在列表中标识连续数字组

more_itertools.consecutive_groups 在4.0版中添加。

演示

import more_itertools as mit


iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
[list(group) for group in mit.consecutive_groups(iterable)]
# [[2, 3, 4, 5], [12, 13, 14, 15, 16, 17], [20]]

代码

应用该工具,我们可以生成一个求连续数字范围的生成器函数。

def find_ranges(iterable):
    """Yield range of consecutive numbers."""
    for group in mit.consecutive_groups(iterable):
        group = list(group)
        if len(group) == 1:
            yield group[0]
        else:
            yield group[0], group[-1]


iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
list(find_ranges(iterable))
# [(2, 5), (12, 17), 20]

这个 source 实现模拟 classic recipe (如@nadia alramli所示)。

注: more_itertools 第三方软件包是否可以通过安装 pip install more_itertools .

编辑2:回答OP新要求

ranges = []
for key, group in groupby(enumerate(data), lambda (index, item): index - item):
    group = map(itemgetter(1), group)
    if len(group) > 1:
        ranges.append(xrange(group[0], group[-1]))
    else:
        ranges.append(group[0])

输出:

[xrange(2, 5), xrange(12, 17), 20]

您可以用range或任何其他自定义类替换xrange。

python文档有一个非常整洁的 recipe 为此:

from operator import itemgetter
from itertools import groupby
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
    print map(itemgetter(1), g)

输出:

[2, 3, 4, 5]
[12, 13, 14, 15, 16, 17]

如果要获得完全相同的输出,可以执行以下操作:

ranges = []
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
    group = map(itemgetter(1), g)
    ranges.append((group[0], group[-1]))

输出:

[(2, 5), (12, 17)]

编辑: 这个例子已经在文档中解释过了,但也许我应该更详细地解释一下:

解决方案的关键是 与范围不同,所以 连续的数字都出现在同一个 组。

如果数据是: [2, 3, 4, 5, 12, 13, 14, 15, 16, 17] 然后 groupby(enumerate(data), lambda (i,x):i-x) 相当于以下内容:

groupby(
    [(0, 2), (1, 3), (2, 4), (3, 5), (4, 12),
    (5, 13), (6, 14), (7, 15), (8, 16), (9, 17)],
    lambda (i,x):i-x
)

lambda函数从元素值中减去元素索引。所以当你在每个项目上应用lambda时。您将获得groupby的以下密钥:

[-2, -2, -2, -2, -8, -8, -8, -8, -8, -8]

groupby按相等的键值对元素进行分组,因此前4个元素将被分组在一起,以此类推。

我希望这能使它更可读。

python 3 版本可能对初学者有帮助

首先导入所需的库

from itertools import groupby
from operator import itemgetter

ranges =[]

for k,g in groupby(enumerate(data),lambda x:x[0]-x[1]):
    group = (map(itemgetter(1),g))
    group = list(map(int,group))
    ranges.append((group[0],group[-1]))

“天真”的解决方案,我觉得至少有点可读性。

x = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 22, 25, 26, 28, 51, 52, 57]

def group(L):
    first = last = L[0]
    for n in L[1:]:
        if n - 1 == last: # Part of the group, bump the end
            last = n
        else: # Not part of the group, yield current group and start a new
            yield first, last
            first = last = n
    yield first, last # Yield the last group


>>>print list(group(x))
[(2, 5), (12, 17), (22, 22), (25, 26), (28, 28), (51, 52), (57, 57)]

假设您的列表已排序:

>>> from itertools import groupby
>>> def ranges(lst):
    pos = (j - i for i, j in enumerate(lst))
    t = 0
    for i, els in groupby(pos):
        l = len(list(els))
        el = lst[t]
        t += l
        yield range(el, el+l)


>>> lst = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
>>> list(ranges(lst))
[range(2, 6), range(12, 18)]

