Large files won't load.

Question

Gilbert Bates 2020-10-28

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https://ww2.mathworks.cn/matlabcentral/answers/629723-large-files-won-t-load

评论： Walter Roberson 2020-10-30

Can you tell me what's wrong with this file?

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madhan ravi 2020-10-28

No files are attached.

Gilbert Bates 2020-10-29

编辑：Walter Roberson 2020-10-30

Unfortunatley the compressed/zip file is still too large to attach.

Anyways, if I suspect running out of memory issues then I found that I can break up the large file by way of 'datastore' so I was going to try that. But I'm stuck and confused by this type name value pair thing with the following error.

Specify the 'Type' name-value pair argument to indicate the type of datastore to create.

ds = datastore('E:\Rise_Times\7436.mat','mat','tall');

tt = tall(ds);

The following doesn't work, how to correct? I know the file location as indicated, but not sure what I need for 'type' it's a matlab file.

ds = datastore('E:\Rise_Times\1000 Hz_10 percent_110 A_start.mat','mat','tall');

tt = tall(ds); %<- Not sure if I need this?

Thank you!

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Answer 1

drummer 2020-10-30

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编辑：drummer 2020-10-30

在 MATLAB Online 中打开

"But I'm stuck and confused by this type name value pair thing with the following error.

Specify the 'Type' name-value pair argument to indicate the type of datastore to create."

ds = datastore('E:\Rise_Times\7436.mat','mat','tall');

The name-value pair argument is a feature of MATLAB so you can inform to the native functions what value you assigning to whatever argument name you want to modify.

In your ds, you should specify the name of the argument, as you correctly specified the values.

ds = datastore('yourPath/blah.mat', 'FileExtensions', '.mat', 'Type', 'tall')

When you type ds in the command window, you will see your variable ds with the attributes of FileExtensions and Type you assigned.

Cheers

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Gilbert Bates 2020-10-30

Hmmm, thank you for the reply, but matlab is fighting back every step of the way. :-(

I followed the example exaclty below, the only difference is that I've added the path that matlab is looking at to a working directory where I have files for testing, so I just have the name of the file...

ds = ('1000 Hz_90 percent_1500W_start.mat', 'FileExtensions', '.mat', 'Type', 'tall')

This is what happens...

>> ds = ('1000 Hz_90 percent_1500W_start.mat', 'FileExtensions', '.mat', 'Type', 'tall');

ds = ('1000 Hz_90 percent_1500W_start.mat', 'FileExtensions', '.mat', 'Type', 'tall');

↑

Error: Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

It points to the first comma???

The second thing I've tried, which should be valid, is to break up the commands with the following...

ds = ('1000 Hz_90 percent_1500W_start.mat');

t = tall(ds);

In this case ds literally becomes the name of the file and t is a 1x33 array of the words/letters in the file name.

What's wrong and thanks!

Gilbert Bates 2020-10-30

Thank you drummer!

I did try the variation you suggested, among many variations, with all resulting in some other error. But I think you are correct with datastore accepting .mat and following a few links, I think it is the filedatastore command that I need to use and that's where .mat is being mentioned. However, it looks like I need to write a custome read function as part of the commend. I'm looking through it now...