Re: Re: finding the position of a pattern in list (Correction)

• To: mathgroup at smc.vnet.net
• Subject: [mg64619] Re: Re: [mg64583] finding the position of a pattern in list (Correction)
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Fri, 24 Feb 2006 00:18:14 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```There is an error in my second method. It will indicate a match for the last
position if the first data element is 1 and the last data element is 0.  For
example,

pos1[data_]:=Position[Partition[data,2,1],{0,1}]//Flatten;

pos2[data_]:=Position[RotateLeft[data]-data,1]//Flatten;

data={1,0,0,1,0,0,1,0};

pos1[data]

{3,6}

pos2[data]

{3,6,8}

The correct method is

pos3[data_]:=Position[Most[RotateLeft[data]]-Most[data],1]//Flatten;

pos3[data]

{3,6}

The revision is still much faster than the original.

data=Table[Random[Integer],{100000}];

pos1[data]==pos3[data]

True

Timing[pos1[data]][[1]]

0.389007 Second

Timing[pos3[data]][[1]]

0.132175 Second

Bob Hanlon

>
> From: Bob Hanlon <hanlonr at cox.net>
To: mathgroup at smc.vnet.net
> Subject: [mg64619] Re: Re: [mg64583] finding the position of a pattern in list
>
> Here is a faster method than the one that I first suggested.
>
> pos1[data_]:=Position[Partition[data,2,1],{0,1}]//Flatten;
>
> pos2[data_]:=Position[RotateLeft[data]-data,1]//Flatten;
>
> data=Table[Random[Integer],{100000}];
>
> pos1[data]==pos2[data]
>
> True
>
> Timing[pos1[data]][[1]]
>
> 0.39032 Second
>
> Timing[pos2[data]][[1]]
>
> 0.128189 Second
>
>
> Bob Hanlon
>
> >
> > From: Bob Hanlon <hanlonr at cox.net>
To: mathgroup at smc.vnet.net
> > Subject: [mg64619] Re: [mg64583] finding the position of a pattern in list
> >
> > data={0,0,1,1,1,0,0,1,1,1,0};
> >
> > Position[Partition[data,2,1],{0,1}]//Flatten
> >
> > {2,7}
> >
> >
> > Bob Hanlon
> >
> > >
> > > From: Gang Ma <contactmagang at gmail.com>
To: mathgroup at smc.vnet.net
> > > Subject: [mg64619] [mg64583] finding the position of a pattern in list
> > >
> > > Hi,
> > > I am working on a program to do the following: My data is a list of 0
> > > and 1. For example, {0,0,1,1,1,0,0,1,1,1,0}. I want to find the
> > > positions of all the pattern of {0,1}.  In my previous example, the
> > > first {0,1} is at 2 and and the second {0,1} appears at 7. I can
> > > write a loop to do this, but I have several thousands such lists,
> > > the computation will be time consuming using loop.
> > >
> > > My question is whether it is possible to use the pattern match to do
> > > this quickly.  If not for the list, do I need to convert the list to
> > > string then use some pattern match for string?  Thank you very much.
> > >
> > > regards,
> > >
> > > Gang Ma
> > >
> > >
> > >
> > >
> >
>

```

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