Re: Length and Drop
- To: mathgroup at smc.vnet.net
- Subject: [mg32193] Re: Length and Drop
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Sat, 5 Jan 2002 00:10:31 -0500 (EST)
- References: <a13vdg$dgm$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Steve,
Drop[timeexp,1] evaluates by the
Drop[{{0,0},{7,8}},1 ]
{{7,8}}
It does nont redefine timeexp.
To get this you would have to evaluate
timeexp = Drop[timeexp,1]
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
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"Steve Gray" <stevebg at adelphia.net> wrote in message
news:a13vdg$dgm$1 at smc.vnet.net...
> The following sequence seems wrong to me, in that
> after dropping some elements from a list, its Length is
> still the same. I copied this directly from the notebook
> but removed the In[] and Out[] statements.
>
>
> timexp={{0,0}} -> {{0,0}} [this is ok]
>
> timexp=Append[timexp,{7,8}] -> {{0,0},{7,8}} [this is ok]
>
> Length[timexp] -> 2 [this is ok]
>
> Drop[timexp,1] -> {{7,8}} [this is ok]
>
> Length[timexp] -> 2 [ this is what I did not expect!]
>
> Should not Drop reduce the Length of the list by the amount
> dropped?
>
>