如何在python中拆分一串数学表达式? [英] How can I split a string of a mathematical expressions in python?

问题描述

我制作了一个程序，可以在 python 中将中缀转换为后缀.问题是当我引入论点时.如果我引入这样的东西:(这将是一个字符串)

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 ) ) )

它将用 .split() 拆分它，程序将正常运行.但我希望用户能够介绍这样的东西:

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) ) )

如您所见，我希望空格可以是微不足道的，但程序会继续按括号、整数(不是数字)和操作数拆分字符串.

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.

我尝试用 for 来解决它，但我不知道如何捕捉整数 (73 , 34 ,72) 而不是一个数字一个数字 (7, 3 , 3 , 4, 7 , 2)

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)

总而言之，我想要的是将像 ((81 * 6)/42+ (3-1)) 这样的字符串拆分为:

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

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

推荐答案

Tree with ast

您可以使用 ast 来获得表达式树:

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)

它输出:

<_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

正如@user2357112 在评论中所写:ast.parse 解释 Python 语法，而不是数学表达式.(1+2)(3+4) 将被解析为函数调用，并且列表推导式将被接受，即使它们可能不应被视为有效的数学表达式.

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.

如果你想要一个扁平的结构，正则表达式可以工作:

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', ')', ')']

它寻找:

• 多位数字
• 或任何不是数字或空格的字符

获得元素列表后，您可以检查语法是否正确，例如使用 stack 来检查括号是否匹配，或者每个元素是否都是已知的.

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.

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