A naive method to create list within a given range is to first create an empty list and append successor of each integer in every iteration of for loop. # list until r2 is reached. We can also use list comprehension for the purpose. Just iterate 'item' in a for loop from r1 to r2 and return all 'item' as list.
In Python, you can generate a series of numbers with the built-in function range() .
The most Pythonic way to convert a list of integers ints to a list of strings is to use the one-liner strings = [str(x) for x in ints] . It iterates over all elements in the list ints using list comprehension and converts each list element x to a string using the str(x) constructor.
Using itertools.groupby()
produces a concise but tricky implementation:
import itertools
def ranges(i):
for a, b in itertools.groupby(enumerate(i), lambda pair: pair[1] - pair[0]):
b = list(b)
yield b[0][1], b[-1][1]
print(list(ranges([0, 1, 2, 3, 4, 7, 8, 9, 11])))
Output:
[(0, 4), (7, 9), (11, 11)]
You can use a list comprehension with a generator expression and a combination of enumerate() and itertools.groupby():
>>> import itertools
>>> l = [0, 1, 2, 3, 4, 7, 8, 9, 11]
>>> [[t[0][1], t[-1][1]] for t in
... (tuple(g[1]) for g in itertools.groupby(enumerate(l), lambda (i, x): i - x))]
[[0, 4], [7, 9], [11, 11]]
First, enumerate()
will build tuples from the list items and their respective index:
>>> [t for t in enumerate(l)]
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 7), (6, 8), (7, 9), (8, 11)]
Then groupby()
will group those tuples using the difference between their index and their value (which will be equal for consecutive values):
>>> [tuple(g[1]) for g in itertools.groupby(enumerate(l), lambda (i, x): i - x)]
[((0, 0), (1, 1), (2, 2), (3, 3), (4, 4)), ((5, 7), (6, 8), (7, 9)), ((8, 11),)]
From there, we only need to build lists from the values of the first and last tuples of each group (which will be the same if the group only contains one item).
You can also use [(t[0][1], t[-1][1]) ...]
to build a list of range tuples instead of nested lists, or even ((t[0][1], t[-1][1]) ...)
to turn the whole expression into a iterable generator
that will lazily build the range tuples on the fly.
This is an improvement over the very elegant answer. This one covers non-unique and non-sorted input and is python3 compatible too:
import itertools
def to_ranges(iterable):
iterable = sorted(set(iterable))
for key, group in itertools.groupby(enumerate(iterable),
lambda t: t[1] - t[0]):
group = list(group)
yield group[0][1], group[-1][1]
Example:
>>> x
[44, 45, 2, 56, 23, 11, 3, 4, 7, 9, 1, 2, 2, 11, 12, 13, 45]
>>> print( list(to_ranges(x)))
[(1, 4), (7, 7), (9, 9), (11, 13), (23, 23), (44, 45), (56, 56)]
Generating range pairs:
def ranges(lst):
s = e = None
r = []
for i in sorted(lst):
if s is None:
s = e = i
elif i == e or i == e + 1:
e = i
else:
r.append((s, e))
s = e = i
if s is not None:
r.append((s, e))
return r
Example:
>>> lst = [1, 5, 6, 7, 12, 15, 16, 17, 18, 30]
>>> print repr(ranges(lst))
[(1, 1), (5, 7), (12, 12), (15, 18), (30, 30)]
As a generator:
def gen_ranges(lst):
s = e = None
for i in sorted(lst):
if s is None:
s = e = i
elif i == e or i == e + 1:
e = i
else:
yield (s, e)
s = e = i
if s is not None:
yield (s, e)
Example:
>>> lst = [1, 5, 6, 7, 12, 15, 16, 17, 18, 30]
>>> print repr(','.join(['%d' % s if s == e else '%d-%d' % (s, e) for (s, e) in gen_ranges(lst)]))
'1,5-7,12,15-18,30'
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