Python 根据值对字典键进行排序
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Sorting dictionary keys based on their values
提问by PDStat
I have a python dictionary setup like so
我有一个像这样的python字典设置
mydict = { 'a1': ['g',6],
'a2': ['e',2],
'a3': ['h',3],
'a4': ['s',2],
'a5': ['j',9],
'a6': ['y',7] }
I need to write a function which returns the ordered keys in a list, depending on which column your sorting on so for example if we're sorting on mydict[key][1] (ascending)
我需要编写一个函数来返回列表中的有序键，这取决于您排序的列，例如，如果我们在 mydict[key][1]（升序）上排序
I should receive a list back like so
我应该像这样收到一份清单
['a2', 'a4', 'a3', 'a1', 'a6', 'a5']
It mostly works, apart from when you have columns of the same value for multiple keys, eg. 'a2': ['e',2] and 'a4': ['s',2]. In this instance it returns the list like so
除了当您有多个键的相同值的列时，它通常有效，例如。'a2': ['e',2] 和 'a4': ['s',2]。在这种情况下，它像这样返回列表
['a4', 'a4', 'a3', 'a1', 'a6', 'a5']
Here's the function I've defined
这是我定义的函数
def itlist(table_dict,column_nb,order="A"):
try:
keys = table_dict.keys()
values = [i[column_nb-1] for i in table_dict.values()]
combo = zip(values,keys)
valkeys = dict(combo)
sortedCols = sorted(values) if order=="A" else sorted(values,reverse=True)
sortedKeys = [valkeys[i] for i in sortedCols]
except (KeyError, IndexError), e:
pass
return sortedKeys
And if I want to sort on the numbers column for example it is called like so
如果我想对数字列进行排序，例如它是这样调用的
sortedkeysasc = itmethods.itlist(table,2)
So any suggestions?
那么有什么建议吗？
Paul
保罗
采纳答案by Sven Marnach
Wouldn't it be much easier to use
使用起来会不会容易很多
sorted(d, key=lambda k: d[k][1])
(with dbeing the dictionary)?
（d作为字典）？
回答by ulidtko
>>> L = sorted(d.items(), key=lambda (k, v): v[1])
>>> L
[('a2', ['e', 2]), ('a4', ['s', 2]), ('a3', ['h', 3]), ('a1', ['g', 6]), ('a6', ['y', 7]), ('a5', ['j', 9])]
>>> map(lambda (k,v): k, L)
['a2', 'a4', 'a3', 'a1', 'a6', 'a5']
Here you sort the dictionary items (key-value pairs) using a key- callable which establishes a total order on the items.
在这里，您使用键对字典项（键值对）进行排序- 可调用，它在项上建立总顺序。
Then, you just filter out needed values using a mapwith a lambdawhich just selects the key. So you get the needed list of keys.
然后，您只需使用 a 过滤掉所需的值map， alambda只选择键。因此，您将获得所需的密钥列表。
EDIT:see this answerfor a much better solution.
编辑：请参阅此答案以获得更好的解决方案。
回答by Amber
def itlist(table_dict, col, desc=False):
return [key for (key,val) in
sorted(
table_dict.iteritems(),
key=lambda x:x[1][col-1],
reverese=desc,
)
]
回答by Kimvais
Although there are multiple working answers above, a slight variation / combination of them is the most pythonicto me:
尽管上面有多个有效的答案，但它们的细微变化/组合对我来说是最pythonic的：
[k for (k,v) in sorted(mydict.items(), key=lambda (k, v): v[1])]
回答by John La Rooy
>>> mydict = { 'a1': ['g',6],
... 'a2': ['e',2],
... 'a3': ['h',3],
... 'a4': ['s',2],
... 'a5': ['j',9],
... 'a6': ['y',7] }
>>> sorted(mydict, key=lambda k:mydict[k][1])
['a2', 'a4', 'a3', 'a1', 'a6', 'a5']
>>> sorted(mydict, key=lambda k:mydict[k][0])
['a2', 'a1', 'a3', 'a5', 'a4', 'a6']
