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Reconciling BinCounts and RangeCounts

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68014] Reconciling BinCounts and RangeCounts
  • From: Gregory Lypny <gregory.lypny at videotron.ca>
  • Date: Thu, 20 Jul 2006 06:04:54 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello Everyone,

I get a discrepancy between the results of BinCounts and RangeCounts 
and can confirm only that RangeCounts is, in fact, counting the 
number of instances where a number is at least as big as the lower 
cut-off and less than the upper cut-off.  Not so for BinCounts, which 
leads me to believe that it is buggy or, more likely, I am.

I have a vector, x, with 7320 observations of real numbers in the 
range .06 to .14 with up to seven decimal places.  Here's what I get 
if I use bins or cut-offs of .01.

First with BinCounts
BinCounts[x, {.06, .14, .01}]
{103, 333, 802, 1266, 997, 662, 611, 2265, 281}

Now with RangeCounts
RangeCounts[x, Range[.07, .14, .01]]
{103, 333, 797, 1270, 997, 663, 611, 2265, 281}

Notice that elements 3, 4, and 6 of the results differ.  So I tried 
to check what was going on by using Select and was able to confirm 
all of the RangeCounts elements.  For example, the third element of 
the RangeCounts results, 797, can be confirmed by using

Length[Select[x, .08 =B2 # < .09 &]] >>>> returns 797

However, the third element of the BinCounts results, 802, can be 
obtained only if I include the upper bound, .09, in the count as

Length[Select[x, .08 =B2 # =B2 .09 &]] >>>>> returns 802,

which of course makes no sense because we need a strong inequality 
for one of them.  But it gets worse.  When I go on to check elements 
4 and 6 of BinCounts, there is no combination of weak or strong 
inequalities that will give me the results 1266 and 662.

Can anyone shed any light on this?  In the meantime, I think it 
safest to use RangeCounts.

Regards,

	Gregory=



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