python 搜索列表中的元素是否是另一个列表的子集，无需内置函数

Question

我正在尝试搜索列表中的元素是否是另一个列表的子集，而不使用“set”或“if item in list”等内置函数。我有以下代码，但我不断收到“索引超出范围”的错误

def letterSearch(sublist,mainlist):
    x = 0
    index = 0
    while x < len(mainlist):
        if sublist[index] == mainlist[x]:
            index = index + 1
            x = x + 1
        else:
            x = x + 1


x = ['d','g']
y = ['d','g','a','b']

letterSearch(x,y)
print(index)

Answer 1

问题:

您的代码将 index 值增加到超过 sublist 的长度。因此，下次比较时，该索引处没有任何项目，从而导致 index out of range 错误。

解决方案:

def letterSearch(sublist,mainlist):
    x = 0
    index = 0
    while x < len(mainlist):
        if len(sublist) != index and sublist[index] == mainlist[x]:
            index = index + 1
        x += 1 
    if len(sublist) == index:
        return index

x = ['d','g']
y = ['d','g','a','b']

index = letterSearch(x,y)
print(index)  # 2

# To display if x is a subset of y or not:
if index:
    print('{} is a subset of {}'.format(x, y))
else:
    print('Not a subset')