尝试使用布尔索引。另外，将metalist从list转换为set以加快搜索速度：

metalist = ['Apple', 'Cake', 'Biscuit']

mask = df['Meta'].notna()
metalist = set(metalist)
df.loc[~mask, 'Available'] = False
df.loc[mask, 'Available'] = df.loc[mask, 'Meta'].apply(lambda x: bool(metalist.intersection(x)))

print(df)

图纸：

Meta Available
0     [Home, Grocery, Cake]      True
1  [Home, Grocery, Biscuit]      True
2                       NaN     False
3    [Home, Grocery, Apple]      True

我将使用一个简单的条件和集合操作：

S = set(metalist)

df['Available'] = [isinstance(x, list) and bool(S.intersection(x))
                   for x in df['Meta']]

使用自定义函数的替代方法：

def overlap(s1, s2):
    try:
        return not s1.isdisjoint(s2)
    except TypeError:
        return False
    
S = set(metalist)

df['Available'] = [overlap(S, x) for x in df['Meta']]

输出：

Meta  Available
0     [Home, Grocery, Cake]       True
1  [Home, Grocery, Biscuit]       True
2                       NaN      False
3    [Home, Grocery, Apple]       True