Compare two numeric cell arrays
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I have two huge cells, a dictionary and a set of possible hands in scrabble. I want to compare the two and find what arrays match
myAletters = ['D','I','R','N','E','E','P'];
myAletters = upper(myAletters);
myAletters = int8(myAletters);
myAletters = myAletters-64;
hand_count=0;
for i=1:length(myAletters)
hand_interim_combination=nchoosek(myAletters,i);
for j=1:length(hand_interim_combination(:,1))
hand_interim_permutation=perms(hand_interim_combination(j,:));
for k=1:length(hand_interim_permutation(:,1))
hand_count=hand_count+1;
hand{hand_count}=hand_interim_permutation(k,:);
end
end
end
end
load('myDict.mat');
word_plays=ismember(hand,myDict);
Ismember won't work because it has to be strings, going one by one won't work because it takes too long. Wat do?
2 件のコメント
Stephen23
2017 年 1 月 17 日
編集済み: Stephen23
2017 年 1 月 17 日
"going one by one won't work because it takes too long. Wat do?"
Stop ignoring mlint messages in the Editor. There is already one message in your code telling you that the way you have written it is going to be slow. The message even tells you how to solve this issue:
Also note that the following code does not create an array of separate characters, because [] is a concatenation operator, not a list operators as many beginners think:
>> myAletters = ['D','I','R','N','E','E','P']
myAletters =
DIRNEEP
That code is therefore equivalent to simply writing this:
>> myAletters = 'DIRNEEP'
myAletters =
DIRNEEP
採用された回答
Stephen23
2017 年 1 月 17 日
編集済み: Stephen23
2017 年 1 月 17 日
Once I got rid of that unnecessary conversion to numeric value then this task becomes a lot easier. I also preallocated the cell array hand to significantly speed the code up.
myAletters = 'DIRNEEP';
hand = cell(1,floor(exp(1)*factorial(numel(myAletters)))-1);
hand_count=0;
for i=1:length(myAletters)
hand_interim_combination=nchoosek(myAletters,i);
for j=1:length(hand_interim_combination(:,1))
hand_interim_permutation=perms(hand_interim_combination(j,:));
for k=1:length(hand_interim_permutation(:,1))
hand_count=hand_count+1;
hand{hand_count}=hand_interim_permutation(k,:);
end
end
end
%load('myDict.mat');
myDict = {'IRE','PEER','XYZ'};
word_plays = ismember(hand,myDict);
and checking the output:
>> hand(word_plays)
ans =
'IRE' 'IRE' 'PEER' 'PEER'
I also tidied up the formatting (crtl+I), replaced unpredictable length with reliable size, and renamed variables to make it more readable (for me):
vec = 'DIRNEEP';
hand = cell(1,floor(exp(1)*factorial(numel(vec)))-1);
itr = 0;
for m = 1:numel(vec)
cmb = nchoosek(vec,m);
for n = 1:size(cmb,1)
prm = perms(cmb(n,:));
for k = 1:size(prm,1)
itr = itr+1;
hand{itr} = prm(k,:);
end
end
end
Perhaps the simplest solution would be to use cellstr instead of the innermost loop:
vec = 'DIRNEEP';
hand = cell(1,2^numel(vec)-1);
k = 0;
for m = 1:numel(vec)
cmb = nchoosek(vec,m);
for n = 1:size(cmb,1)
k = k+1;
hand{k} = cellstr(perms(cmb(n,:)));
end
end
hand = vertcat(hand{:});
Note that these are all just improvements on your algorithm. With a large dictionary a more efficient algorithm might use the small number of available letters to iteratively select smaller sets of the dictionary, until matches are achieved.
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