remove missing entries from cell array

Hello,

I have a 4x20 cell array where each cell in the array is a 40x1 column vector. The data in this cell array is from Excel. I am trying to remove all of the '1x1 missing' entries in this cell array (i.e. from each of these column vector cells). I tried this but it doesn't work:

new_cell_array{1}= cellfun(@rmmissing,old_cell_array{1},'UniformOutput',false)

new_cell_array ends up being a 40x1 cell. I want new_cell_array to be a 4x20 cell array that is basically just old_cell_array but with the 'missing' entries removed. Any suggestions would be much appreciated

Thank you

Answer 1

[Jon]

I think you are close, just use

new_cell_array= cellfun(@rmmissing,old_cell_array,'UniformOutput',false)

[Davindra Usov]

Hi,

Thank you for your reply. When I do this, I get this error:

Error using matlab.internal.math.ismissingKernel/arraySwitch (line 79)

First argument must be numeric, logical, datetime, duration, calendarDuration, string,

categorical, char, cellstr, table, or timetable.

[Jon]

Please save your variable old_cell_array as a .mat file and attach it using the paperclip so I can see what you are actually applying the cellfun to.

[Davindra Usov]

Unfortunately I can't send the data itself cause it's confidential :( But it's basically a 4x20 cell array and each cell within that array is a 40x1 column vector which contains some numbers and some '1x1 missing' entries which I'm trying to remove. I read in the data from Excel

[Jon]

The above works fine for the example I constructed, so your error must have to do either with exactly how you wrote the code, maybe even a typo, or something about the exact contents/types etc in your matrix. Without having access to either of those I can't really help you further. I understand that the data is confidential, but if you could just attach an example cell array that demonstrates the problem, but has random values (non-confidential) and attach a small, generic (non-confidential) piece of code that reproduces the problem I might be able to help.