Re: Re: modified list interval calculation
- To: mathgroup at smc.vnet.net
- Subject: [mg21102] Re: [mg21058] Re: modified list interval calculation
- From: Hartmut Wolf <hwolf at debis.com>
- Date: Fri, 17 Dec 1999 01:21:03 -0500 (EST)
- Organization: debis Systemhaus
- References: <199912130451.XAA16217@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Deborah Leddon schrieb:
>
> Hi,
> I am having trouble with the problem below:
>
> I am trying to remove from list, t, those numbers in list, nt, that
> match and then I want to calculate the differences (intervals )
> between succesive numbers in list, t, that do not contain the
> removed number. For example:
> the two lists are;
> t = {1,4,4.5,6,7,8,8.9,9,10,12,140,16,16.8,17,18,
> 19,19.1,19.3,19.5,20}
> nt = {8,8.9,10,17}
>
> the resulting intervals would look like:
> 4 -1, 4.5 - 4, 6 - 4.5, 7 - 6, 140 -12, 16 - 140, 16.8 - 16, 19 - 18,
> 19.1 -19, 19.3 - 19.1, 19.5 - 19.3, 20 - 19.5
> and of course the final list would be:
> intervals = {3, .5, 1.5, 1, 128, -124, .8, 1, .1, .2, .2, .5};
>
> Note that 9 was omitted from the calculation of intervals because it
> would have been involved in a calculation with a removed number
> from the list, t.
>
Dear Debbie,
The following will do what you want:
In[4]:=
(#2 - #1) & @@@
DeleteCases[Partition[t, 2, 1], Alternatives @@ ({_, #} | {#, _} &) /@
nt]
Out[4]=
{3, 0.5, 1.5, 1, 128, -124, 0.8, 1, 0.1, 0.2, 0.2, 0.5}
Or alternatively:
In[5]:=
(#2 - #1) & @@@
Cases[DeleteCases[Partition[t, 2, 1], Alternatives @@ nt, {2}], {_,
_}]
Out[5]=
{3, 0.5, 1.5, 1, 128, -124, 0.8, 1, 0.1, 0.2, 0.2, 0.5}
Kind regards, Hartmut
- References:
- Re: modified list interval calculation
- From: "Deborah Leddon " <Dleddon@students.cas.unt.edu>
- Re: modified list interval calculation