TypeError: the first argument must be callable

Estimated reading time: 3 minutes

So you may be using Python Classes and have encountered the problem TypeError: First Argument Must be Callable. So what does the problem mean and how can you fix it?

In this article, we are looking to explain how it may occur and the easy fix you can apply to stop the problem in the future.

Let us understand calling Functions/Methods first

Normally in a computer python program, you will have a need to use a function/method as it has the functionality that will save you time, and can continuously be reused.

A classic example is print(“Hello”) that handles all the required logic to show this on a screen. Its output is quickly viewable, and there is not any need to understand what is going on in the background.

Here in this example we have run the function on its own, as a result it does not need to pointed to somehwhere else to run the logic it contains.

For the error we are looking to resolve, this is part of the problem. A function/method can be run on its own, or from within a Class.

The defining differene is that on its own it needs the parenthensis i.e. (), but if if you calling it from within a class, then it has to be handled differently.

Seeing the Type Error First Argument must be callable and fixing it

In the below we have a block of code, that produces the error we are trying to fix.

Following on from what was discussed above, the offending line is highlighted.

In particular the problem lies with printprogress() . This is been called , but in actual fact the problem is that when you have the logic written like this, it gives the error this blog post was setup for.

In essence it is trying to run the program from that exact point, which python does not allow.

Removing the parenthesis then allows the program to go and find the module it is referencing and then run the logic contained within that.

import schedule
import time

class scheduleprint():

    def printprogress(self):
        print("Start of Processing")
        print("Processing Complete")

    def schedule_a_print_job(self, type="Secs", interval=5):

        if type == "Secs": # Fed from the function paramaters
            schedule.every(interval).seconds.do(self.printprogress())===> The problem is here, remove the () after printprogress.
            # Including the parentheses  after printprogess will throw an error as you cant run that method directly from there you can only call it.

        if type == "Mins": # Fed from the function paramaters
            schedule.every(interval).minutes.do(self.printprogress)
            # Including the parentheses  after printprogess will throw an error as you cant run that method directly from there you can only call it.

        while True:
            schedule.run_pending()
            time.sleep(1) # The number of seconds the Python program should pause execution.

run = scheduleprint() # initiating an instance of the class "scheduleprint"
run.schedule_a_print_job() # running the function contained within the class.

In summary to help troubleshoot this problem:

(A) Check your code to see where it is calling a module within a class.

(B) Next make sure that in that call no parenthesis are present, otherwise it wont be able to find the module.

TypeError: ‘NoneType’ object is not iterable

Estimated reading time: 2 minutes

Have you wondered why you cant iterate over a NoneType but never had it properly explained? Read on as here we will fully explain the concept and how you can fix it.

So what is a NoneType anyway?

First of all, to understand a none type, you need to understand what none is.

None is essentially no value, so in other words, nothing is present. It differs from the integer zero, as the integer zero is an actual value and it exists.

If you were to put both of them side by side on a database table, one would be empty the other would have a value of zero.

Also, None is a type in itself:

a = None
print(a)
print(type(a))

Gives the output:

None
<class 'NoneType'>

Whereas zero is an integer datatype:

a = 0
print(a)
print(type(a))

Giving output:
0
<class 'int'>

So how does a none type object is not iterable occur?

Let’s take an example and walk through it. In the below code we have a variable that is empty. In other words, we are asking the program to loop through something that does not exist and does not have any values.

a = None
print(a)
print(type(a))

for i in a:
   print(i)

Giving output:
for i in a:
TypeError: 'NoneType' object is not iterable
None
<class 'NoneType'>

Repeating what we said above, it is not possible to loop through something that does not have a value.

But if we change it to a string value of ‘1’, the loop will work.

a = '1'
print(type(a))

for i in a:
   print(i)

So in summary:

(A) NoneTypes are useful if you just want to return an empty value or check for an empty value.

(B) You cannot loop over NoneTypes as they do not have any values, they are empty!

