如何在 Python 中将字符串转换为元组列表

Question

我有以下格式的字符串，我发现很难将这些字符串转换为元组 -

text = '[(Apple Fruit, 10.88), (Table Top, 1.09), (Kicks, 1.08), (La Liga, 1.05), (Camp Nou, 1.02), (Football Team, 0.82), (, 0.73), (Hattrick, 0.7), (Free kick, 0.68), (Ballon dOr, 0.6), (, 0.53), (Treble, 0.51), (Vinegar, 0.09), (Ronaldo, 0.07)]'

我想将此字符串转换为元组列表 -

output = [('Apple Fruit', 10.88), ('Table Top', 1.09), ('Kicks', 1.08), ('La Liga', 1.05), ('Camp Nou', 1.02), ('Football Team', 0.82), ('', 0.73), ('Hattrick', 0.7), ('Free kick', 0.68), ('Ballon dOr', 0.6), ('', 0.53), ('Treble', 0.51), ('Vinegar', 0.09), ('Ronaldo', 0.07)]

我不知道该怎么做。有人可以帮我解决这个问题吗？

Answer 1

您可以使用convert 函数splits 序列 和构建 元组列表。

text = '[(Apple Fruit, 10.88), (Table Top, 1.09), (Kicks, 1.08), (La Liga, 1.05), (Camp Nou, 1.02), (Football Team, 0.82), (, 0.73), (Hattrick, 0.7), (Free kick, 0.68), (Ballon dOr, 0.6), (, 0.53), (Treble, 0.51), (Vinegar, 0.09), (Ronaldo, 0.07)]'

text = text.replace("[","").replace("]","")

def is_digit(str):
   return str.lstrip('-').replace('.', '').isdigit()

def convert(in_str):
   result = []
   current_tuple = []
   for token in in_str.split(", "):
      chunk = token.replace("(","").replace(")", "")
      if is_digit(chunk):
         chunk = float(chunk)
      current_tuple.append(chunk)
      if ")" in token:
         result.append(tuple(current_tuple))
         current_tuple = []
   return result

输出

[('Apple Fruit', 10.88), ('Table Top', 1.09), ('Kicks', 1.08), ('La Liga', 1.05), ('Camp Nou', 1.02), ('Football Team', 0.82), ('', 0.73), ('Hattrick', 0.7), ('Free kick', 0.68), ('Ballon dOr', 0.6), ('', 0.53), ('Treble', 0.51), ('Vinegar', 0.09), ('Ronaldo', 0.07)]

Answer 2

import re
regex = re.compile(r'\((.*?)\)')

text = '[(Apple Fruit, 10.88), (Table Top, 1.09), (Kicks, 1.08), (La Liga, 1.05), (Camp Nou, 1.02), (Football Team, 0.82), (, 0.73), (Hattrick, 0.7), (Free kick, 0.68), (Ballon dOr, 0.6), (, 0.53), (Treble, 0.51), (Vinegar, 0.09), (Ronaldo, 0.07)]'

pairs = regex.findall(text)

list_of_tuples = [tuple(p.split(',')) for p in pairs]

print(list_of_tuples)

1. 首先我们导入正则表达式模块。
2. 定义要查找的正则表达式模式：两个括号之间的任何内容
3. 在我们的 text 变量中搜索该模式并返回所有匹配项。
4. 使用列表理解创建一个元组列表。

Answer 3

你可以试试这个：

import ast
text = '[(Apple Fruit, 10.88), (Table Top, 1.09), (Kicks, 1.08), (La Liga, 1.05), (Camp Nou, 1.02), (Football Team, 0.82), (, 0.73), (Hattrick, 0.7), (Free kick, 0.68), (Ballon dOr, 0.6), (, 0.53), (Treble, 0.51), (Vinegar, 0.09), (Ronaldo, 0.07)]'

comma_added = True

for char in text:
    if char == '(' and comma_added:
        new_text+='("'
        comma_added = False
        continue
    if char == ',' and not comma_added:
        new_text+='"'
        comma_added = True
    new_text += char
print(ast.literal_eval(new_text))

输出：

[('Apple Fruit', 10.88),
 ('Table Top', 1.09),
 ('Kicks', 1.08),
 ('La Liga', 1.05),
 ('Camp Nou', 1.02),
 ('Football Team', 0.82),
 ('', 0.73),
 ('Hattrick', 0.7),
 ('Free kick', 0.68),
 ('Ballon dOr', 0.6),
 ('', 0.53),
 ('Treble', 0.51),
 ('Vinegar', 0.09),
 ('Ronaldo', 0.07)]

或者（非常丑！！！）：

new_text = text.replace('), ','},').replace('(','("').replace(', ','", ').replace('},','), ')
print(ast.literal_eval(new_text))

Answer 4

使用正则表达式 --> Lookbehind & Lookahead。

例如：

import re
import ast

text = '[(Apple Fruit, 10.88), (Table Top, 1.09), (Kicks, 1.08), (La Liga, 1.05), (Camp Nou, 1.02), (Football Team, 0.82), (, 0.73), (Hattrick, 0.7), (Free kick, 0.68), (Ballon dOr, 0.6), (, 0.53), (Treble, 0.51), (Vinegar, 0.09), (Ronaldo, 0.07)]'
text = re.sub(r"(?<=\()([A-Za-z\s]+)", r'"\1"', text) #Convert letters to string
text = re.sub(r"(?<=\()(?=,)", r'""', text)           #Replace empty space with empty string.        
print(ast.literal_eval(text))                         

输出：

[('Apple Fruit', 10.88),
 ('Table Top', 1.09),
 ('Kicks', 1.08),
 ('La Liga', 1.05),
 ('Camp Nou', 1.02),
 ('Football Team', 0.82),
 ('', 0.73),
 ('Hattrick', 0.7),
 ('Free kick', 0.68),
 ('Ballon dOr', 0.6),
 ('', 0.53),
 ('Treble', 0.51),
 ('Vinegar', 0.09),
 ('Ronaldo', 0.07)]