Bijective relations between ranges and some constants?
Question
Please, move this question to Code Review -area. It is better suited there because I know the code below is junk and I wanted critical feedback to complete rewrite.
How can I write the set-to-constants relations in Python? So if A
in a range, then return its corresponding constant.
[0,10] <-> a
]10,77] <-> b
]77,\inf[ <-> c
Smelling code, bad.
# 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
Solution
If the list was longer than a mere three entries, I would use 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
But this seems a bit overengineered for just three values...
Edit: On second thought, the most readable variant probably is
if sold >= 7700:
rate = 0.2
elif sold >= 2000:
rate = 0.15
elif sold >= 0:
rate = 0.1
else:
# sold is negative
OTHER TIPS
if (sold >=0 & sold <7700):
is equivalent to
if 0 <= sold < 7700:
I'm not aware of a really great way to map ranges but this makes it much nicer looking at least.
You could use your 2nd approach too:
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)
Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow