TypeError: 'dict_keys' object does not support indexing

Question

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 fu         


        

Answer 1

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.

Answer 2

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.

Answer 3

Convert an iterable to a list may have a cost. Instead, to get the the first item, you can use:

next(iter(keys))

Or, if you want to iterate over all items, you can use:

items = iter(keys)
while True:
    try:
        item = next(items)
    except StopIteration as e:
        pass # finish

Answer 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.

Answer 5

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)