可读的尝试除了计算处理

Question

我有一个程序,我需要做很多计算,但输入可能不完整(所以我们不能总是计算所有结果),这本身就很好,但是会给代码的可读性带来问题:

def try_calc():
    a = {'1': 100, '2': 200, '3': 0, '4': -1, '5': None, '6': 'a'}
    try:
        a['10'] = float(a['1'] * a['2'])
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        a['10'] = None
    try:
        a['11'] = float(a['1'] * a['5'])
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        a['11'] = None
    try:
        a['12'] = float(a['1'] * a['6'])
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        a['12'] = None
    try:
        a['13'] = float(a['1'] / a['2'])
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        a['13'] = None
    try:
        a['14'] = float(a['1'] / a['3'])
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        a['14'] = None
    try:
        a['15'] = float((a['1'] * a['2']) / (a['3'] * a['4']))
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        a['15'] = None
    return a

In [39]: %timeit try_calc()
100000 loops, best of 3: 11 µs per loop

所以这很好用,表现很好但是真的不可读.我们提出了另外两种方法来处理这个问题.1:使用内部处理问题的专用函数

import operator
def div(list_of_arguments):
    try:
        result = float(reduce(operator.div, list_of_arguments, 1))
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        result = None
    return result

def mul(list_of_arguments):
    try:
        result = float(reduce(operator.mul, list_of_arguments, 1))
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        result = None
    return result

def add(list_of_arguments):
    try:
        result = float(reduce(operator.add, list_of_arguments, 1))
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        result = None
    return result

def try_calc2():
    a = {'1': 100, '2': 200, '3': 0, '4': -1, '5': None, '6': 'a'}
    a['10'] = mul([a['1'], a['2']])
    a['11'] = mul([a['1'], a['5']])
    a['12'] = mul([a['1'], a['6']])
    a['13'] = div([a['1'], a['2']])
    a['14'] = div([a['1'], a['3']])
    a['15'] = div([
        mul([a['1'], a['2']]), 
        mul([a['3'], a['4']])
        ])
    return a

In [40]: %timeit try_calc2()
10000 loops, best of 3: 20.3 µs per loop

两次虽然缓慢但仍然不那么可读.选项2:封装在eval语句中

def eval_catcher(term):
    try:
        result = float(eval(term))
    except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
        result = None
    return result

def try_calc3():
    a = {'1': 100, '2': 200, '3': 0, '4': -1, '5': None, '6': 'a'}
    a['10'] = eval_catcher("a['1'] * a['2']")
    a['11'] = eval_catcher("a['1'] * a['5']")
    a['12'] = eval_catcher("a['1'] * a['6']")
    a['13'] = eval_catcher("a['1'] / a['2']")
    a['14'] = eval_catcher("a['1'] / a['3']")
    a['15'] = eval_catcher("(a['1'] * a['2']) / (a['3'] * a['4'])")
    return a

In [41]: %timeit try_calc3()
10000 loops, best of 3: 130 µs per loop

所以非常慢(与其他替代方案相比),但同时也是最易读的.我知道一些问题(KeyError,ValueError)也可以通过预处理字典来确保键的可用性来处理,但仍然会留下None(TypeError)和ZeroDivisionErrors,所以我看不到任何优势

我的问题: - 我错过了其他选择吗？ - 我是否因为试图以这种方式解决它而感到非常疯狂？ - 有更多的pythonic方法吗？ - 您认为最好的解决方案是什么？为什么？

Answer 1

如何将计算存储为lambdas？然后你可以循环遍历所有这些,只使用一个try-except块.

def try_calc():
    a = {'1': 100, '2': 200, '3': 0, '4': -1, '5': None, '6': 'a'}
    calculations = {
        '10': lambda: float(a['1'] * a['2']),
        '11': lambda: float(a['1'] * a['5']),
        '12': lambda: float(a['1'] * a['6']),
        '13': lambda: float(a['1'] / a['2']),
        '14': lambda: float(a['1'] / a['3']),
        '15': lambda: float((a['1'] * a['2']) / (a['3'] * a['4']))
    }
    for key, calculation in calculations.iteritems():
        try:
            a[key] = calculation()
        except (ZeroDivisionError, KeyError, TypeError, ValueError) as e:
            a[key] = None

顺便说一句,如果计算顺序很重要,我不建议这样做,就像你在原始代码中有这个一样:

a['3'] = float(a['1'] * a['2'])
a['5'] = float(a['3'] * a['4'])

由于dicts是无序的,因此您无法保证第一个等式在第二个等式之前执行.因此a['5']可能使用新值计算a['3'],或者它可能使用旧值.(这不是问题中计算的问题,因为从不分配一到六个键,并且在计算中从不使用键10到15.)