{"id":128938,"date":"2022-10-16T19:59:58","date_gmt":"2022-10-16T19:59:58","guid":{"rendered":"https:\/\/blog.finxter.com\/?p=796098"},"modified":"2022-10-16T19:59:58","modified_gmt":"2022-10-16T19:59:58","slug":"python-typeerror-dict_keys-not-subscriptable-fix-this-stupid-bug","status":"publish","type":"post","link":"https:\/\/sickgaming.net\/blog\/2022\/10\/16\/python-typeerror-dict_keys-not-subscriptable-fix-this-stupid-bug\/","title":{"rendered":"Python TypeError: \u2018dict_keys\u2019 Not Subscriptable (Fix This Stupid Bug)"},"content":{"rendered":"\n
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5\/5 – (1 vote) <\/div>\n<\/div>\n

Do you encounter the following error message?<\/p>\n

TypeError: 'dict_keys' object is not subscriptable<\/code><\/pre>\n
You’re not alone! This short tutorial will show you why this error occurs, how to fix it, and how to never make the same mistake again. <\/p>\n
So, let’s get started!<\/p>\n
Solution<\/h2>\n
Python raises the “TypeError: 'dict_keys' object is not subscriptable<\/code>” if you use indexing or slicing on the dict_keys<\/code> object obtained with dict.keys()<\/code>. To solve the error, convert the dict_keys<\/code> object to a list such as in list(my_dict.keys())[0]<\/code>.<\/p>\n
print(list(my_dict.keys())[0])<\/pre>\n<\/p>\n
Example<\/h2>\n
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The following minimal example that leads to the error:<\/p>\n
d = {1:'a', 2:'b', 3:'c'}\nprint(d.keys()[0])<\/pre>\n
Output:<\/p>\n
Traceback (most recent call last): File \"C:\\Users\\...\\code.py\", line 2, in <module> print(d.keys()[0])\nTypeError: 'dict_keys' object is not subscriptable<\/pre>\n
Note that the same error message occurs if you use slicing instead of indexing:<\/p>\n
d = {1:'a', 2:'b', 3:'c'}\nprint(d.keys()[:-1]) # <== same error<\/pre>\n<\/p>\n
Fixes<\/h2>\n
The reason this error occurs is that the dictionary.keys()<\/a><\/code> method returns a dict_keys<\/code> object that is not subscriptable. <\/p>\n
You can use the type()<\/a><\/code> function to check it for yourself:<\/p>\n
print(type(d.keys()))\n# <class 'dict_keys'><\/pre>\n
Note<\/strong>: You cannot expect dictionary keys to be ordered<\/a>, so using indexing on a non-ordered type wouldn’t make too much sense, would it? \"\u26a1\"<\/p>\n
You can fix the non-subscriptable TypeError <\/em>by converting the non-indexable dict_keys<\/code> object to an indexable container type such as a list <\/a>in Python using the list()<\/a><\/code> or tuple()<\/code> function.<\/p>\n
Here’s an example fix:<\/p>\n
d = {1:'a', 2:'b', 3:'c'}\nprint(list(d.keys())[0])\n# 1<\/pre>\n
Here’s an other example fix:<\/p>\n
d = {1:'a', 2:'b', 3:'c'}\nprint(tuple(d.keys())[:-1])\n# (1, 2)<\/pre>\n
Both lists and tuples are subscriptable so you can use indexing and slicing after converting the dict_keys<\/code> object to a list or a tuple.<\/p>\n
\"\ud83c\udf0d\" Full Guide<\/strong>: Python Fixing This Subsctiptable Error<\/a> (General)<\/p>\n
Summary<\/h2>\n
Python raises the TypeError: 'dict_keys' object is not subscriptable<\/code> if you try to index x[i]<\/code> or slice x[i:j]<\/code> a dict_keys<\/code> object.<\/p>\n
The dict_keys<\/code> type is not indexable, i.e., it doesn’t define the __getitem__()<\/a><\/code> method. You can fix it by converting the dictionary keys to a list using the list()<\/a><\/code> built-in function.<\/p>\n
Alternatively, you can also fix this by removing the indexing or slicing call, or defining the __getitem__<\/code> method. Although the previous approach is often better.<\/p>\n
What’s Next?<\/h2>\n
I hope you’d be able to fix the bug in your code! Before you go, check out our free Python cheat sheets<\/em><\/strong> that’ll teach you the basics in Python in minimal time:<\/p>\n
If you struggle with indexing in Python, have a look at the following articles on the Finxter blog—especially the third!<\/p>\n
\"\ud83c\udf0d\" Related Articles:<\/strong><\/p>\n