You can use

regexp(str,'-?\d+\.\d+','match')

Or, if you need to also match a + :

regexp(str,'[-+]?\d+\.\d+','match')

If you need to match within whitespace boundaries only:

regexp(str,'(?<!\S)[-+]?\d+\.\d+(?!\S)','match')

If the float value must be in between two whitespace chars:

regexp(str,'(?<=\s)[-+]?\d+\.\d+(?=\s)','match')

• Details*
• (?<=\s) - right before the match, there must be a whitespace
• (?<!\S) - right before the match, there must be start of string, or a whitespace
• [-+]? - an optional + or -
• \d+ - one or more digits
• \. - a dot
• \d+ - one or more digits
• (?!\S) - right after, there must be the end of string or a whitespace
• (?=\s) - right after the match, there must be a whitespace.

And if you need to find both integer or floats, replace \d+\.\d+ with \d+(?:\.\d+)? .

试试这个：

re.compile(r'(?<=\s)\d+(\.\d+)?(?=\s)')

你在问题中没有提到是否/为什么需要使用regexp。如果字符串组合有一致的模式，你可以只使用sscanf或textscan来获取所有非空白元素，而不是regexp。

>> str = '     area                                AMW1  =     93.3 m2 '
str =      area                                AMW1  =     93.3 m2

使用sscanf读取可预测格式的字符串：

>> sscanf(str, " %s %s %s %f %s ")
ans =

    97.000
   114.000
   101.000
    97.000
    65.000
    77.000
    87.000
    49.000
    61.000
    93.300
   109.000
    50.000

(for混合类型，字符作为其数字/ascii值存储在数字数组中）
使用*忽略不需要的元素

> sscanf(str, " %*s %*s %*s %f %*s ")
ans = 93.300

也适用于底片：

>> str = '     area                                AMW1  =     -93.3 m2 '
str =      area                                AMW1  =     -93.3 m2

>> sscanf(str, " %*s %*s %*s %f %*s ")
ans = -93.300

类似地，使用textscan，它将把输出存储在单元格数组中：

>> textscan(str, " %s %s %s %f %s ")
ans =
{
  [1,1] =
  {
    [1,1] = area
  }

  [1,2] =
  {
    [1,1] = AMW1
  }

  [1,3] =
  {
    [1,1] = =
  }

  [1,4] = -93.300
  [1,5] =
  {
    [1,1] = m2
  }

}

>> textscan(str, " %*s %*s %*s %f %*s ")
ans =
{
  [1,1] = -93.300
}

>> cell2mat(textscan(s, " %*s %*s %*s %f %*s "))
ans = -93.300