TypeError: ‘dict_keys’ object does not support indexing

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 is None else int(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

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 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.

Method 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.

Method 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

Method 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.

Method 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)


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

0 0 votes
Article Rating
Subscribe
Notify of
guest

0 Comments
Oldest
Newest Most Voted
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x