REMOVE SPACING IN A STRING
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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 个评论
AYUSH VARSHNEY
2021-5-31
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"??
S = "01110000011100"
regexprep(S, '(.)(?=.)', '$1 ')
S = "01110000011100";
sprintf(" %c",S{1})
Walter Roberson
2021-6-5
But then you have to get rid of the leading space ;-)
采纳的回答
Azzi Abdelmalek
2012-7-15
编辑：Image Analyst
2021-7-6
% 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 个评论
Sadia
2012-7-15
Jan
2012-7-15
The find() is not needed: A= A(~isspace(A)) is sufficient already.
pham quyet
2015-3-18
thank you very much
AYUSH VARSHNEY
2021-5-31
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
2021-5-31
@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
2021-6-4
Image Analyst
2021-6-4
AYUSH VARSHNEY
2021-6-5
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"
S = "01110000011100"
regexprep(S, '(.)(?=.)', '$1 ')
更多回答（3 个）
jwiix
2018-9-6
编辑：Image Analyst
2021-7-6
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 个评论
Shep Bryan
2021-7-6
This answer is much better than the accepted answer
Image Analyst
2012-7-15
编辑：Image Analyst
2012-7-15
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 ~= ' ')
Akansha Saxena
2016-8-31
3 个投票
requiredString = regexprep(theString, '\s+', '')
2 个评论
Alexander Jensen
2018-3-30
编辑：Alexander Jensen
2018-3-30
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
2019-1-10
its works.
Thank you @Akansha Saxena
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