read a text file into an array (file of 0 and 1) - matlab

Question

I have a text file containing randoms of 0 and 1, and I want to read it in Matlab and obtain each element in an array
goal :
I have two text files that I want to compare and see if they are identical and how much difference there is, in fact, the two files are :
1) original file that I send via a communication line
2) the received file should be identical to the send file
Example of my code:
for i=1:1:size
if (send[i] ~= received[i]) error++;
end
but I need to know how to obtain these two arrays from the text files, where all the "0" and "1" are in one line

Answer 1

Since you want to check that the contents of the two files are the same, I do not think you need to worry about the format of their contents or the sequence of zeros and ones, they should be simply identical. You can use the following code to read the entire text file and store it in a char vector:
C = char(join(readlines(filename), ''));
To compare contents of two files and find the error percent you can do the following:
act = char(join(readlines(actualfilename), ''));
exp = char(join(readlines(expectedfilename), ''));
err = (sum(act~=exp))/length(act);
But you should also detect if two files contain different number of characters:
act = char(join(readlines(actualfilename), ''));
exp = char(join(readlines(expectedfilename), ''));
al = length(act); % actual length
el = length(exp); % expected length
dl = abs(al-el);
if (dl>0)
ml = min(al, el); % min length
act = act(1:ml); % shorten act if needed
exp = exp(1:ml); % shorten exp if needed
end
err = (sum(act~=exp)+dl)/al % error
Note that in the second case, if a character is added or lost in the middle of the file, all subsequent characters will be considered as error.

Answer 2

Reading in the Text Files:
If the text file is configured with spaces or line breaks:
Text.txt (line breaks)
0
1
0
1
1
Text.txt (spaces)
1 0 1 0 1 1
Scanning in the data can be done by using the fscanf() function with format specification %d indicated to scan in the file as integers.
File_Name = "Text.txt";
File_ID = fopen(File_Name);
Binary = fscanf(File_ID,'%d');
If the text file has the characters beside/concatenated on the same line without spaces:
Text.txt (single line, no spaces)
01011
Scanning the text file can be done using the format specification, %s indicated to read the file as a string. This string can be split and converted into an array by using split(), cell2mat() and str2num().
split() → Splits the string into a cell array with individual bits/binary
cell2mat() → Converts the cell array to a character array
str2num() → Converts the character array to a numerical double array
File_Name = "Text.txt";
File_ID = fopen(File_Name);
Binary = fscanf(File_ID,'%s');
Binary = split(Binary,'');
Binary = str2num(cell2mat(Binary(2:end-1))).';
Comparing to Evaluate Amount of Errors:
Error checking can be done by comparing the arrays logically in an element-wise fashion. Then by using the nnz() (number of non-zeroes) function we can count the number of times the condition is true, "1". Here the condition is when the two binary signals Binary_1 and Binary_2 not equal to each other.
Code Snippet:
Error = nnz(Binary_1 ~= Binary_2);
Error
Full Script Option 1 (line breaks/spaces text file):
File_Name = "Text_1.txt";
File_ID = fopen(File_Name);
Binary_1 = fscanf(File_ID,'%d');
fclose(File_ID);
File_Name = "Text_2.txt";
File_ID = fopen(File_Name);
Binary_2 = fscanf(File_ID,'%d');
fclose(File_ID);
clearvars -except Binary_1 Binary_2
Error = nnz(Binary_1 ~= Binary_2);
Error
Full Script Option 2 (single line, no spaces text file):
File_Name = "Text_1.txt";
File_ID = fopen(File_Name);
Binary_1 = fscanf(File_ID,'%s');
Binary_1 = split(Binary_1,'');
Binary_1 = str2num(cell2mat(Binary_1(2:end-1))).';
File_Name = "Text_2.txt";
File_ID = fopen(File_Name);
Binary_2 = fscanf(File_ID,'%s');
Binary_2 = split(Binary_2,'');
Binary_2 = str2num(cell2mat(Binary_2(2:end-1))).';
fclose(File_ID);
clearvars -except Binary_1 Binary_2
Error = nnz(Binary_1 ~= Binary_2);
Error
Ran using MATLAB R2019b