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Re: Re: style question

• To: mathgroup at smc.vnet.net
• Subject: [mg79560] Re: [mg79535] Re: style question
• From: Carl Woll <carlw at wolfram.com>
• Date: Sun, 29 Jul 2007 00:13:36 -0400 (EDT)
• References: <200707280945.FAA00305@smc.vnet.net>

Bill Rowe wrote:

>On 7/26/07 at 5:39 AM, carlw at wolfram.com (Carl Woll) wrote:
>
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>>Yaroslav Bulatov wrote:
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>>
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>>>What is the recommended way of counting the number of matches in
>>>two lists?
>>>      
>>>
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>>>The natural way would be to thread over Equal, but Equal will
>>>evaluate before Thread gets to it. The method below works, but
>>>somehow feels wrong
>>>      
>>>
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>>>m1 = RandomInteger[{1}, {10^5, 2, 2}]; m2 = RandomInteger[{1},
>>>{10^5, 2, 2}]; matches = Thread[temporary[m1, m2]] /. temporary ->
>>>Equal; Count[matches, True]
>>>      
>>>
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>>For this particular problem (in version 6), it's faster to use
>>arithmetic and total:
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>>In[5]:= Count[Total[Unitize[m1 - m2], {2, 3}], 0] // Timing
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>>Out[5]= {0.391, 62300}
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>On my machine, it is somewhat faster to make use this idea and
>SparseArray, i.e.,
>
>In[15]:= k = 10^6;
>m1 = RandomInteger[1, {k, 2, 2}];
>m2 = RandomInteger[1, {k, 2, 2}];
>
>In[18]:= Count[Total[Unitize[m1 - m2], {2, 3}], 0] // Timing
>
>Out[18]= {1.12376,62590}
>
>In[19]:= k - (SparseArray[Total[Unitize[m1 - m2], {2, 3}]] /.
>     SparseArray[_, _, _, a_] :> Length[Last@a]) // Timing
>
>Out[19]= {0.929573,62590}
>
>In[20]:= $Version
>
>Out[20]= 6.0 for Mac OS X PowerPC (32-bit) (June 19, 2007)
>--
>To reply via email subtract one hundred and four
>  
>
I used the Count approach for pedagogical reasons. It was already almost 
an order of magnitude faster than the OPs solution, and other solutions 
(that I could think of) were only marginally faster. At any rate, a 
slightly faster solution than your SparseArray approach is to use Total 
and Unitize again:

k = 10^6;
m1 = RandomInteger[1, {k, 2, 2}];
m2 = RandomInteger[1, {k, 2, 2}];

In[62]:= k - (SparseArray[Total[Unitize[m1 - m2], {2, 3}]] /.
    SparseArray[_, _, _, a_] :> Length[Last@a]) // Timing
Out[62]= {0.375,62806}

In[63]:= k - Total[Unitize[Total[Unitize[m1 - m2], {2, 3}]]] // Timing
Out[63]= {0.344,62806}

Carl Woll
Wolfram Research

• References:
  • Re: style question
    • From: Bill Rowe <readnewsciv@sbcglobal.net>