Group consecutive integers and tolerate gaps of 1

问题

In Python, given a list of sorted integers, I would to group them by consecutive values and tolerate gaps of 1.

For instance, given a list my_list:

In [66]: my_list
Out[66]: [0, 1, 2, 3, 5, 6, 10, 11, 15, 16, 18, 19, 20]

I would like the following output:

[[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]]

Now, if I didn't have to tolerate gaps of 1, I could apply the neat solution explained here:

import itertools
import operator
results = []
for k, g in itertools.groupby(enumerate(my_list), lambda (i,x):i-x):
        group = map(operator.itemgetter(1), g)
        results.append(group)

Is there a way to incorporate my extra requirement in the above solution? If not, what's the best way to tackle the problem?

回答1:

When in doubt you can always write your own generator:

def group_runs(li,tolerance=2):
    out = []
    last = li[0]
    for x in li:
        if x-last > tolerance:
            yield out
            out = []
        out.append(x)
        last = x
    yield out

demo:

list(group_runs(my_list))
Out[48]: [[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]]

回答2:

Numpy is a very useful tool, and not very difficult to learn.

This problem is solvable in O(n) with a single line of code (excluding imports, data, and converting to list - if you really need it):

from numpy import array, diff, where, split
l= [0, 1, 2, 3, 5, 6, 10, 11, 15, 16, 18, 19, 20]
result= split(l, where(diff(l)>2)[0]+1 )
print map(list, result)

More importantly, the code is very fast if you need to process large lists, unlike a pure-python solution

回答3:

Remember, groupby in itself, is pretty lame. The strength of itertools.groupby is the key. For this particular problem, you need to create an appropriate key with memory (stateless key will not work here).

Implementation

class Key(object):
    def __init__(self, diff):
        self.diff, self.flag, self.prev = diff, [0,1], None
    def __call__(self, elem):
        if self.prev and abs(self.prev - elem) > self.diff:
            self.flag = self.flag[::-1]
        self.prev= elem
        return self.flag[0]

Object

[list(g) for k, g in groupby(my_list, key = Key(2))]
[[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]]

How it Works

Every time, a new sub-list needs to be created (curr - prev > threshold), you toggle a flag. There are different ways to toggle a flag

flag = 1; flag *= -1
flag = [0, 1 ]; flag = flag[::-1]
flag = 0; flag = 0 if flag else 1

Choose what ever your heart contends

So this generates an accompanying key along with your list

[0, 1, 2, 3, 5, 6, 10, 11, 15, 16, 18, 19, 20]
[0, 0, 0, 0, 0, 0, 1,  1,  0,  0,  0,  0 , 0]
             <------>  <------>
          Toggle flag  Toggle flag
          0 -> 1, as   1 -> 0, as
          10 - 6 > 2   15 - 11 > 2

Now as itertools.groupby, groups consecutive elements with same key, all elements with keys having consecutive 0s or 1s are grouped together

[0, 1, 2, 3, 5, 6, 10, 11, 15, 16, 18, 19, 20]
[0, 0, 0, 0, 0, 0, 1,  1,  0,  0,  0,  0 , 0]

[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]
[0, 0, 0, 0, 0, 0], [1,  1],  [0,  0,  0,  0 , 0]

And your final result would be

[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]

回答4:

An O(nlogn) solution (assuming the input list isn't sorted) is to first the sort the list you're given, then iterate through each value, creating a new group whenever the difference between the current value and the previous value is at least 3.

Demo

>>> my_list = [0, 1, 2, 3, 5, 6, 10, 11, 15, 16, 18, 19, 20]
>>> my_list.sort() # if we can't assume the list is sorted beforehand
>>> groups = [[my_list[0]]] # initialize with the first value in the list
>>> for i, val in enumerate(my_list[1:]):
...     if val - groups[-1][-1] > 2:
...         groups.append( [val] ) # create a new group
...     else:
...         groups[-1].append( val ) # append to the most recent group
... 
>>> groups
[[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]]

回答5:

Here's what I came up with. There's a bit of verbose initialization but it gets the job done. =)

output = []
prev = my_list[0]
temp_list = [my_list[0]]

for num in my_list[1:]:
    if num-2 > prev:
        output += [temp_list]
        temp_list = [num]
    else:
        temp_list.append(num)
    prev = num
output.append(temp_list)

print output

# [[0, 1, 2, 3, 5, 6], [10, 11], [15, 16, 18, 19, 20]]

回答6:

I generally use zip when I want to deal with consecutive elements, and you can use islice you want to avoid building the list slice:

from itertools import islice

def group(lst, tol=1):
    """Group vals in sorted iterable lst, allow tol between consecutive vals."""
    output = [[]]
    for current, next_ in zip(lst, islice(lst, 1, None)):
        output[-1].append(current)
        if next_ > current + tol + 1:
            output.append([])
    output[-1].append(lst[-1])
    return output

Note that in Python 2.x, you need to use itertools.izip to avoid building the list of 2-tuples (current, next_).

来源：https://stackoverflow.com/questions/21142231/group-consecutive-integers-and-tolerate-gaps-of-1

标签

python

list

grouping

itertools