Re: style question
- To: mathgroup at smc.vnet.net
- Subject: [mg79496] Re: style question
- From: Donald DuBois <donabc at comcast.net>
- Date: Sat, 28 Jul 2007 05:24:46 -0400 (EDT)
>Correction to self. I overlooked the three->dimensionality of m1 and >m2. Again I'm without Mathematica; still you might try: > >matches = MapThread[Equal, {m1, m2}, 3]; >Count[matches, True, {3}] > >By the way, your syntax for the first argument of >RandomInteger >appears to be undocumented. > >Vince Virgilio > Hello Vince: I think you were right the first time. {m1, m2} = RandomInteger[1, {2, 10^5, 2, 2}]; (* First time *) matches1 = MapThread[Equal, {m1, m2}]; Count[matches1, True] Out[30]= 6189 (* Second time *) matches2 = MapThread[Equal, {m1, m2}, 3]; Count[matches2, True, {3}] Out[31]= 200218 >From probability, one would expect about 6250 matches. This is because {{a,b},{c,d}} can have 16 different values since a, b , c and d are either 0 or 1 with equal probability. So, {{a,b},{c,d}} will match {{e,f},{g,h}} approx. 1/16-th of the time. 1/16-th of 10^5 = 6250. Best, Don