Re: Replacing 1 to many in a list

• To: mathgroup at smc.vnet.net
• Subject: [mg29276] Re: Replacing 1 to many in a list
• From: Carlos Collier <ccollier at tycho.fciencias.unam.mx>
• Date: Sat, 9 Jun 2001 03:09:05 -0400 (EDT)
• Organization: Facultad de Ciencias, UNAM
• References: <9fq1sf\$qdl\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi!, here's an idea

vector = {6, 6, 5, 4, 4, 4, Null, 6, Null, 44, Null, 4, 4, 5};

pos = Flatten[Position[vector, Null]]

{7, 9, 11}

dat = Map[x, pos]

{x[7], x[9], x[11]}

vector[[pos]] = dat;

vector

{6, 6, 5, 4, 4, 4, x[7], 6, x[9], 44, x[11], 4, 4, 5}

- also -

dat2 = Array[x, Length[dat]];
vector[[pos]] = dat2;
vector
{6, 6, 5, 4, 4, 4, x[1], 6, x[2], 44, x[3], 4, 4, 5}

Saludos
Carlos

On Fri, 8 Jun 2001, Chris Johnson wrote:

> I think this is probably simple, but I can't find a natural way to do it
> yet.  I hope someone out there can help.  I have a vector numbers with
> some Null's scattered throughout.  The number and location of missing data
> varies.  For example,
>
> vector = {6, 6, 5, 4, 4, 4, Null, 6, Null, 44, Null, 4, 4, 5}
>
> Position[vector, Null]
>
> {{7}, {9}, {11}}
>
> Now "/." will let me replace all instances of Null with the same variable
> but I want to replace each instance with a different variable.  I would
> prefer to use x[1], x[2] and x[3], but if it were easier, then x[7], x[9]
> and x[11] would be OK also.
>
> The prefered end result would be:
>
> {6, 6, 5, 4, 4, 4, x[1], 6, x[2], 44, x[3], 4, 4, 5}
>
> Any suggestions?
>