Question

假设您有以下列表：

fruit_types = ["apple","pear","strawberry","mango"]
fruits = ["apple","strawberry","mango"]

我如何编写一个循环来标识“pear”元素缺失 “水果”榜单的第二个指数？

我知道第一步始于：

for element in fruit_types:
    if element not in fruits:

但我完全失去了从这里出发的其他地方。

Answer 1

以上设定的差异有效。但是如果你想要的代码只检查两个列表中的一个是否缺少元素（在你的风格中）。您可以使用枚举函数。

fruit_types = ["apple","pear","strawberry","mango"]
fruits = ["apple","strawberry","mango"]

def find_missing_index():
    for index, element in enumerate(fruit_types):
        if element not in fruits:
            print(fruit_types[index], index)

def main():
    find_missing_index()

if __name__ == '__main__':
    main()

毫无疑问，您需要设置差异代码，但其中有很多其他例子。

为了提高性能，最好提前将fruits转换为set。

fset = set(fruits)
for index, element in enumerate(fruit_types):
    if element not in fset:
        print(fruit_types[index], index)

这会将成员资格测试减少为常量O(1)查找。

Answer 2

您可以在set上使用差异操作来查找此内容。

fruit_types = ["apple","pear","strawberry","mango"]
fruits = ["apple","strawberry","mango"]

missings = list(set(fruits_types) - set(fruit))
print missings

如果这很酷，那么为什么要使用循环？

然后获取缺失项目的索引：

for missing in missings:
    print fruit_types.index(missing)

Answer 3

使用集差异，并找到索引：

fruit_types = ["apple","pear","strawberry","mango"]
fruits = ["apple","strawberry","mango"]

missings = set(fruit_types) - set(fruits)
for missing in missings:
    print(fruit_types.index(missing))

输出：

1  # which is the index of the missing element in fruit_types

Answer 4

你可以这样做：

fruit_types = ["apple","pear","strawberry","mango"]
fruits = ["apple","strawberry","mango"]
missing_elements = []
for element in fruit_types:
    if(element not in fruits):
        missing_elements.append(element)
print(missing_elements)
>>> ["pear"]

创建一个空列表，如果元素不在水果中，则添加到该列表。

如何找到缺失元素的索引？

4 个答案:

输出：