在 Python3 中使用 filter() 从字典值列表中删除列表

Question

我有一本字典，其中 key: str 和 value: list。当键等于某个值并且该列表中的一个元素等于某个值时，我想从其值列表中删除某个列表。

例如，当 d["A"]=[[0, "APPLE", 1202021, "NEW"], [8, "PEAR", 3242413, "NEW"], [1982, "PEAR", 1299021, "OLD"]]" 时，我想从 d["A"] 中删除其第二个索引值等于 3242413 的列表，以便键 A 处的新字典变为：d["A"]=[[0, "APPLE", 1202021, "NEW"], [1982, "PEAR", 1299021, "OLD"]]".

到目前为止，我尝试使用 filter() 和 dict comprehension 但无法想出一个干净的方法来做到这一点。当然总是有循环和删除的方法，但我想知道我们是否可以使用 filter() 来实现同样的事情？像 :

# d is the dictionary key: str value: list of lists
d["A"] = [[0, "APPLE", 1202021, "NEW"], [8, "PEAR", 3242413, "NEW"], [1982, "PEAR", 1299021, "OLD"]]

# 1. use filter
new_dict = dict(filter(lambda elem: elem[1].. != ,d.items()))  # filter out those element inside list value does not equal to 3242413

# 2. use dict comprehension
new_dict = {key: value for (key, value) in d.items() if value ... ==  }

# so the new dict becomes
d["A"]=[[0, "APPLE", 1202021, "NEW"], [1982, "PEAR", 1299021, "OLD"]]

Answer 1

你对字典的理解已经差不多了：

d = {}
d["A"] = [[0, "APPLE", 1202021, "NEW"], [8, "PEAR", 3242413, "NEW"], [1982, "PEAR", 1299021, "OLD"]]
d["B"] = [[0, "APPLE", 1202021, "NEW"], [8, "PEAR", 1, "NEW"], [1982, "PEAR", 3242413, "OLD"]]

d = {k: [i for i in v if i[2] != 3242413] for (k, v) in d.items()}

给出：

{'A': [[0, 'APPLE', 1202021, 'NEW'], [1982, 'PEAR', 1299021, 'OLD']], 'B': [[0, 'APPLE', 1202021, 'NEW'], [8, 'PEAR', 1, 'NEW']]}

在 Python3 中使用 filter() 从字典值列表中删除列表

Remove a list from dictionary value lists using filter() in Python3

python

lambda

dictionary

python-3.x