已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
实在想不出该如何实现,希望高人指点一二.
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·½·¨»¹ÊÇÂù¶àµÄ£¬×î¼òµ¥µÄ¾ÍÊÇת»¯ÎªÇó¶þ½×º¯ÊýµÄ×îСֵÎÊÌâ

ÔÚ07-12-20£¬python Bian <python.bian在gmail.com> дµÀ£º
>
> ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
> ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
>
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ÕâÊǸöÒ»ÔªÏßÐԻعéÎÊÌâ¡£½â·¨²Î¼ûÒÔϵÄÁ´½Ó£º
http://www.core.org.cn/NR/rdonlyres/Sloan-School-of-Management/15-075Applied-StatisticsSpring2003/EEF9C9C0-1AF7-4DD0-A6DE-C7566709CA2C/0/lec12_chap10.pdf

ÔÚ07-12-20£¬python Bian <python.bian在gmail.com> дµÀ£º
>
> ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
> ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
>
> _______________________________________________
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Áîc = Sigma( (b-x/(a+y))^2 ), ½«a,bÁ©×éÊý¾Ý´úÈ룬ÇócµÄ×îСֵ£¬´Ëʱx,y¼«ÎªËùÒªÇóµÄÖµ
µ±È»£¬Õâ¸öÖ»ÊÇ×î´ÖÂÔµÄÒ»ÖÖ½üËÆ·¨...»¹ÓÐÆäËüµÄ·½·¨£¬¿´ÄãÒª¾«È·µ½Ê²Ã´³Ì¶ÈÒÔ¼°Ê¹ÓúÎÖÖ¾àÀ뺯ÊýÁË


ÔÚ07-12-20£¬python Bian <python.bian在gmail.com> дµÀ£º
>
> ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
> ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
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这不是线性回归问题，因为这里要拟合的是a，b，要转换成我们通常习惯的x、y为变元的式子就成了：y=a/(x+b)
假设a=1,b=0则成了y=1/x，成了第一三象限的两条弧形曲线

感觉不大好弄，因为这个距离（任意一点到两条弧形曲线的距离）就不好计算，情况挺复杂

在07-12-20，Li Qingfeng <liqfemail at gmail.com> 写道：
>
> 这是个一元线性回归问题。解法参见以下的链接：
>
>
> <http://www.core.org.cn/NR/rdonlyres/Sloan-School-of-Management/15-075Applied-StatisticsSpring2003/EEF9C9C0-1AF7-4DD0-A6DE-C7566709CA2C/0/lec12_chap10.pdf>
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随便说说
取个倒数，x/b-a=y成直线吗？
反正应该取log之类的直线化
然后好像scipy里有线性拟和的现成函数吧？

Danfeng Lu 写道:
> 令c = Sigma( (b-x/(a+y))^2 ), 将a,b俩组数据代入，求c的最小值，此时x,y
> 极为所要求的值
> 当然，这个只是最粗略的一种近似法...还有其它的方法，看你要精确到什么程
> 度以及使用何种距离函数了
>
>
> 在07-12-20，*python Bian* < python.bian在gmail.com
> python.bian在gmail.com>> 写道：
>
>     已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
>     实在想不出该如何实现,希望高人指点一二.
>
>     _______________________________________________
>     python-chinese
>     Post: send python-chinese在lists.python.cn
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>
>
>
> -- 
> Warmest regards!
>
> 邮箱Email: danfeng.l在gmail.com danfeng.l在gmail.com>
> MSN: hudie_ldf在hotmail.com hudie_ldf在hotmail.com>
> ------------------------------------------------------------------------
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from scipy import stats
再试试stats.linregress
希望对你有用


chosen 写道:
> 随便说说
> 取个倒数，x/b-a=y成直线吗？
> 反正应该取log之类的直线化
> 然后好像scipy里有线性拟和的现成函数吧？
>
> Danfeng Lu 写道:
>   
>> 令c = Sigma( (b-x/(a+y))^2 ), 将a,b俩组数据代入，求c的最小值，此时x,y
>> 极为所要求的值
>> 当然，这个只是最粗略的一种近似法...还有其它的方法，看你要精确到什么程
>> 度以及使用何种距离函数了
>>
>>
>> 在07-12-20，*python Bian* < python.bian在gmail.com
>> python.bian在gmail.com>> 写道：
>>
>>     已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
>>     实在想不出该如何实现,希望高人指点一二.
>>
>>     _______________________________________________
>>     python-chinese
>>     Post: send python-chinese在lists.python.cn
>>     python-chinese在lists.python.cn>
>>     Subscribe: send subscribe to
>>     python-chinese-request在lists.python.cn
>>     python-chinese-request在lists.python.cn>
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>>     python-chinese-request在lists.python.cn
>>     python-chinese-request在lists.python.cn>
>>     Detail Info: http://python.cn/mailman/listinfo/python-chinese
>>
>>
>>
>>
>> -- 
>> Warmest regards!
>>
>> 邮箱Email: danfeng.l在gmail.com danfeng.l在gmail.com>
>> MSN: hudie_ldf在hotmail.com hudie_ldf在hotmail.com>
>> ------------------------------------------------------------------------
>>
>> _______________________________________________
>> python-chinese
>> Post: send python-chinese在lists.python.cn
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>> Unsubscribe: send unsubscribe to  python-chinese-request在lists.python.cn
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>>     
>
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È¡log(x)×îÖ÷ÒªµÄÂé·³Êǵñ£Ö¤x>0...Õâ¸öÔ¼ÊøÌõ¼þÏ൱µÄÂé·³¸öÈËÈÏΪ

ÄâºÏʲôµÄ£¬ÏßÐԵĻ¹ÊÇ·ÇÏßÐԵģ¬ÎÒ¾õµÃÕâЩÆäʵֻÊÇÃû×Ö£¬¹é¸ùµ½µ×£¬Æäʵ¶¼ÊǶ¨ÒåijÖÖÒâÒåÉϵľàÀ룬Ȼºó¶ÔÕâÖÖ¾àÀë½øÐзÖÎöÀ´»ñµÃÐÅÏ¢£¬±ÈÈçÇó¼«ÖµÊ²Ã´µÄ£¬¼«Öµµã»òÕß˵¹Õµã¶¼ÊÇÓÐÖØÒªÐÅÏ¢µÄµØ·½...
±È½ÏÖ±°×ºÇºÇ

ÔÚ07-12-20£¬chosen <chosen在126.com> дµÀ£º
>
> Ëæ±ã˵˵
> È¡¸öµ¹Êý£¬x/b-a=y³ÉÖ±ÏßÂð£¿
> ·´ÕýÓ¦¸ÃÈ¡logÖ®ÀàµÄÖ±Ïß»¯
> È»ºóºÃÏñscipyÀïÓÐÏßÐÔÄâºÍµÄÏֳɺ¯Êý°É£¿
>
> Danfeng Lu дµÀ:
> > Áîc = Sigma( (b-x/(a+y))^2 ), ½«a,bÁ©×éÊý¾Ý´úÈ룬ÇócµÄ×îСֵ£¬´Ëʱx,y
> > ¼«ÎªËùÒªÇóµÄÖµ
> > µ±È»£¬Õâ¸öÖ»ÊÇ×î´ÖÂÔµÄÒ»ÖÖ½üËÆ·¨...»¹ÓÐÆäËüµÄ·½·¨£¬¿´ÄãÒª¾«È·µ½Ê²Ã´³Ì
> > ¶ÈÒÔ¼°Ê¹ÓúÎÖÖ¾àÀ뺯ÊýÁË
> >
> >
> > ÔÚ07-12-20£¬*python Bian* < python.bian在gmail.com
> > python.bian在gmail.com>> дµÀ£º
> >
> >     ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
> >     ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
> >
> >     _______________________________________________
> >     python-chinese
> >     Post: send python-chinese在lists.python.cn
> >     python-chinese在lists.python.cn>
> >     Subscribe: send subscribe to
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> >
> >
> >
> >
> > --
> > Warmest regards!
> >
> > ÓÊÏäEmail: danfeng.l在gmail.com danfeng.l在gmail.com>
> > MSN: hudie_ldf在hotmail.com hudie_ldf在hotmail.com>
> > ------------------------------------------------------------------------
> >
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...ab+by=x
令A=ab
B=b
则有A=x-yB
直接一元线性回归就出来了
反正楼主也没说近似是怎么近似的

