def shuffle(self, x, random=None,int=int):"""x, random=random.random -> shuffle list x in place; return None. Optional arg random is a 0-argument function returning a random float in [0.0, 1.0); by default, the standard random.random. """ randbelow =self._randbelow for i in reversed(range(1, len(x))):# pick an element in x[:i+1] with which to exchange x[i] j = randbelow(i+1)if random isNoneelseint(random()*(i+1)) x[i], x[j]= x[j], x[i]
When I run the shuffle function it raises the following error, why is that?
TypeError:'dict_keys'object does not support indexing
回答1:
Clearly you're passing in d.keys() to your shuffle function. Probably this was written with python2.x (when d.keys() returned a list). With python3.x, d.keys() returns a dict_keys object which behaves a lot more like a set than a list. As such, it can't be indexed.
The solution is to pass list(d.keys()) (or simply list(d)) to shuffle.
回答2:
You're passing the result of somedict.keys() to the function. In Python 3, dict.keys doesn't return a list, but a set-like object that represents a view of the dictionary's keys and (being set-like) doesn't support indexing.
To fix the problem, use list(somedict.keys()) to collect the keys, and work with that.
回答3:
Why you need to implement shuffle when it already exists? Stay on the shoulders of giants.
import random d1 ={0:'zero',1:'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine'} keys = list(d1) random.shuffle(keys) d2 ={}for key in keys: d2[key]= d1[key]print(d1)print(d2)
回答4:
In Python 2 dict.keys() return a list, whereas in Python 3 it returns a generator.
You could only iterate over it's values else you may have to explicitly convert it to a list i.e. pass it to a list function.
回答5:
convert iterative to list may have cost instead of it you can use
next(iter(keys))
for first item or if you want to itrate all items use
翻译