TypeError: ‘list’ object is not an iterator

We have covered off many TypeErrors on this website, here we will go through which using a list with and it is not an iterator gives you errors.

In order to understand this error better, we need to first understand what is an iterator in Python?

An iterator is a Python object that has the following characteristics:

  • You can count the no of values that are contained within it.
  • It also can be iterated through, so you need to apply an iteration method to it.

How does this error occur?

Normally this error occurs when you try to iterate over a list, but you have not made the list iterable.

There are two things required to make this happen:

(A) The iter() returns an iterator object

(B) The next() method moves to the next value.

Without both the code will fail and the error you are about will occur!

In the below code we have a list:

a = ['q', 'w', 'e', 'r', 't', 'y']

with the following:

b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)

As can be seen in the above code we have one component for the iteration , we expect two as per the above.

As a result we get the error:

Traceback (most recent call last):
  File "list_object_is_not_an_iterator.py", line 13, in <module>
    b = next(a)
TypeError: 'list' object is not an iterator

In order to fix this ,all we need to do is apply the iterator to the list as follows:

a = iter(['q', 'w', 'e', 'r', 't', 'y']) ====> We added in the iter() here, enclosing the list within it

b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)
#b = next(a)


print(b)

Giving output:
y

As a result of this, we now have the two required methods that will not give this error.

What is going on within the iterator?

In the above code we have asked to print b. What the iterator is doing is going to the first value of b, in this case q and print.

But because we have a variable b on multiple lines, with the method “next()” in it, the logic is moving through each value of the list till it gets to the end.

What can be done though is , reduce the length of the returned b variables to print as follows:

a = iter(['q', 'w', 'e', 'r', 't', 'y'])
b = next(a)
print(b)
returns:
q

BUT
a = iter(['q', 'w', 'e', 'r', 't', 'y'])
b = next(a)
b = next(a)
print(b)
returns:
w

As can be seen it returns the next value in the list. You can keep adding the b variables.

What happens when you get to the end of the list?

So now we have the below, and we are returning the last value:

a = iter(['q', 'w', 'e', 'r', 't', 'y'])
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)

Returns:
y

The reason for this is that we have the required no of variables with the next method, which equals the length of the list.

If we add in one more b variable:

a = iter(['q', 'w', 'e', 'r', 't', 'y'])
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a)
b = next(a) ===> Additional b variable

Returns: 
Traceback (most recent call last):
  File "list_object_is_not_an_iterator.py", line 19, in <module>
    b = next(a)
StopIteration

The purpose of StopIteration is to not allow a continuous loop and recognise that the end of the list has been reached.

Implementing Iterators

Iterators could be used in the following circumstances:

(A) You have a defined list of object values to work with.

(B) If sequence is important an iterator will help to process values in the order they appear in a list.

TypeError: ‘float’ object is not callable

So you have been presented with the TypeError: ‘float’ object is not callable in Python, but unsure how to fix? Well read on so we can get you a solution.

First things an overview of a Float

A float is any number that can contain a decimal place separating an integer from the fractions to the right of the decimal point.

As an example, in the below code, we have the exact same number. But when a computer comes to read it, it treats them differently, due to the fact that b has a decimal point and a fraction to the right of it.

a = 1
b = 1.0

print (type(a))
print (type(b))

Output:
<class 'int'>
<class 'float'>

So how in a piece of code would this error occur?

In the below code, the error occurs. The reason behind this is that float is an actual function that changes a value to a float data type.

As a result, you cannot assign a variable name as “float” and then call it separately in its code. Python specifically does not allow a programmer to use a function as a variable name.

a= float(10)
b= float(11)
c= float(12)
float = a

d = float(a/b*c)
print(d)

Output:
d = float(a/b*c)
TypeError: 'float' object is not callable

So how is this fixed?

In order to fix this, we rename the variable “float” to something else, and make sure that the new name is not another function name!

