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Dimensions of list used in test

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4

$begingroup$

I would like to reduce the number of brackets in a list for nested lists containing 1 sublist. For instance 1,2,3 should become 1,2,3

However, 1,2,3,4,5 should stay the same.

I used:

DropLevel[x_] := Level[x, -1] /; Dimensions[x][[1]] == 1

Which gives an error on 1,2,3,4,5 but works on 1,2,3

Dimensions[x][[1]]==1 does work as a test for checking whether a list has two sublists or only 1.

Am I missing something about the way /; functions?

expression-manipulation

share|improve this question

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

I would like to reduce the number of brackets in a list for nested lists containing 1 sublist. For instance 1,2,3 should become 1,2,3

However, 1,2,3,4,5 should stay the same.

I used:

DropLevel[x_] := Level[x, -1] /; Dimensions[x][[1]] == 1

Which gives an error on 1,2,3,4,5 but works on 1,2,3

Dimensions[x][[1]]==1 does work as a test for checking whether a list has two sublists or only 1.

Am I missing something about the way /; functions?

expression-manipulation

share|improve this question

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

$endgroup$

add a comment | 

$begingroup$

I would like to reduce the number of brackets in a list for nested lists containing 1 sublist. For instance 1,2,3 should become 1,2,3

However, 1,2,3,4,5 should stay the same.

I used:

DropLevel[x_] := Level[x, -1] /; Dimensions[x][[1]] == 1

Which gives an error on 1,2,3,4,5 but works on 1,2,3

Dimensions[x][[1]]==1 does work as a test for checking whether a list has two sublists or only 1.

Am I missing something about the way /; functions?

expression-manipulation

share|improve this question

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

$endgroup$

DropLevel[x_] := Level[x, -1] /; Dimensions[x][[1]] == 1

share|improve this question

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

share|improve this question

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

edited 8 hours ago

C. E.

52.5k33 gold badges102102 silver badges210210 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

asked 8 hours ago

MikeMike

18766 bronze badges

1 Answer
1

active

oldest

votes

6

$begingroup$

You can use ReplaceAll with the replacement rule x__ :> x:

stripDoubleBraces = # /. x__ :> x &

stripDoubleBraces @ 1, 2, 3

1, 2, 3

stripDoubleBraces @ 1, 2, 3, 4, 5

1, 2, 3,4,5

If you need to the stripping work with arbitrarily deep nesting, you can replace ReplaceAll with ReplaceRepeated:

 1, 5 //. x__ :> x

1, 5

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks again! I will dig into that book soon. Pattern matching is so cool in this language.
  $endgroup$
  – Mike
  6 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

You can use ReplaceAll with the replacement rule x__ :> x:

stripDoubleBraces = # /. x__ :> x &

stripDoubleBraces @ 1, 2, 3

1, 2, 3

stripDoubleBraces @ 1, 2, 3, 4, 5

1, 2, 3,4,5

If you need to the stripping work with arbitrarily deep nesting, you can replace ReplaceAll with ReplaceRepeated:

 1, 5 //. x__ :> x

1, 5

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks again! I will dig into that book soon. Pattern matching is so cool in this language.
  $endgroup$
  – Mike
  6 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

You can use ReplaceAll with the replacement rule x__ :> x:

stripDoubleBraces = # /. x__ :> x &

stripDoubleBraces @ 1, 2, 3

1, 2, 3

stripDoubleBraces @ 1, 2, 3, 4, 5

1, 2, 3,4,5

If you need to the stripping work with arbitrarily deep nesting, you can replace ReplaceAll with ReplaceRepeated:

 1, 5 //. x__ :> x

1, 5

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks again! I will dig into that book soon. Pattern matching is so cool in this language.
  $endgroup$
  – Mike
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use ReplaceAll with the replacement rule x__ :> x:

stripDoubleBraces = # /. x__ :> x &

stripDoubleBraces @ 1, 2, 3

1, 2, 3

stripDoubleBraces @ 1, 2, 3, 4, 5

1, 2, 3,4,5

If you need to the stripping work with arbitrarily deep nesting, you can replace ReplaceAll with ReplaceRepeated:

 1, 5 //. x__ :> x

1, 5

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

$endgroup$

stripDoubleBraces = # /. x__ :> x &

stripDoubleBraces @ 1, 2, 3

stripDoubleBraces @ 1, 2, 3, 4, 5

 1, 5 //. x__ :> x

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges

answered 8 hours ago

kglrkglr

200k1010 gold badges229229 silver badges455455 bronze badges