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Murta
Aug 23, 2012, 2:51:49 AM8/23/12
to
Hi All
I have a simple problem that is:
l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
I want to replace all a,b,a sequence by X to get:
l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
Then I want to group it by X intervals as
l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
If I don't need to put the intermediate X, even better!
I think the with pattern, RaplaceAll and DeleteCases I can do It. Some clue?
Tks
Murta
I have a simple problem that is:
l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
I want to replace all a,b,a sequence by X to get:
l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
Then I want to group it by X intervals as
l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
If I don't need to put the intermediate X, even better!
I think the with pattern, RaplaceAll and DeleteCases I can do It. Some clue?
Tks
Murta
Ulrich Arndt
Aug 23, 2012, 8:52:47 PM8/23/12
to
This one requires no X
ls = ReplaceRepeated[l, {x___, a, b, a, y___} -> {{x}, {y}}]
Extract[ls, Position[ls, {a___?AtomQ}]]
Ulrich
ls = ReplaceRepeated[l, {x___, a, b, a, y___} -> {{x}, {y}}]
Extract[ls, Position[ls, {a___?AtomQ}]]
Ulrich
Ulrich Arndt
Aug 23, 2012, 8:56:21 PM8/23/12
to
Map[Replace[#, {X, y___} -> {y}] &,
Split[ReplaceRepeated[
l, {x___, a, b, a, y___} -> {x, X, y}], #2 =!= X &]]
Map[DeleteCases[#, X] &,
Split[ReplaceRepeated[
l, {x___, a, b, a, y___} -> {x, X, y}], #2 =!= X &]]
Split[ReplaceRepeated[
l, {x___, a, b, a, y___} -> {x, X, y}], #2 =!= X &]]
Map[DeleteCases[#, X] &,
Split[ReplaceRepeated[
l, {x___, a, b, a, y___} -> {x, X, y}], #2 =!= X &]]
Bob Hanlon
Aug 24, 2012, 5:07:23 AM8/24/12
to
Clear[f]
f[list_?VectorQ] := Cases[
(list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
{} ->
Sequence[], _?VectorQ, Infinity]
f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
11, 12}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
a, b, a}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
11, 12, a, b, a}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
Bob Hanlon
f[list_?VectorQ] := Cases[
(list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
{} ->
Sequence[], _?VectorQ, Infinity]
f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
11, 12}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
a, b, a}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
11, 12, a, b, a}]
{{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
Bob Hanlon
Fred Simons
Aug 25, 2012, 4:26:30 AM8/25/12
to
Solutions with pattern matching can be very elegant, such as Bob
Hanlon's solution, but also tend to be very slow when the problem is a
little bit larger.
Here is a solution by finding the position of the triple {a,b,a} and
then taking the parts of the list outside these triples.
g[lst_] := With[{limits=Partition[Flatten[{1,
Position[Partition[lst,3,1], {a,b,a}] /.
{n_Integer}->{n-1,n+3},Length[lst]}],2]}, Take[lst, #]& /@ limits]
In the following commands f is Bob's function:
In[4]:= lst={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} ;
z1=f[lst]; // Timing
z2=g[lst]; // Timing
z1===z2
Out[5]= {0.,Null}
Out[6]= {0.,Null}
Out[7]= True
Both functions are fast enough. Now we make the list 500 times longer:
In[12]:= lstt=Flatten[Table[lst, {500}],1];
z1=f[lstt]; // Timing
z2=g[lstt]; // Timing
z1===z2
Out[13]= {5.007632,Null}
Out[14]= {0.015600,Null}
Out[15]= True
Fred Simons
Eindhoven University of Technology
Op 24-8-2012 11:04, Bob Hanlon schreef:
> -----
> Geen virus gevonden in dit bericht.
> Gecontroleerd door AVG - www.avg.com
> Versie: 2012.0.2197 / Virusdatabase: 2437/5219 - datum van uitgifte: 08/23/12
>
>
>
Hanlon's solution, but also tend to be very slow when the problem is a
little bit larger.
Here is a solution by finding the position of the triple {a,b,a} and
then taking the parts of the list outside these triples.
g[lst_] := With[{limits=Partition[Flatten[{1,
Position[Partition[lst,3,1], {a,b,a}] /.
{n_Integer}->{n-1,n+3},Length[lst]}],2]}, Take[lst, #]& /@ limits]
In the following commands f is Bob's function:
In[4]:= lst={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} ;
z1=f[lst]; // Timing
z2=g[lst]; // Timing
z1===z2
Out[5]= {0.,Null}
Out[6]= {0.,Null}
Out[7]= True
Both functions are fast enough. Now we make the list 500 times longer:
In[12]:= lstt=Flatten[Table[lst, {500}],1];
z1=f[lstt]; // Timing
z2=g[lstt]; // Timing
z1===z2
Out[13]= {5.007632,Null}
Out[14]= {0.015600,Null}
Out[15]= True
Fred Simons
Eindhoven University of Technology
Op 24-8-2012 11:04, Bob Hanlon schreef:
> Geen virus gevonden in dit bericht.
> Gecontroleerd door AVG - www.avg.com
> Versie: 2012.0.2197 / Virusdatabase: 2437/5219 - datum van uitgifte: 08/23/12
>
>
>
Murta
Aug 25, 2012, 4:34:16 AM8/25/12
to
Tks for Bob and Ulrich!
I choosed to work with this combination of both
SplitAt[list_,spliter_]:=Module[{l},
l=(list//.{s___,Sequence@@spliter,r___}->{{s},{r}})/.{}->Sequence[];
Cases[l,_?VectorQ,Infinity]
]
I choosed to work with this combination of both
SplitAt[list_,spliter_]:=Module[{l},
l=(list//.{s___,Sequence@@spliter,r___}->{{s},{r}})/.{}->Sequence[];
Cases[l,_?VectorQ,Infinity]
]
Murta
Aug 26, 2012, 2:51:16 AM8/26/12
to
Good point Fred
You example remembered me I miss in Mathematica the possibility to work with inter row operations in a simpler way (simpler than ListConvolve).
For example, the Position function (or some new one) could looking for not just one element in the list, but for sequence of elements too and retur it position. With this option we could avoid to use partition in your solution.
The problem with Partition in large list is the waste of memory. See the example:
data = RandomReal[1, 10^6]
m1 = ByteCount[data];
m2 = ByteCount[Partition[data, 3, 1]];
N@m2/m1
Out[1]=3
We break the list and then search. The best performance way would be search the sequence (a,b,a) without break it.
Map function to me is a good example of function that could have one "MapOverPartition". It would save a lot of memory.
tks
Murta
You example remembered me I miss in Mathematica the possibility to work with inter row operations in a simpler way (simpler than ListConvolve).
For example, the Position function (or some new one) could looking for not just one element in the list, but for sequence of elements too and retur it position. With this option we could avoid to use partition in your solution.
The problem with Partition in large list is the waste of memory. See the example:
data = RandomReal[1, 10^6]
m1 = ByteCount[data];
m2 = ByteCount[Partition[data, 3, 1]];
N@m2/m1
Out[1]=3
We break the list and then search. The best performance way would be search the sequence (a,b,a) without break it.
Map function to me is a good example of function that could have one "MapOverPartition". It would save a lot of memory.
tks
Murta
> > Clear[f]
>
> >
>
> > f[list_?VectorQ] := Cases[
>
> > (list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
>
> > {} ->
>
> > Sequence[], _?VectorQ, Infinity]
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
>
> > 11, 12}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
>
> > a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
>
> > 11, 12, a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> >
>
> > Bob Hanlon
>
> >
>
> >
>
>
>
> >
>
> > f[list_?VectorQ] := Cases[
>
> > (list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
>
> > {} ->
>
> > Sequence[], _?VectorQ, Infinity]
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
>
> > 11, 12}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
>
> > a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
>
> > 11, 12, a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> >
>
> > Bob Hanlon
>
> >
>
> >
>
>
> >> Hi All
>
> >>
>
> >> I have a simple problem that is:
>
> >>
>
> >> l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
>
> >>
>
> >> I want to replace all a,b,a sequence by X to get:
>
> >>
>
> >> l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
>
> >>
>
> >> Then I want to group it by X intervals as
>
> >> l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
>
> >>
>
> >> If I don't need to put the intermediate X, even better!
>
> >> I think the with pattern, RaplaceAll and DeleteCases I can do It. Some clue?
>
> >> Tks
>
> >> Murta
>
> >>
>
> >>
>
> >>
>
> >
>
> >
>
>
> >>
>
> >> I have a simple problem that is:
>
> >>
>
> >> l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
>
> >>
>
> >> I want to replace all a,b,a sequence by X to get:
>
> >>
>
> >> l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
>
> >>
>
> >> Then I want to group it by X intervals as
>
> >> l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
>
> >>
>
> >> If I don't need to put the intermediate X, even better!
>
> >> I think the with pattern, RaplaceAll and DeleteCases I can do It. Some clue?
>
> >> Tks
>
> >> Murta
>
> >>
>
> >>
>
> >>
>
> >
>
> >
>
Ulrich Arndt
Aug 28, 2012, 4:51:08 AM8/28/12
to
This might come close to a more general position
But is not so efficient in Timing as Partition/Position...
Clear[pos]
pos[lst_?VectorQ,pat_?VectorQ]:=Module[{len=Length[pat]-1},Reap[Scan[If[lst[[#;;#+len]]==pat,Sow[#]]&,Range[Length[lst]-len]]][[2,1]]]
Clear[u]
u[lst_]:=With[{limits=Partition[Flatten[{1,(pos[lst,{a,b,a}]/.n_Integer->{n-1,n+3}),Length[lst]}],2]},Take[lst,#]&/@limits]
Clear[g]
g[lst_] :=
With[{limits =
Partition[
Flatten[{1,
Position[
Partition[lst, 3, 1], {a, b, a}] /. {n_Integer} -> {n - 1,
n + 3}, Length[lst]}], 2]}, Take[lst, #] & /@ limits]
lstt=Flatten[Table[lst,{500}],1];
mmu = MaxMemoryUsed[];
z1=u[lstt];//Timing
MaxMemoryUsed[] - mmu
mmu = MaxMemoryUsed[];
z2=g[lstt];//Timing
MaxMemoryUsed[] - mmu
z1===z2
Out[199]= {0.053322,Null}
Out[200]= 0
Out[202]= {0.025853,Null}
Out[203]= 250064
Out[204]= True
Am 26.08.2012 um 08:51 schrieb Murta:
> Good point Fred
>
> You example remembered me I miss in Mathematica the possibility to =
work with inter row operations in a simpler way (simpler than =
ListConvolve).
> For example, the Position function (or some new one) could looking for =
not just one element in the list, but for sequence of elements too and =
retur it position. With this option we could avoid to use partition in =
your solution.
> The problem with Partition in large list is the waste of memory. See =
a
>>
>> little bit larger.
>>
>>
>>
>> Here is a solution by finding the position of the triple {a,b,a} and=
>>
>> then taking the parts of the list outside these triples.
>>
>>
>>
>> g[lst_] := With[{limits=Partition[Flatten[{1,
>>
>> Position[Partition[lst,3,1], {a,b,a}] /.
>>
>> {n_Integer}->{n-1,n+3},Length[lst]}],2]}, Take[lst, #]& /@ limits]
>>
>>
>>
>> In the following commands f is Bob's function:
>>
>>
>>
>> In[4]:= lst={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} ;
>>
>> z1=f[lst]; // Timing
>>
>> z2=g[lst]; // Timing
>>
>> z1===z2
>>
>> Out[5]= {0.,Null}
>>
>> Out[6]= {0.,Null}
>>
>> Out[7]= True
>>
>>
>>
>> Both functions are fast enough. Now we make the list 500 times =
08/23/12
>>
>>>
>>
>>>
>>
>>>
>
But is not so efficient in Timing as Partition/Position...
Clear[pos]
pos[lst_?VectorQ,pat_?VectorQ]:=Module[{len=Length[pat]-1},Reap[Scan[If[lst[[#;;#+len]]==pat,Sow[#]]&,Range[Length[lst]-len]]][[2,1]]]
Clear[u]
u[lst_]:=With[{limits=Partition[Flatten[{1,(pos[lst,{a,b,a}]/.n_Integer->{n-1,n+3}),Length[lst]}],2]},Take[lst,#]&/@limits]
Clear[g]
g[lst_] :=
With[{limits =
Partition[
Flatten[{1,
Position[
Partition[lst, 3, 1], {a, b, a}] /. {n_Integer} -> {n - 1,
n + 3}, Length[lst]}], 2]}, Take[lst, #] & /@ limits]
lstt=Flatten[Table[lst,{500}],1];
z1=u[lstt];//Timing
MaxMemoryUsed[] - mmu
mmu = MaxMemoryUsed[];
z2=g[lstt];//Timing
MaxMemoryUsed[] - mmu
z1===z2
Out[199]= {0.053322,Null}
Out[200]= 0
Out[202]= {0.025853,Null}
Out[203]= 250064
Out[204]= True
Am 26.08.2012 um 08:51 schrieb Murta:
> Good point Fred
>
work with inter row operations in a simpler way (simpler than =
ListConvolve).
> For example, the Position function (or some new one) could looking for =
not just one element in the list, but for sequence of elements too and =
retur it position. With this option we could avoid to use partition in =
your solution.
> The problem with Partition in large list is the waste of memory. See =
the example:
>
> data = RandomReal[1, 10^6]
> m1 = ByteCount[data];
> m2 = ByteCount[Partition[data, 3, 1]];
> N@m2/m1
> Out[1]=3
>
> We break the list and then search. The best performance way would be =
>
> data = RandomReal[1, 10^6]
> m1 = ByteCount[data];
> m2 = ByteCount[Partition[data, 3, 1]];
> N@m2/m1
> Out[1]=3
>
search the sequence (a,b,a) without break it.
> Map function to me is a good example of function that could have one =
"MapOverPartition". It would save a lot of memory.
>
> tks
> Murta
>
> On Saturday, August 25, 2012 5:26:30 AM UTC-3, Fred Simons wrote:
>> Solutions with pattern matching can be very elegant, such as Bob
>>
>> Hanlon's solution, but also tend to be very slow when the problem is =
>
> tks
> Murta
>
> On Saturday, August 25, 2012 5:26:30 AM UTC-3, Fred Simons wrote:
>> Solutions with pattern matching can be very elegant, such as Bob
>>
a
>>
>> little bit larger.
>>
>>
>>
>> Here is a solution by finding the position of the triple {a,b,a} and=
>>
>> then taking the parts of the list outside these triples.
>>
>>
>>
>> g[lst_] := With[{limits=Partition[Flatten[{1,
>>
>> Position[Partition[lst,3,1], {a,b,a}] /.
>>
>> {n_Integer}->{n-1,n+3},Length[lst]}],2]}, Take[lst, #]& /@ limits]
>>
>>
>>
>> In the following commands f is Bob's function:
>>
>>
>>
>> In[4]:= lst={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} ;
>>
>> z1=f[lst]; // Timing
>>
>> z2=g[lst]; // Timing
>>
>> z1===z2
>>
>> Out[5]= {0.,Null}
>>
>> Out[6]= {0.,Null}
>>
>> Out[7]= True
>>
>>
>>
longer:
>>
>>
>>
>> In[12]:= lstt=Flatten[Table[lst, {500}],1];
>>
>> z1=f[lstt]; // Timing
>>
>> z2=g[lstt]; // Timing
>>
>> z1===z2
>>
>> Out[13]= {5.007632,Null}
>>
>> Out[14]= {0.015600,Null}
>>
>> Out[15]= True
>>
>>
>>
>> Fred Simons
>>
>> Eindhoven University of Technology
>>
>>
>>
>>
>>
>> Op 24-8-2012 11:04, Bob Hanlon schreef:
>>
>>> Clear[f]
>>
>>>
>>
>>> f[list_?VectorQ] := Cases[
>>
>>> (list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
>>
>>> {} ->
>>
>>> Sequence[], _?VectorQ, Infinity]
>>
>>>
>>
>>> f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, =
>>
>>
>>
>> In[12]:= lstt=Flatten[Table[lst, {500}],1];
>>
>> z1=f[lstt]; // Timing
>>
>> z2=g[lstt]; // Timing
>>
>> z1===z2
>>
>> Out[13]= {5.007632,Null}
>>
>> Out[14]= {0.015600,Null}
>>
>> Out[15]= True
>>
>>
>>
>> Fred Simons
>>
>> Eindhoven University of Technology
>>
>>
>>
>>
>>
>> Op 24-8-2012 11:04, Bob Hanlon schreef:
>>
>>> Clear[f]
>>
>>>
>>
>>> f[list_?VectorQ] := Cases[
>>
>>> (list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
>>
>>> {} ->
>>
>>> Sequence[], _?VectorQ, Infinity]
>>
>>>
>>
12}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
>>> f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, =
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
10,
>>
>>> 11, 12}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
>>> f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
>>
>>> a, b, a}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
>>> f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, =
>>
>>> 11, 12}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
>>> f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
>>
>>> a, b, a}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
10,
>>
>>> 11, 12, a, b, a}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
>>>
>>
>>> Bob Hanlon
>>
>>>
>>
>>>
>>
>>
>>>> Hi All
>>
>>>>
>>
>>>> I have a simple problem that is:
>>
>>>>
>>
>>>> l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
>>
>>>>
>>
>>>> I want to replace all a,b,a sequence by X to get:
>>
>>>>
>>
>>>> l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
>>
>>>>
>>
>>>> Then I want to group it by X intervals as
>>
>>>> l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
>>
>>>>
>>
>>>> If I don't need to put the intermediate X, even better!
>>
>>>> I think the with pattern, RaplaceAll and DeleteCases I can do It. =
>>
>>> 11, 12, a, b, a}]
>>
>>>
>>
>>> {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>>
>>>
>>
>>>
>>
>>> Bob Hanlon
>>
>>>
>>
>>>
>>
>>
>>>> Hi All
>>
>>>>
>>
>>>> I have a simple problem that is:
>>
>>>>
>>
>>>> l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
>>
>>>>
>>
>>>> I want to replace all a,b,a sequence by X to get:
>>
>>>>
>>
>>>> l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
>>
>>>>
>>
>>>> Then I want to group it by X intervals as
>>
>>>> l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
>>
>>>>
>>
>>>> If I don't need to put the intermediate X, even better!
>>
Some clue?
>>
>>>> Tks
>>
>>>> Murta
>>
>>>>
>>
>>>>
>>
>>>>
>>
>>>
>>
>>>
>>
>>> -----
>>
>>> Geen virus gevonden in dit bericht.
>>
>>> Gecontroleerd door AVG - www.avg.com
>>
>>> Versie: 2012.0.2197 / Virusdatabase: 2437/5219 - datum van uitgifte: =
>>
>>>> Tks
>>
>>>> Murta
>>
>>>>
>>
>>>>
>>
>>>>
>>
>>>
>>
>>>
>>
>>> -----
>>
>>> Geen virus gevonden in dit bericht.
>>
>>> Gecontroleerd door AVG - www.avg.com
>>
08/23/12
>>
>>>
>>
>>>
>>
>>>
>
Dana DeLouis
Aug 29, 2012, 1:14:35 AM8/29/12
to
Hi. Just an observation on Fred's excellent code.
Just want to post the question on what to do about Overlap. ( ie a,b,a,b,a)
v={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12};
vv=Flatten[Table[v,{500}],1];
// OK when there is no overlap
g[v]
{{1,2,3},{4,5,6,a,b,c,7,8,9},{10,11,12}}
// Error with overlap
g[{a,b,a,b,a,1,2,3}]
Take::take: Cannot take positions 4 through 2 in {a,b,a,b,a,1,2,3}. >>
{{},Take[{a,b,a,b,a,1,2,3},{4,2}],{1,2,3}}
This idea just deletes them all:
Fx[v_List,Patt_List]:=Module[{k,p},
k=Length[Patt];
p=ListConvolve[Patt,v,{-1,1},Null,SameQ,And];
p=Position[p,True];
Do[p=Join[p,p+1],{k-1}];
p=Union[p];
Delete[v,p]
]
// This idea deletes them all
Fx[{a,b,a,b,a,1,2,3},{a,b,a}]
{1,2,3}
// I get the same timing on the larger list:
g[vv]//Timing//First
0.015412
// A negligible speed improvement.
Fx[vv,{a,b,a}]//Timing//First
0.012741
// With same results as your:
Flatten[g[vv]]==Fx[vv,{a,b,a}]
True
= = = = = = = = = =
HTH :>)
Dana DeLouis
Mac & Mathematica 8
= = = = = = = = = =
On Saturday, August 25, 2012 4:26:30 AM UTC-4, Fred Simons wrote:
> Solutions with pattern matching can be very elegant, such as Bob
>
> Hanlon's solution, but also tend to be very slow when the problem is a
>
> >
>
> >
>
> >
Just want to post the question on what to do about Overlap. ( ie a,b,a,b,a)
v={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12};
vv=Flatten[Table[v,{500}],1];
// OK when there is no overlap
g[v]
{{1,2,3},{4,5,6,a,b,c,7,8,9},{10,11,12}}
// Error with overlap
g[{a,b,a,b,a,1,2,3}]
Take::take: Cannot take positions 4 through 2 in {a,b,a,b,a,1,2,3}. >>
{{},Take[{a,b,a,b,a,1,2,3},{4,2}],{1,2,3}}
This idea just deletes them all:
Fx[v_List,Patt_List]:=Module[{k,p},
k=Length[Patt];
p=ListConvolve[Patt,v,{-1,1},Null,SameQ,And];
p=Position[p,True];
Do[p=Join[p,p+1],{k-1}];
p=Union[p];
Delete[v,p]
]
// This idea deletes them all
Fx[{a,b,a,b,a,1,2,3},{a,b,a}]
{1,2,3}
// I get the same timing on the larger list:
g[vv]//Timing//First
0.015412
// A negligible speed improvement.
Fx[vv,{a,b,a}]//Timing//First
0.012741
// With same results as your:
Flatten[g[vv]]==Fx[vv,{a,b,a}]
True
= = = = = = = = = =
HTH :>)
Dana DeLouis
Mac & Mathematica 8
= = = = = = = = = =
On Saturday, August 25, 2012 4:26:30 AM UTC-4, Fred Simons wrote:
> Solutions with pattern matching can be very elegant, such as Bob
>
>
> little bit larger.
>
>
>
> Here is a solution by finding the position of the triple {a,b,a} and
>
> little bit larger.
>
>
>
> Here is a solution by finding the position of the triple {a,b,a} and
>
> then taking the parts of the list outside these triples.
>
> g[lst_] := With[{limits=Partition[Flatten[{1,
>
> Position[Partition[lst,3,1], {a,b,a}] /.
>
> {n_Integer}->{n-1,n+3},Length[lst]}],2]}, Take[lst, #]& /@ limits]
>
>
>
> In the following commands f is Bob's function:
>
>
>
> In[4]:= lst={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} ;
>
> z1=f[lst]; // Timing
>
> z2=g[lst]; // Timing
>
> z1===z2
>
> Out[5]= {0.,Null}
>
> Out[6]= {0.,Null}
>
> Out[7]= True
>
>
>
> Both functions are fast enough. Now we make the list 500 times longer:
>
> g[lst_] := With[{limits=Partition[Flatten[{1,
>
> Position[Partition[lst,3,1], {a,b,a}] /.
