openmedialibrary_platform/Shared/lib/python3.7/site-packages/idna/intranges.py

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2019-01-13 08:01:53 +00:00
"""
Given a list of integers, made up of (hopefully) a small number of long runs
of consecutive integers, compute a representation of the form
((start1, end1), (start2, end2) ...). Then answer the question "was x present
in the original list?" in time O(log(# runs)).
"""
import bisect
def intranges_from_list(list_):
"""Represent a list of integers as a sequence of ranges:
((start_0, end_0), (start_1, end_1), ...), such that the original
integers are exactly those x such that start_i <= x < end_i for some i.
Ranges are encoded as single integers (start << 32 | end), not as tuples.
"""
sorted_list = sorted(list_)
ranges = []
last_write = -1
for i in range(len(sorted_list)):
if i+1 < len(sorted_list):
if sorted_list[i] == sorted_list[i+1]-1:
continue
current_range = sorted_list[last_write+1:i+1]
ranges.append(_encode_range(current_range[0], current_range[-1] + 1))
last_write = i
return tuple(ranges)
def _encode_range(start, end):
return (start << 32) | end
def _decode_range(r):
return (r >> 32), (r & ((1 << 32) - 1))
def intranges_contain(int_, ranges):
"""Determine if `int_` falls into one of the ranges in `ranges`."""
tuple_ = _encode_range(int_, 0)
pos = bisect.bisect_left(ranges, tuple_)
# we could be immediately ahead of a tuple (start, end)
# with start < int_ <= end
if pos > 0:
left, right = _decode_range(ranges[pos-1])
if left <= int_ < right:
return True
# or we could be immediately behind a tuple (int_, end)
if pos < len(ranges):
left, _ = _decode_range(ranges[pos])
if left == int_:
return True
return False