在Matlab中将科学计数形式的数字转换为字符串

Question

I would like to convert a scientific number to a string in Matlab but using a rather specific format. I started with the following:

>> num2str(1E4, '%e')

ans =

    '1.000000e+04'

Then played around with the formatstring to get rid of the digits after the decimal point in the first part

>> num2str(1E4, '%.0e')

ans =

    '1e+04'

The thing is I want it exactly how I am expressing it in numbers, namely I want a string like this '1E4' . I could use strrep to get rid of that plus sign but I refuse to use it to get rid of the leading 0 on the +04 part since I have other instances of the variable which have things like +10 . It it feasible to reproduce the number as a string without resorting to some big complicated algorithm? Preferably using the formatstring?

Answer 1

Solution

According to num2str documentation, you need to use a format parameter of and precision parameter as follows:

num2str(1E4,'%.E')

Result

ans = 1E+04

Answer 2

Read about sprintf . LEt A be your number, to achieve what you want, you can use:

sprintf('%1.0e',A)

Answer 3

Here is a way to convert integers to scientific notation:

function out= scientific(num)
    E = 0;
    if mod(num,10) == 0
        [f n]=factor(num);
        E=min(n(ismember(f,[2 5])));
    end
    out = sprintf('%dE%d',num/10^E,E);
end

>> scientific(134)
ans = 134E0
>> scientific(134000)
ans = 134E3

Another solution that accepts input as vector:

function out= scientific2(num)
    E = sum(cumsum(num2str(num(:))-48,2,'reverse')==0,2);
    out = num2str([num(:)./10.^E,E],'%dE%d\n');
end

Answer 4

You could use a combination of sprintf and regexprep .

my_format = @(x)regexprep(sprintf('%.E',x),'E\+0*','E');

Examples:

>> my_format(1E4)

ans =

1E4

>> my_format(2E12)

ans =

2E12

This is not ideal for all cases:

>> my_format(5) % Expect 5E0

ans =

5E

>> my_format(1E-4) % Expect 1E-4

ans =

1E-04

We can fix the first case with a token :

f2 = @(x)regexprep(sprintf('%.E',x),'E\+0*(\d)','E$1');

>> {f2(1E4), f2(1E20), f2(5)}

ans = 

    '1E4'    '1E20'    '5E0'

And we can fix the second case with tokens and a ? quantifier :

>> f3 = @(x)regexprep(sprintf('%.E',x),'E\+?(-?)0*(\d)','E$1$2');
>> {f3(1E4), f3(1E20), f3(5),f3(1E-1),f3(2E-12)}

ans = 

    '1E4'    '1E20'    '5E0'    '1E-1'    '2E-12'

To explain, sprintf('%.E',x) formats x in scientific notation with E , eg 1E+04 , then it finds

'E\+?(-?)0*(\d)'

 E                 The literal E
  \+?(-?)          Either a + or a -; if - then save to group $1
         0*        As many 0s as it can match, subject to...
           (\d)    At least one digit, saves digit to group $2

Finally, the matched text is replaced with E$1$2 , that is the literal E , then group $1 (a minus sign if found E- , nothing if found E+ ) and the group $2 (a single digit).