Making a list of patterns of arbitrary length

Clash Royale CLAN TAG#URR8PPP

up vote
10
down vote

favorite

1

I would like to create a list of arbitrary length:

x1_, x2_, ...

Where each of the elements of the list has the full form:

Pattern[xi, Blank]

This answer shows how to create a list of symbols:

x1, x2, ...

but I don't know how to adapt that to obtain the above.

I intend to use this list in the definition of a function as in here.

list-manipulation functions pattern-matching

share|improve this question

edited Aug 28 at 2:18

m_goldberg

81.7k869187

asked Aug 28 at 0:45

Winkelried

1634

add a comment | 

up vote
10
down vote

favorite

1

I would like to create a list of arbitrary length:

x1_, x2_, ...

Where each of the elements of the list has the full form:

Pattern[xi, Blank]

This answer shows how to create a list of symbols:

x1, x2, ...

but I don't know how to adapt that to obtain the above.

I intend to use this list in the definition of a function as in here.

list-manipulation functions pattern-matching

share|improve this question

edited Aug 28 at 2:18

m_goldberg

81.7k869187

asked Aug 28 at 0:45

Winkelried

1634

add a comment | 

up vote
10
down vote

favorite

1

up vote
10
down vote

favorite

1

1

I would like to create a list of arbitrary length:

x1_, x2_, ...

Where each of the elements of the list has the full form:

Pattern[xi, Blank]

This answer shows how to create a list of symbols:

x1, x2, ...

but I don't know how to adapt that to obtain the above.

I intend to use this list in the definition of a function as in here.

list-manipulation functions pattern-matching

share|improve this question

edited Aug 28 at 2:18

m_goldberg

81.7k869187

asked Aug 28 at 0:45

Winkelried

1634

I would like to create a list of arbitrary length:

x1_, x2_, ...

Where each of the elements of the list has the full form:

Pattern[xi, Blank]

This answer shows how to create a list of symbols:

x1, x2, ...

but I don't know how to adapt that to obtain the above.

I intend to use this list in the definition of a function as in here.

list-manipulation functions pattern-matching

share|improve this question

edited Aug 28 at 2:18

m_goldberg

81.7k869187

asked Aug 28 at 0:45

Winkelried

1634

share|improve this question

share|improve this question

edited Aug 28 at 2:18

m_goldberg

81.7k869187

edited Aug 28 at 2:18

m_goldberg

81.7k869187

edited Aug 28 at 2:18

m_goldberg

81.7k869187

81.7k869187

asked Aug 28 at 0:45

Winkelried

1634

asked Aug 28 at 0:45

Winkelried

1634

asked Aug 28 at 0:45

Winkelried

1634

1634

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
8
down vote



accepted

Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]

x1_, x2_, x3_, x4_, x5_

FullForm @ %

List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]

Also

Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]

x1_, x2_, x3_, x4_, x5_

and

ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

edited Aug 28 at 4:10

answered Aug 28 at 1:54

kglr

158k8183382

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The middle one will fail if x1=1.
  – Kuba♦
  Aug 28 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba, good point.
  – kglr
  Aug 28 at 8:09
  

  
  
  
  

add a comment | 

up vote
8
down vote

If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:

ToExpression[
 "x1", "x2", "x3",
 StandardForm,
 Pattern[#,Blank]&
]

x1_, x2_, x3_

Or, creating the list and converting:

ToExpression[
 Table["x" <> ToString@i, i, 5],
 StandardForm,
 Pattern[#, Blank]&
]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

answered Aug 28 at 1:31

Carl Woll

56k272146

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It won't work if x1 = 1.
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
  – Anton.Sakovich
  Aug 28 at 7:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba what exactly do you mean by x1 = 1 ?
  – Winkelried
  Aug 28 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
  – Kuba♦
  Aug 28 at 8:58
  

  
  
  
  

add a comment | 

up vote
4
down vote

Nothing new but shorter:

StringTemplate["x``_"] /@ Range[10] // ToExpression

x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_

share|improve this answer

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

add a comment | 

up vote
3
down vote

Here is an example of how to make the solution in the link work for this case:

patt = Table[
 With[
 s = Symbol["x" <> ToString[i]],
 Pattern[s, Blank]
 ], i, 10];

Range[10] /. patt :> x5, x8

5, 8

Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.

