Python: Mapping from intervals to values

Question

I'm refactoring a function that, given a series of endpoints that implicitly define intervals, checks if a number is included in the interval, and then return a corresponding (not related in any computable way). The code that is now handling the work is:

if p <= 100:
    return 0
elif p > 100 and p <= 300:
    return 1
elif p > 300 and p <= 500:
    return 2
elif p > 500 and p <= 800:
    return 3
elif p > 800 and p <= 1000:
    return 4
elif p > 1000:
    return 5

Which is IMO quite horrible, and lacks in that both the intervals and the return values are hardcoded. Any use of any data structure is of course possible.

Solution 1

import bisect
bisect.bisect_left([100,300,500,800,1000], p)

here the docs: bisect

Solution 2

You could try a take on this:

def check_mapping(p):
    mapping = [(100, 0), (300, 1), (500, 2)] # Add all your values and returns here

    for check, value in mapping:
        if p <= check:
            return value

print check_mapping(12)
print check_mapping(101)
print check_mapping(303)

produces:

0
1
2

As always in Python, there will be any better ways to do it.

Solution 3

It is indeed quite horrible. Without a requirement to have no hardcoding, it should have been written like this:

if p <= 100:
    return 0
elif p <= 300:
    return 1
elif p <= 500:
    return 2
elif p <= 800:
    return 3
elif p <= 1000:
    return 4
else:
    return 5

Here are examples of creating a lookup function, both linear and using binary search, with the no-hardcodings requirement fulfilled, and a couple of sanity checks on the two tables:

def make_linear_lookup(keys, values):
    assert sorted(keys) == keys
    assert len(values) == len(keys) + 1
    def f(query):
        return values[sum(1 for key in keys if query > key)]
    return f

import bisect
def make_bisect_lookup(keys, values):
    assert sorted(keys) == keys
    assert len(values) == len(keys) + 1
    def f(query):
        return values[bisect.bisect_left(keys, query)]
    return f

Comments

[stefanw]

Does not consider the case of p > 1000!

[kjfletch]

That is why I specified: "You could try a take on this"

[sykora]

That last sentence is ironic, considering the python philosophy of having preferably only one obvious way to do something.

[John Machin]

BUG: It produces None if p is greater than the last endpoint.

[JAB]

I like this one better than the one that has the most votes because of its more generalized/non-hardcoded form and because it is more in-depth.

[John Machin]

Lose the "previous" caper; it's quite redundant.

[Steef]

Yeah, you're right, I guess the original code "inspired" me to use it. BTW, your use of the imperative might sound a bit gruff to some.

[Agos]

Truly impressive. Super clean, and I believe very fast too. It can also be easily extended in case one does need a non-natural ordering or something else in return, like a string: import bisect n = bisect.bisect_left([100,300,500,800,1000], p) a=["absent","low","average","high", "very high", "extreme"] a[n]

[John Machin]

@Steef: You may wish to consider a humble suggestion that you might at your leisure revist your answer, note that your answer still includes a redundant line of code, and in the fullness of time, excise the same.

[Charlie Parker]

hmmm but this doesn't return "arbitrary values" it returns the index. How do I have it return the arbitrary value? I tried p = 10 x = bisect.bisect_left(OrderedDict({10: 'a', 11: 'b'}), p) print() but it didn't work.