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Re: Thread::tdlen: Objects of unequal length in

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  • Subject: [mg127765] Re: Thread::tdlen: Objects of unequal length in
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Mon, 20 Aug 2012 21:28:43 -0400 (EDT)
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You give an example of something that doesn't do what you want but you
never say what it is that you do want done.

Since Last[testdata] is a vector I assume that testdata is an m x n array,

testdata = Array[a, {3, 5}]

{{a[1, 1], a[1, 2], a[1, 3], a[1, 4], a[1, 5]}, {a[2, 1], a[2, 2], a[2, 3],
  a[2, 4], a[2, 5]}, {a[3, 1], a[3, 2], a[3, 3], a[3, 4], a[3, 5]}}

Your first function is

f1 = MapThread[Mean, Sequence[{{testdata}}]]

{{(1/3)*(a[1, 1] + a[2, 1] + a[3, 1]),
     (1/3)*(a[1, 2] + a[2, 2] + a[3, 2]),
     (1/3)*(a[1, 3] + a[2, 3] + a[3, 3]),
     (1/3)*(a[1, 4] + a[2, 4] + a[3, 4]),
     (1/3)*(a[1, 5] + a[2, 5] + a[3, 5])}}

What you have implemented is

f1 ==
 {Mean[testdata]} ==
   {Mean /@ Transpose[testdata]} ==
   {Total[testdata]/Length[testdata]}

True

Length[f1]

1

If what you want is the Mean of each row of testdata then use

meanByRows = Mean /@ testdata

{(1/5)*(a[1, 1] + a[1, 2] + a[1, 3] + a[1, 4] + a[1, 5]),
   (1/5)*(a[2, 1] + a[2, 2] + a[2, 3] + a[2, 4] + a[2, 5]),
   (1/5)*(a[3, 1] + a[3, 2] + a[3, 3] + a[3, 4] + a[3, 5])}

Length[meanByRows]

3

If you want the mean of each column then use

meanByColumns = Mean[testdata]

{(1/3)*(a[1, 1] + a[2, 1] + a[3, 1]),
   (1/3)*(a[1, 2] + a[2, 2] + a[3, 2]),
   (1/3)*(a[1, 3] + a[2, 3] + a[3, 3]),
   (1/3)*(a[1, 4] + a[2, 4] + a[3, 4]),
   (1/3)*(a[1, 5] + a[2, 5] + a[3, 5])}

Length[meanByColumns]

5

meanByColumns ==
 Mean /@ Transpose[testdata] ==
 Total[testdata]/Length[testdata] ==
 f1[[1]]

True

If what you want is the mean of all of the elements of testdata then use

meanOfAll = testdata // Flatten // Mean

(1/15)*(a[1, 1] + a[1, 2] + a[1, 3] + a[1, 4] + a[1, 5] +
      a[2, 1] + a[2, 2] + a[2, 3] + a[2, 4] + a[2, 5] +
      a[3, 1] + a[3, 2] + a[3, 3] + a[3, 4] + a[3, 5])

meanOfAll ==
  Mean[meanByRows] ==
  Mean[meanByColumns] //
 Simplify

True


Bob Hanlon


On Mon, Aug 20, 2012 at 4:14 AM, Aaron <aaron.sokolik at gmail.com> wrote:
> I'm new to Mathematica and trying to run rebuild some programs from other=
 systems in Mathematica.  I'm operating on large data lists and receiving t=
he unequal length error.  However, if I simply paste the output without the=
 extra curly bracket into an operation, everything works.  Obviously, copyi=
ng and pasting wont work for functions... How can I get around this?
>
> Below are the functions I'm using to generate two lists from a single dat=
aset:
>
> ln[13]:= MapThread[Mean, Sequence[{{testdata}}]]
> Out[13]= {{15.0059, 14.9897, 15.0248,....}}
>
> ln[14]:= Last[testdata]
> Out[14]= {14.9602, 14.8624, 15.3364, 15.0231,....}
>
> When I run Length on each of the outputs, I receive the proper number of =
datapoints for the Last function but receive an output of "1" for the Mean =
function.  If I paste the output with only 1 "{" though and run Length, I r=
eceive the proper output.
>
> I know this must be simple but I'm stuck.
>



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