TypeError: 'dict_keys' object is not subscriptable (Python)

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Borislav Hadzhiev

Last updated: Jan 31, 2023
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# Table of Contents

  1. TypeError: 'dict_keys' object is not subscriptable
  2. TypeError: 'generator' object is not subscriptable in Python
  3. TypeError: 'map' object is not subscriptable in Python

# TypeError: 'dict_keys' object is not subscriptable (Python)

The Python "TypeError: 'dict_keys' object is not subscriptable" occurs when we try to access a dict_keys object at a specific index.

To solve the error, convert the dict_keys object to a list, before accessing an index, e.g. list(my_dict.keys())[0].

typeerror dict keys object is not subscriptable

Here is an example of how the error occurs.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30} keys = my_dict.keys() print(type(keys)) # ๐Ÿ‘‰๏ธ <class 'dict_keys'> # โ›”๏ธ TypeError: 'dict_keys' object is not subscriptable print(keys[0])

Notice that the dict.keys() method returns a dict_keys object, not a list.

# Convert the dict_keys object to a list

To solve the error, pass the dict_keys object to the list constructor to convert it to a list before accessing a list item at a specific index.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30} # โœ… convert to list keys = list(my_dict.keys()) print(keys[0]) # ๐Ÿ‘‰๏ธ "name" print(keys[1]) # ๐Ÿ‘‰๏ธ "age" print(keys[0:2]) # ๐Ÿ‘‰๏ธ ['name', 'age']

convert dict keys object to list

The dict.keys method returns a new view of the dictionary's keys.

Since dict_keys objects are not subscriptable, we can't access them at a specific index.

Once you convert the dict_keys object to a list, you can also use list slicing.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30, 'country': 'Belgium'} keys = list(my_dict.keys()) print(keys[0:2]) # ๐Ÿ‘‰๏ธ ['name', 'age'] print(keys[1:3]) # ๐Ÿ‘‰๏ธ ['age', 'country']

The syntax for list slicing is a_list[start:stop:step].

The start index is inclusive and the stop index is exclusive (up to, but not including).

If the start index is omitted, it is considered to be 0, if the stop index is omitted, the slice goes to the end of the list.

Python indexes are zero-based, so the first item in a list has an index of 0, and the last item has an index of -1 or len(a_list) - 1.

# TypeError: 'dict_values' or 'dict_items' object is not subscriptable

The solution is the same if you got one of the following errors:

  • TypeError: 'dict_values' object is not subscriptable
  • TypeError: 'dict_items' object is not subscriptable

You have to convert the dict_values or dict_items object to a list before accessing it at an index.

Here is an example that does that for dict_values.

main.p
my_dict = {'name': 'Bobby Hadz', 'age': 30} # โœ… convert to list values = list(my_dict.values()) print(values[0]) # ๐Ÿ‘‰๏ธ "Bobby Hadz" print(values[1]) # ๐Ÿ‘‰๏ธ 30 print(values[0:2]) # ๐Ÿ‘‰๏ธ ['Bobby Hadz', 30]

And here is an example that does that for dict_items.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30} # โœ… convert to list items = list(my_dict.items()) print(items[0]) # ๐Ÿ‘‰๏ธ ('name', 'Bobby Hadz') print(items[1]) # ๐Ÿ‘‰๏ธ ('age', 30) print(items[0:2]) # ๐Ÿ‘‰๏ธ [('name', 'Bobby Hadz'), ('age', 30)]

# Iterating over a dictionary's keys

You can iterate over a dictionary's keys without having to convert the dict_keys object to a list.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30, 'country': 'Belgium'} for key in my_dict.keys(): # name # age # country print(key)

iterate over dictionary keys

On each iteration, we get access to the current key and print the result.

If you need to access the value of the current iteration, use the dict.items() method instead.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30, 'country': 'Belgium'} for key, value in my_dict.items(): # name Bobby Hadz # age 30 # country Belgium print(key, value)

# Handling out-of-range exceptions with a try/except

Note that if you try to access a list index that is out of bounds, you would get an error.

You can use a try/except statement if you need to handle that.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30} keys = list(my_dict.keys()) try: print(keys[100]) except IndexError: print('index out of bounds') # ๐Ÿ‘‰๏ธ this runs

handle out of range exception in try except

The example catches the IndexError that is thrown if the index is out of bounds.

Note that an IndexError is not raised when you use list slicing, even if the specified indices do not exist.

main.py
my_dict = {'name': 'Bobby Hadz', 'age': 30, 'country': 'Belgium'} keys = list(my_dict.keys()) print(keys[100:200]) # ๐Ÿ‘‰๏ธ [] print(keys[1:100]) # ๐Ÿ‘‰๏ธ ['age', 'country']

The list constructor takes an iterable, such as a dict_keys object, and converts it to a list.

The list constructor builds a list whose items are the same and in the same order as the iterable's items.

The list constructor takes an iterable that may be a sequence, a container that supports iteration or an iterator object.

You should only use square brackets to access subscriptable objects.

The subscriptable objects in Python are:

  • list
  • tuple
  • dictionary
  • string

All other objects have to be converted to a subscriptable object by using the list(), tuple(), dict() or str() classes to be able to use bracket notation.

Subscriptable objects implement the __getitem__ method whereas non-subscriptable objects do not.

main.py
a_list = [2, 4, 6, 8] # ๐Ÿ‘‡๏ธ <built-in method __getitem__ of list object at 0x7f71f3252640> print(a_list.__getitem__)

# Table of Contents

  1. TypeError: 'generator' object is not subscriptable in Python
  2. TypeError: 'map' object is not subscriptable in Python

# TypeError: 'generator' object is not subscriptable in Python

The Python "TypeError: 'generator' object is not subscriptable" occurs when we try to access a generator object at a specific index.

To solve the error, pass the generator to the list constructor, e.g. list(gen)[0].

typeerror generator object is not subscriptable

Here is an example of how the error occurs

main.py
def g(): yield 1 yield 2 yield 3 gen = g() print(type(gen)) # โ›”๏ธ TypeError: 'generator' object is not subscriptable print(gen[0])

We used square brackets to try to access the element at index 0, but the generator object is not subscriptable.

We can get around this by converting the generator object to a list.

main.py
def g(): yield 1 yield 2 yield 3 # โœ… convert generator object to list my_list = list(g()) print(my_list) # ๐Ÿ‘‰๏ธ [1, 2, 3] print(my_list[0]) # ๐Ÿ‘‰๏ธ 1 print(my_list[1]) # ๐Ÿ‘‰๏ธ 2

List objects are subscriptable, so we can access the element at a specific index after converting the generator object to a list.

You can iterate over a generator with a simple for loop.

main.py
def g(): yield 1 yield 2 yield 3 gen = g() for i in gen: print(i) # ๐Ÿ‘‰๏ธ 1, 2, 3

Alternatively, you can pass the generator to the next() function to retrieve the next item.

main.py
def g(): yield 1 yield 2 yield 3 gen = g() print(next(gen)) # ๐Ÿ‘‰๏ธ 1 print(next(gen)) # ๐Ÿ‘‰๏ธ 2 print(next(gen)) # ๐Ÿ‘‰๏ธ 3

Note that if you try to access a list index that is out of bounds, you would get an error. You can use a try/except statement if you need to handle that.

main.py
def g(): yield 1 yield 2 yield 3 my_list = list(g()) try: print(my_list[100]) except IndexError: print('index out of bounds') # ๐Ÿ‘‰๏ธ this runs

The example catches the IndexError that is thrown if the index is out of bounds.

The list constructor takes an iterable, such as a generator object, and converts it to a list.

The list constructor builds a list whose items are the same and in the same order as the iterable's items.

The list constructor takes an iterable that may be a sequence, a container that supports iteration or an iterator object.

You should only use square brackets to access subscriptable objects.

The subscriptable objects in Python are:

  • list
  • tuple
  • dictionary
  • string

All other objects have to be converted to a subscriptable object by using the list(), tuple(), dict() or str() classes to be able to use bracket notation.

