RE: Replacing 1 to many in a list

• To: mathgroup at smc.vnet.net
• Subject: [mg29268] RE: [mg29250] Replacing 1 to many in a list
• From: "David Park" <djmp at earthlink.net>
• Date: Sat, 9 Jun 2001 03:08:56 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Chris,

One possibility:

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

i = 0;
If[# === Null, x[++i], #] & /@ vector
{6, 6, 5, 4, 4, 4, x[1], 6, x[2], 44, x[3], 4, 4, 5}

David Park

> From: Chris Johnson [mailto:cjohnson at shell.faradic.net]
To: mathgroup at smc.vnet.net
>
> 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?
>
> Thanks in advance,
> Chris
>
>

```

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