Reputation: 3022

TypeError: 'int' object is not callable

Given the following:

a = 23
b = 45
c = 16

round((a/b)*0.9*c)

Running the above outputs an error:

TypeError: 'int' object is not callable.

How can I round the output to an integer?

Upvotes: 94

Answers (11)

w. Patrick Gale

Reputation: 2347

FYI: the 'int' object is not callable error also appears if you accidentally use .size() as a method (which does not exist) instead of the .size property in Pandas dataframes as demonstrated below. Thus the error can appear in unexpected places.

import pandas as pd
s = pd.Series([35, 52, 63, 52])
print("unique numbers: ",s.unique())
print("number of unique values: ",s.unique().size())

Upvotes: 0

Code on the Rocks

Reputation: 17854

I encountered this error because I was calling a function inside my model that used the @property decorator.

@property
def volume_range(self):
    return self.max_oz - self.min_oz

When I tried to call this method in my serializer, I hit the error "TypeError: 'int' object is not callable".

    def get_oz_range(self, obj):
      return obj.volume_range()

In short, the issue was that the @property decorator turns a function into a getter. You can read more about property() in this SO response.

The solution for me was to access volume_range like a variable and not call it as a function:

    def get_oz_range(self, obj):
      return obj.volume_range # No more parenthesis

Upvotes: 0

Nithish Thomas

Reputation: 1565

You can always use the below method to disambiguate the function.

__import__('__builtin__').round((a/b)*0.9*c)

__builtin__ is the module name for all the built in functions like round, min, max etc. Use the appropriate module name for functions from other modules.

Upvotes: 0

Shambhav Agrawal

Reputation: 151

There are two reasons for this error "TypeError: 'int' object is not callable"

Function Has an Integer Value

Consider

a = [5, 10, 15, 20]
max = 0
max = max(a)
print(max)

This will produce TypeError: 'int' object is not callable.

Just change the variable name "max" to var(say).

a = [5, 10, 15, 20]
var = 0
var = max(a)
print(var)

The above code will run perfectly without any error!!

Missing a Mathematical Operator

Consider

a = 5
b = a(a+1)
print(b)

This will also produce TypeError: 'int' object is not callable.

You might have forgotten to put the operator in between ( '*' in this case )

Upvotes: 3

Mebatsion Sahle

Reputation: 429

Sometimes the problem would be forgetting an operator while calculation.
Example:
print(n-(-1+(math.sqrt(1-4(2*(-n))))/2)) rather
it has to be
print(n-(-1+(math.sqrt(1-4*(2*(-n))))/2))

HTH

Upvotes: 2

Sugandha Jain

Reputation: 353

I was also facing this issue but in a little different scenario.

Scenario:

param = 1

def param():
    .....
def func():
    if param:
        var = {passing a dict here}
        param(var)

It looks simple and a stupid mistake here, but due to multiple lines of codes in the actual code, it took some time for me to figure out that the variable name I was using was same as my function name because of which I was getting this error.

Changed function name to something else and it worked.

So, basically, according to what I understood, this error means that you are trying to use an integer as a function or in more simple terms, the called function name is also used as an integer somewhere in the code. So, just try to find out all occurrences of the called function name and look if that is being used as an integer somewhere.

I struggled to find this, so, sharing it here so that someone else may save their time, in case if they get into this issue.

Upvotes: 5

Tomate

Reputation: 305

I got the same error (TypeError: 'int' object is not callable)

def xlim(i,k,s1,s2):
   x=i/(2*k)
   xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x(1-x))
   return xl 
... ... ... ... 

>>> xlim(1,100,0,0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in xlim
TypeError: 'int' object is not callable

after reading this post I realized that I forgot a multiplication sign * so

def xlim(i,k,s1,s2):
   x=i/(2*k)
   xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x * (1-x))
   return xl 

xlim(1.0,100.0,0.0,0.0)
0.005

tanks

Upvotes: 17

krishnakeshan

Reputation: 1280

As mentioned you might have a variable named round (of type int) in your code and removing that should get rid of the error. For Jupyter notebooks however, simply clearing a cell or deleting it might not take the variable out of scope. In such a case, you can restart your notebook to start afresh after deleting the variable.

Upvotes: 0