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Delete strings: {name, John, John Doe, Doe} to {name, John Doe}

5

$begingroup$

I am not very familiar with coding and I am trying to solve the following:

Input:

{name, John, John Doe, Doe}

Output:

{name, John Doe}

So, it should delete {John, Doe} only if there is a string which contains already both of them.

I would be very grateful for any help! Thank you very much!

string-manipulation

share|improve this question

edited 6 hours ago

user64494

3,61411122

asked 15 hours ago

Coffee_09Coffee_09

312

New contributor

Coffee_09 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to the forum! You should watch the TOUR ;-)
  $endgroup$
  – Vitaliy Kaurov
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

I am not very familiar with coding and I am trying to solve the following:

Input:

{name, John, John Doe, Doe}

Output:

{name, John Doe}

So, it should delete {John, Doe} only if there is a string which contains already both of them.

I would be very grateful for any help! Thank you very much!

string-manipulation

share|improve this question

edited 6 hours ago

user64494

3,61411122

asked 15 hours ago

Coffee_09Coffee_09

312

New contributor

Coffee_09 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to the forum! You should watch the TOUR ;-)
  $endgroup$
  – Vitaliy Kaurov
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

I am not very familiar with coding and I am trying to solve the following:

Input:

{name, John, John Doe, Doe}

Output:

{name, John Doe}

So, it should delete {John, Doe} only if there is a string which contains already both of them.

I would be very grateful for any help! Thank you very much!

string-manipulation

share|improve this question

edited 6 hours ago

user64494

3,61411122

asked 15 hours ago

Coffee_09Coffee_09

312

New contributor

Coffee_09 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

{name, John, John Doe, Doe}

{name, John Doe}

share|improve this question

edited 6 hours ago

user64494

3,61411122

asked 15 hours ago

Coffee_09Coffee_09

312

New contributor

Coffee_09 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 6 hours ago

user64494

3,61411122

asked 15 hours ago

Coffee_09Coffee_09

312

New contributor

Coffee_09 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 6 hours ago

user64494

3,61411122

edited 6 hours ago

user64494

3,61411122

asked 15 hours ago

Coffee_09Coffee_09

312

New contributor

Coffee_09 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 15 hours ago

Coffee_09Coffee_09

312

4 Answers
4

active

oldest

votes

4

$begingroup$

strings = {"name", "John", "John Doe", "Doe"};



Pick[strings,

 BitAnd @@ (IdentityMatrix[Length[strings]] +

    Boole[Outer[StringFreeQ, strings, strings]]), 1]

{"name", "John Doe"}

Or equivalently:

Pick[strings,

 MapIndexed[StringFreeQ[Delete[strings, #2], #] &, strings],

 ConstantArray[True, Length[strings] - 1]]

share|improve this answer

edited 13 hours ago

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! thank you very much for your help, Coolwater!
  $endgroup$
  – Coffee_09
  13 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I'll assume that the list will contain strings, not variables:

L = {"name", "John", "John Doe", "Doe"};

Delete element sequences like {"John", "Doe"} that are together contained in a single string "John Doe":

ClearAll[f];

SetAttributes[f, Orderless];

(f[Sequence @@ #, x___] = f[x]) & /@ Select[StringSplit /@ L, Length[#] > 1 &];

f[x___] = {x};

f @@ L

{"John Doe", "name"}

This method does not keep the original sorting order of the list. If this order needs to be kept, we can replace the last line with

SortBy[f @@ L, Position[L, #][[1, 1]] &]

{"name", "John Doe"}

share|improve this answer

edited 14 hours ago

answered 14 hours ago

RomanRoman

6,03611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much, Roman! Is there a possibility to keep the original sorting order?
  $endgroup$
  – Coffee_09
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be honest this is probably not the most efficient solution, but I wrote it down as an exercise in using pattern-matching on Orderless functions.
  $endgroup$
  – Roman
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your solution, Roman!
  $endgroup$
  – Coffee_09
  13 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

# /. Flatten[Cases[#, 

      x_ /; StringContainsQ[x, " "] :> 

       Thread[StringSplit[x] -> Nothing]]] &[{"name", "John", 

  "John Doe", "Doe"}]

