将数字列表转换为字符串范围

Question

我想知道是否有一种简单的（或已经创建的）与此相反的方法：Generate List of Numbers from Hyphenated...。这个链接可以用来做：

>> list(hyphen_range('1-9,12,15-20,23'))
[1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 16, 17, 18, 19, 20, 23]:

我想做相反的事情（请注意，包括 10 和 21，因此它与 range 函数兼容，其中 range(1,10)=[1,2,3,4,5,6, 7,8,9]):

>> list_to_ranges([1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 16, 17, 18, 19, 20, 23])
'1-10,12,15-21,23'

最后，我希望输出也包含一个步骤，其中输出的最后一个数字表示步骤：

>> list_to_ranges([1, 3, 5, 7, 8, 9, 10, 11])
'1-13:2,8,10'

基本上，这最终会有点像“逆”范围函数

>> tmp = list_to_ranges([1, 3, 5])
>> print tmp
'1-7:2'
>> range(1, 7, 2)
[1, 3, 5]

我的猜测是没有真正简单/简单的方法可以做到这一点，但我想我会在这里问一下，然后再做一些蛮力，长方法。

编辑

以this post 的回答中的代码为例，我想出了一个简单的方法来完成第一部分。但我认为确定执行步骤的模式会有点困难。

from itertools import groupby
from operator import itemgetter

data = [ 1,  4,5,6, 10, 15,16,17,18, 22, 25,26,27,28]
print data, '\n'

str_list = []
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
   ilist = map(itemgetter(1), g)
   print ilist
   if len(ilist) > 1:
      str_list.append('%d-%d' % (ilist[0], ilist[-1]+1))
   else:
      str_list.append('%d' % ilist[0])
print '\n', ','.join(str_list)

编辑 2

这是我尝试包含步长大小的尝试......它非常接近，但第一个数字会重复。我认为只要稍微调整一下，它就会接近我想要的——或者至少足够好。

import numpy as np
from itertools import groupby

def list_to_ranges(data):
   data = sorted(data)
   diff_data = np.diff(data).tolist()
   ranges = []
   i = 0
   for k, iterable in groupby(diff_data, None):
      rng = list(iterable)
      step = rng[0]
      if len(rng) == 1:
         ranges.append('%d' % data[i])
      elif step == 1:
         ranges.append('%d-%d' % (data[i], data[i+len(rng)]+step))
      else:
         ranges.append('%d-%d:%d' % (data[i], data[i+len(rng)]+step, step))
      i += len(rng)
   return ','.join(ranges)

data = [1, 3, 5, 6, 7, 11, 13, 15, 16, 17, 18, 19, 22, 25, 28]
print data
data_str = list_to_ranges(data)
print data_str

_list = []
for r in data_str.replace('-',':').split(','):
   r = [int(a) for a in r.split(':')]
   if len(r) == 1:
      _list.extend(r)
   elif len(r) == 2:
      _list.extend(range(r[0], r[1]))
   else:
      _list.extend(range(r[0], r[1], r[2]))
print _list
print list(set(_list))

Answer 1

一种方法可以是逐个“吃掉”输入序列并存储部分范围结果，直到你得到所有结果：

def formatter(start, end, step):
    return '{}-{}:{}'.format(start, end, step)
    # return '{}-{}:{}'.format(start, end + step, step)

def helper(lst):
    if len(lst) == 1:
        return str(lst[0]), []
    if len(lst) == 2:
        return ','.join(map(str,lst)), []

    step = lst[1] - lst[0]
    for i,x,y in zip(itertools.count(1), lst[1:], lst[2:]):
        if y-x != step:
            if i > 1:
                return formatter(lst[0], lst[i], step), lst[i+1:]
            else:
                return str(lst[0]), lst[1:]
    return formatter(lst[0], lst[-1], step), []

def re_range(lst):
    result = []
    while lst:
        partial,lst = helper(lst)
        result.append(partial)
    return ','.join(result)

我用一堆单元测试对其进行了测试，它都通过了，它也可以处理负数，但它们看起来有点难看（这真的是任何人的错）。

例子：

>>> re_range([1,  4,5,6, 10, 15,16,17,18, 22, 25,26,27,28])
'1,4-6:1,10,15-18:1,22,25-28:1'
>>> re_range([1, 3, 5, 7, 8, 9, 10, 11, 13, 15, 17])
'1-7:2,8-11:1,13-17:2'

注意：我为 Python 3 编写了代码。

---

性能

我没有在上述解决方案中投入任何性能努力。特别是，每次使用切片重新构建列表时，如果输入列表具有特定形状，则可能需要一些时间。因此，第一个简单的改进是尽可能使用itertools.islice()。

