       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

> 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:= l1 /. {x_, y_} -> x
>
> Out= {1, 3, 5}
>
> This is what I would expect. Now consider the following list
>
> In:= l2 = {{3, 4}, {5, 6}};
>
> Then I use the same pattern/replacement rule
>
> In:=l2 /. {x_, y_} -> x
>
> Out= {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