[英]creating lists from nested dictionaries
I am new to coding and I am trying to extract 2 lists from the following nested dictionary:我是编码新手,我正在尝试从以下嵌套字典中提取 2 个列表:
dict : {'player2': {'player': 'player2', 'firstname': 'Ross', 'lastname': 'Estrada'}, 'player3': {'player': 'player3', 'firstname': 'Melvin', 'lastname': 'Graves'}, 'player0': {'player': 'player0', 'firstname': 'Alfredo', 'lastname': 'Hopkins'}, 'player1': {'player': 'player1', 'firstname': 'Jay', 'lastname': 'Ramos'}, 'player6': {'player': 'player6', 'firstname': 'Claudia', 'lastname': 'Brown'}, 'player7': {'player': 'player7', 'firstname': 'Ben', 'lastname': 'Carson'}, 'player4': {'player': 'player4', 'firstname': 'Virginia', 'lastname': 'Harmon'}, 'player5': {'player': 'player5', 'firstname': 'Caleb', 'lastname': 'Reese'}, 'player8': {'player': 'player8', 'firstname': 'Olivia', 'lastname': 'Olson'}, 'player9': {'player': 'player9', 'firstname': 'Amy', 'lastname': 'Valdez'}}
The first list should contain all the first names, whereas the second list should contain all the second names.第一个列表应该包含所有的名字,而第二个列表应该包含所有的第二个名字。
The problem is that I do not know what key to use to signal all outer keys rather than just one.问题是我不知道用什么键来发出所有外部键的信号,而不仅仅是一个。
So, for example I know dict["player2"]["firstname"]
will return the player2's firstname, but how can I write this to extract all the firstnames at once.因此,例如我知道
dict["player2"]["firstname"]
将返回 player2 的名字,但是我如何编写它以一次提取所有名字。
Thanks for your help!感谢您的帮助!
>>> list(zip(*map(operator.itemgetter('firstname', 'lastname'), D.values())))
[('Ross', 'Melvin', 'Alfredo', 'Jay', 'Claudia', 'Ben', 'Virginia', 'Caleb', 'Olivia', 'Amy'), ('Estrada', 'Graves', 'Hopkins', 'Ramos', 'Brown', 'Carson', 'Harmon', 'Reese', 'Olson', 'Valdez')]
>>> fnames, lnames = zip(*map(operator.itemgetter('firstname', 'lastname'), D.values()))
>>> fnames
('Ross', 'Melvin', 'Alfredo', 'Jay', 'Claudia', 'Ben', 'Virginia', 'Caleb', 'Olivia', 'Amy')
>>> lnames
('Estrada', 'Graves', 'Hopkins', 'Ramos', 'Brown', 'Carson', 'Harmon', 'Reese', 'Olson', 'Valdez')
You can use a list comprehension and iterate over all outer keys:您可以使用列表理解并迭代所有外部键:
d = {
'player2': {'player': 'player2', 'firstname': 'Ross', 'lastname': 'Estrada'},
'player3': {'player': 'player3', 'firstname': 'Melvin', 'lastname': 'Graves'},
'player0': {'player': 'player0', 'firstname': 'Alfredo', 'lastname': 'Hopkins'},
'player1': {'player': 'player1', 'firstname': 'Jay', 'lastname': 'Ramos'},
'player6': {'player': 'player6', 'firstname': 'Claudia', 'lastname': 'Brown'},
'player7': {'player': 'player7', 'firstname': 'Ben', 'lastname': 'Carson'},
'player4': {'player': 'player4', 'firstname': 'Virginia', 'lastname': 'Harmon'},
'player5': {'player': 'player5', 'firstname': 'Caleb', 'lastname': 'Reese'},
'player8': {'player': 'player8', 'firstname': 'Olivia', 'lastname': 'Olson'},
'player9': {'player': 'player9', 'firstname': 'Amy', 'lastname': 'Valdez'}}
firstNames = [ d[pKey].get('firstname','') for pKey in d]
lastNames = [ d[pKey].get('lastname','') for pKey in d]
print(firstNames)
print(lastNames)
By using dict.get(key,default)
you can avoid getting key-errors on the inner dictionaries in case one has the key you look for not inside it - and provide a default.通过使用
dict.get(key,default)
您可以避免在内部字典中出现键错误,以防万一您查找的键不在其中 - 并提供默认值。
See fe Why dict.get(key) instead of dict[key]?请参阅 fe 为什么使用 dict.get(key) 而不是 dict[key]? for answers on why its smart.