>
> {n_Integer}->{n-1,n+3},Length[lst]}],2]}, Take[lst, #]& /@ limits]
>
>
>
> In the following commands f is Bob's function:
>
>
>
> In[4]:= lst={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} ;
>
> z1=f[lst]; // Timing
>
> z2=g[lst]; // Timing
>
> z1===z2
>
> Out[5]= {0.,Null}
>
> Out[6]= {0.,Null}
>
> Out[7]= True
>
>
>
>
>
>
> In[12]:= lstt=Flatten[Table[lst, {500}],1];
>
> z1=f[lstt]; // Timing
>
> z2=g[lstt]; // Timing
>
> z1===z2
>
> Out[13]= {5.007632,Null}
>
> Out[14]= {0.015600,Null}
>
> Out[15]= True
>
>
>
> Fred Simons
>
> Eindhoven University of Technology
>
>
>
>
>
> Op 24-8-2012 11:04, Bob Hanlon schreef:
>
> > Clear[f]
>
> >
>
> > f[list_?VectorQ] := Cases[
>
> > (list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
>
> > {} ->
>
> > Sequence[], _?VectorQ, Infinity]
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12}]
>
>
> In[12]:= lstt=Flatten[Table[lst, {500}],1];
>
> z1=f[lstt]; // Timing
>
> z2=g[lstt]; // Timing
>
> z1===z2
>
> Out[13]= {5.007632,Null}
>
> Out[14]= {0.015600,Null}
>
> Out[15]= True
>
>
>
> Fred Simons
>
> Eindhoven University of Technology
>
>
>
>
>
> Op 24-8-2012 11:04, Bob Hanlon schreef:
>
> > Clear[f]
>
> >
>
> > f[list_?VectorQ] := Cases[
>
> > (list //. {s___, a, b, a, r___} -> {{s}, {r}}) /.
>
> > {} ->
>
> > Sequence[], _?VectorQ, Infinity]
>
> >
>
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
>
> > 11, 12}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
>
> > a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10,
> > 11, 12}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12,
>
> > a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
>
> > 11, 12, a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> >
>
> > Bob Hanlon
>
> >
>
> >
>
> > 11, 12, a, b, a}]
>
> >
>
> > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}}
>
> >
>
> >
>
> > Bob Hanlon
>
> >
>
> >
>
> >> Hi All
>
> >>
>
> >> I have a simple problem that is:
>
> >>
>
> >> l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
>
> >>
>
> >> I want to replace all a,b,a sequence by X to get:
>
> >>
>
> >> l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
>
> >>
>
> >> Then I want to group it by X intervals as
>
> >> l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
>
> >>
>
> >> If I don't need to put the intermediate X, even better!
>
> >> I think the with pattern, RaplaceAll and DeleteCases I can do It. Some clue?
>
> >>
>
> >> I have a simple problem that is:
>
> >>
>
> >> l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12}
>
> >>
>
> >> I want to replace all a,b,a sequence by X to get:
>
> >>
>
> >> l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12}
>
> >>
>
> >> Then I want to group it by X intervals as
>
> >> l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}}
>
> >>
>
> >> If I don't need to put the intermediate X, even better!
>
>
> >> Tks
>
> >> Murta
>
> >>
>
> >>
>
> >>
>
> >
>
> >
>
> > -----
>
> > Geen virus gevonden in dit bericht.
>
> > Gecontroleerd door AVG - www.avg.com
>
> > Versie: 2012.0.2197 / Virusdatabase: 2437/5219 - datum van uitgifte: 08/23/12
> >> Tks
>
> >> Murta
>
> >>
>
> >>
>
> >>
>
> >
>
> >
>
> > -----
>
> > Geen virus gevonden in dit bericht.
>
> > Gecontroleerd door AVG - www.avg.com
>
>
> >
>
> >
>
> >
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