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3

$begingroup$

I searched and tried different solutions on SE but nothing worked.

I have two lists and would like to remove items from list1 that have a PARTIAL match with items in list2. I'm new to this.

I tried this but it didn't filter anything out

Select[List1, ! MemberQ[List2, #] &]

I also tried with DeleteCases and StringMatchQ but only got various errors. I tried a few solutions from SE but none worked for my specific case.

list1 = "http://www.domain.com/research", "http://www.domain.com/signup", "tel:5559986644", "mailto:info@domain.com"

list2 = "tel:", "mailto:"

results = "http://www.domain.com/research", "http://www.domain.com/signup"

Thank you in advance

pattern-matching filtering

share|improve this question

asked 3 hours ago

Sam BSam B

485

$endgroup$

add a comment | 

3

$begingroup$

I searched and tried different solutions on SE but nothing worked.

I have two lists and would like to remove items from list1 that have a PARTIAL match with items in list2. I'm new to this.

I tried this but it didn't filter anything out

Select[List1, ! MemberQ[List2, #] &]

I also tried with DeleteCases and StringMatchQ but only got various errors. I tried a few solutions from SE but none worked for my specific case.

list1 = "http://www.domain.com/research", "http://www.domain.com/signup", "tel:5559986644", "mailto:info@domain.com"

list2 = "tel:", "mailto:"

results = "http://www.domain.com/research", "http://www.domain.com/signup"

Thank you in advance

pattern-matching filtering

share|improve this question

asked 3 hours ago

Sam BSam B

485

$endgroup$

add a comment | 

$begingroup$

I searched and tried different solutions on SE but nothing worked.

I have two lists and would like to remove items from list1 that have a PARTIAL match with items in list2. I'm new to this.

I tried this but it didn't filter anything out

Select[List1, ! MemberQ[List2, #] &]

I also tried with DeleteCases and StringMatchQ but only got various errors. I tried a few solutions from SE but none worked for my specific case.

list1 = "http://www.domain.com/research", "http://www.domain.com/signup", "tel:5559986644", "mailto:info@domain.com"

list2 = "tel:", "mailto:"

results = "http://www.domain.com/research", "http://www.domain.com/signup"

Thank you in advance

pattern-matching filtering

share|improve this question

asked 3 hours ago

Sam BSam B

485

$endgroup$

Select[List1, ! MemberQ[List2, #] &]

share|improve this question

asked 3 hours ago

Sam BSam B

485

share|improve this question

asked 3 hours ago

Sam BSam B

485

asked 3 hours ago

Sam BSam B

485

asked 3 hours ago

Sam BSam B

485

1 Answer
1

active

oldest

votes

3

$begingroup$

Cases[_?(Not@*StringStartsQ[Alternatives @@ list2])]@list1
Cases[Except[_?(StringStartsQ[Alternatives@@list2])]]@list1
DeleteCases[_?(StringStartsQ[Alternatives @@ list2])]@list1
Select[list1, Not @* StringStartsQ[Alternatives @@ list2]]

all give

"http://www.domain.com/research", "http://www.domain.com/signup"

You can also use the pattern _?(StringMatchQ[#, Alternatives @@ list2 ~~ "*"]&), e.g.,

DeleteCases[list1, _?(StringMatchQ[#,Alternatives @@ list2~~"*"]&)]

same result

share|improve this answer

edited 26 mins ago

answered 3 hours ago

kglrkglr

193k10214434

$endgroup$

add a comment | 

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3

$begingroup$

Cases[_?(Not@*StringStartsQ[Alternatives @@ list2])]@list1
Cases[Except[_?(StringStartsQ[Alternatives@@list2])]]@list1
DeleteCases[_?(StringStartsQ[Alternatives @@ list2])]@list1
Select[list1, Not @* StringStartsQ[Alternatives @@ list2]]

all give

"http://www.domain.com/research", "http://www.domain.com/signup"

You can also use the pattern _?(StringMatchQ[#, Alternatives @@ list2 ~~ "*"]&), e.g.,

DeleteCases[list1, _?(StringMatchQ[#,Alternatives @@ list2~~"*"]&)]

same result

share|improve this answer

edited 26 mins ago

answered 3 hours ago

kglrkglr

193k10214434

$endgroup$

add a comment | 

3

$begingroup$

Cases[_?(Not@*StringStartsQ[Alternatives @@ list2])]@list1
Cases[Except[_?(StringStartsQ[Alternatives@@list2])]]@list1
DeleteCases[_?(StringStartsQ[Alternatives @@ list2])]@list1
Select[list1, Not @* StringStartsQ[Alternatives @@ list2]]

all give

"http://www.domain.com/research", "http://www.domain.com/signup"

You can also use the pattern _?(StringMatchQ[#, Alternatives @@ list2 ~~ "*"]&), e.g.,

DeleteCases[list1, _?(StringMatchQ[#,Alternatives @@ list2~~"*"]&)]

same result

share|improve this answer

edited 26 mins ago

answered 3 hours ago

kglrkglr

193k10214434

$endgroup$

add a comment | 

$begingroup$

Cases[_?(Not@*StringStartsQ[Alternatives @@ list2])]@list1
Cases[Except[_?(StringStartsQ[Alternatives@@list2])]]@list1
DeleteCases[_?(StringStartsQ[Alternatives @@ list2])]@list1
Select[list1, Not @* StringStartsQ[Alternatives @@ list2]]

all give

"http://www.domain.com/research", "http://www.domain.com/signup"

You can also use the pattern _?(StringMatchQ[#, Alternatives @@ list2 ~~ "*"]&), e.g.,

DeleteCases[list1, _?(StringMatchQ[#,Alternatives @@ list2~~"*"]&)]

same result

share|improve this answer

edited 26 mins ago

answered 3 hours ago

kglrkglr

193k10214434

$endgroup$

Cases[_?(Not@*StringStartsQ[Alternatives @@ list2])]@list1
Cases[Except[_?(StringStartsQ[Alternatives@@list2])]]@list1
DeleteCases[_?(StringStartsQ[Alternatives @@ list2])]@list1
Select[list1, Not @* StringStartsQ[Alternatives @@ list2]]

DeleteCases[list1, _?(StringMatchQ[#,Alternatives @@ list2~~"*"]&)]

share|improve this answer

edited 26 mins ago

answered 3 hours ago

kglrkglr

193k10214434

answered 3 hours ago

kglrkglr

193k10214434

answered 3 hours ago

kglrkglr

193k10214434