检测列表中的连续整数

From the docs:

>>> from itertools import groupby
>>> from operator import itemgetter
>>> data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28]
>>> for k, g in groupby(enumerate(data), lambda (i, x): i-x):
...     print map(itemgetter(1), g)
...
[1]
[4, 5, 6]
[10]
[15, 16, 17, 18]
[22]
[25, 26, 27, 28]

You can adapt this fairly easily to get a printed set of ranges.

Built-In: No, as far as I'm aware.

You have to run through the array. Start off with putting the first value in a variable and print it, then as long as you keep hitting the next number do nothing but remember the last number in another variable. If the next number is not in line, check the last number remembered versus the first number. If it's the same, do nothing. If it's different, print "-" and the last number. Then put the current value in the first variable and start over. At the end of the array you run the same routine as if you had hit a number out of line.

I could have written the code, of course, but I don't want to spoil your homework :-)

This will print exactly as you specified:

>>> nums = [1, 2, 3, 4, 7, 8, 10, 11, 12, 13, 14]
>>> ranges = sum((list(t) for t in zip(nums, nums[1:]) if t[0]+1 != t[1]), [])
>>> iranges = iter(nums[0:1] + ranges + nums[-1:])
>>> print ', '.join([str(n) + '-' + str(next(iranges)) for n in iranges])
1-4, 7-8, 10-14

If the list has any single number ranges, they would be shown as n-n:

>>> nums = [1, 2, 3, 4, 5, 7, 8, 9, 12, 15, 16, 17, 18]
>>> ranges = sum((list(t) for t in zip(nums, nums[1:]) if t[0]+1 != t[1]), [])
>>> iranges = iter(nums[0:1] + ranges + nums[-1:])
>>> print ', '.join([str(n) + '-' + str(next(iranges)) for n in iranges])
1-5, 7-9, 12-12, 15-18

Here is another basic solution without using any module, which is good for interview, generally in the interview they asked without using any modules:

#!/usr/bin/python


def split_list(n):
"""will return the list index"""
return [(x+1) for x,y in zip(n, n[1:]) if y-x != 1]


def get_sub_list(my_list):
"""will split the list base on the index"""
my_index = split_list(my_list)
output = list()
prev = 0
for index in my_index:
new_list = [ x for x in my_list[prev:] if x < index]
output.append(new_list)
prev += len(new_list)
output.append([ x for x in my_list[prev:]])
return output


my_list = [1, 3, 4, 7, 8, 10, 11, 13, 14]
print get_sub_list(my_list)

Output:

[[1], [3, 4], [7, 8], [10, 11], [13, 14]]

I had a similar problem and am using the following for a sorted list. It outputs a dictionary with ranges of values listed in a dictionary. The keys separate each run of consecutive numbers and are also the running total of non-sequential items between numbers in sequence.

Your list gives me an output of {0: [1, 4], 1: [7, 8], 2: [10, 14]}

def series_dictf(index_list):
from collections import defaultdict
series_dict = defaultdict(list)
sequence_dict = dict()


list_len = len(index_list)
series_interrupts = 0


for i in range(list_len):
if i == (list_len - 1):
break


position_a = index_list[i]
position_b = index_list[i + 1]


if position_b == (position_a + 1):
sequence_dict[position_a] = (series_interrupts)
sequence_dict[position_b] = (series_interrupts)


if position_b != (position_a + 1):
series_interrupts += 1


for position, series in sequence_dict.items():
series_dict[series].append(position)
for series, position in series_dict.items():
series_dict[series] = [position[0], position[-1]]


return series_dict

Using set operation, the following algorithm can be executed

def get_consecutive_integer_series(integer_list):
integer_list = sorted(integer_list)
start_item = integer_list[0]
end_item = integer_list[-1]


a = set(integer_list)  # Set a
b = range(start_item, end_item+1)


# Pick items that are not in range.
c = set(b) - a  # Set operation b-a


li = []
start = 0
for i in sorted(c):
end = b.index(i)  # Get end point of the list slicing
li.append(b[start:end])  # Slice list using values
start = end + 1  # Increment the start point for next slicing
li.append(b[start:])  # Add the last series


for sliced_list in li:
if not sliced_list:
# list is empty
continue
if len(sliced_list) == 1:
# If only one item found in list
yield sliced_list[0]
else:
yield "{0}-{1}".format(sliced_list[0], sliced_list[-1])




a = [1, 2, 3, 6, 7, 8, 4, 14, 15, 21]
for series in get_consecutive_integer_series(a):
print series

Output for the above list "a"
1-4
6-8
14-15
21

You can use collections library which has a class called Counter. Counter can come in handy if trying to poll the no of distinct elements in any iterable

from collections import Counter
data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28]
cnt=Counter(data)
print(cnt)

the output for this looks like

Counter({1: 1, 4: 1, 5: 1, 6: 1, 10: 1, 15: 1, 16: 1, 17: 1, 18: 1, 22: 1, 25: 1, 26: 1, 27: 1, 28: 1})

which just like any other dictionary can be polled for key values

A short solution that works without additional imports. It accepts any iterable, sorts unsorted inputs, and removes duplicate items:

def ranges(nums):
nums = sorted(set(nums))
gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s+1 < e]
edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])
return list(zip(edges, edges))

Example:

>>> ranges([2, 3, 4, 7, 8, 9, 15])
[(2, 4), (7, 9), (15, 15)]


>>> ranges([-1, 0, 1, 2, 3, 12, 13, 15, 100])
[(-1, 3), (12, 13), (15, 15), (100, 100)]


>>> ranges(range(100))
[(0, 99)]


>>> ranges([0])
[(0, 0)]


>>> ranges([])
[]

This is the same as @dansalmo's solution which I found amazing, albeit a bit hard to read and apply (as it's not given as a function).

Note that it could easily be modified to spit out "traditional" open ranges [start, end), by e.g. altering the return statement:

    return [(s, e+1) for s, e in zip(edges, edges)]