Comments (2)
You can already achieve the same result with portion.Interval(*[portion.open(i*10, i*10+1) for i in range(5)])
.
More generally, you can rely on the operator
module and the reduce
function to achieve the same result. For example,
>>> from operator import or_
>>> from functools import reduce
>>> intervals = [portion.open(i*10, i*10+1) for i in range(5)]
>>> result = reduce(or_, intervals)
from portion.
Got it! Thank you for the answer
from portion.
Related Issues (20)
- Iteration of empty intervals is inconsistent. in general "empty := (+inf,-inf)" is problematic HOT 15
- Is this library performant enough to work with (non-atomic) intervals which span integers between 1 and 1 billion? HOT 1
- Using an external comparator? HOT 3
- iterate is broken with subclasses of Interval HOT 12
- Add __format__ method to Interval (improvement) HOT 16
- Thoughts about text-annotation use case and Pandas Ext. API HOT 5
- interval diameter (length, width, measure, range, or size) HOT 8
- "compatible version" specifier in setup.py confuses poetry HOT 2
- AttributeError: module 'portion.interval' has no attribute 'empty' HOT 3
- IntervalMultiDict HOT 3
- Add example for pandas in README HOT 5
- importlib error with create_api HOT 2
- Error to import interval, inf, imath from interval HOT 1
- importlib.machinery error with create_api HOT 6
- Enclosure Calculation Bug HOT 1
- Empty Calculation Bug HOT 3
- The performance issues of interval calculations in large quantities. HOT 1
- mass/Lebesgue measure? HOT 7
- Value out of range HOT 4
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from portion.