Question

说我有一个清单，

l = [1, 2, 3, 4, 5, 6, 7, 8]

我想获取任意元素的索引及其邻居的值。例如，

i = l.index(n)
j = l[i-1]
k = l[i+1]

但是，对于i == len(l) - 1失败的边缘情况。所以我以为我只是把它包起来，

if i == len(l) - 1:
    k = l[0]
else:
    k = l[i+1]

有没有pythonic的方法来做到这一点？

Answer 1

您可以使用模运算符！

i = len(l) - 1
jIndex = (i - 1) % len(l)
kIndex = (i + 1) % len(l)

j = l[jIndex]
k = l[kIndex]

或者，不那么冗长：

k = l[(i + 1) % len(l)]

Answer 2

包装固定长度列表的最简单方法是使用％（modulo）运算符

list_element = my_list[idx % len(my_list)]

但无论如何看 http://docs.python.org/library/itertools.html

from itertools import cycle

for p in cycle([1,2,3]):
  print "endless cycle:", p

Answer 3

将值拟合到特定范围的典型方法是使用%运算符：

k = l[(i + 1) % len(l)]

Answer 4

如果你想把它作为一个类，我就掀起了这个快速的循环列表：

import operator

class CircularList(list):
    def __getitem__(self, x):
        if isinstance(x, slice):
            return [self[x] for x in self._rangeify(x)]

        index = operator.index(x)
        try:
            return super().__getitem__(index % len(self))
        except ZeroDivisionError:
            raise IndexError('list index out of range')

    def _rangeify(self, slice):
        start, stop, step = slice.start, slice.stop, slice.step
        if start is None:
            start = 0
        if stop is None:
            stop = len(self)
        if step is None:
            step = 1
        return range(start, stop, step)

它支持切片，所以

CircularList(range(5))[1:10] == [1, 2, 3, 4, 0, 1, 2, 3, 4]

Answer 5

如果你不想环绕，最Pythonic的答案是使用切片。缺少用无替换的邻居。 E.g：

def nbrs(l, e):
   i = l.index(e)
   return (l[i-1:i] + [None])[0], (l[i+1:i+2] + [None])[0]

这是函数的工作方式：

>>> nbrs([2,3,4,1], 1)
(4, None)
>>> nbrs([1,2,3], 1)
(None, 2)
>>> nbrs([2,3,4,1,5,6], 1)
(4, 5)
>>> nbrs([], 1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in nbrs
ValueError: 1 is not in list

Answer 6

a = [2,3,5,7,11,13]

def env (l, n, count):
    from itertools import cycle, islice
    index = l.index(n) + len(l)
    aux = islice (cycle (l), index - count, index + count + 1)
    return list(aux)

表现如下

>>> env (a, 2,1)
[13, 2, 3]
>>> env (a,13,2)
[7, 11, 13, 2, 3]
>>> env (a,7,0)
[7]

Pythonic循环清单

6 个答案: