Question

我想根据主字典中的属性中是否存在任何字符串数组，将字典拆分为两个。目前我可以通过两个单独的字典理解来实现这一点（下面），但是只有一个行/字典理解有更有效的方法吗？

included = {k:v for k,v in users.items() if not any(x.lower() in v["name"].lower() for x in EXCLUDED_USERS)}
excluded = {k:v for k,v in users.items() if any(x.lower() in v["name"].lower() for x in EXCLUDED_USERS)}

修改

EXCLUDED_USERS包含模式列表。

Answer 1

这个解决方案更加冗长，但它应该更高效，可能更具可读性：

included = {}
excluded = {}

lower_excluded_users = [x.lower() for x in EXCLUDED_USERS]

for k,v in users.items():
    if any(x in v["name"].lower() for x in lower_excluded_users):
        excluded[k] = v
    else:
        included[k] = v

我认为一次理解不能做到这一点。可以在k:v语句中使用a ternary operator，在else模式中if之后无法使用{k:v for k,v in users.items() if k ...}。

Answer 2

一行解决方案（lower_excluded_users我无法抗拒）

included, excluded = dict(), dict()

# ssly, you don't have to do this everytime
lower_excluded_users = [x.lower() for x in EXCLUDED_USERS]

# and now the one-line answer using if-else-for construct with
# v substituted by D[k]. And instead of using `for k, v in dicn.items()`
# I have used [... for aKey in dicn.keys()]

[ excluded.update({aKey: users[aKey]}) \
    if any(x in users[aKey]["name"].lower() for x in lower_excluded_users) \
    else \
    included.update({aKey: users[aKey]}) \
    for aKey in users.keys()
]

或者没有美化的人：

[excluded.update({aKey: users[aKey]}) if any(x in users[aKey]["name"].lower() for x in lower_excluded_users) else included.update({aKey: users[aKey]}) for aKey in users.keys()]

（Python）基于属性将字典拆分为两个

2 个答案: