题
请把这个问题移至 代码评论 - 区域. 。它更适合在那里,因为我知道下面的代码是垃圾,我希望关键的反馈以完成重写。
如何在Python中编写固定关系的关系?因此,如果 A
在范围内,然后返回其相应的常数。
[0,10] <-> a
]10,77] <-> b
]77,\inf[ <-> c
闻到代码,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
解决方案
如果列表比仅三个条目更长,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
但这似乎只有三个值过度设计...
编辑:第二个思考,最可读的变体可能是
if sold >= 7700:
rate = 0.2
elif sold >= 2000:
rate = 0.15
elif sold >= 0:
rate = 0.1
else:
# sold is negative
其他提示
if (sold >=0 & sold <7700):
等同于
if 0 <= sold < 7700:
我不知道一个 真的很棒 要映射范围,但这至少使看起来更好。
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)
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