Re: Re: Mathematica 6.0 easier for me ... (small review)

• To: mathgroup at smc.vnet.net
• Subject: [mg76639] Re: [mg76556] Re: [mg76457] Mathematica 6.0 easier for me ... (small review)
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Fri, 25 May 2007 06:28:21 -0400 (EDT)
• References: <200705230859.EAA23180@smc.vnet.net> <2889719.1180010222826.JavaMail.root@m35>

```Twenty minutes search of the documentation yields no clue what \$1 and \$2

Care to share the secret?

Bobby

On Thu, 24 May 2007 04:53:53 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl>

wrote:

>
> On 23 May 2007, at 17:59, Paul at desinc.com wrote:
>
>> If I have
>> lis={{1,10},{2,10},...{9,10},{11,20},{12,20}...{19,20}
>>
>> How do I use functional and/or rule to determine where the second
>> number (lis[[i,2]]) jumped from 10 to 20 to 30 and save the pair.
>> Assuming there was noise, I only want to store the first 10->20, then
>> look for 20->30 and so on.  So in time, I want my search to change as
>> I progress through the list.  Any input appreciated!
>
> I am not quite sure if I really understand what you wish to do. But
> if what I think is correct, then it seems easy to do it using rule
> based programming:
>
> FirstJump[lis_, n_, m_] := Flatten[({
>      lis} /. {___, {x_, n}, {y_, m}, ___} :> Position[lis, {x, n}])]
>
> For example:
>
>
> ls={{1,10},{2,10},{9,10},{11,20},{12,20},{19,20},{20,30}};
>
>
> FirstJump[ls,10,20]
>
> {3}
>
>
> FirstJump[ls,20,30]
>
> {6}
>
> The above assumes that all the pairs in your list are distinct. If
> they are not, then something more complicated is needed, e.g.:
>
> Clear[FirstJump]
>
> FirstJump[lis_, n_, m_] := Position[Flatten[({lis} /. {a___, {x_, n}, {
>      y_, m}, b___} :> {a, \$1, \$2, b}), 1], \$1, 1]
>
> ls = {{1,10}, {2, 10}, {9, 10}, {9, 10}, {11, 20}, {12, 20}, {19,
> 20}, {20, 30}};
>
>
> FirstJump[ls, 10, 20]
>
> {{4}}
>
>
> Andrzej Kozlowski
>
>

--

DrMajorBob at bigfoot.com

```

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