在一行中构建 for 循环并收到错误的输出，即 Nonetype

Question

我正在尝试在列表理解中编写代码。但是，当我这样做时，我收到了 nonetype。代码：a = ["I", "have", "something", "to", "buy"]delete = ["I", "have"]aa = [a.remove(x) for x in delete]print(aa)输出：[None, None]预期输出：["something", "to", "buy"]

Answer 1

list.remove就地更改列表并返回 None。例如：

a = ["I", "have", "something", "to", "buy"]

print(a.remove("I")) # ['have', 'something', 'to', 'buy']

print(a) # None

如果你真的想在一行中做到这一点，你可以保留你的代码。虽然您不需要 aa，但只需打印 a。

Answer 2

您需要使用理解 if 子句，例如：

代码：

aa = [x for x in a if x not in delete]

测试代码：

a = ["I", "have", "something", "to", "buy"]

delete = ["I", "have"]

aa = [x for x in a if x not in delete]

print(aa)

结果：

['something', 'to', 'buy']

Answer 3

尝试打印（a）而不是打印（aa）

当您执行 a.remove(x) 时，它会将其从数组 'a' 中删除。

  a = ["I", "have", "something", "to", "buy"]

  delete = ["I", "have"]

  aa = [a.remove(x) for x in delete]

  print(a)