Question

lst = [
8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8,             #0-19
49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0,       #20-39
81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65,      #40-59
52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91,
22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80,
24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50,
32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70,
67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21,
24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72,
21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95,
78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92,
16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57,
86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58,
19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40,
4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66,
88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69,
4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36,     #320-339
20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16,    #340-359
20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54,     #360-379
1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48]     #380-399

prodsum = 1

def prod(iter):
   p = 1
    for n in iter:
         p *= n
     return p

for n in range(0,5000,20):    #NOT OUT OF RANGE???
    for i in range(0,17):
        if prod(lst[n+i:n+i+4]) > prodsum:
            prodsum = prod(lst[n+i:n+i+4])

我试图学习/提高我在Python中的基本技能，所以我一直在经历欧拉计划的挑战。挑战问题较为复杂，但我基本上有一个20x20的网格，并找到4个具有最大乘积的相邻数字。

我基本上将网格变成了一个列表（具有400个值），并且要扫描行索引。我不小心为for循环输入了很多数字，并发现我没有出现超出范围的错误。为什么是这样？

Answer 1

您不会超出范围，因为您永远不会直接根据索引（n）访问列表。

您使用lst[n+i:n+i+4]来获取lst的一个切片...如果索引超出范围，则该切片为空，因此prod(...)被[]调用并返回1。

Answer 2

使用普通索引会出现超出范围的错误。例如，如果您有一个由10个元素组成的数组，并且您要求my_list[20]。但是，使用切片 my_array[a: b]，您可以从a到b-1或列表的末尾获取元素。那只是语言的设计决定。

Answer 3

在序列范围之外进行切片不会导致错误。如果您尝试为单个项目建立索引，则会出现错误。

为什么没有索引超出范围错误？

3 个答案: