get 0 in the end of a regexp function

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Adrian Brown el 11 de Jun. de 2021
Editada: Stephen23 el 11 de Jun. de 2021
Hello,
I have a number x= 1.25690e-15 in the first step I extract only the the first part of the x which mean firstPart = 1.25690 using the program bellow, then I used the fuction regexp to split the firstPart varibale to elements. in the result of the program I expect to get : [1 2 5 6 9 0] but Unfortunatly I got only [1 2 5 6 9].
x = 1.25690e-15;
string = sprintf('%.8e', x); % Convert number to a scientific notation string with 8 decimal places of precision
stringParts = strsplit(string,'e'); % Split the string where 'e' is
firstPart = str2double(stringParts(1)); % Get the 1st part of stringParts which is the first part of standard form.
k=str2double(regexp(num2str(firstPart),'\d','match')); % slpit the firstPart variable to elements
I really appreciate any help and suggestion.
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Sebastiano Marinelli el 11 de Jun. de 2021
I think you should try to work with string untill the end since when you convert the string to a number:
firstPart = str2double(stringParts(1));
Matlab does automatically remove the not usefull 0 since 1.25690 = 1.2569
something like this should work (yep it is not very elegant, but you can base on this code to solve your problem)
x = 1.25690e-15;
%Note that I had to change .8 to .5!!!
string = sprintf('%.5e', x); % Convert number to a scientific notation string with 8 decimal places of precision
stringParts = strsplit(string,'e'); % Split the string where 'e' is
firstPart = stringParts(1); % Get the 1st part of stringParts which is the first part of standard form.
%split the number into two parts: 1.25690 -> 1 and 25690
strSplit = split((firstPart{1}),'.');
%extract the decimal
decimals = strSplit(2);
%Reshape the decimal
for i = 1:length(decimals{1})
k(i) = str2double(decimals{1}(i));
end
Sebastiano Marinelli el 11 de Jun. de 2021
of course you can still mantain the 8 decimal, you only have to change a bit the code

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Stephen23 el 11 de Jun. de 2021
Editada: Stephen23 el 11 de Jun. de 2021
x = 1.2569e-15;
v = regexprep(sprintf('%#.6g',x),{'\.','e.+'},'')-'0'
v = 1×6
1 2 5 6 9 0
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