REMOVE SPACING IN A STRING

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Sadia
Sadia on 15 Jul 2012
Edited: Image Analyst on 6 Jul 2021
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'
  4 Comments
Walter Roberson
Walter Roberson on 5 Jun 2021
But then you have to get rid of the leading space ;-)

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Accepted Answer

Azzi Abdelmalek
Azzi Abdelmalek on 15 Jul 2012
Edited: Image Analyst on 6 Jul 2021
% 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'
  9 Comments
Walter Roberson
Walter Roberson on 5 Jun 2021
S = "01110000011100"
S = "01110000011100"
regexprep(S, '(.)(?=.)', '$1 ')
ans = "0 1 1 1 0 0 0 0 0 1 1 1 0 0"

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More Answers (3)

Image Analyst
Image Analyst on 15 Jul 2012
Edited: Image Analyst on 15 Jul 2012
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 ~= ' ')

jwiix
jwiix on 6 Sep 2018
Edited: Image Analyst on 6 Jul 2021
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.
:)
  1 Comment
Shep Bryan
Shep Bryan on 6 Jul 2021
This answer is much better than the accepted answer

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Akansha Saxena
Akansha Saxena on 31 Aug 2016
requiredString = regexprep(theString, '\s+', '')
  2 Comments
Rajbir Singh
Rajbir Singh on 10 Jan 2019
its works.
Thank you @Akansha Saxena

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