How do I compare a cell array containing string arrays against a string array without using loops?
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I'm comparing two variables (data attached):
- tableOfTextByTime.("tweetUniqueMentions"), which is a 500x1 cell array. The content of each cell is a string array that may contain 0 or more words. Each cell can contain a different number of words. See screenshot:
- tableOfUsers{:,1}, which is a 334x1 string array
The code below works using a for loop, an anonymous function, and cellfun, but it's slow.
It's ok for a small test dataset, but when running on a real data set (20,000 x 1 cell array) and (5,000 x 1 string array) it takes way too long.
for i = height(tableOfUsers): -1: 1
% create a wrapped strcmp anon fcn that takes each cell element and
% each string element
wStrcmp = @(anonInp1) any(strcmp(anonInp1, tableOfUsers{i,1}));
% create matrix of indices for the entries that match the criteria (500x334)
idxMat(:,i) = cellfun( wStrcmp, tableOfTextByTime.("tweetUniqueMentions"),'UniformOutput',false);
% grab the relevant text that match the criteria
correspondingText{i,1} = tableOfTextByTime(cell2mat(idxMat(:,i)),:);
end
How can I get an equivalent result while drastically speeding up the code? Is there a way to do this in a vectorized or element-wise manner? bsxfun and arrayfun seem to have limitations when working with strings. Parallel computing toolbox not an option : )
3 个评论
Walter Roberson
2022-12-21
Please describe in words what the desired outcome is.
- for each cell, you need to know whether at least one string in the cell appears anywhere in the string array?
- for each string in each cell, you need to know of the string appears anywhere in the string array?
- for each string in the string array, you need to know which cells it appears in?
采纳的回答
Stephen23
2022-12-21
编辑:Stephen23
2022-12-21
S = load('answersData.mat');
A = S.tableOfTextByTime.("tweetUniqueMentions")
B = S.tableOfUsers{:,1}
tic
T = vertcat(A{:});
X = repelem(1:numel(A),cellfun(@numel,A)).';
[Y,Z] = ismember(T,B);
F = @(x) {S.tableOfTextByTime(x,:)};
C = accumarray(Z(Y),X(Y),[],F);
toc
C
Lets compare against a loop (as the cyclist mentioned, faster due to moving indexing before the loop):
tic
for kk = numel(B):-1: 1
% create a wrapped strcmp anon fcn that takes each cell element and
% each string element
wStrcmp = @(anonInp1) any(strcmp(anonInp1, B{kk}));
% create matrix of indices for the entries that match the criteria (500x334)
idxMat(:,kk) = cellfun( wStrcmp, A,'UniformOutput',false);
%idx = cell2mat(idxMat(:,kk))
% grab the relevant text that match the criteria
correspondingText{kk,1} = S.tableOfTextByTime(cell2mat(idxMat(:,kk)),:);
end
toc
correspondingText
isequal(C,correspondingText)
更多回答(1 个)
the cyclist
2022-12-21
I think more can be done, but here a couple improvements that make the small test case faster. Hopefully it is an ever larger speed-up on your real problem.
load("answersData.mat")
tic
for i = height(tableOfUsers): -1: 1
% create a wrapped strcmp anon fcn that takes each cell element and
% each string element
wStrcmp = @(anonInp1) any(strcmp(anonInp1, tableOfUsers{i,1}));
% create matrix of indices for the entries that match the criteria (500x334)
idxMat(:,i) = cellfun( wStrcmp, tableOfTextByTime.("tweetUniqueMentions"),'UniformOutput',false);
% grab the relevant text that match the criteria
correspondingText{i,1} = tableOfTextByTime(cell2mat(idxMat(:,i)),:);
end
toc
tic
% Preallocate, and pull out desired subset of data (so indexing doesn't need to be done repeatedly)
idxMat2 = false(height(tableOfTextByTime),height(tableOfUsers));
C = tableOfTextByTime.("tweetUniqueMentions");
T = tableOfUsers{:,1};
for i = height(tableOfUsers): -1: 1
% create a wrapped strcmp anon fcn that takes each cell element and
% each string element
wStrcmp2 = @(anonInp1) any(strcmp(anonInp1, T(i)));
% create matrix of indices for the entries that match the criteria (500x334)
idxMat2(:,i) = cellfun( wStrcmp2, C);
% grab the relevant text that match the criteria
correspondingText2{i,1} = tableOfTextByTime((idxMat2(:,i)),:);
end
toc
% Test that the two methods result in the same output
isequal(correspondingText,correspondingText2)
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