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Re: Is there any efficient easy way to compare two lists with the same length with Mathematica?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg124469] Re: Is there any efficient easy way to compare two lists with the same length with Mathematica?
  • From: Rex <Aoirex at gmail.com>
  • Date: Sat, 21 Jan 2012 05:11:25 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

Great. Thanks a lot for all of you.

Hope you all have a nice day.


On Tue, Jan 17, 2012 at 7:37 AM, Andy Ross <andyr at wolfram.com> wrote:
> On 1/17/2012 2:34 AM, Rex wrote:
>>
>> Given two lists `A={a1,a2,a3,...an}` and `B={b1,b2,b3,...bn}`, I would
>> say `A>=B` if and only if all `ai>=bi`.
>>
>> There is a built-in logical comparison of two lists, `A==B`, but no
>> `A>B`.
>> Do we need to compare each element like this
>>
>> And@@Table[A[[i]]>=B[[i]],{i,n}]
>>
>> Any better tricks to do this?
>>
>
> If the vectors are very long and there isn't a good reason to expect that
> most false cases will occur early you might consider UnitStep.
>
> a=RandomReal[{0,1},10^6];
> b=a+1;
> c=RandomReal[{0,1},10^6];
>
> In[311]:= And@@Thread[b>a]//AbsoluteTiming
> Out[311]= {0.858011,True}
>
> In[312]:= Total[UnitStep[b-a]]==Length[b]//AbsoluteTiming
> Out[312]= {0.046801,True}
>
> In[313]:= And@@Thread[c>a]//AbsoluteTiming
> Out[313]= {0.858011,False}
>
> In[314]:= Total[UnitStep[c-a]]==Length[c]//AbsoluteTiming
> Out[314]= {0.046801,False}
>
> Andy Ross
> Wolfram Research



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