Question

考虑以下列表：

>>> circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']  
>>> list(enumerate(circle))  
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g'), (7, 'h')]

如果circle被概念化为循环列表，即circle[0]已连接到circle[7]，则给定start索引和end索引{{1}我想构建两个表示顺时针和逆时针方向的线性遍历顺序的列表。

根据start != end和start的值，这里有我提出的内容：

案例1：end

start < end

案例2：>>> circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'] >>> start, end = 1, 6 >>> clockwise = circle[start:end+1] >>> clockwise ['b', 'c', 'd', 'e', 'f', 'g'] >>> counter_clockwise = circle[start::-1] + circle[:end-1:-1] >>> counter_clockwise ['b', 'a', 'h', 'g']

start > end

是否有一种pythonic /更高效/更简单的方法来构建这两个列表？

Answer 1

您可以使用itertools.cycle：

import itertools

circle = ['a', 'b', 'c', 'd', 'e', 'f', 
  'g', 'h']

def clockwise(start, end):
    endless = itertools.cycle(circle)
    if start > end:
        end = start + (len(circle)-(start
          -end))
    return [next(endless) for i in range
      (end+1)][start:]

def counter_clockwise(start, end):
    endless = itertools.cycle(circle)
    if end > start:
        start = end + (len(circle)-(end
          -start))    
    return [next(endless) for i in range
      (start+1)][end:][::-1]

# start < end:
forward=clockwise(1, 6)
b1=counter_clockwise(1, 6)
#start > end:
f1=clockwise(6, 1)
backward=counter_clockwise(6, 1)

print(forward)
print(b1)
print(f1)
print(backward)

输出：

['b', 'c', 'd', 'e', 'f', 'g']
['b', 'a', 'h', 'g']
['g', 'h', 'a', 'b']
['g', 'f', 'e', 'd', 'c', 'b']

Answer 2

import itertools

def slice_it(circle,start,end,step=1):
    if end < start and step > 0:
        end = len(circle)+end
    if step < 0:
        return list(itertools.islice(itertools.cycle(circle),end,start,-1*step))[::-1]
    return list(itertools.islice(itertools.cycle(circle),start,end,step))

circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']

print slice_it(circle,6,1)
print slice_it(circle,6,1,-1)

可能？

Answer 3

您可以使用lambda表达式创建顺时针和逆时针方法并列出切片！

>>> clockwise = lambda circle, start, end: circle[start:] + circle[:end+1] if start > end else circle[start:end+1]
>>> counter_clockwise = lambda circle, start, end : clockwise(circle, end, start)[::-1]

这有点类似于你所尝试过的，但是采用更加传播的方式和广泛的方式！

>>> clockwise(circle,1,6)
['b', 'c', 'd', 'e', 'f', 'g']
>>> counter_clockwise(circle,1,6)
['b', 'a', 'h', 'g']
>>> clockwise(circle,6,1)
['g', 'h', 'a', 'b']
>>> counter_clockwise(circle,6,1)
['g', 'f', 'e', 'd', 'c', 'b']

Answer 4

您可以使用deque模块中的collections来处理counter_clockwise条件的某些部分，例如此示例（如果需要，您可以对其进行修改或改进）：

from collections import deque

circle = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']


clockwise = lambda x,start,end: x[start:end+1] if start < end else x[start:] + x[:end+1]

def counter_clockwise(itertable, start, end):
    if end > start:
        a = deque(itertable[:start+1])
        b = deque(itertable[end:])
        a.rotate()
        b.rotate()
        # Or:
        # return list(a.__iadd__(b))
        return list(a) + list(b)

    if end < start:
        return itertable[start:end-1:-1]


print("start > end:")
start, end = 6,1
print(clockwise(circle, start, end))
print(counter_clockwise(circle, start, end))

print("start < end:")
start, end = 1,6
print(clockwise(circle, start, end))
print(counter_clockwise(circle, start, end))

输出：

start > end:
['g', 'h', 'a', 'b']
['g', 'f', 'e', 'd', 'c', 'b']
start < end:
['b', 'c', 'd', 'e', 'f', 'g']
['b', 'a', 'h', 'g']

从python中的概念循环列表创建两个线性列表

4 个答案: