delete words from list based on logical statement
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say I have a cell array x={'abatable';'abate';'abatement';'abater';'dog';'cat';'make';'abator';'see'}
and word='abatis' letter='a'
how would I delete all words from x that do not have the letter in the respective positions of the word
For example with the letter a it appears in word abatis as the 1st letter and the 3rd letter. I would then like to delete all words from x that do not have 'a' as their 1st and 3rd positions.
I have tried this
if ismember(letter,word)
x=letter==word % returns a 1 row vector of logical statements
%.....
end
but im stuck there what else should I try.
Thanks
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Plus de réponses (2)
List = {'abatable';'abate';'abatement';'abater';'dog';'cat'; ...
'make';'abator';'see'}
Word = 'abatis'
C = 'a';
pattern = (Word == C);
match = false(1, numel(List));
for k = 1:numel(List)
if isequal(pattern, (List{k} == C))
match(k) = true;
end
end
List(match) = [];
[EDITED, consider word with different lengths:]
...
pattern = strfind(Word, C);
match = false(1, numel(List));
for k = 1:numel(List)
match(k) = isequal(pattern, strfind(List{k}, C))
end
List(match) = [];
In addition I've replaced the IF branching by a direct assignment.
Thorsten
le 12 Nov 2015
Another way would be to use strvcat
List = {'abatable';'abate';'abatement';'abater';'dog';'cat'; ...
'make';'abator';'see'}
Word = 'abatis'
C = 'a';
pattern = (Word == C);
L = strvcat(List) == 'a'; n = size(L, 2);
if length(pattern) < n, pattern(n) = 0; end
P = repmat(pattern, size(L, 1), 1);
ind = ~any(abs(P - L), 2);
List = List(ind);
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