REMOVE SPACING IN A STRING

I want to convert my binary data into hex, the function that does so only that string as an input. but when I convert my 64 bit binary matrix into a string, it doesn't remove the spaces, which is messing my solution, any idea how to get rid of these

here is wat im talking about;

what i want: '0111001101100001011001000'

what i get: '0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'

[AYUSH VARSHNEY]

And what about vice versa???

what i want : ''0 1 1 1 1 0 0 0 0 0 1 1 0 0 0" like this from "01110000011100"??

[Walter Roberson]

S = "01110000011100"

S = "01110000011100"

regexprep(S, '(.)(?=.)', '$1 ')

ans = "0 1 1 1 0 0 0 0 0 1 1 1 0 0"

[Stephen23]

S = "01110000011100";
sprintf(" %c",S{1})

ans = " 0 1 1 1 0 0 0 0 0 1 1 1 0 0"

[Walter Roberson]

But then you have to get rid of the leading space ;-)

Answer 1

[Azzi Abdelmalek]

% Add this code
 A = '0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0' % Has spaces
 A = A(find(~isspace(A)))

You get a string with no spaces:

A =

'0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'

A =

'0111001101100001011001000'

[Sadia]

thx alot

[Jan]

The find() is not needed: A= A(~isspace(A)) is sufficient already.

[pham quyet]

thank you very much

[AYUSH VARSHNEY]

And what about vice versa???

what i want : ''0 1 1 1 1 0 0 0 0 0 1 1 0 0 0" like this from "01110000011100"??

[Image Analyst]

@AYUSH VARSHNEY, try this:

str = '0 1 1 1 1 0 0 0 0 0 1 1  0 0 0'
% Make it like this form "01110000011100"
str(str == ' ') = []

[AYUSH VARSHNEY]

[Image Analyst]

@AYUSH VARSHNEY, it looks like all the spaces are now gone so I'm glad my code worked for you.

[AYUSH VARSHNEY]

no no no... it won't work for me....the pic i showed to you is the string is without space...

which is like this = "000011101010001010011"

and what i want is the spaces between them .

like this = "0 0 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0"

[Walter Roberson]

S = "01110000011100"

S = "01110000011100"

regexprep(S, '(.)(?=.)', '$1 ')

ans = "0 1 1 1 0 0 0 0 0 1 1 1 0 0"

Answer 2

[jwiix]

As an alternative

A = '0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'
A = strrep(A,' ','') % Replace space with null.

It's slightly faster than the current logical indexing answer I think.

-------------------------------------------------------------------------------------------

K>> A= '0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'

A =

'0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'

K>> tic; A= A(~isspace(A)); toc

Elapsed time is 0.000653 seconds.

-------------------------------------------------------------------------------------------

K>> A= '0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'

A =

'0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'

K>> tic; A = strrep(A,' ',''); toc

Elapsed time is 0.000098 seconds.

:)

[Shep Bryan]

This answer is much better than the accepted answer

Answer 3

[Image Analyst]

Just set locations with spaces equal to null:

% Generate sample string.
theString = '0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0'
% Now change existing string by setting locations with spaces equal to null.
theString(theString == ' ') = []

% Alternative method.
% Create a brand new string with a different name
% by extracting non-space elements.
stringWithoutSpaces = theString(theString ~= ' ')

Answer 4

[Akansha Saxena]

requiredString = regexprep(theString, '\s+', '')

[Alexander Jensen]

This also works on cell arrays containing strings! (at least as of version R2017a)

Example:

str = {'1 GC 2 H M', 'food nam nam';'hello world','meh bleb'};
requiredString = regexprep(str, '\s+', '')

requiredString =

    2×2 cell array

      '1GC2HM'        'foodnamnam'
      'helloworld'    'mehbleb'

Thank you

[Rajbir Singh]

its works.

Thank you @Akansha Saxena