在这里,它应该可以工作,而不需要任何导入:

def myfunc(lst):
    ret = []
    a = b = lst[0]                           # a and b are range's bounds

    for el in lst[1:]:
        if el == b+1: 
            b = el                           # range grows
        else:                                # range ended
            ret.append(a if a==b else (a,b)) # is a single or a range?
            a = b = el                       # let's start again with a single
    ret.append(a if a==b else (a,b))         # corner case for last single/range
    return ret

请注意,代码使用 groupby 不能像Python3中给出的那样工作,所以使用这个。

for k, g in groupby(enumerate(data), lambda x:x[0]-x[1]):
    group = list(map(itemgetter(1), g))
    ranges.append((group[0], group[-1]))

这不使用标准函数-它只是对输入进行迭代,但它应该可以工作:

def myfunc(l):
    r = []
    p = q = None
    for x in l + [-1]:
        if x - 1 == q:
            q += 1
        else:
            if p:
               if q > p:
                   r.append('%s-%s' % (p, q))
               else:
                   r.append(str(p))
            p = q = x
    return '(%s)' % ', '.join(r)

注意,它要求输入只包含按升序排列的正数。您应该验证输入,但为了清晰起见,省略了此代码。

这是我想出来的答案。我编写代码是为了让其他人理解,所以我对变量名和注释相当冗长。

首先是一个快速助手函数:

def getpreviousitem(mylist,myitem):
    '''Given a list and an item, return previous item in list'''
    for position, item in enumerate(mylist):
        if item == myitem:
            # First item has no previous item
            if position == 0:
                return None
            # Return previous item    
            return mylist[position-1] 

然后是实际代码:

def getranges(cpulist):
    '''Given a sorted list of numbers, return a list of ranges'''
    rangelist = []
    inrange = False
    for item in cpulist:
        previousitem = getpreviousitem(cpulist,item)
        if previousitem == item - 1:
            # We're in a range
            if inrange == True:
                # It's an existing range - change the end to the current item
                newrange[1] = item
            else:    
                # We've found a new range.
                newrange = [item-1,item]
            # Update to show we are now in a range    
            inrange = True    
        else:   
            # We were in a range but now it just ended
            if inrange == True:
                # Save the old range
                rangelist.append(newrange)
            # Update to show we're no longer in a range    
            inrange = False 
    # Add the final range found to our list
    if inrange == True:
        rangelist.append(newrange)
    return rangelist

实例运行:

getranges([2, 3, 4, 5, 12, 13, 14, 15, 16, 17])

返回:

[[2, 5], [12, 17]]

import numpy as np

myarray = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
sequences = np.split(myarray, np.array(np.where(np.diff(myarray) > 1)[0]) + 1)
l = []
for s in sequences:
    if len(s) > 1:
        l.append((np.min(s), np.max(s)))
    else:
        l.append(s[0])
print(l)

输出:

[(2, 5), (12, 17), 20]

使用numpy+理解列表:
使用numpy diff函数,可以识别出其差不等于一的后续输入向量项。需要考虑输入向量的开始和结束。

import numpy as np
data = np.array([2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20])

d = [i for i, df in enumerate(np.diff(data)) if df!= 1] 
d = np.hstack([-1, d, len(data)-1])  # add first and last elements 
d = np.vstack([d[:-1]+1, d[1:]]).T

print(data[d])

输出:

 [[ 2  5]   
  [12 17]   
  [20 20]]

注:省略了个别数字应被区别对待的请求(作为个别数字返回,而不是范围)。这可以通过进一步的后处理结果来实现。通常情况下,这会使事情变得更复杂,而不会获得任何好处。

不需要额外导入就可以工作的简短解决方案。它接受任何iterable,对未排序的输入进行排序,并删除重复项:

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))

例子:

>>> 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([])
[]

这和@dansalmo的一样 solution 我发现这很神奇,虽然有点难阅读和应用(因为它不是作为函数给出的)。

注意,它可以很容易地修改为吐出“传统”的开放范围。 [start, end) ,例如,通过修改返回语句:

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

我把这个答案抄过来 another question 它被标记为这个主题的副本,目的是让它更容易被发现(在我刚刚再次搜索这个主题之后,一开始只在这里找到问题,对给出的答案不满意)。