Pattern[Symbol["x" <> ToString[i]], Blank]

because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.

share|improve this answer

edited Aug 28 at 9:49

answered Aug 28 at 1:19

C. E.

47.6k390192

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It won't work if x1 = 1
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba You mean if the symbols have values?
  – C. E.
  Aug 28 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry for not being clear.
  – Kuba♦
  Aug 28 at 9:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
  – C. E.
  Aug 28 at 10:40
  

  
  
  
  

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
8
down vote



accepted

Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]

x1_, x2_, x3_, x4_, x5_

FullForm @ %

List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]

Also

Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]

x1_, x2_, x3_, x4_, x5_

and

ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

edited Aug 28 at 4:10

answered Aug 28 at 1:54

kglr

158k8183382

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The middle one will fail if x1=1.
  – Kuba♦
  Aug 28 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba, good point.
  – kglr
  Aug 28 at 8:09
  

  
  
  
  

add a comment | 

up vote
8
down vote



accepted

Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]

x1_, x2_, x3_, x4_, x5_

FullForm @ %

List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]

Also

Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]

x1_, x2_, x3_, x4_, x5_

and

ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

edited Aug 28 at 4:10

answered Aug 28 at 1:54

kglr

158k8183382

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The middle one will fail if x1=1.
  – Kuba♦
  Aug 28 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba, good point.
  – kglr
  Aug 28 at 8:09
  

  
  
  
  

add a comment | 

up vote
8
down vote



accepted

up vote
8
down vote



accepted

Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]

x1_, x2_, x3_, x4_, x5_

FullForm @ %

List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]

Also

Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]

x1_, x2_, x3_, x4_, x5_

and

ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

edited Aug 28 at 4:10

answered Aug 28 at 1:54

kglr

158k8183382

Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]

x1_, x2_, x3_, x4_, x5_

FullForm @ %

List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]

Also

Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]

x1_, x2_, x3_, x4_, x5_

and

ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

edited Aug 28 at 4:10

answered Aug 28 at 1:54

kglr

158k8183382

share|improve this answer

share|improve this answer

edited Aug 28 at 4:10

edited Aug 28 at 4:10

edited Aug 28 at 4:10

answered Aug 28 at 1:54

kglr

158k8183382

answered Aug 28 at 1:54

kglr

158k8183382

answered Aug 28 at 1:54

kglr

158k8183382

158k8183382

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The middle one will fail if x1=1.
  – Kuba♦
  Aug 28 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba, good point.
  – kglr
  Aug 28 at 8:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The middle one will fail if x1=1.
  – Kuba♦
  Aug 28 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba, good point.
  – kglr
  Aug 28 at 8:09
  

  
  
  
  

1

1

The middle one will fail if x1=1.
– Kuba♦
Aug 28 at 7:37

The middle one will fail if x1=1.
– Kuba♦
Aug 28 at 7:37

@Kuba, good point.
– kglr
Aug 28 at 8:09

@Kuba, good point.
– kglr
Aug 28 at 8:09

add a comment | 

up vote
8
down vote

If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:

ToExpression[
 "x1", "x2", "x3",
 StandardForm,
 Pattern[#,Blank]&
]

x1_, x2_, x3_

Or, creating the list and converting:

ToExpression[
 Table["x" <> ToString@i, i, 5],
 StandardForm,
 Pattern[#, Blank]&
]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

answered Aug 28 at 1:31

Carl Woll

56k272146

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It won't work if x1 = 1.
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
  – Anton.Sakovich
  Aug 28 at 7:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba what exactly do you mean by x1 = 1 ?
  – Winkelried
  Aug 28 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
  – Kuba♦
  Aug 28 at 8:58
  

  
  
  
  

add a comment | 

up vote
8
down vote

If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:

ToExpression[
 "x1", "x2", "x3",
 StandardForm,
 Pattern[#,Blank]&
]

x1_, x2_, x3_

Or, creating the list and converting:

ToExpression[
 Table["x" <> ToString@i, i, 5],
 StandardForm,
 Pattern[#, Blank]&
]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

answered Aug 28 at 1:31

Carl Woll

56k272146

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It won't work if x1 = 1.
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
  – Anton.Sakovich
  Aug 28 at 7:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba what exactly do you mean by x1 = 1 ?
  – Winkelried
  Aug 28 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
  – Kuba♦
  Aug 28 at 8:58
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:

ToExpression[
 "x1", "x2", "x3",
 StandardForm,
 Pattern[#,Blank]&
]

x1_, x2_, x3_

Or, creating the list and converting:

ToExpression[
 Table["x" <> ToString@i, i, 5],
 StandardForm,
 Pattern[#, Blank]&
]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

answered Aug 28 at 1:31

Carl Woll

56k272146

If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:

ToExpression[
 "x1", "x2", "x3",
 StandardForm,
 Pattern[#,Blank]&
]

x1_, x2_, x3_

Or, creating the list and converting:

ToExpression[
 Table["x" <> ToString@i, i, 5],
 StandardForm,
 Pattern[#, Blank]&
]

x1_, x2_, x3_, x4_, x5_

share|improve this answer

answered Aug 28 at 1:31

Carl Woll

56k272146

share|improve this answer

share|improve this answer

answered Aug 28 at 1:31

Carl Woll

56k272146

answered Aug 28 at 1:31

Carl Woll

56k272146

answered Aug 28 at 1:31

Carl Woll

56k272146

56k272146

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It won't work if x1 = 1.
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
  – Anton.Sakovich
  Aug 28 at 7:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba what exactly do you mean by x1 = 1 ?
  – Winkelried
  Aug 28 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
  – Kuba♦
  Aug 28 at 8:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It won't work if x1 = 1.
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
  – Anton.Sakovich
  Aug 28 at 7:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba what exactly do you mean by x1 = 1 ?
  – Winkelried
  Aug 28 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
  – Kuba♦
  Aug 28 at 8:58
  

  
  
  
  

2

2

It won't work if x1 = 1.
– Kuba♦
Aug 28 at 7:36

It won't work if x1 = 1.
– Kuba♦
Aug 28 at 7:36

Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
– Anton.Sakovich
Aug 28 at 7:44

Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
– Anton.Sakovich
Aug 28 at 7:44

@Kuba what exactly do you mean by x1 = 1 ?
– Winkelried
Aug 28 at 8:57

@Kuba what exactly do you mean by x1 = 1 ?
– Winkelried
Aug 28 at 8:57

@Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
– Kuba♦
Aug 28 at 8:58

@Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
– Kuba♦
Aug 28 at 8:58

add a comment | 

up vote
4
down vote

Nothing new but shorter:

StringTemplate["x``_"] /@ Range[10] // ToExpression

x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_

share|improve this answer

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

add a comment | 

up vote
4
down vote

Nothing new but shorter:

StringTemplate["x``_"] /@ Range[10] // ToExpression

x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_

share|improve this answer

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

add a comment | 

up vote
4
down vote

up vote
4
down vote

Nothing new but shorter:

StringTemplate["x``_"] /@ Range[10] // ToExpression

x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_

share|improve this answer

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

Nothing new but shorter:

StringTemplate["x``_"] /@ Range[10] // ToExpression

x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_

share|improve this answer

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

share|improve this answer

share|improve this answer

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

answered Aug 28 at 7:38

Kuba♦

99.3k11194491

99.3k11194491

add a comment | 

add a comment | 

up vote
3
down vote

Here is an example of how to make the solution in the link work for this case:

patt = Table[
 With[
 s = Symbol["x" <> ToString[i]],
 Pattern[s, Blank]
 ], i, 10];

Range[10] /. patt :> x5, x8

5, 8

Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.

Pattern[Symbol["x" <> ToString[i]], Blank]

because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.

share|improve this answer

edited Aug 28 at 9:49

answered Aug 28 at 1:19

C. E.

47.6k390192

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It won't work if x1 = 1
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba You mean if the symbols have values?
  – C. E.
  Aug 28 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry for not being clear.
  – Kuba♦
  Aug 28 at 9:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
  – C. E.
  Aug 28 at 10:40
  

  
  
  
  

add a comment | 

up vote
3
down vote

Here is an example of how to make the solution in the link work for this case:

patt = Table[
 With[
 s = Symbol["x" <> ToString[i]],
 Pattern[s, Blank]
 ], i, 10];

Range[10] /. patt :> x5, x8

5, 8

Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.

Pattern[Symbol["x" <> ToString[i]], Blank]

because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.

share|improve this answer

edited Aug 28 at 9:49

answered Aug 28 at 1:19

C. E.

47.6k390192

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It won't work if x1 = 1
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba You mean if the symbols have values?
  – C. E.
  Aug 28 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry for not being clear.
  – Kuba♦
  Aug 28 at 9:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
  – C. E.
  Aug 28 at 10:40
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Here is an example of how to make the solution in the link work for this case:

patt = Table[
 With[
 s = Symbol["x" <> ToString[i]],
 Pattern[s, Blank]
 ], i, 10];

Range[10] /. patt :> x5, x8

5, 8

Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.

Pattern[Symbol["x" <> ToString[i]], Blank]

because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.

share|improve this answer

edited Aug 28 at 9:49

answered Aug 28 at 1:19

C. E.

47.6k390192

Here is an example of how to make the solution in the link work for this case:

patt = Table[
 With[
 s = Symbol["x" <> ToString[i]],
 Pattern[s, Blank]
 ], i, 10];

Range[10] /. patt :> x5, x8

5, 8

Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.

Pattern[Symbol["x" <> ToString[i]], Blank]

because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.

share|improve this answer

edited Aug 28 at 9:49

answered Aug 28 at 1:19

C. E.

47.6k390192

share|improve this answer

share|improve this answer

edited Aug 28 at 9:49

edited Aug 28 at 9:49

edited Aug 28 at 9:49

answered Aug 28 at 1:19

C. E.

47.6k390192

answered Aug 28 at 1:19

C. E.

47.6k390192

answered Aug 28 at 1:19

C. E.

47.6k390192

47.6k390192

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It won't work if x1 = 1
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba You mean if the symbols have values?
  – C. E.
  Aug 28 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry for not being clear.
  – Kuba♦
  Aug 28 at 9:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
  – C. E.
  Aug 28 at 10:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It won't work if x1 = 1
  – Kuba♦
  Aug 28 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba You mean if the symbols have values?
  – C. E.
  Aug 28 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry for not being clear.
  – Kuba♦
  Aug 28 at 9:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
  – C. E.
  Aug 28 at 10:40
  

  
  
  
  

It won't work if x1 = 1
– Kuba♦
Aug 28 at 7:36

It won't work if x1 = 1
– Kuba♦
Aug 28 at 7:36

@Kuba You mean if the symbols have values?
– C. E.
Aug 28 at 9:48

@Kuba You mean if the symbols have values?
– C. E.
Aug 28 at 9:48

Yes, sorry for not being clear.
– Kuba♦
Aug 28 at 9:54

Yes, sorry for not being clear.
– Kuba♦
Aug 28 at 9:54

@Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
– C. E.
Aug 28 at 10:40

@Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
– C. E.
Aug 28 at 10:40

add a comment | 

 

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