Subscriptable objects implement the __getitem__ method whereas non-subscriptable objects do not.

main.py
a_list = ['bobby', 'hadz', 'com'] # ๐Ÿ‘‡๏ธ <built-in method __getitem__ of list object at 0x7f71f3252640> print(a_list.__getitem__)

# TypeError: 'map' object is not subscriptable in Python

The Python "TypeError: 'map' object is not subscriptable" occurs when we try to access a map object at a specific index.

To solve the error, convert the map object to a list before accessing it at an index, e.g. new_list = list(map(int, a_list)).

typeerror map object is not subscriptable

Here is an example of how the error occurs.

main.py
a_list = ['2', '4', '6', '8'] new_list = map(int, a_list) print(new_list) # ๐Ÿ‘‰๏ธ <map object at 0x7fde832679a0> # โ›”๏ธ TypeError: 'map' object is not subscriptable print(new_list[0])

In Python 3, the map() function returns a map object, not a list.

To solve the error, use the list() class to convert the map object to a list.
main.py
a_list = ['2', '4', '6', '8'] new_list = list(map(int, a_list)) print(new_list) # ๐Ÿ‘‰๏ธ [2, 4, 6, 8] print(new_list[0]) # ๐Ÿ‘‰๏ธ 2

The map() function takes a function and an iterable as arguments and calls the function with each item of the iterable.

The function returns a map object and map objects are not subscriptable (cannot be accessed at an index).

You can use square brackets if you need to access:

  • a list at a specific index
  • a tuple at a specific index
  • a string at a specific index
  • a specific key in a dictionary
An alternative approach to solving the error is to use a for loop to iterate over the map object without having to convert it to a list.
main.py
a_list = ['2', '4', '6', '8'] new_list = [] map_obj = map(int, a_list) for item in map_obj: print(int(item)) new_list.append(int(item)) print(new_list) # ๐Ÿ‘‰๏ธ [2, 4, 6, 8] print(new_list[0]) # ๐Ÿ‘‰๏ธ 2 print(new_list[:2]) # ๐Ÿ‘‰๏ธ [2, 4]

Map objects are evaluated lazily, so we used a for loop to iterate over the object.

You can optionally append the result of calling a function with each item to a list.

A better approach than having to convert a map object to a list is to use a list comprehension.

main.py
a_list = ['2', '4', '6', '8'] new_list = [int(item) for item in a_list] print(new_list) # ๐Ÿ‘‰๏ธ [2, 4, 6, 8]

We used a list comprehension to iterate over the original list.

List comprehensions are used to perform some operation for every element or select a subset of elements that meet a condition.

On each iteration, we call the int() class with the current item and return the result.

The list comprehension returns a new list containing the results.

A list comprehension can be used as a replacement for the map function. All you have to do is call a function with each list item and return the result.

You can also use a list comprehension or the map function with user-defined functions.

main.py
def do_math(a): return a + 100 a_list = [2, 4, 6, 8] result = list(map(do_math, a_list)) print(result) # ๐Ÿ‘‰๏ธ [102, 104, 106, 108] print(result[0]) # ๐Ÿ‘‰๏ธ 102 result = [do_math(item) for item in a_list] print(result) # ๐Ÿ‘‰๏ธ [102, 104, 106, 108] print(result[0]) # ๐Ÿ‘‰๏ธ 102

The error basically means that the object cannot be accessed using square brackets.

You should only use square brackets to access subscriptable objects.

The subscriptable objects in Python are:

  • list
  • tuple
  • dictionary
  • string

All other objects have to be converted to a subscriptable object by using the list(), tuple(), dict() or str() classes to be able to use bracket notation.

Subscriptable objects implement the __getitem__ method whereas non-subscriptable objects do not.

main.py
a_list = [1, 2, 3] # ๐Ÿ‘‡๏ธ <built-in method __getitem__ of list object at 0x7f71f3252640> print(a_list.__getitem__)

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