在列表中查找最长的运行

Question

给定一个数据列表，我试图创建一个新列表，其中位置i的值是从原始列表中的位置i开始的最长运行的长度。例如，给定

x_list = [1, 1, 2, 3, 3, 3]

应该返回：

run_list = [2, 1, 1, 3, 2, 1]

我的解决方案：

freq_list = []
current = x_list[0]
count = 0
for num in x_list:
    if num == current:
        count += 1
    else:
        freq_list.append((current,count))
        current = num
        count = 1
freq_list.append((current,count))

run_list = []
for i in freq_list:
    z = i[1]
    while z > 0:
        run_list.append(z)
        z -= 1 

首先，我创建一个元组列表freq_list，其中每个元组的第一个元素是来自x_list的元素，其中第二个元素是总运行的数量。

在这种情况下：

freq_list = [(1, 2), (2, 1), (3, 3)]

有了这个，我创建了一个新列表并附加了适当的值。

然而，我想知道是否有更短的方式/另一种方式来做到这一点？

Answer 1

Here's a simple solution that iterates over the list backwards and increments a counter each time a number is repeated:

last_num = None
result = []
for num in reversed(x_list):
    if num != last_num:
        # if the number changed, reset the counter to 1
        counter = 1
        last_num = num
    else:
        # if the number is the same, increment the counter
        counter += 1

    result.append(counter)

# reverse the result
result = list(reversed(result))

Result:

[2, 1, 1, 3, 2, 1]

Answer 2

This is possible using itertools:

from itertools import groupby, chain

x_list = [1, 1, 2, 3, 3, 3]

gen = (range(len(list(j)), 0, -1) for _, j in groupby(x_list))
res = list(chain.from_iterable(gen))

Result

[2, 1, 1, 3, 2, 1]

Explanation

• First use itertools.groupby to group identical items in your list.
• For each item in your groupby, create a range object which counts backwards from the length of the number of consecutive items to 1.
• Turn this all into a generator to avoid building a list of lists.
• Use itertools.chain to chain the ranges from the generator.

Performance note

Performance will be inferior to @Aran-Fey's solution. Although itertools.groupby is O(n), it makes heavy use of expensive __next__ calls. These do not scale as well as iteration in simple for loops. See itertools docs for groupby pseudo-code.

If performance is your main concern, stick with the for loop.

Answer 3

您正在对连续组执行反向累积计数。我们可以使用

创建Numpy累积计数功能

import numpy as np

def cumcount(a):
    a = np.asarray(a)
    b = np.append(False, a[:-1] != a[1:])
    c = b.cumsum()
    r = np.arange(len(a))
    return r - np.append(0, np.flatnonzero(b))[c] + 1

然后使用

生成我们的结果

a = np.array(x_list)

cumcount(a[::-1])[::-1]

array([2, 1, 1, 3, 2, 1])

Answer 4

我会使用生成器来完成这种任务，因为它可以避免逐步构建结果列表，如果需要，可以懒得使用：

def gen(iterable):  # you have to think about a better name :-)
    iterable = iter(iterable)
    # Get the first element, in case that fails
    # we can stop right now.
    try:
        last_seen = next(iterable)
    except StopIteration:
        return
    count = 1

    # Go through the remaining items
    for item in iterable:
        if item == last_seen:
            count += 1
        else:
            # The consecutive run finished, return the
            # desired values for the run and then reset
            # counter and the new item for the next run.
            yield from range(count, 0, -1)
            count = 1
            last_seen = item
    # Return the result for the last run
    yield from range(count, 0, -1)

如果输入不能为reversed（某些生成器/迭代器无法反转），这也会有效：

>>> x_list = (i for i in range(10))  # it's a generator despite the variable name :-)
>>> ... arans solution ...
TypeError: 'generator' object is not reversible

>>> list(gen((i for i in range(10))))
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

它适合您的输入：

>>> x_list = [1, 1, 2, 3, 3, 3]
>>> list(gen(x_list))
[2, 1, 1, 3, 2, 1]

使用itertools.groupby：

实际上可以更简单

import itertools

def gen(iterable):
    for _, group in itertools.groupby(iterable):
        length = sum(1 for _ in group)  # or len(list(group))
        yield from range(length, 0, -1)

>>> x_list = [1, 1, 2, 3, 3, 3]
>>> list(gen(x_list))
[2, 1, 1, 3, 2, 1]

我也做了一些基准测试，根据这些Aran-Feys解决方案是最快的，除了piRSquareds解决方案获胜的长列表：

如果您想确认结果，这是我的基准测试设置：

from itertools import groupby, chain
import numpy as np

def gen1(iterable):
    iterable = iter(iterable)
    try:
        last_seen = next(iterable)
    except StopIteration:
        return
    count = 1
    for item in iterable:
        if item == last_seen:
            count += 1
        else:
            yield from range(count, 0, -1)
            count = 1
            last_seen = item
    yield from range(count, 0, -1)

def gen2(iterable):
    for _, group in groupby(iterable):
        length = sum(1 for _ in group)
        yield from range(length, 0, -1)

def mseifert1(iterable):
    return list(gen1(iterable))

def mseifert2(iterable):
    return list(gen2(iterable))

def aran(x_list):
    last_num = None
    result = []
    for num in reversed(x_list):
        if num != last_num:
            counter = 1
            last_num = num
        else:
            counter += 1
        result.append(counter)
    return list(reversed(result))

def jpp(x_list):
    gen = (range(len(list(j)), 0, -1) for _, j in groupby(x_list))
    res = list(chain.from_iterable(gen))
    return res

def cumcount(a):
    a = np.asarray(a)
    b = np.append(False, a[:-1] != a[1:])
    c = b.cumsum()
    r = np.arange(len(a))
    return r - np.append(0, np.flatnonzero(b))[c] + 1

def pirsquared(x_list):
    a = np.array(x_list)
    return cumcount(a[::-1])[::-1]

from simple_benchmark import benchmark
import random

funcs = [mseifert1, mseifert2, aran, jpp, pirsquared]
args = {2**i: [random.randint(0, 5) for _ in range(2**i)] for i in range(1, 20)}

bench = benchmark(funcs, args, "list size")

%matplotlib notebook
bench.plot()

Python 3.6.5，NumPy 1.14

Answer 5

这是使用collections.Counter实现它的简单迭代方法：

from collections import Counter

x_list = [1, 1, 2, 3, 3, 3]
x_counter, run_list = Counter(x_list), []

for x in x_list:
    run_list.append(x_counter[x])
    x_counter[x] -= 1

将返回run_list作为：

[2, 1, 1, 3, 2, 1]

作为替代方案，这里使用使用列表理解 与enumerate一起实现，但由于{{的迭代使用，因此效率不高1}}：

list.index(..)

Answer 6

您可以计算连续的相等项目，然后将项目数量的倒计时加1到结果：

def runs(p):
    old = p[0]
    n = 0
    q = []
    for x in p:
        if x == old:
            n += 1
        else:
            q.extend(range(n, 0, -1))
            n = 1
            old = x

    q.extend(range(n, 0, -1))

    return q

（几分钟后）哦，这与MSeifert's code相同，但没有可迭代的方面。此版本似乎与method shown by Aran-Fey几乎一样快。