# [PD] list operation

Matt Barber brbrofsvl at gmail.com
Wed Apr 28 02:50:53 CEST 2010

```One thing to watch out for;  if you get two items in the list that are
equally as far as the test value, mine outputs one list with the value
replacement at each appropriate index, while Jack's outputs several
lists with the value replacement at only one index, but each
replacement sequentially, if that makes any sense.

So if the test value is 2.02, and the input list is <1 0 3 2 4 3 3 2 0
4>, mine will output:

<1 0 3 2.02 4 3 3 2.02 0 4> as one list, while Jack's gives two lists:

<1 0 3 2.02 4 3 3 2 0 4>
<1 0 3 2 4 3 3 2.02 0 4>

Matt

On Tue, Apr 27, 2010 at 8:16 PM, Jack <jack at rybn.org> wrote:
> An alternative. Also full pure-pd.
> ++
>
> Jack
>
>
>
> Le mardi 27 avril 2010 à 18:49 -0400, Matt Barber a écrit :
>> Try the attached (threw together using list-abs) -- right inlet is the
>> value (12 in your example) and left inlet is the list of floats; I
>> think this follows established list-abs syntax.
>>
>> It's possible that I overlooked a list-abs that already does this, and
>> there is probably a step or two you could skip.
>>
>> You might get poor results if the list is a bang or contains symbols.
>>
>> Matt
>>
>>
>> > Hello everyone,
>> >
>> > I would like to compare all the values in a list with a value of input and
>> > then decide which is the closest value and replace that value closer to the
>> > amount of input.
>> >
>> > For example, I have a list <2, 10, 35> and have an input value of <12> I
>> > would like the list as output <2, 12, 35>. I'm almost getting, but the patch
>> > is so ugly that I will not show here, i'm sure there is a more elegant
>> > solution. Does anyone have an idea?
>> > tnx ;)
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>

```