.html
>>
>> ------------------------------
>>
>> Message: 5
>> Date: Sun, 8 Jan 2006 11:36:29 +0000
>> From: ??? <leexiaofeng at gmail.com>
>> Subject: Re: [python-chinese] Re: ??????????
>> To: python-chinese at lists.python.cn
>> Message-ID: <87af76540601080336t535eef15w at mail.gmail.com>
>> Content-Type: text/plain; charset=GB2312
>>
>> 我想他的代码是不是直接算出比这个数小的数中所有1的个数，而不是一个一个加起来。
>>
>> 在 06-1-8，魏忠<weizhong2004 at gmail.com> 写道：
>> > 老兄的代码速度很快,但俺没有看懂.能再加点注释么?
>> > 在俺的机器上,你的代码
>> > D:\py>python -m profile f112.py
>> > 199981
>> >          8557 function calls (7799 primitive calls) in 0.118 CPU seconds
>> > 当俺将你代码中的 range函数改为 xrange 之后,性能又提高了将近一倍
>> >
>> >
>> >  在06-1-8，yzhh <yezonghui at gmail.com> 写道：
>> > > 魏忠 wrote:
>> > >
>> > > > 据说这道题来自google的面试：
>> > > > 已知：
>> > > > f(1)=1
>> > > > f(10)=2
>> > > > f(11)=4
>> > > > f(n)表示从1到n的全部整数中1这个数字出现的次数
>> > > > 写一个尽可能短的程序，计算出下一次f(n)=n时，n等于多少
>> > > > 分析清楚之后，用python实现很容易。忙累了的高手也可以休息一下试试哦！
>> > > >
>> > > > 这里给出答案：199981
>> > > > D:\py>python -m profile f11.py
>> > > > 答案是:199981
>> > > >          199985 function calls in 1.582 CPU seconds
>> > > >
>> > > > 开飞机的舒克
>> > > > http://www.lvye.org/shuke
>> > > > msn:weizhong at netease.com
>> > >
>> > > 这是我的代码，比较快，请指教
>> > >
>> > > def fac(n):
>> > >     """Comput factorial of n"""
>> > >     r = 1
>> > >     i = 1
>> > >     while (i<=n):
>> > >         r *= i
>> > >         i += 1
>> > >     return r
>> > >
>> > > def C(n,m):
>> > >     """Comput combinatory: select m from n"""
>> > >     return fac(n)/(fac(m)*fac(n-m))
>> > >
>> > > def f(n):
>> > >     if n == 0:
>> > >         return 0
>> > >
>> > >     d = 0
>> > >     m = 1
>> > >     while m <= n :
>> > >         m *= 10
>> > >         d += 1
>> > >     m /= 10
>> > >     d -= 1  #now m == 10**d <= n < 10**(d+1)
>> > >
>> > >     count = 0
>> > >     for i in range(1,d+1):
>> > >         #count += No. of '1's where there is i occurrence of '1' in a
>> > number
>> > >         count += i * C(d,i) * (9 **(d-i))
>> > >     r = (n / m) * count
>> > >     if (n / m ) == 1:
>> > >         r += n - m + 1
>> > >     else:
>> > >         r += m
>> > >     r += f(n%m)
>> > >     return r
>> > >
>> > > #import pdb
>> > > #pdb.set_trace()
>> > >
>> > > n2 = 2
>> > > while n2 > f(n2):
>> > >     n2 += n2 - f(n2)
>> > > #    print "f(" + str(n2) + ") = " + str(f(n2))
>> > > n1 = n2 / 2
>> > > n = n2
>> > > while (n != f(n)):
>> > >     n = (n1 + n2) / 2
>> > >     if n > f(n):
>> > >         n1 = n
>> > >     else:
>> > >         n2 = n
>> > > #    print "f(" + str(n) + ") = " + str(f(n))
>> > >
>> > > print n
>> > > --
>> > >    regards,
>> > > yzhh
>> > >
>> > > _______________________________________________
>> > > python-chinese
>> > > Post: send python-chinese at lists.python.cn
>> > > Subscribe: send subscribe to
>> > python-chinese-request at lists.python.cn
>> > > Unsubscribe: send unsubscribe to
>> > python-chinese-request at lists.python.cn
>> > > Detail Info:
>> > http://python.cn/mailman/listinfo/python-chinese
>> > >
>> >
>> >
>> >
>> > --
>> >
>> >
>> > 开飞机的舒克
>> >  http://www.lvye.org/shuke
>> > msn:weizhong at netease.com
>> > _______________________________________________
>> > python-chinese
>> > Post: send python-chinese at lists.python.cn
>> > Subscribe: send subscribe to
>> > python-chinese-request at lists.python.cn
>> > Unsubscribe: send unsubscribe to
>> > python-chinese-request at lists.python.cn
>> > Detail Info:
>> > http://python.cn/mailman/listinfo/python-chinese
>> >
>> >
>>
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Best Wishes.^^

　　　　　　　　Beisi Xu
　　　　　　　　　　2006-01-09
               Email:xubeisi at ustc.edu
　　　　　　　　QQ   :2+3=0+5 14=2×7
               MSN  :xubeisi at hotmail.com
               Blog :http://xubeisi.blog.edu.cn Chinese only...T_T

标题：[python-chinese] Re: python-chinese Digest, Vol 25, Issue 39 Message: 5