{"name", "John Doe"}

share|improve this answer

answered 3 hours ago

halmirhalmir

10.8k2544

$endgroup$

add a comment | 

0

$begingroup$

If you have exact matches this is as short as it gonna get:

DeleteCases[{"name","John","John Doe","Doe"},"John"|"Doe"]

{name,John Doe}

share|improve this answer

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285

$endgroup$

add a comment | 

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4

$begingroup$

strings = {"name", "John", "John Doe", "Doe"};



Pick[strings,

 BitAnd @@ (IdentityMatrix[Length[strings]] +

    Boole[Outer[StringFreeQ, strings, strings]]), 1]

{"name", "John Doe"}

Or equivalently:

Pick[strings,

 MapIndexed[StringFreeQ[Delete[strings, #2], #] &, strings],

 ConstantArray[True, Length[strings] - 1]]

share|improve this answer

edited 13 hours ago

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! thank you very much for your help, Coolwater!
  $endgroup$
  – Coffee_09
  13 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

strings = {"name", "John", "John Doe", "Doe"};



Pick[strings,

 BitAnd @@ (IdentityMatrix[Length[strings]] +

    Boole[Outer[StringFreeQ, strings, strings]]), 1]

{"name", "John Doe"}

Or equivalently:

Pick[strings,

 MapIndexed[StringFreeQ[Delete[strings, #2], #] &, strings],

 ConstantArray[True, Length[strings] - 1]]

share|improve this answer

edited 13 hours ago

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! thank you very much for your help, Coolwater!
  $endgroup$
  – Coffee_09
  13 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

strings = {"name", "John", "John Doe", "Doe"};



Pick[strings,

 BitAnd @@ (IdentityMatrix[Length[strings]] +

    Boole[Outer[StringFreeQ, strings, strings]]), 1]

{"name", "John Doe"}

Or equivalently:

Pick[strings,

 MapIndexed[StringFreeQ[Delete[strings, #2], #] &, strings],

 ConstantArray[True, Length[strings] - 1]]

share|improve this answer

edited 13 hours ago

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

$endgroup$

strings = {"name", "John", "John Doe", "Doe"};



Pick[strings,

 BitAnd @@ (IdentityMatrix[Length[strings]] +

    Boole[Outer[StringFreeQ, strings, strings]]), 1]

Pick[strings,

 MapIndexed[StringFreeQ[Delete[strings, #2], #] &, strings],

 ConstantArray[True, Length[strings] - 1]]

share|improve this answer

edited 13 hours ago

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

answered 14 hours ago

CoolwaterCoolwater

15.5k32553

3

$begingroup$

I'll assume that the list will contain strings, not variables:

L = {"name", "John", "John Doe", "Doe"};

Delete element sequences like {"John", "Doe"} that are together contained in a single string "John Doe":

ClearAll[f];

SetAttributes[f, Orderless];

(f[Sequence @@ #, x___] = f[x]) & /@ Select[StringSplit /@ L, Length[#] > 1 &];

f[x___] = {x};

f @@ L

{"John Doe", "name"}

This method does not keep the original sorting order of the list. If this order needs to be kept, we can replace the last line with

SortBy[f @@ L, Position[L, #][[1, 1]] &]

{"name", "John Doe"}

share|improve this answer

edited 14 hours ago

answered 14 hours ago

RomanRoman

6,03611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much, Roman! Is there a possibility to keep the original sorting order?
  $endgroup$
  – Coffee_09
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be honest this is probably not the most efficient solution, but I wrote it down as an exercise in using pattern-matching on Orderless functions.
  $endgroup$
  – Roman
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your solution, Roman!
  $endgroup$
  – Coffee_09
  13 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I'll assume that the list will contain strings, not variables:

L = {"name", "John", "John Doe", "Doe"};

Delete element sequences like {"John", "Doe"} that are together contained in a single string "John Doe":

ClearAll[f];

SetAttributes[f, Orderless];

(f[Sequence @@ #, x___] = f[x]) & /@ Select[StringSplit /@ L, Length[#] > 1 &];

f[x___] = {x};

f @@ L

{"John Doe", "name"}

This method does not keep the original sorting order of the list. If this order needs to be kept, we can replace the last line with

SortBy[f @@ L, Position[L, #][[1, 1]] &]

