Replacing unassigned values in a vector with nans instead of zeros
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Morning everyone,
I'm resizing vectors in a loop. Not good I know and probably why it takes 30 minutes to run, but beside from reading through all the files a first time to determine the correct length of the vectors and then reading through a second time to fill in the vectors I can't see a better way. As there is no easy method of determining the length I need.
However, currently if a file can't be read - some of the files are dodgy and I put a try-catch statement around them - then unassigned values in the vector automatically equal zero. Is there a way to change that behaviour so that an unassigned vector value is a nan?
For example
r(1) = 2;
r(3) = 4;
>> r
r =
1 0 2
is there a way to make r(2) a nan without explicitly assigning it?
2 个评论
Chad Greene
2016-9-20
Is preallocating r an option? I'm thinking something like this:
r = nan(1,3);
r(1) = 2;
r(3) = 4
回答(2 个)
Steven Lord
2016-9-20
Preallocate r to be a NaN vector with a number of elements equal to some upper bound on the number of elements that it could possibly contain (for instance, if you have 10 files and each could add 7 elements to the vector preallocate it to be nan(1, 70)) then fill it in and trim off the excess elements after the for loop is complete.
5 个评论
Guillaume
2016-9-21
The reading is efficient, but I don't see how that portion of code links to your question. Assuming you're concatenating the tables vertically and you want nans for the missing files, this is how I'd do it:
%...
ML = cell(numel(files));
for filenumber = 1:numel(files)
try
ML{filenumber} = readtable(...);
catch
...
end
end
failedread = cellfun(@isempty, ML);
maxheight = max(cellfun(@height, ML(~failedread)));
width = cellfun(@width, ML(~failedread));
assert(numel(width) > 0 && all(width == width(1)), 'table concatenation will fail')
ML{failedread} = array2table(nan(maxheight, width(1)));
allML = vertcat(ML{:});
Jonathan Avesar
2019-9-17
I also have this question. I can't pre-allocate because I don't know how large the matrix will be in advance.
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