y = imread(file1, 'BackgroundColor', [1 1 1]); WHAT DOES IT DO?

I am getting an error in this line as:

Error using strfind
Input strings must have one row.

Error in getFileFromURL (line 8)
if (strfind(filenameIn, '://'))

Error in imread (line 327)
[isUrl, filename] = getFileFromURL(filename);

Error in oooo (line 63)
y = imread(I, 'BackgroundColor', [1 1 1]);

Answer 1

[dpb]

doc imread

Read image from graphics file

Syntax
A = imread(filename)
A = imread(filename,fmt)
A = imread(___,idx)
A = imread(___,Name,Value)
...

Description

A = imread(filename) reads the image from the file specified by filename, inferring the format of the file 
from its contents. If filename is a multi-image file, then imread reads the first image in the file.

A = imread(filename,fmt) additionally specifies the format of the file with the standard file extension indicated 
by fmt. If imread cannot find a file with the name specified by filename, it looks for a file named filename.fmt.

 ...

A = imread(___,Name,Value) specifies format-specific options using one or more name-value pair arguments, 
in addition to any of the input arguments in the previous syntaxes.

...

Ergo, the first argument must be a valid filename string; you've passed what looks to be an array, I.

[dpb]

[OP's comment from Answer--dpb]

I is the name of the image file.. this image is extracted from a video and then saved in my computer. so I wrote

I = imread('Frame 0001.png')

And later in the code used as...

y = imread(file1, 'BackgroundColor', [1 1 1]);

Is something wrong with it?

[dpb]

I = imread('Frame 0001.png');

is OK, yes. Now I is the immge array.

"And later in the code used as..."

y = imread(file1, 'BackgroundColor', [1 1 1]);

Well, whether that would be ok or not depends upon what variable file1 happened to hold at the time IF that were the actual code used.

But, what the error message says you used was

Error in oooo (line 63)
y = imread(I, 'BackgroundColor', [1 1 1]);

as taken directly from the error message, so you didn't use file1, you did, in fact, use I and you had, it appears, already read the image into I.

We don't have the preceding lines in the function or script oooo.m (a remarkably uninformative script name) so we can't see all the code, but clearly I was NOT a file name when that line was executed.

y = imread(I, 'BackgroundColor', [1 1 1]);

[Walter Roberson]

If I were a file name you would still have a problem, because BackgroundColor is not a valid option to imread.

[Ameer Hamza]

It is a valid option in case of a .png file.

[Walter Roberson]

Huh, so it is!

[SURILA GUGLANI]

Sorry, for the typing error.

Answer 2

[SURILA GUGLANI]

I is the name of the image file.. this image is extracted from a video and then saved in my computer. so I wrote

I = imread('Frame 0001.png') And later in the code used as...

y = imread(I, 'BackgroundColor', [1 1 1]); Is something wrong with it?

[Walter Roberson]

After you do

I = imread('Frame 0001.png') And later in the code used as...

then I is the content of Frame 0001.png , and is not a file name.

You would need something like

filename = 'Frame 0001.png';
I = imread(filename);
y = imread(filename, 'BackgroundColor', [1 1 1]);