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XQuery compare order of two sequences

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0

$begingroup$

I have written a function in XQuery 3.0 (using BaseX) which tests a base sequence against a test sequence, with the goal of determining if the items in the test sequence appear in the same order as the base sequence.

This is best illustrated with an example

E.G.
let $baseSequence := ('a','b','c'..'x','y','z')

Tested against

('a','b','c') => true appears in the same order in $baseSequence

('a','b','a') => false is not in the same order in $baseSequence

('c','h','x','y') => true appears in the same order in $baseSequence

I would like to know if there are any improvements on the function I have written for this, or is there a better way of doing this?

declare function local:testOrder($baseSequence, $testSequence, $position) {

    let $itemPosition := 

    if(count($testSequence) = 0) then (32768) else (index-of($baseSequence, head($testSequence)))

  return

    if ($itemPosition > $position) then (

      local:testOrder($baseSequence, tail($testSequence), $itemPosition)

    ) else (

      count($testSequence) = 0

    )

};



(:Test cases:)

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

let $testAgainst := ('a','c','e','h','w')

return local:testOrder($baseSequence, $testAgainst, 0)

Would return true

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

    let $testAgainst := ('f','d','l','m','n')

    return local:testOrder($baseSequence, $testAgainst, 0)

Would return false

xml xquery

share|improve this question

edited 10 hours ago

200_success

130k16153417

asked 10 hours ago

swshaunswshaun

486

$endgroup$

add a comment | 

0

$begingroup$

I have written a function in XQuery 3.0 (using BaseX) which tests a base sequence against a test sequence, with the goal of determining if the items in the test sequence appear in the same order as the base sequence.

This is best illustrated with an example

E.G.
let $baseSequence := ('a','b','c'..'x','y','z')

Tested against

('a','b','c') => true appears in the same order in $baseSequence

('a','b','a') => false is not in the same order in $baseSequence

('c','h','x','y') => true appears in the same order in $baseSequence

I would like to know if there are any improvements on the function I have written for this, or is there a better way of doing this?

declare function local:testOrder($baseSequence, $testSequence, $position) {

    let $itemPosition := 

    if(count($testSequence) = 0) then (32768) else (index-of($baseSequence, head($testSequence)))

  return

    if ($itemPosition > $position) then (

      local:testOrder($baseSequence, tail($testSequence), $itemPosition)

    ) else (

      count($testSequence) = 0

    )

};



(:Test cases:)

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

let $testAgainst := ('a','c','e','h','w')

return local:testOrder($baseSequence, $testAgainst, 0)

Would return true

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

    let $testAgainst := ('f','d','l','m','n')

    return local:testOrder($baseSequence, $testAgainst, 0)

Would return false

xml xquery

share|improve this question

edited 10 hours ago

200_success

130k16153417

asked 10 hours ago

swshaunswshaun

486

$endgroup$

add a comment | 

$begingroup$

I have written a function in XQuery 3.0 (using BaseX) which tests a base sequence against a test sequence, with the goal of determining if the items in the test sequence appear in the same order as the base sequence.

This is best illustrated with an example

E.G.
let $baseSequence := ('a','b','c'..'x','y','z')

Tested against

('a','b','c') => true appears in the same order in $baseSequence

('a','b','a') => false is not in the same order in $baseSequence

('c','h','x','y') => true appears in the same order in $baseSequence

I would like to know if there are any improvements on the function I have written for this, or is there a better way of doing this?

declare function local:testOrder($baseSequence, $testSequence, $position) {

    let $itemPosition := 

    if(count($testSequence) = 0) then (32768) else (index-of($baseSequence, head($testSequence)))

  return

    if ($itemPosition > $position) then (

      local:testOrder($baseSequence, tail($testSequence), $itemPosition)

    ) else (

      count($testSequence) = 0

    )

};



(:Test cases:)

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

let $testAgainst := ('a','c','e','h','w')

return local:testOrder($baseSequence, $testAgainst, 0)

Would return true

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

    let $testAgainst := ('f','d','l','m','n')

    return local:testOrder($baseSequence, $testAgainst, 0)

