How can I split a string of a mathematical express

问题:

I made a program which convert infix to postfix in python. The problem is when I introduce the arguments. If i introduce something like this: (this will be a string)

( ( 73 + ( ( 34 - 72 ) / ( 33 - 3 ) ) ) + ( 56 + ( 95 - 28 ) ) )

it will split it with .split() and the program will work correctly. But I want the user to be able to introduce something like this:

((73 + ( (34- 72 ) / ( 33 -3) )) + (56 +(95 - 28) ) )

As you can see I want that the blank spaces can be trivial but the program continue splitting the string by parentheses, integers (not digits) and operands.

I try to solve it with a for but I don't know how to catch the whole number (73 , 34 ,72) instead one digit by digit (7, 3 , 3 , 4 , 7 , 2)

To sum up, what I want is split a string like ((81 * 6) /42+ (3-1)) into:

[(, (, 81, *, 6, ), /, 42, +, (, 3, -, 1, ), )]

回答1:

Tree with ast

You could use ast to get a tree of the expression :

import ast

source = '((81 * 6) /42+ (3-1))'
node = ast.parse(source) 

def show_children(node, level=0):
    if isinstance(node, ast.Num):
        print(' ' * level + str(node.n))
    else:
        print(' ' * level + str(node))
    for child in ast.iter_child_nodes(node):
        show_children(child, level+1)

show_children(node)

It outputs :

<_ast.Module object at 0x7f56abbc5490>
 <_ast.Expr object at 0x7f56abbc5350>
  <_ast.BinOp object at 0x7f56abbc5450>
   <_ast.BinOp object at 0x7f56abbc5390>
    <_ast.BinOp object at 0x7f56abb57cd0>
     81
     <_ast.Mult object at 0x7f56abbd0dd0>
     6
    <_ast.Div object at 0x7f56abbd0e50>
    42
   <_ast.Add object at 0x7f56abbd0cd0>
   <_ast.BinOp object at 0x7f56abb57dd0>
    3
    <_ast.Sub object at 0x7f56abbd0d50>
    1

As @user2357112 wrote in the comments : ast.parse interprets Python syntax, not mathematical expressions. (1+2)(3+4) would be parsed as a function call and list comprehensions would be accepted even though they probably shouldn't be considered a valid mathematical expression.

List with a regex

If you want a flat structure, a regex could work :

import re

number_or_symbol = re.compile('(\d+|[^ 0-9])')
print(re.findall(number_or_symbol, source))
# ['(', '(', '81', '*', '6', ')', '/', '42', '+', '(', '3', '-', '1', ')', ')']

It looks for either :

• multiple digits
• or any character which isn't a digit or a space

Once you have a list of elements, you could check if the syntax is correct, for example with a stack to check if parentheses are matching, or if every element is a known one.

回答2:

You need to implement a very simple tokenizer for your input. You have the following types of tokens:

• (
• )
• +
• -
• *
• /
• \d+

You can find them in your input string separated by all sorts of white space.

So a first step is to process the string from start to finish, and extract these tokens, and then do your parsing on the tokens, rather than on the string itself.

A nifty way to do this is to use the following regular expression: '\s*([()+*/-]|\d+)'. You can then:

import re

the_input='(3+(2*5))'
tokens = []
tokenizer = re.compile(r'\s*([()+*/-]|\d+)')
current_pos = 0
while current_pos < len(the_input):
  match = tokenizer.match(the_input, current_pos)
  if match is None:
     raise Error('Syntax error')
  tokens.append(match.group(1))
  current_pos = match.end()
print(tokens)

This will print ['(', '3', '+', '(', '2', '*', '5', ')', ')']

You could also use re.findall or re.finditer, but then you'd be skipping non-matches, which are syntax errors in this case.

回答3:

It actual would be pretty trivial to hand-roll a simple expression tokenizer. And I'd think you'd learn more that way as well.

So for the sake of education and learning, Here is a trivial expression tokenizer implementation which can be extended. It works based upon the "maximal-much" rule. This means it acts "greedy", trying to consume as many characters as it can to construct each token.

Without further ado, here is the tokenizer:

class ExpressionTokenizer:
    def __init__(self, expression, operators):
        self.buffer = expression
        self.pos = 0
        self.operators = operators

    def _next_token(self):
        atom = self._get_atom()

        while atom and atom.isspace():
            self._skip_whitespace()
            atom = self._get_atom()

        if atom is None:
            return None
        elif atom.isdigit():
            return self._tokenize_number()
        elif atom in self.operators:
            return self._tokenize_operator()
        else:
            raise SyntaxError()

    def _skip_whitespace(self):
        while self._get_atom():
            if self._get_atom().isspace():
                self.pos += 1
            else:
                break

    def _tokenize_number(self):
        endpos = self.pos + 1
        while self._get_atom(endpos) and self._get_atom(endpos).isdigit():
            endpos += 1
        number = self.buffer[self.pos:endpos]
        self.pos = endpos
        return number

    def _tokenize_operator(self):
        operator = self.buffer[self.pos]
        self.pos += 1
        return operator

    def _get_atom(self, pos=None):
        pos = pos or self.pos
        try:
            return self.buffer[pos]
        except IndexError:
            return None

    def tokenize(self):
        while True:
            token = self._next_token()
            if token is None:
                break
            else:
                yield token

Here is a demo the usage:

tokenizer = ExpressionTokenizer('((81 * 6) /42+ (3-1))', {'+', '-', '*', '/', '(', ')'})
for token in tokenizer.tokenize():
    print(token)

Which produces the output:

(
(
81
*
6
)
/
42
+
(
3
-
1
)
)

回答4:

Quick regex answer: re.findall(r"\d+|[()+\-*\/]", str_in)

Demonstration:

>>> import re
>>> str_in = "((81 * 6) /42+ (3-1))"
>>> re.findall(r"\d+|[()+\-*\/]", str_in)
['(', '(', '81', '*', '6', ')', '/', '42', '+', '(', '3', '-', '1', 
')', ')']

For the nested parentheses part, you can use a stack to keep track of the level.

回答5:

If you don't want to use re module, you can try this:

s="((81 * 6) /42+ (3-1))"

r=[""]

for i in s.replace(" ",""):
    if i.isdigit() and r[-1].isdigit():
        r[-1]=r[-1]+i
    else:
        r.append(i)
print(r[1:])

Output:

['(', '(', '81', '*', '6', ')', '/', '42', '+', '(', '3', '-', '1', ')', ')']

回答6:

This does not provide quite the result you want but might be of interest to others who view this question. It makes use of the pyparsing library.

# Stolen from http://pyparsing.wikispaces.com/file/view/simpleArith.py/30268305/simpleArith.py
# Copyright 2006, by Paul McGuire
# ... and slightly altered

from pyparsing import *

integer = Word(nums).setParseAction(lambda t:int(t[0]))
variable = Word(alphas,exact=1)
operand = integer | variable

expop = Literal('^')
signop = oneOf('+ -')
multop = oneOf('* /')
plusop = oneOf('+ -')
factop = Literal('!')

expr = operatorPrecedence( operand,
    [("!", 1, opAssoc.LEFT),
     ("^", 2, opAssoc.RIGHT),
     (signop, 1, opAssoc.RIGHT),
     (multop, 2, opAssoc.LEFT),
     (plusop, 2, opAssoc.LEFT),]
    )

print (expr.parseString('((81 * 6) /42+ (3-1))'))

Output:

[[[[81, '*', 6], '/', 42], '+', [3, '-', 1]]]

回答7:

Using grako:

start = expr $;
expr = calc | value;
calc = value operator value;
value = integer | "(" @:expr ")" ;
operator = "+" | "-" | "*" | "/";
integer = /\d+/;

grako transpiles to python.

For this example, the return value looks like this:

['73', '+', ['34', '-', '72', '/', ['33', '-', '3']], '+', ['56', '+', ['95', '-', '28']]]

Normally you'd use the generated semantics class as a template for further processing.

回答8:

To provide a more verbose regex approach that you could easily extend:

import re

solution = []
pattern = re.compile('([\d\.]+)')

s = '((73 + ( (34- 72 ) / ( 33 -3) )) + (56 +(95 - 28) ) )'

for token in re.split(pattern, s):
    token = token.strip()
    if re.match(pattern, token):
        solution.append(float(token))
        continue
    for character in re.sub(' ', '', token):
        solution.append(character)

Which will give you the result:

 solution = ['(', '(', 73, '+', '(', '(', 34, '-', 72, ')', '/', '(', 33, '-', 3, ')', ')', ')', '+', '(', 56, '+', '(', 95, '-', 28, ')', ')', ')']