如何在Python中用括号替换文本？(How to replace text between parentheses in Python?)

我有一个包含以下键值对的字典：d = {'Alice'：'x'，'Bob'：'y'，'Chloe'：'z'}

我想用任何给定字符串中的常量（键）替换小写变量（值）。

例如，如果我的字符串是：

A（x）的B（Y）C（X，Z）

如何替换字符以获得结果字符串：

A（爱丽丝）B（鲍勃）C（爱丽丝，小克）

我应该使用正则表达式吗？

I have a dictionary containing the following key-value pairs: d={'Alice':'x','Bob':'y','Chloe':'z'}

I want to replace the lower case variables(values) by the constants(keys) in any given string.

For example, if my string is:

A(x)B(y)C(x,z)

how do I replace the characters in order to get a resultant string of :

A(Alice)B(Bob)C(Alice,Chloe)

Should I use regular expressions?

最满意答案

具有替换功能的re.sub()解决方案：

import re d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') for k in m.group().strip('()').split(','))), s) print(result)

输出：

A(Alice)B(Bob)C(Alice,Chloe)

扩大的视野：

import re def repl(m): val = m.group().strip('()') d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) if ',' in val: return '({})'.format(','.join(flipped.get(k,'') for k in val.split(','))) else: return '({})'.format(flipped.get(val,'')) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', repl, s) print(result)

特定输入案例A(x)B(y)C(Alice,z) 奖励方法：

... s = 'A(x)B(y)C(Alice,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') or k for k in m.group().strip('()').split(','))), s) print(result)

re.sub() solution with replacement function:

import re d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') for k in m.group().strip('()').split(','))), s) print(result)

The output:

A(Alice)B(Bob)C(Alice,Chloe)

Extended version:

import re def repl(m): val = m.group().strip('()') d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) if ',' in val: return '({})'.format(','.join(flipped.get(k,'') for k in val.split(','))) else: return '({})'.format(flipped.get(val,'')) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', repl, s) print(result)

Bonus approach for particular input case A(x)B(y)C(Alice,z):

... s = 'A(x)B(y)C(Alice,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') or k for k in m.group().strip('()').split(','))), s) print(result)如何在Python中用括号替换文本？(How to replace text between parentheses in Python?)

我有一个包含以下键值对的字典：d = {'Alice'：'x'，'Bob'：'y'，'Chloe'：'z'}

我想用任何给定字符串中的常量（键）替换小写变量（值）。

例如，如果我的字符串是：

A（x）的B（Y）C（X，Z）

如何替换字符以获得结果字符串：

A（爱丽丝）B（鲍勃）C（爱丽丝，小克）

我应该使用正则表达式吗？

I have a dictionary containing the following key-value pairs: d={'Alice':'x','Bob':'y','Chloe':'z'}

I want to replace the lower case variables(values) by the constants(keys) in any given string.

For example, if my string is:

A(x)B(y)C(x,z)

how do I replace the characters in order to get a resultant string of :

A(Alice)B(Bob)C(Alice,Chloe)

Should I use regular expressions?

最满意答案

具有替换功能的re.sub()解决方案：

import re d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') for k in m.group().strip('()').split(','))), s) print(result)

输出：

A(Alice)B(Bob)C(Alice,Chloe)

扩大的视野：

import re def repl(m): val = m.group().strip('()') d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) if ',' in val: return '({})'.format(','.join(flipped.get(k,'') for k in val.split(','))) else: return '({})'.format(flipped.get(val,'')) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', repl, s) print(result)

特定输入案例A(x)B(y)C(Alice,z) 奖励方法：

... s = 'A(x)B(y)C(Alice,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') or k for k in m.group().strip('()').split(','))), s) print(result)

re.sub() solution with replacement function:

import re d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') for k in m.group().strip('()').split(','))), s) print(result)

The output:

A(Alice)B(Bob)C(Alice,Chloe)

Extended version:

import re def repl(m): val = m.group().strip('()') d = {'Alice':'x','Bob':'y','Chloe':'z'} flipped = dict(zip(d.values(), d.keys())) if ',' in val: return '({})'.format(','.join(flipped.get(k,'') for k in val.split(','))) else: return '({})'.format(flipped.get(val,'')) s = 'A(x)B(y)C(x,z)' result = re.sub(r'\([^()]+\)', repl, s) print(result)

Bonus approach for particular input case A(x)B(y)C(Alice,z):

... s = 'A(x)B(y)C(Alice,z)' result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') or k for k in m.group().strip('()').split(','))), s) print(result)