delete words from list based on logical statement

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Max
Max am 10 Nov. 2015
Beantwortet: Thorsten am 12 Nov. 2015
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

Akzeptierte Antwort

Adam Barber
Adam Barber am 10 Nov. 2015
I slightly modified Jan's answer:
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)
currentWord = List{k};
lenToCheck = min(numel(pattern), numel(currentWord));
if isequal(pattern(1:lenToCheck), currentWord(1:lenToCheck) == C)
match(k) = true;
end
end
List(~match) = [];
Note that I am using the smaller of the two lengths between the words. As such, "dog" and "cat" won't work, but just the word "as" would.
Of course you could modify this to make your application work.
  1 Kommentar
Max
Max am 11 Nov. 2015
Is it possible to have this work for all words like dog and cat

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Weitere Antworten (2)

Jan
Jan am 10 Nov. 2015
Bearbeitet: Jan am 12 Nov. 2015
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.
  2 Kommentare
Max
Max am 10 Nov. 2015
When I type that in it returns List={'abatable';'abate';'abatement';'dog';'cat';'make';'see'}
Jan
Jan am 12 Nov. 2015
@Max: Correct, the code considers words of the same length only. See EDITED.

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Thorsten
Thorsten am 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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