如何用括號替換美元符號？

Question

我編寫了以下源代碼來用方括號替換美元符號：

text= r"mu-the mean direction $\mu \in [-\pi,\pi)$.   kappa-a concentration parameter $\kappa > 0$.  "

    def replace_dollar(self, text):
        new_text = ""
        flag = False
        for ch in text:
            if ch == '$':
                if flag is False:
                    new_text += "["
                    flag = True
                else:
                    new_text += "]"
                    flag = False
            else:
                new_text += ch
        return new_text


if __name__ == '__main__':
    new_text = replace_dollar(text)
    print(new_text)

Output

mu-the mean direction [\mu \in [-\pi,\pi)].   kappa-a concentration parameter [\kappa > 0].  

有沒有更有效的技術？

Answer 1

常用表達。 也許效率不高（你必須測量），但肯定更整潔，更不容易出錯。

>>> import re
>>> re.sub(r'\$(.*?)\$', r'[\1]', text)
'mu-the mean direction [\\mu \\in [-\\pi,\\pi)].   kappa-a concentration parameter [\\kappa > 0].  '

編輯：如果您要重復使用它，請預編譯 RE 以提高效率：

>>> dollar_to_brackets = re.compile(r'\$(.*?)\$')
>>> dollar_to_brackets.sub(r'[\1]', text)
'mu-the mean direction [\\mu \\in [-\\pi,\\pi)].   kappa-a concentration parameter [\\kappa > 0].  '

Answer 2

這是使用您當前邏輯的版本，但更具可讀性。

text= r"mu-the mean direction $\mu \in [-\pi,\pi)$.   kappa-a concentration parameter $\kappa > 0$.  "

bracket = "[]"    # Index this string with values True/False, which are 0/1
right = False     # Indicates to use right bracket (or not)

while '$' in text:    # In each iteration, replace the left-most occurrence
    text = text.replace('$', bracket[right], 1)
    right = not right    # Switch between left and right brackets 
    print(text)          # This is for tracing only

Output：

mu-the mean direction [\mu \in [-\pi,\pi)$.   kappa-a concentration parameter $\kappa > 0$.  
mu-the mean direction [\mu \in [-\pi,\pi)].   kappa-a concentration parameter $\kappa > 0$.  
mu-the mean direction [\mu \in [-\pi,\pi)].   kappa-a concentration parameter [\kappa > 0$.  
mu-the mean direction [\mu \in [-\pi,\pi)].   kappa-a concentration parameter [\kappa > 0].  

Answer 3

另一種正則表達式方式：

import re
from itertools import cycle

def replace_dollar(text):
    return re.sub(r'\$', lambda _, c=cycle('[]'): next(c), text)

使用問題的示例文本對結果進行基准測試：

8.41 us  original
0.82 us  Prune
4.66 us  Woodford
4.09 us  Woodford2
2.12 us  Manuel

並且文本乘以 10（因此它有 40 個美元符號 - OP 說他們有“不超過 50 個美元符號” ，所以也許這是一個現實的測試）：

91.60 us  original
30.75 us  Prune
27.33 us  Woodford
27.41 us  Woodford2
11.79 us  Manuel

基准代碼， 在線試用！ ：

from timeit import repeat
from functools import partial
import re
from itertools import cycle

def original(text):
        new_text = ""
        flag = False
        for ch in text:
            if ch == '$':
                if flag is False:
                    new_text += "["
                    flag = True
                else:
                    new_text += "]"
                    flag = False
            else:
                new_text += ch
        return new_text

def Prune(text):
    bracket = "[]"
    right = False
    while '$' in text:
        text = text.replace('$', bracket[right], 1)
        right = not right
    return text

def Woodford(text):
    return re.sub(r'\$(.*?)\$', r'[\1]', text)

def Woodford2(text, dollar_to_brackets=re.compile(r'\$(.*?)\$')):
    return dollar_to_brackets.sub(r'[\1]', text)

def Manuel(text):
    return re.sub(r'\$', lambda _, c=cycle('[]'): next(c), text)

def benchmark():
    text = r"mu-the mean direction $\mu \in [-\pi,\pi)$.   kappa-a concentration parameter $\kappa > 0$.  "
    funcs = original, Prune, Woodford, Woodford2, Manuel

    # Correctness
    expect = original(text)
    for func in funcs:
        assert func(text) == expect

    # Speed
    number = 10 ** 4
    for _ in range(3):
        for func in funcs:
            t = min(repeat(partial(func, text), number=number)) / number
            print('%.2f us ' % (t * 1e6), func.__name__)
        print()

benchmark()