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

*To*: mathgroup at smc.vnet.net*Subject*: [mg124429] Re: Is there any efficient easy way to compare two lists with the same length with Mathematica?*From*: Armand Tamzarian <mike.honeychurch at gmail.com>*Date*: Thu, 19 Jan 2012 05:13:04 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

On Jan 18, 9:58 pm, Andy Ross <an... 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 Min[UnitStep[c - a]] == 1 and Min[Sign[c - a]] == 1 are also pretty quick. Mike