Question

我定义了以下函数：

def makeAnagram(a, b):
    c = list(a)
    d = list(b)
    x = 0
    for i in c:
        if i not in d:
            c = c.remove(i)
            x += 1
    for i in d:
        if i not in c:
            d = d.remove(i)
            x += 1
    return int(x)

测试时，它在第 15 行（上面的斜体字）返回以下错误：

  File "/Users/ob/untitled0.py", line 15, in ma

AttributeError: 'NoneType' object has no attribute 'remove'

它如何不将 c 识别为列表？

a 和 b 是两个字符列表

先谢谢大家！

Answer 1

list.remove() 将返回 None。如果你想从列表中删除元素，你可以使用索引或就地删除。

#inplace remove
def makeAnagram(a, b):
    c = list(a)
    d = list(b)
    x = 0
    for i in c:
        if i not in d:
            c.remove(i)
            x += 1
    for i in d:
        if i not in c:
            d.remove(i)
            x += 1
    return int(x)


makeAnagram([5, 9], [6, 0])

# using index
def makeAnagram(a, b):
    c = list(a)
    d = list(b)
    x = 0
    for i in c:
        if index,i not in enumerate(d):
            del c[index]
            x += 1
    for index,i in enumerate(d):
        if i not in c:
            del d[index]
            x += 1
    return int(x)

Answer 2

我想我们的目标是获得两个列表中都没有出现的元素总数，你可以用更好更短的方式来做到这一点。

c = ['a','b','e','d']
d = ['w','b','c','d']

c_not_d = len(set(c).difference(set(d)))
# Output of above without len(), will be {'a', 'e'}

d_not_c = len(set(d).difference(set(c)))
# Output of above without len(), will be {'w', 'c'}

# Total Count required
total_count = c_not_d + d_not_c

引用列表时出现“NoneType”错误

2 个答案: