Re: Partition a list into sublists of unequal size

• To: mathgroup at smc.vnet.net
• Subject: [mg83946] Re: Partition a list into sublists of unequal size
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Wed, 5 Dec 2007 07:11:13 -0500 (EST)
• Organization: The Open University, Milton Keynes, UK
• References: <fj37cv\$hq0\$1@smc.vnet.net>

```antf wrote:

> I have a list that looks like the following:
>
> list={1,0,0,1,0,0,1,1,0,1,0,0,0,0,1,0}
>
> I then find the positions of the 1s in the list:
>
> positions = Flatten@Position[list,1]
>
> {1,4,7,8,10,15}
>
> I then want to partition a third list into sublists so that the first
> sublist is from index 1;;1, the second sublist is from index 2;;4, the
> third sublist is from index 5;;7, the fourth sublist is just element
> 8, etc.
>
> Currently, I use Partition on the positions sublist with overlap as
> in:
>
> Partition[Join[{1},positions],2,1]
>
> {{1,1},{1,4},{4,7},{7,8},{8,10},{10,15}}
>
> I then use this list to partition the target list by Mapping Take to
> each element of this list and adjusting the first index appropriately.
>
> I am sure there must be a simpler way to do this.  Can anyone tell me
> how to do this operation?

The following might not be the simpler way you were expected; still, it
might be a source of inspiration :)

lst = {1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0};

myPart[target_List, source_List] /; Length@target == Length@source :=
Take[target, #] & /@
Flatten@Position[source, 1]]]

myPart[lst, lst]

returns {{1}, {0, 0, 1}, {0, 0, 1}, {1}, {0, 1}, {0, 0, 0, 0, 1}}

(Note that in the case where the source and target lists are the same
Split[lst, (#1 != 1) &] would be enough to get the desired result.)

Also, in the example you provided, the last sequence of running zeros is
not taken into account. Is this desired?

Regards,
--
Jean-Marc

```

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