i have this output:
output = dial ('1FUNDOG4YOU')
output =
1 3 8 6 3 6 4 4 9 6 8
(1 X 11) cell
but this is what i want
output :13863644968
(1X 1)
I've tried all sort of ridiculous stuff, str2num and the likes but nothing works . any idea on what to do to make the conversion happen? thanks

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Stephen23
Stephen23 2016년 6월 14일
편집: Stephen23 2016년 6월 14일

1 개 추천

If the elements of the cell array are numeric scalars:
>> C = {1,3,8,6,3,6,4,4,9,6,8};
>> sprintf('%d',C{:})
ans = 13863644968
Or if they are strings:
>> C = {'1','3','8','6','3','6','4','4','9','6','8'};
>> sprintf('%s',C{:})
ans = 13863644968
Note that this string has size 1x11 (strings are not size 1x1 in MATLAB). If you need it to be stored in a 1x1 array, then you can put it inside a cell.

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OLUBUKOLA ogunsola
OLUBUKOLA ogunsola 2016년 6월 14일
works but its comes out as char , but my objective is to get a uint64 class. when i use uint64(ans), it gives me a 1X11
C = {1,3,8,6,3,6,4,4,9,6,8}
C=sprintf('%d',C{:})
C=uint64(str2num(C))
OLUBUKOLA ogunsola
OLUBUKOLA ogunsola 2016년 6월 14일
편집: OLUBUKOLA ogunsola 2016년 6월 14일
perfect ! thanks

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추가 답변 (1개)

Walter Roberson
Walter Roberson 2016년 6월 14일

1 개 추천

Be careful, str2num() will create double precision with only 53 bits of precision. If you need the full 64 bits of precision you will need a different calculation.
C = {1,3,8,6,3,6,4,4,9,6,8};
Cv = uint64([C{:}]);
output = sum( Cv .* (uint64(10).^uint64(length(Cv)-1:-1:0)), 'native' );

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OLUBUKOLA ogunsola
OLUBUKOLA ogunsola 2016년 6월 15일
it actually works , problem now is testing for conditions. input should be only numbers and letters . I've tried ~isnumber , if ~isletter , isstrprop but nothing works
Stephen23
Stephen23 2016년 6월 15일
편집: Stephen23 2016년 6월 15일
@OLUBUKOLA ogunsola: read the isstrprop documentation:
>> inp = '1FUNDOG4YOU';
>> all(isstrprop(inp,'alphanum'))
ans = 1
>> inp = '!%()="';
>> all(isstrprop(inp,'alphanum'))
ans = 0
And read about assert.
OLUBUKOLA ogunsola
OLUBUKOLA ogunsola 2016년 6월 15일
thanks for the info . since all(isstrprop(inp,'alphanum'))returns ans=1 or 0, i did this
x=all(isstrprop(inp,'alphanum'))
then
if x=1
the code else
output =0
but it still doesnt work. Debugging shows that the program continues with the program even when x==0. maybe its something in my code that is causing that. meanwhile I'm studying the assert ,hopefully that will work
Stephen23
Stephen23 2016년 6월 15일
편집: Stephen23 2016년 6월 15일
The function isstrprop and all both return logical values, so you don't need to compare them to anything. They are already boolean values, that can be used directly in an if statement:
if all(isstrprop(...))
...
else
...
end
No testing for "==0" is required.
OLUBUKOLA ogunsola
OLUBUKOLA ogunsola 2016년 6월 15일
problem with assert is it outputs an error , i don't want that i want output =0 . so i wrote the code this way
assert(isa(testStr,'alphanum'),'output=0.') but gets an error :
Error using dial (line 6) output=0
OLUBUKOLA ogunsola
OLUBUKOLA ogunsola 2016년 6월 15일
편집: Walter Roberson 2016년 6월 15일
this is the code i wrote :
function output = dial (testStr)
c = containers.Map({'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'}, ...
[0,1,2,3,4,5,6,7,8,9,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9]) ;
testOut = NaN(1,numel(testStr)) ;
x=all(isstrprop('testStr','alphanum'));
if x==1
for ii = 1:numel(testStr);
testOut(ii) = c(strcat(testStr(ii)));
C=sprintf('%d',testOut(:));
C=uint64(str2double(C));
end
output=C ;
else
output=0;
end
end

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도움말 센터File Exchange에서 Data Type Identification에 대해 자세히 알아보기

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