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Re: Partition a list based on columns
*To*: mathgroup at smc.vnet.net
*Subject*: [mg119707] Re: Partition a list based on columns
*From*: Ray Koopman <koopman at sfu.ca>
*Date*: Sat, 18 Jun 2011 06:13:20 -0400 (EDT)
*References*: <itcdai$d13$1@smc.vnet.net> <itek2d$s5q$1@smc.vnet.net>
On Jun 16, 9:09 pm, Ray Koopman <koop... at sfu.ca> wrote:
> On Jun 16, 1:02 am, StatsMath <stats.ma... at gmail.com> wrote:
>>
>> [...]
>>
>> What would be a good way to line up elements in a list in a diagonal
>> fashion, for ex: from Range[9] want to create the follwing matrix:
>>
>> 1 2 4
>> 3 5 7
>> 6 8 9
>>
>> Thanks in advance!
>
> This will give an r x c matrix whose elements are
> the integers 1...r*c in the positions you asked for:
>
> m[r_,c_] := With[{d = Which[
> r < c-1, Join[Range[ r ],Table[r,{c-2-r}],Range[ r ,1,-1]],
> r > c-1, Join[Range[c-1],Table[c-1,{r-c}],Range[c-1,1,-1]],
> True , Join[Range[c-2], Range[c-1,1,-1]]]},
> Partition[FoldList[Plus,0,d],c,1] + Range@r]
>
> m[3,5] //TableForm
>
> 1 2 4 7 10
> 3 5 8 11 13
> 6 9 12 14 15
Here's a better way to get an r x c matrix whose elements are the
integers 1...r*c in the desired positions. To put the elements of
a vector v in those positions, just change Range[r*c] to v.
m[r_,c_] := Partition[Range[r*c][[ Ordering@SortBy[
Tuples@{Range@r,Range@c}, Tr] ]], c]
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