{"name", "John Doe"}

share|improve this answer

edited 14 hours ago

answered 14 hours ago

RomanRoman

6,03611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much, Roman! Is there a possibility to keep the original sorting order?
  $endgroup$
  – Coffee_09
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be honest this is probably not the most efficient solution, but I wrote it down as an exercise in using pattern-matching on Orderless functions.
  $endgroup$
  – Roman
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your solution, Roman!
  $endgroup$
  – Coffee_09
  13 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I'll assume that the list will contain strings, not variables:

L = {"name", "John", "John Doe", "Doe"};

Delete element sequences like {"John", "Doe"} that are together contained in a single string "John Doe":

ClearAll[f];

SetAttributes[f, Orderless];

(f[Sequence @@ #, x___] = f[x]) & /@ Select[StringSplit /@ L, Length[#] > 1 &];

f[x___] = {x};

f @@ L

{"John Doe", "name"}

This method does not keep the original sorting order of the list. If this order needs to be kept, we can replace the last line with

SortBy[f @@ L, Position[L, #][[1, 1]] &]

{"name", "John Doe"}

share|improve this answer

edited 14 hours ago

answered 14 hours ago

RomanRoman

6,03611132

$endgroup$

L = {"name", "John", "John Doe", "Doe"};

ClearAll[f];

SetAttributes[f, Orderless];

(f[Sequence @@ #, x___] = f[x]) & /@ Select[StringSplit /@ L, Length[#] > 1 &];

f[x___] = {x};

f @@ L

SortBy[f @@ L, Position[L, #][[1, 1]] &]

share|improve this answer

edited 14 hours ago

answered 14 hours ago

RomanRoman

6,03611132

answered 14 hours ago

RomanRoman

6,03611132

answered 14 hours ago

RomanRoman

6,03611132

0

$begingroup$

# /. Flatten[Cases[#, 

      x_ /; StringContainsQ[x, " "] :> 

       Thread[StringSplit[x] -> Nothing]]] &[{"name", "John", 

  "John Doe", "Doe"}]

{"name", "John Doe"}

share|improve this answer

answered 3 hours ago

halmirhalmir

10.8k2544

$endgroup$

add a comment | 

0

$begingroup$

# /. Flatten[Cases[#, 

      x_ /; StringContainsQ[x, " "] :> 

       Thread[StringSplit[x] -> Nothing]]] &[{"name", "John", 

  "John Doe", "Doe"}]

{"name", "John Doe"}

share|improve this answer

answered 3 hours ago

halmirhalmir

10.8k2544

$endgroup$

add a comment | 

$begingroup$

# /. Flatten[Cases[#, 

      x_ /; StringContainsQ[x, " "] :> 

       Thread[StringSplit[x] -> Nothing]]] &[{"name", "John", 

  "John Doe", "Doe"}]

{"name", "John Doe"}

share|improve this answer

answered 3 hours ago

halmirhalmir

10.8k2544

$endgroup$

# /. Flatten[Cases[#, 

      x_ /; StringContainsQ[x, " "] :> 

       Thread[StringSplit[x] -> Nothing]]] &[{"name", "John", 

  "John Doe", "Doe"}]

share|improve this answer

answered 3 hours ago

halmirhalmir

10.8k2544

answered 3 hours ago

halmirhalmir

10.8k2544

answered 3 hours ago

halmirhalmir

10.8k2544

0

$begingroup$

If you have exact matches this is as short as it gonna get:

DeleteCases[{"name","John","John Doe","Doe"},"John"|"Doe"]

{name,John Doe}

share|improve this answer

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285

$endgroup$

add a comment | 

0

$begingroup$

If you have exact matches this is as short as it gonna get:

DeleteCases[{"name","John","John Doe","Doe"},"John"|"Doe"]

{name,John Doe}

share|improve this answer

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285

$endgroup$

add a comment | 

$begingroup$

If you have exact matches this is as short as it gonna get:

DeleteCases[{"name","John","John Doe","Doe"},"John"|"Doe"]

{name,John Doe}

share|improve this answer

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285

$endgroup$

DeleteCases[{"name","John","John Doe","Doe"},"John"|"Doe"]

share|improve this answer

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285

answered 1 hour ago

Vitaliy KaurovVitaliy Kaurov

58.2k6163285