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Whalefall126

问题：
Write a function defined as:

def linear_interpolate_index (slist, match_value, low_index, high_index):
that takes an ordered list of ascending numbers and a desired 'match' value to search for in the list.

The partition of interest in the list is defined by the index of the low and high values in the list. The function uses linear interpolation between slist[high] and slist[low] to return the index to try interpolation searching for the match value or None if the desired match falls outside the range slist[low] to slist[high-1)].

For example, a call to linear_interpolate_index(list(range(0, 20, 2)), 6, 0, 9) would return an index value of 3.

我的答案：
def linear_interpolate_index (slist, match_value, low_index, high_index):
   
    # YOUR CODE HERE
    low_value = slist[low_index]
    high_value = slist[high_index]

    if match_value < low_value or match_value > high_value:
        return None  

    if low_value == high_value:
        return low_index  
   
    interpolated_index = low_index + ((match_value - low_value) / (high_value - low_value)) * (high_index - low_index)

    return int(interpolated_index)


slist = list(range(0, 20, 2))
result = linear_interpolate_index(slist, 6, 0, 9)
print(result)

但是系统提醒我list index out of range
求大神帮我瞅瞅

isdkz

你的代码出现了问题，当low_index和high_index的值过大时，slist[high_index]会超出列表的范围，导致list index out of range错误。因此，需要在函数中添加一些条件来检查low_index和high_index是否超出列表的范围。下面是修改后的代码：

def linear_interpolate_index(slist, match_value, low_index, high_index):
    # 检查low_index和high_index是否超出列表的范围
    if low_index < 0 or high_index >= len(slist) or low_index >= high_index:
        return None
   
    low_value = slist[low_index]
    high_value = slist[high_index]

    if match_value < low_value or match_value > high_value:
        return None

    if low_value == high_value:
        return low_index

    interpolated_index = low_index + ((match_value - low_value) / (high_value - low_value)) * (high_index - low_index)

    return int(interpolated_index)

slist = list(range(0, 20, 2))
result = linear_interpolate_index(slist, 6, 0, 9)
print(result)

这样就可以避免出现list index out of range错误了。

Mike_python小

问题：函数linear_interpolate_index的代码分析

你的linear_interpolate_index函数的逻辑大致是正确的，但可能在索引计算上出现了问题。让我们一起来检查一下代码：

2. def linear_interpolate_index(slist, match_value, low_index, high_index):
3.     low_value = slist[low_index]
4.     high_value = slist[high_index]
6.     if match_value < low_value or match_value > high_value:
7.         return None
9.     if low_value == high_value:
10.         return low_index
12.     interpolated_index = low_index + ((match_value - low_value) / (high_value - low_value)) * (high_index - low_index)
14.     return int(interpolated_index)
16. slist = list(range(0, 20, 2))
17. result = linear_interpolate_index(slist, 6, 0, 9)
18. print(result)

复制代码

我认为导致 "list index out of range" 错误的原因是在计算插值索引时使用了错误的公式。根据题目描述，插值公式应该是 low_index + ((match_value - low_value) / (high_value - low_value)) * (high_index - low_index - 1)，而不是 low_index + ((match_value - low_value) / (high_value - low_value)) * (high_index - low_index)。

修正后的代码如下：

2. def linear_interpolate_index(slist, match_value, low_index, high_index):
3.     low_value = slist[low_index]
4.     high_value = slist[high_index]
6.     if match_value < low_value or match_value > high_value:
7.         return None
9.     if low_value == high_value:
10.         return low_index
12.     interpolated_index = low_index + ((match_value - low_value) / (high_value - low_value)) * (high_index - low_index - 1)
14.     return int(interpolated_index)
16. slist = list(range(0, 20, 2))
17. result = linear_interpolate_index(slist, 6, 0, 9)
18. print(result)

复制代码

请尝试运行修正后的代码，看看问题是否得到解决。如果还有其他问题，请提供更多相关信息以便我能够帮助你进一步调试代码。

球一个最佳答案谢谢啦！这对我非常重要！