[ACCEPTED]-Python: Mapping from intervals to values-intervals

Accepted answer
Score: 51
import bisect
bisect.bisect_left([100,300,500,800,1000], p)

here the docs: bisect

0

Score: 3

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 2 always in Python, there will be any better 1 ways to do it.

Score: 3

It is indeed quite horrible. Without a requirement 6 to have no hardcoding, it should have been 5 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 4 a lookup function, both linear and using 3 binary search, with the no-hardcodings requirement 2 fulfilled, and a couple of sanity checks 1 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
Score: 0

Try something along the lines of:

d = {(None,100): 0, 
    (100,200): 1,
    ...
    (1000, None): 5}
value = 300 # example value
for k,v in d.items():
    if (k[0] is None or value > k[0]) and (k[1] is None or value <= k[1]):
        return v

0

Score: 0
def which_interval(endpoints, number):
    for n, endpoint in enumerate(endpoints):
        if number <= endpoint:
            return n
        previous = endpoint
    return n + 1

Pass your endpoints as a list in endpoints, like 3 this:

which_interval([100, 300, 500, 800, 1000], 5)

Edit:

The above is a linear search. Glenn 2 Maynard's answer will have better performance, since 1 it uses a bisection algorithm.

Score: 0

Another way ...

def which(lst, p): 
    return len([1 for el in lst if p > el])

lst = [100, 300, 500, 800, 1000]
which(lst, 2)
which(lst, 101)
which(lst, 1001)

0

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