How to use RE to solve this, I did it with other method.
Here is Question:

'''Parse Stock prices. Create a function that will
decode the old-style fractional stock price. The price
can be a simple floating point number or it can be a
fraction, for example, 4 5/8.

Develop two patterns, one for numbers with optional
decimal places and another for a number with a space
and a fraction. Write a function that accepts a string
and checks both patterns, returning the correct decimal
price for whole numbers (e.g., 14), decimal prices
(e.g., 5.28) and fractional prices (27 1/4).

For the fractional prices (27 1/4), only whole numbers
are allowed in it, and the fractional part must be proper,
that is, it must in the (0,1) range, excluding 0 and 1.
Otherwise, your fuction should return float('nan'),
that is, Not-A-Number.

return a float type object representing the parsed price.

For example,

stock_price('27 1/4') returns float('27.25')
stock_price('27 4/2') returns float('nan')
stock_price('1/2') returns float('0.5')
stock_price('half') returns float('nan')
stock_price('0.1/0.2') returns float('nan')
stock_price('0.2') returns float('0.2')
stock_price('2') returns float('2')


def stock_price(price):
 if price.count('.')==1:
        while price.count('/')==0:# 如果有'.'和'/',那么pretest将其输出为'nan'
           return float(price) #如果有'.'没有'/',那么就是直接转为float
        return float('nan')
    elif price.count('.')==0: 
        if  price.count('/')==1:#如果没有'.',有'/'
            b=price.find('/')  #找到'/'和 ' '所在的位置(如果有' '的话)
            c=price.find(' ')
            if price[b-1]<price[b+1] and price.count(' ')==1:#前面的数字小于后面的数字
                return float(int(price[:c])+int(price[c+1:b])/int(price[b+1:]))
            #找到位置后,将string分开并且转成int,此时有' ',那么将' '前面的数字和后面'/'
            elif price[b-1]<price[b+1] and price.count(' ')==0:#前面的数字小于后面的数字
                return float(int(price[:b])/int(price[b+1:]))
            #如果没有' ',那么直接将'/'前后的数字相除即可
                return float('nan')
        elif price.count('/')==0: #既没有'.'也没有'/'
            if price.isdigit()==True:#如果是数字,没有' '
                return float(price)
            else: #如果不是数字,包括其中包含' '
                return float('nan')
    else: # 多于两个'.的输入一定无效
        return float('nan')

Here is a pattern which seems to work

from __future__ import print_function
import re

pat = re.compile(
    r"^(?P<X>\d+)(?:(?: (?P<Y>\d+)/(?P<Z>\d+))|[.](?P<T>\d*))?$"
def test_pat(data):
    print("TEST_PAT", repr(data))
    match = pat.match(data)
    if not match:
        print("no match")
    for symbol in "XYZT":
        print("%s: %s," % (symbol, repr(, end = " ")
test_pat("12 2/5")

""" my output -->
TEST_PAT '3.35'
X: '3', Y: None, Z: None, T: '35', 
TEST_PAT '12 2/5'
X: '12', Y: '2', Z: '5', T: None, 
TEST_PAT 'two'
no match
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