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Re: List element replacement.

• To: mathgroup at smc.vnet.net
• Subject: [mg52031] Re: [mg51980] List element replacement.
• From: DrBob <drbob at bigfoot.com>
• Date: Sun, 7 Nov 2004 01:04:52 -0500 (EST)
• References: <200411060707.CAA25986@smc.vnet.net> <opsg2daylkiz9bcq@monster.cox-internet.com>
• Reply-to: drbob at bigfoot.com
• Sender: owner-wri-mathgroup at wolfram.com

Here's a more readable routine (I think):

ClearAll@removeAndReplace
removeAndReplace::usage = "
removeAndReplace[a,b,c] checks whether List a is
     contained in List b (counting multiplicities). If no, b is returned. If \
yes, the return value is b with the elements of a removed and the elements of \
c added (counting multiplicities)."
removeAndReplace[a_List, b_List, c_List] := Block[{aFreq, bFreq, aa, bb, f},
     If[
       Or[
         Length@a > Length@b,
         {aFreq, bFreq} = Frequencies /@ {a, b};
         Length@aFreq > Length@bFreq,
         {aa, bb} = {aFreq, bFreq}[[All, All, 2]];
         Union[aa, bb] =!= Union[bb],
         f[{n_, x_}] :=
            n > Extract[bFreq, {Position[bb, x][[1, 1]], 1}] &&
             Throw@True;
         Catch[Scan[f, aFreq]; False]
       ],
       b,
       Clear@f; f[lst_, {n_, x_}] := DeleteCases[lst, x, {1}, n];
       Fold[f, Join[b, c], aFreq]]
     ]

I removed Sort and added a check with Union, among other things.

Bobby

On Sat, 06 Nov 2004 16:12:24 -0600, DrBob <drbob at bigfoot.com> wrote:

> Does this do the trick? (My testing wasn't extensive.)
>
> ClearAll@removeAndReplace
> removeAndReplace::usage = "
> removeAndReplace[a,b,c] checks whether List a is
>      contained in List b (counting multiplicities). If no, b is returned. If \
> yes, the return value is b with the elements of a removed and the elements of \
> c added (counting multiplicities and sorted)."
> removeAndReplace[a_List, b_List, c_List] := Block[{aFreq, bFreq, f, pos},
>      Which[
>        Length@a > Length@b, b,
>        Length@(aFreq = Frequencies@a) > Length@(bFreq = Frequencies@b), b,
>        f[{n_, x_}] := ((pos = Position[bFreq[[All, 2]],
>        x]) === {} || (n > Extract[bFreq, {pos[[1, 1]], 1}])) && Throw[False];
>        Catch[Scan[f, aFreq]; Throw@True],
>        f[lst_, {n_, x_}] := DeleteCases[lst, x, {1}, n];
>        Sort@Fold[f, Join[b, c], aFreq],
>        True, b]
>      ]
>
> This could be useful without the Sort, so I'd probably omit it and enter Sort@removeAndReplace[a,b,c], if I needed a sorted result.
>
> In that case, when it removes elements, it removes them starting at the front of Join[b,c].
>
> Bobby
>
> On Sat, 6 Nov 2004 02:07:45 -0500 (EST), Robert G. Wilson v <rgwv at rgwv.com> wrote:
>
>> Et al,
>>
>> 	HELP, I tried this once before but what I received back I could not make it work.
>> Therefore I am following the adage that if at first you do not succeed then wait
>> awhile, rethink, restate and resubmit.
>>
>> 	I have three lists of objects, sorted but not unioned, that is there are elements
>> which will be repeated. Let us label them 'initial' list, 'compare' list, and
>> 'replace' list. If all of the elements in the 'compare' list, including repeats,
>> are in the 'initial' list, then remove those elements and put in the elements from
>> the 'replace' list.
>>
>> Thank you in advance,
>>
>> Bob.
>>
>>
>>
>>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net

• References:
  • List element replacement.
    • From: "Robert G. Wilson v" <rgwv@rgwv.com>