问题
I know there's tonnes of questions on python sorting lists/dictionaries already, but I can't seem to find one which helps in my case, and i'm looking for the most efficient solution as I'm going to be sorting a rather large dataset.
My data basically looks like this at the moment:
a = {'a': (1, 2, 3), 'b': (3, 2, 1)}
I'm basically creating a word list in which I store each word along with some stats about it (n, Sigma(x), Sigma(x^2) )
I want to sort it based on a particular stat. So far I've been trying something along the lines of:
b = a.items()
b.sort(key = itemgetter(1), reverse=True)
I'm not sure how to control which index it is sorted based on when its effectively a list of tuples of tuples? I guess I effectively need to nest two itemgetter operations but not really sure how to do this.
If there's a better data structure I should be using instead please let me know. Should I perhaps create a small class/struct and then use a lambda function to access a member of the class?
Many Thanks
回答1:
Something like this?
>>> a = {'a': (1, 2, 3), 'b': (3, 2, 1)}
>>> b = a.items()
>>> b
[('a', (1, 2, 3)), ('b', (3, 2, 1))]
>>> b.sort(key=lambda x:x[1][2]) # sorting by the third item in the tuple
>>> b
[('b', (3, 2, 1)), ('a', (1, 2, 3))]
回答2:
Names are easier to work with and remember that indices, so I would go with a class:
class Word(object): # don't need `object` in Python 3
def __init__(self, word):
self.word = word
self.sigma = (some calculation)
self.sigma_sq = (some other calculation)
def __repr__(self):
return "Word(%r)" % self.word
def __str__(self):
return self.word
@property
def sigma(self):
return self._sigma
@sigma.setter # requires python 2.6+
def sigma(self, value):
if not value:
raise ValueError("sigma must be ...")
self._sigma = value
word_list = [Word('python'), Word('totally'), Word('rocks')]
word_list.sort(key=lambda w: w.sigma_sq)
来源:https://stackoverflow.com/questions/7349646/sorting-a-dictionary-of-tuples-in-python