Reading an image sequence: "Error: The filename or url argument must be a string."

Hello, I'm trying to work over a sequence of images, I have used dir to obtain a list of all the files:

dirData = dir([dirName, '\*.png']);      %# Get the data for the current directory
dirIndex = [dirData.isdir];  %# Find the index for directories
fileList = {dirData(~dirIndex).name}';  %'# Get a list of the files
if ~isempty(fileList)
    imgList = cellfun(@(x) fullfile(dirName,x),...  %# Prepend path to files
        fileList,'UniformOutput',false);
end
 imgst=cellstr(imgList);

I tried looping over the list but I got an error so I then tried imread over only one member of the array: imread(imgst(1))

and again I got the error saying argument wasn't a string.

I thought cellstr would convert the members in the list to strings? If they're not strings what are they? how can I reference each file in my list so I can work with them over a loop? I know similar questions have been asked before but I didn't find an answer for an error like mine.

Thanks a lot

[Walter Roberson]

The line

imgList = cellfun(@(x) fullfile(dirName,x),...  %# Prepend path to files
        fileList,'UniformOutput',false);

can be replaced with

imgList = fullfile(dirName, fileList);

fullfile() is happy to take a cell array of character vectors as the second argument.

Answer 1

[Fer]

Okay, so I found the answer to my own question. Not deleting in case someone has the same problem. The only thing I had to so was calling

 imread(imgst{1})

so basically just using {} instead of ().

I had not used cells before so, my bad.

[Virgil of the Matrix]

THANK YOU! YOU ARE A S A I N T!

[Nathaniel Conrad]

Thank you for pointing this out! I was using a cell data container to hold a string character for a file path for the first time and having the same issue, too.

I do find it odd that if a cell only has one element in it, then you must index the container variable with {1} in order to extract the string that is labeling the relevant image file path.

For example, if you store the string in a structure array, then you don't need to index the structure array data container variable at all. Weird!

Answer 2

[Tess Meier]

Yes thank you!!!!!!

Answer 3

[hossam mahmoud]

I found the solution

just try this

s=convertStringsToChars(s);