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Re: Creating sub-list based on a given criterion
*To*: mathgroup at smc.vnet.net
*Subject*: [mg110592] Re: Creating sub-list based on a given criterion
*From*: Peter Pein <petsie at dordos.net>
*Date*: Sat, 26 Jun 2010 08:42:22 -0400 (EDT)
*References*: <i0495h$5lm$1@smc.vnet.net>
Am Sat, 26 Jun 2010 07:10:41 +0000 (UTC)
schrieb Dimitris Emmanoulopoulos <demmanoulopoulos at hotmail.com>:
> Hello
>
> I was wondering if there is any straightforward way to bin a list
> into sub-lists containing those elements that full-fill a given
> criterion. Consider the following list
>
> ds=
> {{{0,0,1},{1,0,-1},{2,0,-1},{3,0,-1},{4,0,-1},{5,588,-1},{6,2345,1},{7,2418,1},{8,2323,1},{9,2179,1}},{{480,23,1},{487,23,1},{470,29,1},{478,32,1},{468,25,1},{498,19,-1},{494,15,1},{496,20,1},{455,33,1},{462,27,1}},{{781,1,1},{779,1,1},{784,0,1},{778,0,1},{766,1,-1},{735,1,-1},{758,1,-1},{733,2,-1},{716,0,-1},{794,1,1}}}
>
> The criterion is that the sum of the second elements in each new
> sub-list should be at least 25?
>
> After the re-binning we should get something like:
>
> {{0,0,1},{1,0,-1},{2,0,-1},{3,0,-1},{4,0,-1},{5,588,-1}}
> {6,2345,1}
> {7,2418,1}
> {8,2323,1}
> {9,2179,1}
> {{480,23,1},{487,23,1}}
> {470,29,1}
> {478,32,1}
> {468,25,1},
> {498,19,-1},{494,15,1},
> {496,20,1},{455,33,1}
> {{462,27,1},{781,1,1},{779,1,1},{784,0,1},{778,0,1},{766,1,-1},{735,1,-1},{758,1,-1},{733,2,-1},{716,0,-1},{794,1,1}}}
>
> The first bin has a total of the second elements 588, the second
> 2345, the third 2418,...,the sixth 46, the seventh 29, =C3=A2=E2=82=AC=C2=
=A6,the
> eleventh 53, and the last one 35.
>
> In other words, imagine that these are the results of a counting
> experiment and the second element in every entry (in ds) corresponds
> to counts. The question is how can we re-bin the data in order to
> have at least 25 counts per new-bin? Of course it is straightforward
> to do it using a Do-loop, but I was wondering if there is any simpler
> way to do it using the efficient Mathematica syntax in an
> =C3=A2=E2=82=AC=C5=93one-liner expression=C3=A2=E2=82=AC_.
>
> Thanks a lot
> Dimitris
>
Hi Dimitris,
I did not manage it in one line but at least looks like Mathematica ;-)
In[2]:= d2 = Block[
{sum = 0, lst = Flatten[ds, 1]},
lst = Map[First,
Split[
Map[{#, If[(sum += #[[2]]) >= 25, sum=0; False, True]}&, lst],
Last[#1]&],
{2}];
If[Length[lst] >= 2 && Total[Last[lst][[All, 2]]] < 25,
Join[Drop[lst, -2], {Join @@ Take[lst, -2]}],
lst]
]
Out[2]=
{
{{0,0,1},{1,0,-1},{2,0,-1},{3,0,-1},{4,0,-1},{5,588,-1}},
{{6,2345,1}},
{{7,2418,1}},
{{8,2323,1}},
{{9,2179,1}},
{{480,23,1},{487,23,1}},
{{470,29,1}},
{{478,32,1}},
{{468,25,1}},
{{498,19,-1},{494,15,1}},
{{496,20,1},{455,33,1}},
{{462,27,1},{781,1,1},{779,1,1},{784,0,1},{778,0,1},
{766,1,-1},{735,1,-1},{758,1,-1},{733,2,-1},{716,0,-1},
{794,1,1}}
}
I guess a loop based approach would be faster in execution but I did
not try it.
Cheers,
Peter
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