Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: Replacement Rule

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32088] RE: [mg32064] Replacement Rule
  • From: "David Park" <djmp at earthlink.net>
  • Date: Sat, 22 Dec 2001 04:23:09 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Brian,

A more efficient solution to your problem would be

l1 = {{1, 2}, {3, 4}, {5, 6}};
l2 = {{3, 4}, {5, 6}};

First /@ l1
{1, 3, 5}

First /@ l2
{3, 5}

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/ 


> From: Brian Higgins [mailto:bghiggins at ucdavis.edu]
To: mathgroup at smc.vnet.net
> 
> I have the following list
> 
> l1={{1,2},{3,4},{5,6}};
> 
> and then I operate on it with the following pattern/replacement rule
> 
> In[3]:= l1 /. {x_, y_} -> x
> 
> Out[3]= {1, 3, 5}
> 
> This is what I would expect. Now consider the following list
> 
> In[4]:= l2 = {{3, 4}, {5, 6}};
> 
> Then I use the same pattern/replacement rule 
> 
> In[5]:=l2 /. {x_, y_} -> x
> 
> Out[5]= {3, 4}
> 
> I was hoping to get
> 
> {3,5}
> 
> What am I missing? The FullForm of l2 is basically the same structure
> as l1, as far as I can tell....
> 
> Brian
> 


  • Prev by Date: Re: Displaying Mathematica's Global rules
  • Next by Date: Re: Solutions that are not solutions
  • Previous by thread: Re: Replacement Rule
  • Next by thread: Re: Replacement Rule