无论如何，这是相同算法的另一种实现，它使用scan 索引而不是切片来扫描输入列表：

def re_range(lst):
    n = len(lst)
    result = []
    scan = 0
    while n - scan > 2:
        step = lst[scan + 1] - lst[scan]
        if lst[scan + 2] - lst[scan + 1] != step:
            result.append(str(lst[scan]))
            scan += 1
            continue

        for j in range(scan+2, n-1):
            if lst[j+1] - lst[j] != step:
                result.append(formatter(lst[scan], lst[j], step))
                scan = j+1
                break
        else:
            result.append(formatter(lst[scan], lst[-1], step))
            return ','.join(result)

    if n - scan == 1:
        result.append(str(lst[scan]))
    elif n - scan == 2:
        result.append(','.join(map(str, lst[scan:])))

    return ','.join(result)

一旦它比以前的顶级解决方案快约 65%，我就停止了它的工作，这似乎足够了 :)

无论如何，我认为可能还有改进的空间（特别是在中间的 for 循环中）。

Answer 2

这是三种方法的比较。通过以下值更改数据量和密度......无论我使用什么值，第一个解决方案对我来说似乎是最快的。对于非常大的数据集，第三种解决方案变得非常慢。

已编辑

已编辑以包含下面的 cmets 并添加新的解决方案。最后一个解决方案现在似乎是最快的。

import numpy as np
import itertools
import random
import timeit

# --- My Solution --------------------------------------------------------------
def list_to_ranges1(data):
   data = sorted(data)
   diff_data = np.diff(data)
   ranges = []
   i = 0
   skip_next = False
   for k, iterable in itertools.groupby(diff_data, None):
      rng = list(iterable)
      step = rng[0]
      if skip_next:
         skip_next = False
         rng.pop()

      if len(rng) == 0:
         continue
      elif len(rng) == 1:
         ranges.append('%d' % data[i])
      elif step == 1:
         ranges.append('%d-%d' % (data[i], data[i+len(rng)]+step))
         i += 1
         skip_next = True
      else:
         ranges.append('%d-%d:%d' % (data[i], data[i+len(rng)]+step, step))
         i += 1
         skip_next = True
      i += len(rng)

   if len(rng) == 0 or len(rng) == 1:
      ranges.append('%d' % data[i])
   return ','.join(ranges)

# --- Kaidence Solution --------------------------------------------------------
# With a minor edit for use in range function
def list_to_ranges2(data):
   onediff = np.diff(data)
   twodiff = np.diff(onediff)
   increments, breakingindices = [], []
   for i in range(len(twodiff)):
       if twodiff[i] != 0:
           breakingindices.append(i+2)  # Correct index because of the two diffs
           increments.append(onediff[i]) # Record the increment for this section

  # Increments and breakingindices should be the same size
   str_list = []
   start = data[0]
   for i in range(len(breakingindices)):
       str_list.append("%d-%d:%d" % (start,
                                     data[breakingindices[i]-1] + increments[i],
                                     increments[i]))
       start = data[breakingindices[i]]
   str_list.append("%d-%d:%d" % (start,
                                 data[len(data)-1] + onediff[len(onediff)-1],
                                 onediff[len(onediff)-1]))
   return ','.join(str_list)

# --- Rik Poggi Solution -------------------------------------------------------
# With a minor edit for use in range function
def helper(lst):
    if len(lst) == 1:
        return str(lst[0]), []
    if len(lst) == 2:
        return ','.join(map(str,lst)), []

    step = lst[1] - lst[0]
    #for i,x,y in itertools.izip(itertools.count(1), lst[1:], lst[2:]):
    for i,x,y in itertools.izip(itertools.count(1),
                                itertools.islice(lst, 1, None, 1),
                                itertools.islice(lst, 2, None, 1)):
        if y-x != step:
            if i > 1:
                return '{}-{}:{}'.format(lst[0], lst[i]+step, step), lst[i+1:]
            else:
                return str(lst[0]), lst[1:]
    return '{}-{}:{}'.format(lst[0], lst[-1]+step, step), []

def list_to_ranges3(lst):
    result = []
    while lst:
        partial,lst = helper(lst)
        result.append(partial)
    return ','.join(result)

# --- Rik Poggi Solution 2 -----------------------------------------------------
def formatter(start, end, step):
    #return '{}-{}:{}'.format(start, end, step)
    return '{}-{}:{}'.format(start, end + step, step)

def list_to_ranges4(lst):
    n = len(lst)
    result = []
    scan = 0
    while n - scan > 2:
        step = lst[scan + 1] - lst[scan]
        if lst[scan + 2] - lst[scan + 1] != step:
            result.append(str(lst[scan]))
            scan += 1
            continue

        for j in xrange(scan+2, n-1):
            if lst[j+1] - lst[j] != step:
                result.append(formatter(lst[scan], lst[j], step))
                scan = j+1
                break
        else:
            result.append(formatter(lst[scan], lst[-1], step))
            return ','.join(result)

    if n - scan == 1:
        result.append(str(lst[scan]))
    elif n - scan == 2:
        result.append(','.join(itertools.imap(str, lst[scan:])))

    return ','.join(result)

# --- Test Function ------------------------------------------------------------
def test_data(data, f_to_test):
   data_str = f_to_test(data)
   _list = []
   for r in data_str.replace('-',':').split(','):
      r = [int(a) for a in r.split(':')]
      if len(r) == 1:
         _list.extend(r)
      elif len(r) == 2:
         _list.extend(range(r[0], r[1]))
      else:
         _list.extend(range(r[0], r[1], r[2]))
   return _list

# --- Timing Tests -------------------------------------------------------------
# Generate some sample data...
data_list = []
for i in range(5):
   # Note: using the "4000" and "5000" values below, the relative density of
   # the data can be changed.  This has a huge effect on the results
   # (particularly on the results for list_to_ranges3 which uses recursion).
   data_list.append(sorted(list(set([random.randint(1,4000) for a in \
                                      range(random.randint(5,5000))]))))

testfuncs = list_to_ranges1, list_to_ranges2, list_to_ranges3, list_to_ranges4
for f in testfuncs:
   print '\n', f.__name__
   for i, data in enumerate(data_list):
      t = timeit.Timer('f(data)', 'from __main__ import data, f')
      #print f(data)
      print i, data==test_data(data, f), round(t.timeit(200), 3)

Answer 3

This 很可能是您要查找的内容。

编辑：我看到你已经找到了帖子。我很抱歉。

为了帮助完成第二部分，我自己做了一些修改。这是我想出的：

from numpy import diff

data = [ 1, 3, 5, 7, 8, 9, 10, 11, 13, 15, 17 ]
onediff, twodiff = diff(data), diff(diff(data))
increments, breakingindices = [], []
for i in range(len(twodiff)):
    if twodiff[i] != 0:
        breakingindices.append(i+2) # Correct index because of the two diffs
        increments.append(onediff[i]) # Record the increment for this section

# Increments and breakingindices should be the same size
str_list = []
start = data[0]
for i in range(len(breakingindices)):
    str_list.append("%d-%d:%d" % (start, data[breakingindices[i]-1], increments[i]))
    start = data[breakingindices[i]]
str_list.append("%d-%d:%d" % (start, data[len(data)-1], onediff[len(onediff)-1]))
print str_list

对于给定的输入列表，这给出：['1-7:2', '8-11:1', '13-17:2']。该代码可以进行一些清理，但这与您的问题有关，假设分组可以按顺序完成。

{注意：对于 [1,2,3,5,6,7] 这给出 ['1-3:1', '5-5:2', '6-7:1'] 而不是 [ '1-3:1', '5-7:1']}

Answer 4

这类似于处理步长为一的情况enumerated here 的版本，但也处理单例（序列中不超过 2 个元素的元素或重复元素）和非单一步长（包括负步长）。它也不会删除 [1, 2, 3, 3, 4, 5] 等列表的重复项。

至于运行时间：在你眨眼之前就完成了。

def ranges(L):
    """return a list of singletons or ranges of integers, (first, last, step)
    as they occur sequentially in the list of integers, L.

    Examples
    ========

    >>> list(ranges([1, 2, 4, 6, 7, 8, 10, 12, 13]))
    [1, (2, 6, 2), 7, (8, 12, 2), 13]
    >>> list(ranges([1,2,3,4,3,2,1,3,5,7,11,1,2,3]))
    [(1, 4, 1), (3, 1, -1), (3, 7, 2), 11, (1, 3, 1)]

    """
    if not L:
        return []
    r = []
    for i in L:
        if len(r) < 2:
            r.append(i)
            if len(r) == 2:
                d = r[1] - r[0]
        else:
            if i - r[1] == d:
                r[1] = i
            else:
                if r[1] - r[0] == d:
                    yield(r.pop(0))
                    r.append(i)
                    d = r[1] - r[0]
                else:
                    yield(tuple(r+[d]))
                    r[:] = [i]
    if len(r) == 1:
        yield(r.pop())
    elif r[1] - r[0] == d:
        for i in r:
            yield i
    else:
        yield(tuple(r+[d]))

可以根据需要修改原始输出，例如可以创建实际的range 实例。

def sranges(i):
    """return pretty string for output of ranges.

    Examples
    ========

    >>> sranges([1,2,4,6,7,8,10,12,13,15,16,17])
    '1, range(2, 8, 2), 7, range(8, 14, 2), 13, range(15, 18)'

    """
    out = []
    for i in ranges(i):
        if type(i) is int:
            out.append(str(i))
        elif i[-1] == 1:
            if i[0] == 0:
                out.append('range(%s)'%(i[1] + 1))
            else:
                out.append('range(%s, %s)'%(i[0], i[1] + 1))
        else:
            out.append('range(%s, %s, %s)'%(i[0], i[1] + i[2], i[2]))
    return ', '.join(out)

Answer 5

这个函数应该做你需要的，而不需要任何导入。

def listToRanges(self, intList):
    ret = []
    for val in sorted(intList):
        if not ret or ret[-1][-1]+1 != val:
            ret.append([val])
        else:
            ret[-1].append(val)
    return ",".join([str(x[0]) if len(x)==1 else str(x[0])+"-"+str(x[-1]) for x in ret])