python - 从子字符串中提取的映射运算符

Question

我有字典列表:

print (L)
[{0: 'x==1', 1: 'y==2', 2: 'z!=1'}, {0: 'x==1', 1: 'y<=3', 2: 'z>1'}]

我想创建元组，其值在运算符之前、运算符和之后的值:

#first step
wanted = [[('x', '==', '1'), ('y', '==', '2'), ('z', '!=', '1')], 
          [('x', '==', '1'), ('y', '<=', '3'), ('z', '>', '1')]]

然后通过运算符映射第二个值:

import operator

ops = {'>': operator.gt,
        '<': operator.lt,
       '>=': operator.ge,
       '<=': operator.le,
       '==': operator.eq,
        '!=': operator.ne}

#expected final output
wanted = [[('x', <built-in function eq>, '1'), 
           ('y', <built-in function eq>, '2'), 
           ('z', <built-in function ne>, '1')], 
          [('x', <built-in function eq>, '1'), 
           ('y', <built-in function le>, '3'), 
           ('z', <built-in function gt>, '1')]]

我尝试:

L = [[re.findall(r'(.*)([<>=!]+)(.*)', v)[0] for k, v in x.items()] for x in L]
print (L)
[[('x=', '=', '1'), ('y=', '=', '2'), ('z!', '=', '1')], 
 [('x=', '=', '1'), ('y<', '=', '3'), ('z', '>', '1')]]

L = [[ops[y[1]] for y in x] for x in L]

但问题是错误匹配的中间子串 - 运算符，然后错误匹配运算符的值。

正确匹配的正确正则表达式是什么？或者这是另一种可能的解决方案。例如通过 string.partition ？我开放所有可能的解决方案。

Answer 1

如果您的输入确实如此简单，我认为最直接的方法是在运算符字符上拆分:

In [1]: import re

In [2]: data = [{0: 'x==1', 1: 'y==2', 2: 'z!=1'}, {0: 'x==1', 1: 'y<=3', 2: 'z>1'}]

In [3]: rgx = re.compile(r'([<>=!]+)')

In [4]: [[rgx.split(v) for v in d.values()] for d in data]
Out[4]:
[[['x', '==', '1'], ['y', '==', '2'], ['z', '!=', '1']],
 [['x', '==', '1'], ['y', '<=', '3'], ['z', '>', '1']]]

请注意，如果您将捕获组添加到拆分器正则表达式，它会被包含在内!

然后，完成它:

In [11]: ops = {'>': operator.gt,
    ...:         '<': operator.lt,
    ...:        '>=': operator.ge,
    ...:        '<=': operator.le,
    ...:        '==': operator.eq,
    ...:         '!=': operator.ne}
    ...:

In [12]: parsed = [[rgx.split(v) for v in d.values()] for d in data]

In [13]: [[(x, ops[op], y) for x,op,y in ps] for ps in parsed]
Out[13]:
[[('x', <function _operator.eq>, '1'),
  ('y', <function _operator.eq>, '2'),
  ('z', <function _operator.ne>, '1')],
 [('x', <function _operator.eq>, '1'),
  ('y', <function _operator.le>, '3'),
  ('z', <function _operator.gt>, '1')]]