Re: split again

• To: mathgroup at smc.vnet.net
• Subject: [mg73671] Re: [mg73622] split again
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sat, 24 Feb 2007 02:18:13 -0500 (EST)

```Your zSplit is missing a & and even then would never result in lists longer than one whose elements are all less than or equal to 0.7

Using a different zSplit

z=Table[Random[],{1000}];

zSplit=Split[z,#1<=0.7 === #2<=0.7&];

With this Split, lists longer than one can only contain elements less than or equal to 0.7 so it is unnecessary to test the elements' values.

Select[zSplit,Length[#]>5&][[1]]

{0.465823,0.397768,0.520191,0.0301858,0.516125,0.616719,0.133392,0.33706}

Creating mixed lists

zSplit=Split[z,#1<=0.7 && #2<=0.7||#1>0.7 && #2>0.7&];

{0.465823,0.397768,0.520191,0.0301858,0.516125,0.616719,0.133392,0.33706}

Bob Hanlon

> Hi,
>
> Thank you very much for your help in my provious post. Now, consider
>
> z=Table[Random[],{1000}];
>
> zSplit=Split[z,#1<=0.7 === #2>=0.7];           (bulit thanks to your
> help)
>
> I want to pick the first sublist of zSplit that consists of elements
> <= 0.7 and whose length is greater than a certain number (say 5). I
> think that a good candidate would be using Cases and variations of _ ,
> but I don't know how.
>
> What I want to do (and what my both posts are about) is to find the
> first sublist of z with all elements less than A and the length of
> this sublist must be greater than B. Maybe there exists better
> solution than to Split z in advance since what I need to do in my next
> step is to find ONLY the FIRST sublist of splitted z.
>
> Thanks again,
>
> Arek

```

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