CODEWITHSUNDEEP
pythonpython-3.xrange
mr.wolle
mr.wolle

Reputation: 1336

How to check if a range is a part of another range in Python 3.x

How can I simply check if a range is a subrange of another ?

range1 in range2 will not work as expected.

Upvotes: 17

Views: 15431

Answers (4)

Álvaro H.G
Álvaro H.G

Reputation: 602

I was working on a similar problem: how to find if two ranges overlap. Hence, the order of ranges introduced in the function was a limitation for me (the order should not affect). This is the solution I propose (inspired by @jonrsharpe answer):

def range_overlap(range1, range2):
  """Whether range1 and range2 overlap."""
  x1, x2 = range1.start, range1.stop
  y1, y2 = range2.start, range2.stop
  return x1 <= y2 and y1 <= x2

a = range(20, 25, 2)
b = range(23, 25, 3) # different step does not matter
print(list(a), list(b)) # [20, 22, 24] [23]
range_overlap(a, b) # True

In case any range is empty:

# if range1 is non-empty and range2 is empty
a = range(20, 25, 1)
b = range(22, 22, 3) # empty but the range start and end overlap

assert range_overlap(a, b) == range_overlap(a, b)
range_overlap(a, b)
# True

Upvotes: 4

user2804197
user2804197

Reputation: 402

This is the most readable imo, but inefficient:

all(e in range2 for e in range1)

I think this would be the most efficient if both ranges have the same step:

range1[0] in range2 and range1[-1] in range2

Upvotes: 3

jonrsharpe
jonrsharpe

Reputation: 122096

You can do it in O(1), as follows:

def range_subset(range1, range2):
    """Whether range1 is a subset of range2."""
    if not range1:
        return True  # empty range is subset of anything
    if not range2:
        return False  # non-empty range can't be subset of empty range
    if len(range1) > 1 and range1.step % range2.step:
        return False  # must have a single value or integer multiple step
    return range1.start in range2 and range1[-1] in range2

In use:

>>> range_subset(range(0, 1), range(0, 4))
True

Upvotes: 22

mr.wolle
mr.wolle

Reputation: 1336

set((range(0,1))).issubset(range(0,4)) will do it.

Upvotes: 5

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