TypeError: 'list' object is not callable in python

Seems like you've shadowed the builtin name list pointing at a class by the same name pointing at its instance. Here is an example:

>>> example = list('easyhoss')  # here `list` refers to the builtin class
>>> list = list('abc')  # we create a variable `list` referencing an instance of `list`
>>> example = list('easyhoss')  # here `list` refers to the instance
Traceback (most recent call last):
  File "<string>", line 1, in <module>
TypeError: 'list' object is not callable

I believe this is fairly obvious. Python stores object names (functions and classes are objects, too) in namespaces (which are implemented as dictionaries), hence you can rewrite pretty much any name in any scope. It won't show up as an error of some sort. As you might know, Python emphasizes that "special cases aren't special enough to break the rules". And there are two major rules behind the problem you've faced:

1. Namespaces. Python supports nested namespaces. Theoretically you can endlessly nest namespaces. As I've already mentioned, namespaces are basically dictionaries of names and references to corresponding objects. Any module you create gets its own "global" namespace. In fact it's just a local namespace with respect to that particular module.

2. Scoping. When you reference a name, the Python runtime looks it up in the local namespace (with respect to the reference) and, if such name does not exist, it repeats the attempt in a higher-level namespace. This process continues until there are no higher namespaces left. In that case you get a NameError. Builtin functions and classes reside in a special high-order namespace __builtins__. If you declare a variable named list in your module's global namespace, the interpreter will never search for that name in a higher-level namespace (that is __builtins__). Similarly, suppose you create a variable var inside a function in your module, and another variable var in the module. Then, if you reference var inside the function, you will never get the global var, because there is a var in the local namespace - the interpreter has no need to search it elsewhere.

Here is a simple illustration.

>>> example = list("abc")  # Works fine
>>> 
>>> # Creating name "list" in the global namespace of the module
>>> list = list("abc")
>>> 
>>> example = list("abc")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'list' object is not callable
>>> # Python looks for "list" and finds it in the global namespace,
>>> # but it's not the proper "list".
>>> 
>>> # Let's remove "list" from the global namespace
>>> del list
>>> # Since there is no "list" in the global namespace of the module,
>>> # Python goes to a higher-level namespace to find the name. 
>>> example = list("abc")  # It works.

So, as you see there is nothing special about Python builtins. And your case is a mere example of universal rules. You'd better use an IDE (e.g. a free version of PyCharm, or Atom with Python plugins) that highlights name shadowing to avoid such errors.

You might as well be wondering what is a "callable", in which case you can read this post. list, being a class, is callable. Calling a class triggers instance construction and initialisation. An instance might as well be callable, but list instances are not. If you are even more puzzled by the distinction between classes and instances, then you might want to read the documentation (quite conveniently, the same page covers namespaces and scoping).

If you want to know more about builtins, please read the answer by Christian Dean.

P.S. When you start an interactive Python session, you create a temporary module.

If you are in a interactive session and don't want to restart you can remove the shadowing with

del list

In the league of stupid Monday morning mistakes, using round brackets instead of square brackets when trying to access an item in the list will also give you the same error message:

l=[1,2,3]

print(l[2])#GOOD
print(l(2))#BAD

TypeError: 'list' object is not callable

Before you can fully understand what the error means and how to solve, it is important to understand what a built-in name is in Python.

What is a built-in name?

In Python, a built-in name is a name that the Python interpreter already has assigned a predefined value. The value can be either a function or class object. These names are always made available by default, no matter the scope. Some of the values assigned to these names represent fundamental types of the Python language, while others are simple useful.

As of the latest version of Python - 3.6.2 - there are currently 61 built-in names. A full list of the names and how they should be used, can be found in the documentation section Built-in Functions.

An important point to note however, is that Python will not stop you from re-assigning builtin names. Built-in names are not reserved, and Python allows them to be used as variable names as well.

Here is an example using the dict built-in:

>>> dict = {}
>>> dict
{}
>>>

As you can see, Python allowed us to assign the dict name, to reference a dictionary object.

What does "TypeError: 'list' object is not callable" mean?

To put it simply, the reason the error is occurring is because you re-assigned the builtin name list in the script:

list = [1, 2, 3, 4, 5]

When you did this, you overwrote the predefined value of the built-in name. This means you can no longer use the predefined value of list, which is a class object representing Python list.

Thus, when you tried to use the list class to create a new list from a range object:

myrange = list(range(1, 10))

Python raised an error. The reason the error says "'list' object is not callable", is because as said above, the name list was referring to a list object. So the above would be the equivalent of doing:

[1, 2, 3, 4, 5](range(1, 10))

Which of course makes no sense. You cannot call a list object.

How can I fix the error?

Suppose you have code such as the following:

list = [1, 2, 3, 4, 5]
myrange = list(range(1, 10))

for number in list:
    if number in myrange:
        print(number, 'is between 1 and 10')

Running the above code produces the following error:

Traceback (most recent call last):
  File "python", line 2, in <module>
TypeError: 'list' object is not callable

If you are getting a similar error such as the one above saying an "object is not callable", chances are you used a builtin name as a variable in your code. In this case and other cases the fix is as simple as renaming the offending variable. For example, to fix the above code, we could rename our list variable to ints:

ints = [1, 2, 3, 4, 5] # Rename "list" to "ints"
myrange = list(range(1, 10))

for number in ints: # Renamed "list" to "ints"
    if number in myrange:
        print(number, 'is between 1 and 10')

PEP8 - the official Python style guide - includes many recommendations on naming variables.

This is a very common error new and old Python users make. This is why it's important to always avoid using built-in names as variables such as str, dict, list, range, etc.

Many linters and IDEs will warn you when you attempt to use a built-in name as a variable. If your frequently make this mistake, it may be worth your time to invest in one of these programs.

I didn't rename a built-in name, but I'm still getting "TypeError: 'list' object is not callable". What gives?

Another common cause for the above error is attempting to index a list using parenthesis (()) rather than square brackets ([]). For example:

>>> lst = [1, 2]
>>> lst(0)

Traceback (most recent call last):
  File "<pyshell#32>", line 1, in <module>
    lst(0)
TypeError: 'list' object is not callable

For an explanation of the full problem and what can be done to fix it, see TypeError: 'list' object is not callable while trying to access a list.