检查两个列表在同一索引处是否具有相同的值，如果是，则返回索引。如果不是，则返回 -1

我认为一个更简洁的答案使用内置和函数：enumeratezip

Dlist = [17,13,10,6,2]
Ilist = [5,9,10,15,18]

def seqsearch(DS,IS):
    for idx, (d, s) in enumerate(zip(DS, IS)):
        if d == s:
            return f"Yes! Found at index = {idx}"

    return "No!\n-1"


print(seqsearch(Dlist,Ilist))

目前尚不清楚是要只返回第一个索引，还是要返回所有具有匹配元素的索引。无论哪种情况，最好只返回所需的值，然后在函数范围之外添加任何格式和打印语句。

您可以像这样遍历两个列表：

Dlist = [17, 13, 10, 6, 2]
Ilist = [5, 9, 10, 15, 18]


def seqsearch(DS, IS):

    for index_1, element_1 in enumerate(DS):
        for index_2, element_2 in enumerate(IS):

            if (element_1 == element_2) and (index_1 == index_2):
                print(f"Yes! Found at index ={index_1}")
                return index_1
    print("No!")
    return -1


print(seqsearch(Dlist, Ilist))

但是，您可以进行更多改进。 确实是一个更好的选择，但理解起来稍微复杂一些。zip()

另外，请注意，这与 .你没有回来;你正在打印它。returnprint-1

我尝试使用您的代码来更好地使用 undrestand（但根据您的使用，我们有很多选择）

Dlist = [17, 13, 10, 6, 2]
Ilist = [5, 9, 10, 15, 18]


def seqsearch_with_print(DS, IS):
    """this function only print not return!!!
    if you use return your function ended!"""
    for i in range(len(DS) - 1):
        if DS[i] == IS[i]:
            print(f"Yes! Found at index {i}")
        else:
            print(f"No not equal in {i} index")


def seqsearch_with_list(DS, IS):
    """this function save history of (equal index or not: as 1, -1)"""
    temp_list = []
    for i in range(len(DS) - 1):
        if DS[i] == IS[i]:
            temp_list.append(1)
        else:
            temp_list.append(-1)
    return temp_list


def seqsearch_with_list_dict(DS, IS):
    """this function save history of (equal index or not: as 1, -1)
    as key, value (key== index, value==list item)"""
    temp_list = []
    for i in range(len(DS) - 1):
        if DS[i] == IS[i]:
            temp_list.append({i: 1})
        else:
            temp_list.append({i: -1})
    return temp_list


# no need print(in function we used print)
seqsearch_with_print(Dlist, Ilist)
"""
return of function: 
    No not equal in 0 index
    No not equal in 1 index
    Yes! Found at index 2
    No not equal in 3 index"""

# save history of search index in list
print(seqsearch_with_list(Dlist, Ilist))
"""
return of function with print:
    [-1, -1, 1, -1]
"""
# save history as dict key value
print(seqsearch_with_list_dict(Dlist, Ilist))
"""
return of function with print
    [{0: -1}, {1: -1}, {2: 1}, {3: -1}]
"""