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请将此问题移至Code Review -area。它更适合那里，因为我知道下面的代码是垃圾，我想要关键的反馈来完成重写。

如何在 Python 中编写集合到常量的关系？所以如果A在一个范围内，则返回其对应的常数。

[0,10]    <-> a
]10,77]   <-> b
]77,\inf[ <-> c

闻代码，不好。

    # Bad style

    provSum=0


    # TRIAL 1: messy if-clauses
    for sold in getSelling():
            if (sold >=0 & sold <7700):
                    rate =0.1 
            else if (sold>=7700 & sold <7700):   
            #won't even correct mistakes here because it shows how not to do things
                    rate =0.15
            else if (sold>=7700):
                    rate =0.20


    # TRIAL 2: messy, broke it because it is getting too hard to read
    provisions= {"0|2000":0.1, "2000|7700":0.15, "7700|99999999999999":0.20}


    if int(sold) >= int(border.split("|")[0]) & int(sold) < int(border.split("|")[1]):
            print sold, rate
            provSum = provSum + sold*rate
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2 回答 2

3

如果列表长于三个条目，我会使用bisect.bisect()：

limits = [0, 2000, 7700]
rates = [0.1, 0.15, 0.2]
index = bisect.bisect(limits, sold) - 1
if index >= 0:
    rate = rates[index]
else:
    # sold is negative

但这对于仅三个值似乎有点过度设计......

编辑：再想一想，最易读的变体可能是

if sold >= 7700:
    rate = 0.2
elif sold >= 2000:
    rate = 0.15
elif sold >= 0:
    rate = 0.1
else:
    # sold is negative
于 2011-01-25T16:33:30.577 回答
1 
if (sold >=0 & sold <7700):

相当于

if 0 <= sold < 7700:

我不知道映射范围的真正好方法，但这至少使它看起来更好看。

您也可以使用第二种方法：

provisions = {(0, 2000) : 0.1, (2000,7700):0.15, (7700, float("inf")):0.20}

# loop though the items and find the first that's in range
for (lower, upper), rate in provisions.iteritems():
    if lower <= sold < upper:
        break # `rate` remains set after the loop ..

# which pretty similar (see comments) to
rate = next(rate for (lower, upper), rate in 
                 provisions.iteritems() if lower <= sold < upper)
于 2011-01-25T16:27:12.110 回答