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

*To*: mathgroup at smc.vnet.net*Subject*: [mg127773] Re: Thread::tdlen: Objects of unequal length in*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Tue, 21 Aug 2012 04:59:32 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

On 8/20/12 at 4:14 AM, aaron.sokolik at gmail.com (Aaron) 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 the unequal length error. However, if I simply paste the >output without the extra curly bracket into an operation, everything >works. Obviously, copying 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 dataset: >ln[13]:= MapThread[Mean, Sequence[{{testdata}}]] It is not entirely clear as to what is being assigned to testdata. So, I will assume you have two simple (1D) lists of equal length. For example, starting with a = RandomInteger[{0, 10}, 5]; b = RandomInteger[{11, 20}, 5]; Then I can set testData to testData = {a, b}; With this I can do In[4]:= MapThread[Mean[{##}] &, testData] Out[4]= {13/2,13,11,27/2,29/2} note the difference in syntax and note I get a list of 5 elements. That is I am averaging the first element of list a with the first element of list b and so forth. Notice In[5]:= Dimensions[testData] Out[5]= {2,5} That is testData consists of two lists of equal length. Notice In[6]:= Dimensions[Sequence[{{testData}}]] Out[6]= {1,1,2,5} additionally In[7]:= Dimensions[{{testData}}] Out[7]= {1,1,2,5} That is Sequence isn't doing what you think and if you have defined testData as I have above the result then {{testData}} will not have the dimensional structure needed by MapThread. In fact, MapThread is not needed at all since In[8]:= Mean[testData] Out[8]= {13/2,13,11,27/2,29/2} produces exactly the same result. And if the your desire was to have return the mean of list a along with the mean of list b rather than averaging an element of list a with the corresponding element of list b, you could do either In[9]:= Mean /@ testData Out[9]= {36/5,81/5} or In[9]:= Mean /@ testData Out[9]= {36/5,81/5}