Re: Question about Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg106571] Re: Question about Mathematica
- From: "Peter.Pein" <petsie at dordos.net>
- Date: Sat, 16 Jan 2010 06:14:35 -0500 (EST)
- References: <hipld8$7po$1@smc.vnet.net>
- Reply-to: "Peter.Pein" <petsie at dordos.net>
"Dominic" <miliotodc at rtconline.com> schrieb im Newsbeitrag news:hipld8$7po$1 at smc.vnet.net... > Hi. Can someone help me with the following question: > > I have a table of a table of number pairs: > > {{{1,2),(3,-1),{2,-4)},{{1,2},{4,-5},{6,-8}},{{2,-1},{-2,-3},{-4,6}}} > > How may I find the minimum second element in each sub-table? For > example, the first sub-table is: > > {{1,2},{3,-1},{2,-4}} > > I would like to then extract the {2,-4} element from this item. Then > the {6,-8} from the second sub-table, then the {-2,-3} element from the > last. > > Thank you, > Dominic > Hi Domonic, test3 = Table[row[[Ordering[row[[All, 2]]][[1]]]], {row, #}] & is the fastest solution I found so far. The others have been test1 = (Cases[#1, {_, Min[#1[[All, 2]]]}, 1, 1][[1]] &) /@ # & and the slow test2 = Fold[If[#2[[2]] < #1[[2]], #2, #1] &, {foo, Infinity}, #] & /@ # & All three give the desired result for your example and with bigdat = RandomInteger[{-9, 9}, {10^6, 10, 2}]; I get In[12]:= Timing[result1=test1[bigdat];][[1]] Out[12]= 19.406 In[13]:= Timing[result2=test2[bigdat];][[1]] Out[13]= 46.738 In[14]:= Timing[result3=test3[bigdat];][[1]] Out[14]= 8.112 In[15]:= SameQ[result1,result2,result3] Out[15]= True Peter
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