使用负向后查找：(?<!\d)(\.\d+)

你可以做一个正则表达式替换与负查找一个数字。
正则表达式-(?<!\d)(\.\d+)

• (?<!\d)-检查正则表达式前面没有数字
• (\.\d+)-捕获点和一个或多个连续数字

替代-0\1

• 0-为捕获添加一个零
• \1-对捕获组的反向引用，该捕获组是以点开始的浮点数

Regex101 Demo for python regex flavour

Python代码示例

import re

string = "The dimensions of the object-1 is: 1.4 meters wide, .5 meters long, 5.6 meters high \nThe dimensions of the object-2 is: .8 meters wide, .11 meters long, 0.6 meters high"
pattern = r"(?<!\d)(\.\d+)"

match = re.search(pattern, string)
new_string = re.sub(pattern, "Number: 0\\1", string)

print(new_string)

按照你的意图做似乎很奇怪。为什么不捕获所有的浮点数并格式化它们呢？re.sub将接受一个函数作为repl参数。该函数应该接受一个match作为其参数，并返回一个str。你可以使用这种方法来格式化所有的浮点数。

import re

data = """
The dimensions of the object-1 is: 1.4 meters wide, .5 meters long, 5.6 meters high
The dimensions of the object-2 is: .8 meters wide, .11 meters long, 0.6 meters high
"""

reg   = re.compile(r'\d*\.\d+')
frepl = lambda m: f'{float(m.group(0))}'
data  = reg.sub(frepl, data)

print(data)

The dimensions of the object-1 is: 1.4 meters wide, 0.5 meters long, 5.6 meters high
The dimensions of the object-2 is: 0.8 meters wide, 0.11 meters long, 0.6 meters high