I have an interval of integers that comprises some inner intervals. Given these intervals I want to compute a list including the intervals between
Question:
Inner intervals are always inside the global one.
All intervals are integer, left-closed, right-open intervals.
Let’s take this example.
The "global" interval is [0, 22[.
"Inner" intervals are [3, 6[ and [12, 15[.
For this example I expect :
[0, 3[ U [3, 6[ U [6, 12[ U [12, 15[ U [15, 22[
I’ve tried to define a function but then messed up with indices while iterating over intervals.
def allspans(r, spans):
pass
allspans((0, 22), [(3,6), (12,15)]) # expected : [(0, 3), (3, 6), (6, 12), (12, 15), (15, 22)]
Answers:
Yes you have to iterate over your spans but take care of maintaining a position to correctly fill the spaces between.
from typing import Generator
def allspans(r, spans) -> Generator:
pos = 0
for lower, upper in spans:
if pos < lower:
yield pos, lower
yield lower, upper
pos = upper
if pos <= r[1]:
yield pos, r[1]
I find it easier to use a Generator.
Just use list() to convert to a List.
list(allspans((0, 22), [(3,6), (12,15)])) # [(0, 3), (3, 6), (6, 12), (12, 15), (15, 22)]
Using a normal loop:
def allspans(r, spans):
intervals = []
intervals.append((r[0], spans[0][0]))
for i in range(len(spans)):
current_span = spans[i]
if i != 0:
intervals.append((spans[i - 1][1], current_span[0]))
intervals.append((current_span[0], current_span[1]))
intervals.append((spans[-1][1], r[1]))
return intervals
print(allspans((0, 22), [(3, 6), (12, 15)]))
# [(0, 3), (3, 6), (6, 12), (12, 15), (15, 22)]
Using itertools.chain and itertools.pairwise (Python 3.10+):
from itertools import chain, pairwise
def all_spans(r, spans):
start, end = r
it = chain((start,), chain.from_iterable(spans), (end,))
return [t for t in pairwise(it) if t[0] != t[1]]
First, we construct an iterator it over all of the interval endpoints in order; then the sub-ranges are all the pairs of consecutive endpoints, excluding the empty sub-ranges where two consecutive endpoints are equal.
Inner intervals are always inside the global one.
All intervals are integer, left-closed, right-open intervals.
Let’s take this example.
The "global" interval is [0, 22[.
"Inner" intervals are [3, 6[ and [12, 15[.
For this example I expect :
[0, 3[ U [3, 6[ U [6, 12[ U [12, 15[ U [15, 22[
I’ve tried to define a function but then messed up with indices while iterating over intervals.
def allspans(r, spans):
pass
allspans((0, 22), [(3,6), (12,15)]) # expected : [(0, 3), (3, 6), (6, 12), (12, 15), (15, 22)]
Yes you have to iterate over your spans but take care of maintaining a position to correctly fill the spaces between.
from typing import Generator
def allspans(r, spans) -> Generator:
pos = 0
for lower, upper in spans:
if pos < lower:
yield pos, lower
yield lower, upper
pos = upper
if pos <= r[1]:
yield pos, r[1]
I find it easier to use a Generator.
Just use list() to convert to a List.
list(allspans((0, 22), [(3,6), (12,15)])) # [(0, 3), (3, 6), (6, 12), (12, 15), (15, 22)]
Using a normal loop:
def allspans(r, spans):
intervals = []
intervals.append((r[0], spans[0][0]))
for i in range(len(spans)):
current_span = spans[i]
if i != 0:
intervals.append((spans[i - 1][1], current_span[0]))
intervals.append((current_span[0], current_span[1]))
intervals.append((spans[-1][1], r[1]))
return intervals
print(allspans((0, 22), [(3, 6), (12, 15)]))
# [(0, 3), (3, 6), (6, 12), (12, 15), (15, 22)]
Using itertools.chain and itertools.pairwise (Python 3.10+):
from itertools import chain, pairwise
def all_spans(r, spans):
start, end = r
it = chain((start,), chain.from_iterable(spans), (end,))
return [t for t in pairwise(it) if t[0] != t[1]]
First, we construct an iterator it over all of the interval endpoints in order; then the sub-ranges are all the pairs of consecutive endpoints, excluding the empty sub-ranges where two consecutive endpoints are equal.