Question

我有一个整数列表......

[1,2,3,4,5,8,9,10,11,200,201,202]

我想将它们分组到一个列表列表中，其中每个子列表包含其序列尚未被破坏的整数。像这样......

[[1,5],[8,11],[200,202]]

我有一个相当笨重的工作...

lSequenceOfNum = [1,2,3,4,5,8,9,10,11,200,201,202]

lGrouped = []
start = 0
for x in range(0,len(lSequenceOfNum)):
    if x != len(lSequenceOfNum)-1:
        if(lSequenceOfNum[x+1] - lSequenceOfNum[x]) > 1:
            lGrouped.append([lSequenceOfNum[start],lSequenceOfNum[x]])
            start = x+1

    else:
        lGrouped.append([lSequenceOfNum[start],lSequenceOfNum[x]])
print lGrouped

这是我能做的最好的事情。是否有更“pythonic”的方式来做到这一点？感谢..

Answer 1

假设列表将始终按升序排列：

from itertools import groupby, count

numberlist = [1,2,3,4,5,8,9,10,11,200,201,202]

def as_range(g):
    l = list(g)
    return l[0], l[-1]

print [as_range(g) for _, g in groupby(numberlist, key=lambda n, c=count(): n-next(c))]

Answer 2

我意识到我有点过于复杂，比使用稍微复杂的发电机更容易手动计算：

def ranges(seq):
    start, end = seq[0], seq[0]
    count = start
    for item in seq:
        if not count == item:
            yield start, end
            start, end = item, item
            count = item
        end = item
        count += 1
    yield start, end

print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))

产：

[(1, 5), (8, 11), (200, 202)]

这种方法非常快：

这种方法（和旧方法一样，它们的表现几乎完全相同）：

python -m timeit -s "from test import ranges" "ranges([1,2,3,4,5,8,9,10,11,200,201,202])"
1000000 loops, best of 3: 0.47 usec per loop

Jeff Mercado's Method：

python -m timeit -s "from test import as_range; from itertools import groupby, count" "[as_range(g) for _, g in groupby([1,2,3,4,5,8,9,10,11,200,201,202], key=lambda n, c=count(): n-next(c))]"
100000 loops, best of 3: 11.1 usec per loop

速度提高了20倍 - 当然，除非速度很重要，否则这不是真正的问题。

我使用生成器的旧解决方案：

import itertools

def resetable_counter(start):
    while True:
        for i in itertools.count(start):
            reset = yield i
            if reset:
                start = reset
                break

def ranges(seq):
    start, end = seq[0], seq[0]
    counter = resetable_counter(start)
    for count, item in zip(counter, seq): #In 2.x: itertools.izip(counter, seq)
        if not count == item:
            yield start, end
            start, end = item, item
            counter.send(item)
        end = item
    yield start, end

print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))

产：

[(1, 5), (8, 11), (200, 202)]

Answer 3

您可以通过三个步骤有效地完成此任务

给定的

list1=[1,2,3,4,5,8,9,10,11,200,201,202]

计算不连续性

     [1,2,3,4,5,8,9,10,11 ,200,201,202]
-      [1,2,3,4,5,8,9 ,10 ,11 ,200,201,202]
----------------------------------------
       [1,1,1,1,3,1,1 ,1  ,189,1  ,1]
(index) 1 2 3 4 5 6 7  8   9   10  11 
                *          *
rng = [i+1 for i,e in enumerate((x-y for x,y in zip(list1[1:],list1))) if e!=1] 
>>> rng
[5, 9]

添加边界

rng = [0] + rng + [len(list1)]
>>> rng
[0, 5, 9,12]

现在计算实际的连续性范围

[(list1[i],list1[j-1]) for i,j in zip(list2,list2[1:])]
[(1, 5), (8, 11), (200, 202)]

LB                [0,   5,    9,  12]
UB             [0, 5,   9,    12]
     -----------------------
indexes (LB,UB-1) (0,4) (5,8) (9,11)

Answer 4

这个问题已经很老了，但我还是认为我会分享我的解决方案

假设$(data).each(function(){ console.log(this.chart_data); console.log(this.chart_data.pop()); var end = (this.chart_data).pop(); ret.push({ name: this.name, y: end['y'], link: this.link, color: chart_colours[i] }); i++; });

import numpy as np

Answer 5

伪代码（需要修复一个错误）：

jumps = new array;
for idx from 0 to len(array)
if array[idx] != array[idx+1] then jumps.push(idx);

我认为这实际上是一种情况，使用索引（如在C中，在java / python / perl / etc之前改进之前）而不是数组中的对象是有意义的。

Answer 6

这是一个易于阅读的版本：

def close_range(el, it):
    while True:
        el1 = next(it, None)
        if el1 != el + 1:
            return el, el1
        el = el1

def compress_ranges(seq):
    iterator = iter(seq)
    left = next(iterator, None)
    while left is not None:
        right, left1 = close_range(left, iterator)
        yield (left, right)
        left = left1

list(compress_ranges([1, 2, 3, 4, 5, 8, 9, 10, 11, 200, 201, 202]))

Answer 7

类似的问题：
Python - find incremental numbered sequences with a list comprehension
Pythonic way to convert a list of integers into a string of comma-separated ranges

input = [1, 2, 3, 4, 8, 10, 11, 12, 17]

i, ii, result = iter(input), iter(input[1:]), [[input[0]]]
for x, y in zip(i,ii):
    if y-x != 1:
        result.append([y])
    else:
        result[-1].append(y)

>>> result
[[1, 2, 3, 4], [8], [10, 11, 12], [17]]

>>> print ", ".join("-".join(map(str,(g[0],g[-1])[:len(g)])) for g in result)
1-4, 8, 10-12, 17

>>> [(g[0],g[-1])[:len(g)] for g in result]
[(1, 4), (8,), (10, 12), (17,)]

按连续顺序排列的组列表

7 个答案: