Question

所以我想知道是否有更“美丽”的方式来做到这一点。目前，我有超过一千个列表list_of_lists，其中每个列表看起来都像这样：

list_of_items = ["dog", "mouse", "cow", "goat", "fish"]

其中一些列表包含其他动物/字符串，但其中一些没有。这取决于。

我现在想做一个if语句，说：

list_of_items = ["dog", "mouse", "cow", "goat", "fish"]
for x in list_of_items:
    if "cow" not in list_of_items and "cat" not in list_of_items:
       print("Cat or Cow could not be found in list {}".format(x))

这完全符合预期。如果在当前列表中找到“ cat”或“ cow”，则不会打印任何内容。但是，如果没有找到，就会出现打印语句。

我的问题是我有几个“牛”，“猫”，因此需要在if语句中包括它们。以我为例，如果有10个，那么看起来会很长很丑。因此，有什么方法可以说：if list_of_animals not in list_of_items:，其中list_of_animals只是应该包含在and语句中的字符串的列表？

Answer 1

您可以将列表转换为set，并使用issubset

例如：

list_of_items = set(["dog", "mouse", "cow", "goat", "fish", "cat"])
toCheck = set(["cow", "cat"])

if toCheck.issubset(list_of_items):
    print("Ok")

根据评论进行编辑

if any(i in list_of_items for i in toCheck):
    print("Ok")

Answer 2

如果您希望其中任何一个匹配，也许是这样的事情？

a = ["dog", "mouse", "goat", "fish"]
b = ["cat", "cow"]
if(any(x in a for x in b)):
    print("True")
else:
    print("False")

返回错误

a = ["dog", "mouse", "cow", "goat", "fish"]
b = ["cat", "cow"]
if(any(x in a for x in b)):
    print("True")
else:
    print("False")

返回True

如果您希望两者都匹配，则：

a = ["dog", "mouse", "cow", "goat", "fish"]
b = ["cat", "cow"]
if(all(x in a for x in b)):
    print("True")
else:
    print("False")

返回错误

a = ["dog", "mouse", "cow", "cat", "goat", "fish"]
b = ["cat", "cow"]
if(all(x in a for x in b)):
    print("True")
else:
    print("False")

返回True

Answer 3

尝试一下：

list_of_items = ["dog", "mouse", "cow", "goat", "fish"]
another_list = ["cow", "cat"]
for x in list_of_items:
    if x in another_list:
        string_text = "String found"
    else:
        string_text = "String not found"
print(string_text)

Answer 4

for x in list_of_lists:
    if not any(animal in list_of_items for animal in ("cow", "cat"))
        print("Cat or Cow could not be found in list {}".format(x)

如果列表项不在列表中，则

4 个答案: