Searching word list for key letters

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user86753
user86753 2022 年 1 月 24 日
コメント済み: Walter Roberson 2022 年 1 月 24 日
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?

採用された回答

Walter Roberson
Walter Roberson 2022 年 1 月 24 日
編集済み: Walter Roberson 2022 年 1 月 24 日
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 件のコメント
Walter Roberson
Walter Roberson 2022 年 1 月 24 日
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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その他の回答 (1 件)

David Hill
David Hill 2022 年 1 月 24 日
wordList = {'apples','orange','banana','show','unit'};
ContainList = 'shu';
NotContainList = 'br';
r=wordList(contains(wordList,num2cell(ContainList))&~contains(wordList,num2cell(NotContainList)));

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