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Re: list equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121940] Re: list equation
  • From: Adriano Pascoletti <adriano.pascoletti at uniud.it>
  • Date: Fri, 7 Oct 2011 04:44:52 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201110060824.EAA22708@smc.vnet.net>

In[1]:= lst = Tuples[Range[100], 2];
In[9]:= Timing[lstnew = Union[lst, SameTest -> (#1 == Reverse[#2] & )]; ]
Out[9]= {39.09032700000002, Null}


Faster


In[10]:= Timing[lstnew1 = DeleteDuplicates[Sort /@ lst]; ]
Out[10]= {0.008493999999998891, Null}
In[11]:= lstnew === lstnew1
Out[11]= True


Much faster


In[12]:= Timing[lstnew1 = (Sequence @@ Thread[{#1, Range[#1, 100]}] & ) /@
    Range[100]; ]
Out[12]= {0.003968000000000416, Null}
In[13]:= lstnew1 === lstnew
Out[13]= True


Adriano Pascoletti

2011/10/6 dimitris <dimmechan at yahoo.com>

> Hello to all.
>
> Let
>
> lst = Tuples[Range[100], 2];
>
> In the previous list appear elements such us {x,y} and {y,x}. (e.g.
> {3,4} and {4,3}).
> I want to create a new list with {y,x} dropped (that is, in the new
> list appears only {3,4}
> and not {4,3}).
>
> I use
>
> lstnew = Union[lst, SameTest -> (#1 == Reverse[#2] &)]
>
> However it is needed almost 150 sec for this procedure ($Version->5.2
> for Windows)
> I know that my laptop is too old but I guess there is a more efficient
> way to
> create lstnew.
>
> Any ideas?
>
> Thanks a lot
>
> Dimitris
>
>




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