有关为什么它很聪明的答案。
Using for k in dictName
will iterate overall possible keys of this dictionary, the rest is a simple list comprehension to generate the wanted lists. for k in dictName
使用for k in dictName
将迭代这个字典的所有可能的键,剩下的就是一个简单的列表理解来生成想要的列表。
That one-liner will do那个单线就行
first_names, last_names = zip(*((d[k]['firstname'], d[k]['lastname']) for k in d))
Output:输出:
>>> first_names
('Virginia', 'Amy', 'Melvin', 'Claudia', 'Jay', 'Alfredo', 'Ross',
'Caleb', 'Ben', 'Olivia')
>>> last_names
('Harmon', 'Valdez', 'Graves', 'Brown', 'Ramos', 'Hopkins', 'Estrada',
'Reese', 'Carson', 'Olson')
Note 1: Maybe you would prefer dict.get(key)
to dict[key]
, as suggested by Patrick Artner Note 2: Those variables will be tuples.注 1:也许您更喜欢
dict.get(key)
到dict[key]
,正如 Patrick Artner 所建议的那样 注 2:这些变量将是元组。 Convert them to list
if you want to.如果需要,将它们转换为
list
。
Another way of doing it is this (naming your dictionary d
instead of dict
, as dict
is a Python built-in type and reserved keyword):另一种方法是这样的(将字典命名为
d
而不是dict
,因为dict
是 Python 内置类型和保留关键字):
firstNames = list(map(lambda i: d[i]['firstname'], d.keys()))
lastNames = list(map(lambda i: d[i]['lastname'], d.keys()))
print(firstNames)
print(lastNames)
Output:输出:
['Ross', 'Melvin', 'Alfredo', 'Jay', 'Claudia', 'Ben', 'Virginia', 'Caleb', 'Olivia', 'Amy']
['Estrada', 'Graves', 'Hopkins', 'Ramos', 'Brown', 'Carson', 'Harmon', 'Reese', 'Olson', 'Valdez']
You can use zip
and unpacking:您可以使用
zip
和解包:
d = {'player2': {'player': 'player2', 'firstname': 'Ross', 'lastname': 'Estrada'}, 'player3': {'player': 'player3', 'firstname': 'Melvin', 'lastname': 'Graves'}, 'player0': {'player': 'player0', 'firstname': 'Alfredo', 'lastname': 'Hopkins'}, 'player1': {'player': 'player1', 'firstname': 'Jay', 'lastname': 'Ramos'}, 'player6': {'player': 'player6', 'firstname': 'Claudia', 'lastname': 'Brown'}, 'player7': {'player': 'player7', 'firstname': 'Ben', 'lastname': 'Carson'}, 'player4': {'player': 'player4', 'firstname': 'Virginia', 'lastname': 'Harmon'}, 'player5': {'player': 'player5', 'firstname': 'Caleb', 'lastname': 'Reese'}, 'player8': {'player': 'player8', 'firstname': 'Olivia', 'lastname': 'Olson'}, 'player9': {'player': 'player9', 'firstname': 'Amy', 'lastname': 'Valdez'}}
first, last = zip(*[[b['firstname'], b['lastname']] for _, b in d.items()])
Output:输出:
('Ross', 'Melvin', 'Alfredo', 'Jay', 'Claudia', 'Ben', 'Virginia', 'Caleb', 'Olivia', 'Amy')
('Estrada', 'Graves', 'Hopkins', 'Ramos', 'Brown', 'Carson', 'Harmon', 'Reese', 'Olson', 'Valdez')
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