在 07-12-21，Danfeng Lu<danfeng.l在gmail.com> 写道：
> 取log(x)最主要的麻烦是得保证x>0...这个约束条件相当的麻烦个人认为
>
> 拟合什么的，线性的还是非线性的，我觉得这些其实只是名字，归根到底，其实都是定义某种意义上的距离，然后对这种距离进行分析来获得信息，比如求极值什么的，极值点或者说拐点都是有重要信息的地方...
> 比较直白呵呵
>
> 在07-12-20， chosen <chosen在126.com> 写道：
> > 随便说说
> > 取个倒数，x/b-a=y成直线吗？
> > 反正应该取log之类的直线化
> > 然后好像scipy里有线性拟和的现成函数吧？
> >
> > Danfeng Lu 写道:
> > > 令c = Sigma( (b-x/(a+y))^2 ), 将a,b俩组数据代入，求c的最小值，此时x,y
> > > 极为所要求的值
> > > 当然，这个只是最粗略的一种近似法...还有其它的方法，看你要精确到什么程
> > > 度以及使用何种距离函数了
> > >
> > >
> > > 在07-12-20，*python Bian* < python.bian在gmail.com
> > > python.bian在gmail.com>> 写道：
> > >
> > >     已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
> > >     实在想不出该如何实现,希望高人指点一二.
> > >
> > >     _______________________________________________
> > >     python-chinese
> > >     Post: send python-chinese在lists.python.cn
> > >     python-chinese在lists.python.cn>
> > >     Subscribe: send subscribe to
> > >     python-chinese-request在lists.python.cn
> > >     python-chinese-request在lists.python.cn>
> > >     Unsubscribe: send unsubscribe to
> > >     python-chinese-request在lists.python.cn
> > >     python-chinese-request在lists.python.cn>
> > >     Detail Info:
> http://python.cn/mailman/listinfo/python-chinese
> > >
> > >
> > >
> > >
> > > --
> > > Warmest regards!
> > >
> > > 邮箱Email: danfeng.l在gmail.com danfeng.l在gmail.com>
> > > MSN: hudie_ldf在hotmail.com hudie_ldf在hotmail.com>
> > >
> ------------------------------------------------------------------------
> > >
> > > _______________________________________________
> > > python-chinese
> > > Post: send python-chinese在lists.python.cn
> > > Subscribe: send subscribe to
> python-chinese-request在lists.python.cn
> > > Unsubscribe: send unsubscribe to
> python-chinese-request在lists.python.cn
> > > Detail Info:
> http://python.cn/mailman/listinfo/python-chinese
> >
> >
> > _______________________________________________
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> > Post: send python-chinese在lists.python.cn
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>
>
>
> --
> Warmest regards!
>
> 邮箱Email: danfeng.l在gmail.com
> MSN: hudie_ldf在hotmail.com
> _______________________________________________
> python-chinese
> Post: send python-chinese在lists.python.cn
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bingo~ÓеÀÀí£¬Õâ¸ö·½·¨ºÃ£¡²»¹ý£¬Ã»ÆÕÊÊÐÔ£¡^^

ÔÚ07-12-21£¬yuting cui <yutingcui在gmail.com> дµÀ£º
>
> ...ab+by=x
> ÁîA=ab
> B=b
> ÔòÓÐA=x-yB
> Ö±½ÓÒ»ÔªÏßÐԻعé¾Í³öÀ´ÁË
> ·´ÕýÂ¥Ö÷Ҳû˵½üËÆÊÇÔõô½üËƵÄ
>
> ÔÚ 07-12-21£¬Danfeng Lu<danfeng.l在gmail.com> дµÀ£º
> > È¡log(x)×îÖ÷ÒªµÄÂé·³Êǵñ£Ö¤x>0...Õâ¸öÔ¼ÊøÌõ¼þÏ൱µÄÂé·³¸öÈËÈÏΪ
> >
> >
> ÄâºÏʲôµÄ£¬ÏßÐԵĻ¹ÊÇ·ÇÏßÐԵģ¬ÎÒ¾õµÃÕâЩÆäʵֻÊÇÃû×Ö£¬¹é¸ùµ½µ×£¬Æäʵ¶¼ÊǶ¨ÒåijÖÖÒâÒåÉϵľàÀ룬Ȼºó¶ÔÕâÖÖ¾àÀë½øÐзÖÎöÀ´»ñµÃÐÅÏ¢£¬±ÈÈçÇó¼«ÖµÊ²Ã´µÄ£¬¼«Öµµã»òÕß˵¹Õµã¶¼ÊÇÓÐÖØÒªÐÅÏ¢µÄµØ·½...
> > ±È½ÏÖ±°×ºÇºÇ
> >
> > ÔÚ07-12-20£¬ chosen <chosen在126.com> дµÀ£º
> > > Ëæ±ã˵˵
> > > È¡¸öµ¹Êý£¬x/b-a=y³ÉÖ±ÏßÂð£¿
> > > ·´ÕýÓ¦¸ÃÈ¡logÖ®ÀàµÄÖ±Ïß»¯
> > > È»ºóºÃÏñscipyÀïÓÐÏßÐÔÄâºÍµÄÏֳɺ¯Êý°É£¿
> > >
> > > Danfeng Lu дµÀ:
> > > > Áîc = Sigma( (b-x/(a+y))^2 ), ½«a,bÁ©×éÊý¾Ý´úÈ룬ÇócµÄ×îСֵ£¬´Ëʱx,y
> > > > ¼«ÎªËùÒªÇóµÄÖµ
> > > > µ±È»£¬Õâ¸öÖ»ÊÇ×î´ÖÂÔµÄÒ»ÖÖ½üËÆ·¨...»¹ÓÐÆäËüµÄ·½·¨£¬¿´ÄãÒª¾«È·µ½Ê²Ã´³Ì
> > > > ¶ÈÒÔ¼°Ê¹ÓúÎÖÖ¾àÀ뺯ÊýÁË
> > > >
> > > >
> > > > ÔÚ07-12-20£¬*python Bian* < python.bian在gmail.com
> > > > python.bian在gmail.com>> дµÀ£º
> > > >
> > > >     ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
> > > >     ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
> > > >
> > > >     _______________________________________________
> > > >     python-chinese
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> > > > --
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> > > >
> > > > ÓÊÏäEmail: danfeng.l在gmail.com danfeng.l在gmail.com>
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> > > >
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Warmest regards!

ÓÊÏäEmail: danfeng.l在gmail.com
MSN: hudie_ldf在hotmail.com
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这个纯粹是捣乱算法...
正常的应该认为误差是正态分布的...
结论就是你那个

在 07-12-21，Danfeng Lu<danfeng.l在gmail.com> 写道：
> bingo~有道理，这个方法好！不过，没普适性！^^
>
> 在07-12-21，yuting cui <yutingcui在gmail.com> 写道：
> > ...ab+by=x
> > 令A=ab
> > B=b
> > 则有A=x-yB
> > 直接一元线性回归就出来了
> > 反正楼主也没说近似是怎么近似的
> >
> > 在 07-12-21，Danfeng Lu<danfeng.l在gmail.com> 写道：
> > > 取log(x)最主要的麻烦是得保证x>0...这个约束条件相当的麻烦个人认为
> > >
> > >
> 拟合什么的，线性的还是非线性的，我觉得这些其实只是名字，归根到底，其实都是定义某种意义上的距离，然后对这种距离进行分析来获得信息，比如求极值什么的，极值点或者说拐点都是有重要信息的地方...
> > > 比较直白呵呵
> > >
> > > 在07-12-20， chosen <chosen在126.com> 写道：
> > > > 随便说说
> > > > 取个倒数，x/b-a=y成直线吗？
> > > > 反正应该取log之类的直线化
> > > > 然后好像scipy里有线性拟和的现成函数吧？
> > > >
> > > > Danfeng Lu 写道:
> > > > > 令c = Sigma( (b-x/(a+y))^2 ), 将a,b俩组数据代入，求c的最小值，此时x,y
> > > > > 极为所要求的值
> > > > > 当然，这个只是最粗略的一种近似法...还有其它的方法，看你要精确到什么程
> > > > > 度以及使用何种距离函数了
> > > > >
> > > > >
> > > > > 在07-12-20，*python Bian* < python.bian在gmail.com
> > > > > python.bian在gmail.com>> 写道：
> > > > >
> > > > >     已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
> > > > >     实在想不出该如何实现,希望高人指点一二.
> > > > >
> > > > >     _______________________________________________
> > > > >     python-chinese
> > > > >     Post: send python-chinese在lists.python.cn
> > > > >     python-chinese在lists.python.cn>
> > > > >     Subscribe: send subscribe to
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> > > > >     python-chinese-request在lists.python.cn>
> > > > >     Unsubscribe: send unsubscribe to
> > > > >     python-chinese-request在lists.python.cn
> > > > >     python-chinese-request在lists.python.cn
> >
> > > > >     Detail Info:
> > > http://python.cn/mailman/listinfo/python-chinese
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > --
> > > > > Warmest regards!
> > > > >
> > > > > 邮箱Email: danfeng.l在gmail.com danfeng.l在gmail.com>
> > > > > MSN: hudie_ldf在hotmail.com hudie_ldf在hotmail.com>
> > > > >
> > >
> ------------------------------------------------------------------------
> > > > >
> > > > > _______________________________________________
> > > > > python-chinese
> > > > > Post: send python-chinese在lists.python.cn
> > > > > Subscribe: send subscribe to
> > > python-chinese-request在lists.python.cn
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> > > >
> > > >
> > > > _______________________________________________
> > > > python-chinese
> > > > Post: send python-chinese在lists.python.cn
> > > > Subscribe: send subscribe to
> > > python-chinese-request在lists.python.cn
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> > >
> > >
> > >
> > > --
> > > Warmest regards!
> > >
> > > 邮箱Email: danfeng.l在gmail.com
> > > MSN: hudie_ldf在hotmail.com
> > > _______________________________________________
> > > python-chinese
> > > Post: send python-chinese在lists.python.cn
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> > > Unsubscribe: send unsubscribe to
> > > python-chinese-request在lists.python.cn
> > > Detail Info:
> > > http://python.cn/mailman/listinfo/python-chinese
> > >
> > _______________________________________________
> > python-chinese
> > Post: send python-chinese在lists.python.cn
> > Subscribe: send subscribe to
> python-chinese-request在lists.python.cn
> > Unsubscribe: send unsubscribe to
> python-chinese-request在lists.python.cn
> > Detail Info:
> http://python.cn/mailman/listinfo/python-chinese
>
>
>
> --
> Warmest regards!
>
> 邮箱Email: danfeng.l在gmail.com
> MSN: hudie_ldf在hotmail.com
> _______________________________________________
> python-chinese
> Post: send python-chinese在lists.python.cn
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>

ÀÏʵ½²£¬ÎÒ¾õµÃÊ×Ïȶ¨Òå¾ÍÓе㲻Çå³þ¡£
ʲô½Ð×ö×îºÏÊÊ£¿ Õâ¸öÐèÒª¶¨Òå¡£
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d = |x-by-ab|/sqrt(1+b^2)
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D = ¡Æ(x-by-ab)^2/(1+b^2)
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ÒÔºó¾ÍÊÇÊýѧ·ÖÎöµÄÄÚÈÝÁË

¸Ð¾õÕâ¸öÎÊÌâûË㷨ʲôʶù

2007-12-21 



snake 



·¢¼þÈË£º python Bian 
·¢ËÍʱ¼ä£º 2007-12-20  21:43:30 
ÊÕ¼þÈË£º python china 
³­ËÍ£º 
Ö÷Ì⣺ [python-chinese]ÇóÖú:Ëã·¨ÎÊÌâ 
 
ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
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偶认错...偶那个纯粹是为了偷懒在误差足够小的情况下的一个近似算法
标准的回归分析是这样的
鉴于楼主给了个公式b=f(a)
两组数据(a0,b0),...,(an,bn)
那么一般不加说明的情况下认为对任意的bi有bi=f(ai)+εi
其中εi满足均值为0的正态分布N(0,σ^2 )
就是求使得方差σ^2最小x,y...

在 07-12-21，snake<deleteobject在gmail.com> 写道：
>
>
> 老实讲，我觉得首先定义就有点不清楚。
> 什么叫做最合适？ 这个需要定义。
> 这里我尝试自己发挥一下：
> ab+by=x，坐标系里就是一条直线，(x,y)就是直线上的点。
> 所谓合适，就是尽量接近这条直线的点，点到直线的距离最小。
> 也就是求到所有直线的距离之和最小的点
> d = |x-by-ab|/sqrt(1+b^2)
> 要求所有的a和b的组合所形成的d之和最小
> 为了去除绝对值，可以转化为求d的平方之和最小
> 于是
> D = ∑(x-by-ab)^2/(1+b^2)
> 也即对D进行二元函数求极值
> 于是对D求关于x和y的偏导，然后联立方程组求出所有驻点 ....
> 以后就是数学分析的内容了
>
> 感觉这个问题没算法什么事儿
>
> 2007-12-21 ________________________________
>
> snake ________________________________
>
> 发件人： python Bian
> 发送时间： 2007-12-20  21:43:30
> 收件人： python china
> 抄送：
> 主题： [python-chinese]求助:算法问题
>
>
> 已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
> 实在想不出该如何实现,希望高人指点一二.
> _______________________________________________
> python-chinese
> Post: send python-chinese在lists.python.cn
> Subscribe: send subscribe to
> python-chinese-request在lists.python.cn
> Unsubscribe: send unsubscribe to
> python-chinese-request在lists.python.cn
> Detail Info:
> http://python.cn/mailman/listinfo/python-chinese
>

呵呵,谢谢大家回复,发现数学高人不少啊.
承认标题误导大家了,现在发现纯属数学问题.
"任意的bi有bi=f(ai)+εi
其中εi满足均值为0的正态分布N(0,σ^2 )
就是求使得方差σ^2最小x,y..." 正是我想说的.
好好学习回归分析中


在07-12-21，yuting cui <yutingcui at gmail.com> 写道：
>
> 偶认错...偶那个纯粹是为了偷懒在误差足够小的情况下的一个近似算法
> 标准的回归分析是这样的
> 鉴于楼主给了个公式b=f(a)
> 两组数据(a0,b0),...,(an,bn)
> 那么一般不加说明的情况下认为对任意的bi有bi=f(ai)+εi
> 其中εi满足均值为0的正态分布N(0,σ^2 )
> 就是求使得方差σ^2最小x,y...
>
> 在 07-12-21，snake<deleteobject at gmail.com> 写道：
> >
> >
> > 老实讲，我觉得首先定义就有点不清楚。
> > 什么叫做最合适？ 这个需要定义。
> > 这里我尝试自己发挥一下：
> > ab+by=x，坐标系里就是一条直线，(x,y)就是直线上的点。
> > 所谓合适，就是尽量接近这条直线的点，点到直线的距离最小。
> > 也就是求到所有直线的距离之和最小的点
> > d = |x-by-ab|/sqrt(1+b^2)
> > 要求所有的a和b的组合所形成的d之和最小
> > 为了去除绝对值，可以转化为求d的平方之和最小
> > 于是
> > D = ∑(x-by-ab)^2/(1+b^2)
> > 也即对D进行二元函数求极值
> > 于是对D求关于x和y的偏导，然后联立方程组求出所有驻点 ....
> > 以后就是数学分析的内容了
> >
> > 感觉这个问题没算法什么事儿
> >
> > 2007-12-21 ________________________________
> >
> > snake ________________________________
> >
> > 发件人： python Bian
> > 发送时间： 2007-12-20  21:43:30
> > 收件人： python china
> > 抄送：
> > 主题： [python-chinese]求助:算法问题
> >
> >
> > 已知两组数据a,b,试求出最合适的x,y是所有的a,b都近似满足b=x/(a+y)
> > 实在想不出该如何实现,希望高人指点一二.
> > _______________________________________________
> > 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
> >
> _______________________________________________
> 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
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ÊÇ°¡,,ºÃ¶àÊýѧ´ïÈË,,,,ĬĬѧϰing...

ÔÚ07-12-22£¬python Bian <python.bian在gmail.com> дµÀ£º
>
> ºÇºÇ,лл´ó¼Ò»Ø¸´,·¢ÏÖÊýѧ¸ßÈ˲»ÉÙ°¡.
> ³ÐÈϱêÌâÎóµ¼´ó¼ÒÁË,ÏÖÔÚ·¢ÏÖ´¿ÊôÊýѧÎÊÌâ.
> "ÈÎÒâµÄbiÓÐbi=f(ai)+¦Åi
> ÆäÖЦÅiÂú×ã¾ùֵΪ0µÄÕý̬·Ö²¼N(0,¦Ò^2 )
> ¾ÍÊÇÇóʹµÃ·½²î¦Ò^2×îСx,y..." ÕýÊÇÎÒÏë˵µÄ.
> ºÃºÃѧϰ»Ø¹é·ÖÎöÖÐ
>
>
> ÔÚ07-12-21£¬yuting cui <yutingcui在gmail.com> дµÀ£º
> >
> > żÈÏ´í...żÄǸö´¿´âÊÇΪÁË͵ÀÁÔÚÎó²î×㹻СµÄÇé¿öϵÄÒ»¸ö½üËÆËã·¨
> > ±ê×¼µÄ»Ø¹é·ÖÎöÊÇÕâÑùµÄ
> > ¼øÓÚÂ¥Ö÷¸øÁ˸ö¹«Ê½b=f(a)
> > Á½×éÊý¾Ý(a0,b0),...,(an,bn)
> > ÄÇôһ°ã²»¼Ó˵Ã÷µÄÇé¿öÏÂÈÏΪ¶ÔÈÎÒâµÄbiÓÐbi=f(ai)+¦Åi
> > ÆäÖЦÅiÂú×ã¾ùֵΪ0µÄÕý̬·Ö²¼N(0,¦Ò^2 )
> > ¾ÍÊÇÇóʹµÃ·½²î¦Ò^2×îСx,y...
> >
> > ÔÚ 07-12-21£¬snake<deleteobject在gmail.com> дµÀ£º
> > >
> > >
> > > ÀÏʵ½²£¬ÎÒ¾õµÃÊ×Ïȶ¨Òå¾ÍÓе㲻Çå³þ¡£
> > > ʲô½Ð×ö×îºÏÊÊ£¿ Õâ¸öÐèÒª¶¨Òå¡£
> > > ÕâÀïÎÒ³¢ÊÔ×Ô¼º·¢»Óһϣº
> > > ab+by=x£¬×ø±êϵÀï¾ÍÊÇÒ»ÌõÖ±Ïߣ¬(x,y)¾ÍÊÇÖ±ÏßÉϵĵ㡣
> > > ËùνºÏÊÊ£¬¾ÍÊǾ¡Á¿½Ó½üÕâÌõÖ±Ïߵĵ㣬µãµ½Ö±ÏߵľàÀë×îС¡£
> > > Ò²¾ÍÊÇÇóµ½ËùÓÐÖ±ÏߵľàÀëÖ®ºÍ×îСµÄµã
> > > d = |x-by-ab|/sqrt(1+b^2)
> > > ÒªÇóËùÓеÄaºÍbµÄ×éºÏËùÐγɵÄdÖ®ºÍ×îС
> > > ΪÁËÈ¥³ý¾ø¶ÔÖµ£¬¿ÉÒÔת»¯ÎªÇódµÄƽ·½Ö®ºÍ×îС
> > > ÓÚÊÇ
> > > D = ¡Æ(x-by-ab)^2/(1+b^2)
> > > Ò²¼´¶ÔD½øÐжþÔªº¯ÊýÇó¼«Öµ
> > > ÓÚÊǶÔDÇó¹ØÓÚxºÍyµÄÆ«µ¼£¬È»ºóÁªÁ¢·½³Ì×éÇó³öËùÓÐפµã ....
> > > ÒÔºó¾ÍÊÇÊýѧ·ÖÎöµÄÄÚÈÝÁË
> > >
> > > ¸Ð¾õÕâ¸öÎÊÌâûË㷨ʲôʶù
> > >
> > > 2007-12-21 ________________________________
> > >
> > > snake ________________________________
> > >
> > > ·¢¼þÈË£º python Bian
> > > ·¢ËÍʱ¼ä£º 2007-12-20  21:43:30
> > > ÊÕ¼þÈË£º python china
> > > ³­ËÍ£º
> > > Ö÷Ì⣺ [python-chinese]ÇóÖú:Ëã·¨ÎÊÌâ
> > >
> > >
> > > ÒÑÖªÁ½×éÊý¾Ýa,b,ÊÔÇó³ö×îºÏÊʵÄx,yÊÇËùÓеÄa,b¶¼½üËÆÂú×ãb=x/(a+y)
> > > ʵÔÚÏë²»³ö¸ÃÈçºÎʵÏÖ,Ï£Íû¸ßÈËÖ¸µãÒ»¶þ.
> > > _______________________________________________
> > > python-chinese
> > > Post: send python-chinese在lists.python.cn
> > > Subscribe: send subscribe to
> > > python-chinese-request在lists.python.cn
> > > Unsubscribe: send unsubscribe to
> > > python-chinese-request在lists.python.cn
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> > > http://python.cn/mailman/listinfo/python-chinese
> > >
> > _______________________________________________
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> > Post: send python-chinese在lists.python.cn
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> > Unsubscribe: send unsubscribe to  python-chinese-request在lists.python.cn
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>
>
>
> _______________________________________________
> python-chinese
> Post: send python-chinese在lists.python.cn
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·¢¼þÈË: python-chinese-bounces在lists.python.cn
[mailto:python-chinese-bounces在lists.python.cn] ´ú±í beck917
·¢ËÍʱ¼ä: 2007Äê12ÔÂ22ÈÕ 15:58
ÊÕ¼þÈË: python-chinese在lists.python.cn
Ö÷Ìâ: Re: [python-chinese] ÇóÖú:Ëã·¨ÎÊÌâ

 

ÊÇ°¡,,ºÃ¶àÊýѧ´ïÈË,,,,ĬĬѧϰing...

ÔÚ07-12-22£¬python Bian <python.bian在gmail.com> дµÀ£º

ºÇºÇ,лл´ó¼Ò»Ø¸´,·¢ÏÖÊýѧ¸ßÈ˲»ÉÙ°¡.

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ÔÚ07-12-21£¬yuting cui <yutingcui在gmail.com  yutingcui在gmail.com> >
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