我建议对动态替换表达式使用regexprep:

I suggest using regexprep with dynamic replacement expression:

regexprep(inputString,'([0-9\.E-]+)','${sprintf(''%8.6f'',str2double($0))}')

这是第一个字符串的结果(注意:它是正确地"四舍五入，而不是四舍五入的数字)

Here's the result for the first string (note: it's rounding "correctly", rather than chopping off digits)

\left(\begin{array}{ccc} 0.000000 & 0.002325 & 0.002322 & 0.002324 \end{array}\right)

这是第二个更难输入的字符串的结果

Here is the result for the second, harder, input string

\left(\begin{array}{ccc} 0.000000 & 0.002324 & 0.002321 & 0.002323 \end{array}\right)

编辑

刚注意到它说更难的 output 字符串"-这是实现该目标的方法(尽管它确实总是带有两位数的指数).当然，如果您不想匹配指数符号，则可以在match表达式中省略E-.

Just noticed that it says "harder output string" - here's how to achieve that (though it does come with an always two-digit exponent). You can, of course, leave out the E- in the match expression if you don't want to match exponential notations.

out = regexprep(instr,'([0-9\.E-]+)','${sprintf(''%5.3e'',str2double($0))}')
ans =
\left(\begin{array}{ccc} 1.100e-12 & 2.325e-03 & 2.322e-03 & 2.324e-03 \end{array}\right)