Searching word list for key letters

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user86753
user86753 el 24 de En. de 2022
Comentada: Walter Roberson el 24 de En. de 2022
wordList = {'apples'
'orange'
'banana'};
ContainList = 'shu';
NotContainList = 'br';
I'd like to be able to find all words that contain the letters S,H or U and do not have the letters B or R. I'm using 'contains' for each letter but is there a way to loop through the ContainList/NotContainList for each word?

Respuesta aceptada

Walter Roberson
Walter Roberson el 24 de En. de 2022
Editada: Walter Roberson el 24 de En. de 2022
wordList = {'apples'
'orange'
'banana'};
ContainList = 'shu';
NotContainList = 'br';
mask1 = contains(wordList, num2cell(ContainList))
mask1 = 3×1 logical array
1 0 0
mask2 = contains(wordList, num2cell(NotContainList))
mask2 = 3×1 logical array
0 1 1
filtered_words = wordList(mask1 & ~mask2)
filtered_words = 1×1 cell array
{'apples'}
%another approach
mask3 = ~cellfun(@isempty, regexp(wordList, "[" + ContainList + "]"))
mask3 = 3×1 logical array
1 0 0
mask4 = ~cellfun(@isempty, regexp(wordList, "[" + NotContainList + "]"))
mask4 = 3×1 logical array
0 1 1
filtered_words2 = wordList(mask3 & ~mask4)
filtered_words2 = 1×1 cell array
{'apples'}
%another approach
pattern = "^[^" + NotContainList + ContainList + "]*[" + ContainList + "][^" + NotContainList + "]*$"
pattern = "^[^brshu]*[shu][^br]*$"
mask5 = ~cellfun(@isempty, regexp(wordList, pattern))
mask5 = 3×1 logical array
1 0 0
filtered_words3 = wordList(mask5)
filtered_words3 = 1×1 cell array
{'apples'}
  1 comentario
Walter Roberson
Walter Roberson el 24 de En. de 2022
In some cases you might want to work iteratively
wordList = {'apples'
'orange'
'banana'};
ContainList = 'shu';
NotContainList = 'br';
words_containing = wordList(contains(wordList, num2cell(ContainList)));
filtered_words = words_containing(~contains(wordList, num2cell(NotContainList)));
In situations where you are doing stepwise refinement, it is most efficient to start from the steps that are expected to make the most difference, each step discarding as much as practical, so that each step is searching less and less information.

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Más respuestas (1)

David Hill
David Hill el 24 de En. de 2022
wordList = {'apples','orange','banana','show','unit'};
ContainList = 'shu';
NotContainList = 'br';
r=wordList(contains(wordList,num2cell(ContainList))&~contains(wordList,num2cell(NotContainList)));

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