Dimensions of arrays being concatenated are not consistent error.

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Leonidas Daedalus
Leonidas Daedalus 2018-12-15
编辑： dpb 2018-12-15
How to fix this error?
Error using HelperTestKNNClassifier (line 35)
Could not concatenate the table variable 'ActualSpeaker' using VERTCAT.
Caused by:
Error using vertcat
Dimensions of arrays being concatenated are not consistent.
for idx = 1:length(Filenames)
T = featuresTest(Fidx==idx,2:end); % Rows that correspond to one file
predictedLabels = string(predict(trainedClassifier,T(:,1:15))); % Predict
totalVals = size(predictedLabels,1);
[predictedLabel, freq] = mode(categorical(predictedLabels)); % Find most frequently predicted label
match = freq/totalVals*100;
result_file.Filename = Filenames(idx);
result_file.ActualSpeaker = T.Label{1};
result_file.PredictedSpeaker = char(predictedLabel);
result_file.ConfidencePercentage = match;
result = [result; struct2table(result_file)]; %#ok
end

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回答（1 个）

dpb
dpb 2018-12-15
编辑：dpb 2018-12-15
result_file.ActualSpeaker = T.Label{1};
...
result = [result; struct2table(result_file)];
You're trying to catenate char() arrays that may not be the same length (and, in fact, aren't). This happened because you first dereference a cell string to char, then tried to build a table. Illustration of barebones problem:
>> T.Label='Label 1'; % first char label (did the {} dereference manually)
>> res=struct2table(T); % first entry into a table is ok...
>> T.Label='Label 1001'; % later on the string is longer...
>> res=[res;struct2table(T)] % and, BOOM!
Could not concatenate the table variable 'Label' using VERTCAT.
Caused by:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
>>
The solution is to not dereference the label but leave as cell strings (or, if appropriate, convert to categorical )
result_file.ActualSpeaker = T.Label(1);

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