Re: options Transpose[] ?
- To: mathgroup at smc.vnet.net
- Subject: [mg48997] Re: options Transpose[] ?
- From: koopman at sfu.ca (Ray Koopman)
- Date: Mon, 28 Jun 2004 04:13:40 -0400 (EDT)
- References: <cbi6v2$lb6$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Petr Kujan <kujanp at fel.cvut.cz> wrote in message
news:<cbi6v2$lb6$1 at smc.vnet.net>...
> Hello Mathematica User Group.
>
> If I have an multidimensional array A:
>
> In[1] :=
> A = Array[Random[]&, dims = {3,2,4}]
>
> Than I expect that dimension of transposed A with options per = {3,1,2}
> is equal {4,3,2} === dims[[per]] , but it is not equal!
>
> Dimensions of transposed A with options per = {3,1,2} is:
> {2,4,3}.
>
> In[2] :=
> Dimensions[Transpose[A, per = {3, 1, 2}]]
> dims[[per]]
>
> It' s correct or it's bug?
> How can I get the same dimensions?
> [...]
Mathematica treats per as a "TO" list, in which the i-th value
specifies which new dimension the old i-th dimension goes TO.
It interprets {3,1,2} as meaning that
old dimension 1 --> new dimension 3,
old dimension 2 --> new dimension 1,
old dimension 3 --> new dimension 2.
You're treating per as a "FROM" list, in which the i-th value
specifies which old dimension the new i-th dimension comes FROM:
you want {3,1,2} to mean that
new dimension 1 <-- old dimension 3,
new dimension 2 <-- old dimension 1,
new dimension 3 <-- old dimension 2.
In this case, per = {2,3,1} will get you what you want. More generally,
convert[per_] := Last@Transpose@Sort@Transpose@{per,Range@Length@per}
will convert one form of permutation list to the other.
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