如何在Python中反向迭代列表?


请参见:如何获得列表的反向副本(避免在.reverse后链接方法时使用单独的语句)?


当前回答

如果你想在其他变量中存储反转列表的元素,那么你可以使用revArray = array[::-1]或revArray = list(reversed(array))。

但是第一个变体稍微快一点:

z = range(1000000)
startTimeTic = time.time()
y = z[::-1]
print("Time: %s s" % (time.time() - startTimeTic))

f = range(1000000)
startTimeTic = time.time()
g = list(reversed(f))
print("Time: %s s" % (time.time() - startTimeTic))

输出:

Time: 0.00489711761475 s
Time: 0.00609302520752 s

其他回答

另一种方法使用列表理解和abs

array = [0, 10, 20, 40]

reversed_array = [array[abs(indx)] for indx in range(abs(len(array)-1),1)]
            
reversed_array
[40, 20, 10, 0]

一个干净简单的类对象来解决您的问题。

class lister():
    def reverse(self):
        return  [self[len(self)-e]for e,x in enumerate(self,start=1)]
print(lister.reverse([0, 10, 20, 40]))
for x in array[::-1]:
    do stuff

使用列表推导式:

[array[n] for n in range(len(array)-1, -1, -1)]

内置功能最少,假设是面试设置

array = [1, 2, 3, 4, 5, 6,7, 8]
inverse = [] #create container for inverse array
length = len(array)  #to iterate later, returns 8 
counter = length - 1  #because the 8th element is on position 7 (as python starts from 0)

for i in range(length): 
   inverse.append(array[counter])
   counter -= 1
print(inverse)