Would return false

xml xquery

share|improve this question

edited 10 hours ago

200_success

130k16153417

asked 10 hours ago

swshaunswshaun

486

$endgroup$

declare function local:testOrder($baseSequence, $testSequence, $position) {

    let $itemPosition := 

    if(count($testSequence) = 0) then (32768) else (index-of($baseSequence, head($testSequence)))

  return

    if ($itemPosition > $position) then (

      local:testOrder($baseSequence, tail($testSequence), $itemPosition)

    ) else (

      count($testSequence) = 0

    )

};



(:Test cases:)

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

let $testAgainst := ('a','c','e','h','w')

return local:testOrder($baseSequence, $testAgainst, 0)

let $baseSequence := ('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z')

    let $testAgainst := ('f','d','l','m','n')

    return local:testOrder($baseSequence, $testAgainst, 0)

share|improve this question

edited 10 hours ago

200_success

130k16153417

asked 10 hours ago

swshaunswshaun

486

share|improve this question

edited 10 hours ago

200_success

130k16153417

asked 10 hours ago

swshaunswshaun

486

edited 10 hours ago

200_success

130k16153417

edited 10 hours ago

200_success

130k16153417

asked 10 hours ago

swshaunswshaun

486

asked 10 hours ago

swshaunswshaun

486

1 Answer
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0

$begingroup$

There is a more concise possible approach if you, first, convert the test sequence into the corresponding positions in the base sequence, then, substract each position to the next one (replace "ge 0" with "gt 0" if duplicates are not allowed):

declare function local:testOrder2($baseSequence, $testSequence) {

    let $positions := $testSequence ! index-of($baseSequence, .)

    return min(for $pos at $index in tail($positions) return $pos - $positions[$index]) ge 0

};

Your recursive approach will stop at the first disorder detection so, depending on inputs, it might be faster!

share|improve this answer

edited 15 mins ago

answered 4 hours ago

Alain CouthuresAlain Couthures

1011

New contributor

Alain Couthures is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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0

$begingroup$

There is a more concise possible approach if you, first, convert the test sequence into the corresponding positions in the base sequence, then, substract each position to the next one (replace "ge 0" with "gt 0" if duplicates are not allowed):

declare function local:testOrder2($baseSequence, $testSequence) {

    let $positions := $testSequence ! index-of($baseSequence, .)

    return min(for $pos at $index in tail($positions) return $pos - $positions[$index]) ge 0

};

Your recursive approach will stop at the first disorder detection so, depending on inputs, it might be faster!

share|improve this answer

edited 15 mins ago

answered 4 hours ago

Alain CouthuresAlain Couthures

1011

New contributor

Alain Couthures is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

There is a more concise possible approach if you, first, convert the test sequence into the corresponding positions in the base sequence, then, substract each position to the next one (replace "ge 0" with "gt 0" if duplicates are not allowed):

declare function local:testOrder2($baseSequence, $testSequence) {

    let $positions := $testSequence ! index-of($baseSequence, .)

    return min(for $pos at $index in tail($positions) return $pos - $positions[$index]) ge 0

};

Your recursive approach will stop at the first disorder detection so, depending on inputs, it might be faster!

share|improve this answer

edited 15 mins ago

answered 4 hours ago

Alain CouthuresAlain Couthures

1011

New contributor

Alain Couthures is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

There is a more concise possible approach if you, first, convert the test sequence into the corresponding positions in the base sequence, then, substract each position to the next one (replace "ge 0" with "gt 0" if duplicates are not allowed):

declare function local:testOrder2($baseSequence, $testSequence) {

    let $positions := $testSequence ! index-of($baseSequence, .)

    return min(for $pos at $index in tail($positions) return $pos - $positions[$index]) ge 0

};

Your recursive approach will stop at the first disorder detection so, depending on inputs, it might be faster!

share|improve this answer

edited 15 mins ago

answered 4 hours ago

Alain CouthuresAlain Couthures

1011

New contributor

Alain Couthures is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

declare function local:testOrder2($baseSequence, $testSequence) {

    let $positions := $testSequence ! index-of($baseSequence, .)

    return min(for $pos at $index in tail($positions) return $pos - $positions[$index]) ge 0

};

share|improve this answer

edited 15 mins ago

answered 4 hours ago

Alain CouthuresAlain Couthures

1011

New contributor

Alain Couthures is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 4 hours ago

Alain CouthuresAlain Couthures

1011

New contributor

Alain Couthures is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 4 hours ago

Alain CouthuresAlain Couthures

1011