Python 按照元素的特定顺序对列表进行排序

发布于05月27日

假设我需要对列表进行如下排序:

A=[1,1,1,1,1,1,1,0,0,0]

按照1和0之间4:1的比率，获得so A=[1,1,1,1,0,1,1,1,0,0].

这可能吗？

scheme=[1,1,1,1,0,1,1,1,0,0]
       
for k, number in enumerate(scheme):
    visited.append(number)
    scheme[k] += visited.count(number)/len(scheme)

for z in scheme:
    new = sorted(scheme).index(z)
    final.append(sorted(que)[new])

但这并不是一个舒适的方法，因为scheme，指南列表，强烈依赖于初始列表的长度.

提前感谢您！

使用简单算法

假设序列只包含0和1.

from collections import Counter

def reorder_4_1(seq):
    c = Counter(seq)
    q1, r1 = divmod(c[1], 4)
    diff = q1 - c[0]
    if diff > 0:
        return [1,1,1,1,0] * c[0] + [1] * (diff + r1)
    else:
        return [1,1,1,1,0] * q1 + [1] * r1 + [0] * (-diff)

print( reorder_4_1([1,1,1,1,1,1,1,0,0,0]) )
# [1, 1, 1, 1, 0, 1, 1, 1, 0, 0]

Using module `itertools`

使用the itertools documentation中的配方roundrobin:

假设有两组元素交错4:1

from itertools import cycle, islice

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            # Remove the iterator we just exhausted from the cycle.
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

def interleave_4_1(a, b):
    a = iter(a)
    b = iter(b)
    return roundrobin(a, a, a, a, b)

print(list( interleave_4_1([1,1,1,1,1,1,1],[0,0,0]) ))
# [1, 1, 1, 1, 0, 1, 1, 1, 0, 0]

假设序列保证为1和0的列表

from collections import Counter
from itertools import repeat

# def roundrobin...

def reorder_4_1(seq):
    c = Counter(seq)
    a = repeat(1, c[1])
    b = repeat(0, c[0])
    return roundrobin(a, a, a, a, b)

print(list( reorder_4_1([1,1,1,1,1,1,1,0,0,0]) ))
# [1, 1, 1, 1, 0, 1, 1, 1, 0, 0]