如何分解这个列表推导式，或者换句话说，如何合并列表中的子集?

Question

我遵循了一些技巧，但仍然不能分解这个列表理解，这在中间列表中意味着什么?它主要用于合并列表中的一些子集，例如

l1 = ['员工:张三', '实习员工:张三', '职员:李四']

因为'员工:张三'是'实习员工:张三'的子集，所以结果将去掉'员工:张三'

l1_res = [word for word in l1 if not any(word in other != word for other in l1)]

我知道外在可能会喜欢，但内在呢?

l1_res = []
for word in l1:
if not any(word in other!=word for other in l1): #how to decompose this line?
l1_res.append(word)

或者其他实现这个函数的方法?

Answer 1

可以分解为:

l1_res = []
for word in l1:
for other in l1:
if word in other != word:
break
else:
l1_res.append(word)

这里使用了两个Python容易引起误解的结构:

• word in other != word的链式比较，即word in other and other != word
• for..else，其中else只在循环不是broken，错误，break没有被调用时才会被执行

或者你也可以这样做:

l1_res = []
for word in l1:
found = False
for other in l1:
if word in other != word:
found = True
break
if not found:
l1_res.append(word)

Answer 2

下面是基于标准循环的等效代码:

l1_res = []
for word in l1:
tmp = []
for other in l1:
tmp.append(word in other != word)
if not any(tmp):
l1_res.append(word)
print(l1_res)

Answer 3

l1 = ['a', 'ab','c','abc']
l1_res = []
for word in l1:
tmp = []
for other in l1:
tmp.append( word in other and other != word)  # word in other != word
if not any(tmp):
l1_res.append(word)
print(l1_res)