       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 06:20:37 -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 want to create the follwing matrix:
>>
>> 1 2 4
>> 3 5 7
>> 6 8 9
>>
>
> 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]

```

• Prev by Date: Re: capacitor equation solution not working
• Next by Date: Re: Partition a list based on columns
• Previous by thread: Re: Partition a list based on columns
• Next by thread: Re: Partition a list based on columns