As can be seen when we do this, the below output runs as expected with no errors:

a= float(10)
b= float(11)
c= float(12)
float_value = a   ===> This line was corrected from above from "float" to "float_value"

d = float(a/b*c)
print(d)

Output: 
10.909090909090908

In summary the crux of this problem is that when creating a variable, it cannot be a function name.

You can see similar posts here on TypeErrors.

Have you seen these posts?

TypeError: ‘float’ object is not callable

TypeError: ‘int’ object is not callable

TypeError: ‘int’ object is not callable

Estimated reading time: 1 minute

We have posted several python solutions to TypeErrors here on the website. Here is another one we hope you will find useful in solving a programming issue you may have.

So what does the error mean?

In the below code, we have four variables that have been assigned an integer.

a= 10
b= 11
c= 12
int= 13

d = int(a/b*c)
print(d)

As can be seen we also have a variable d that is assigned to a function int , that is using the variables a,b,c.

As int is a function it cannot be assigned as a variable, and for this reason the below error will appear:

d = int(a/b*c)
TypeError: 'int' object is not callable

So how can this be fixed?

The solution to this is quite straight forward, it is important not to assign a function as a variable. The fix you would apply is as follows:

Incorrect code:

a= 10
b= 11
c= 12
int = 13  ===> change this line to fix the error

d = float(a/b*c)
print(d)

Corrected code:
a= 10
b= 11
c= 12
int_value = 13  ===> corrected line of code
d = float(a/b*c)
print(d)

Giving you the result:
10

As can also be seen with TypeError: ‘str’ object is not callable assign variables to functions should be avoided at all cost.

TypeError: ‘str’ object is not callable

Estimated reading time: 2 minutes

This is a common TypeError that you will come across in Python, it is actually easy to diagnose how it occurred.

To start off we need to understand “str” is a function in python, and it converts any value into a string.

As a result , because it is a function your ability to call it has limitations.

So in essence it has parenthesis () beside it, and allows parameters to be passed to it.

So lets first look at how the string function works:

x = str("10")
y = 10

print(type(x))
print(type(y))
print(x)
print(y)

With output:
<class 'str'>
<class 'int'>
10
10

As you will see the above the value 10, on its own is an integer, but when you call the string function, it now becomes a string.

For this reason this , calling a string function , completes a conversion to string of ten, but what if the variable is called str?

Lets take an example below from an input:

str = input("what year where you born?")
print(str(str))

Output:

what year where you born?2021
Traceback (most recent call last):
  File "str obj is not callable.py", line 2, in <module>
    print(str(str))
TypeError: 'str' object is not callable

Process finished with exit code 1

The reason for this error above , is that we have named the variable “str”.

As can be seen the programme is trying to use the first str in the print statement as a function.

As we know by now string variables are not callable.

Accordingly the function str() which the programme is trying to run, fails with the TypeError identified.

For this reason to fix this problem we would change the variable called str to “year”, the error then disappears.

The updated code will work as when calling str(), it is not conflicted with a variable name.

year = input("what year where you born?")
print(str(year))

Output:
"str obj is not callable.py"
what year where you born?2021
2021

Process finished with exit code 0

So to summarise:

  • Strings are not callable
  • Don’t name a variable as str, or any function name, these are reserved words by the system.
  • Functions are callable, strings are not.

TypeError: cannot unpack non-iterable int object

Often when working on a data analytics project it requires you to split data out into its constituent parts.

There are a number of reasons for this, it can be confusing when you get errors as with the title of this post.

Before we explain this error and what it means, lets us first explain unpacking

Unpacking basically means splitting something up into a number of its parts that make it up.

To demonstrate if you take a,b,c = 123, and look to unpack it, it throws out the error, but why?

Well pure and simple, we have three values on the left “a,b,c”, looking for three values on the right.

a,b,c = 123
print(a)

Output:
 a,b,c = 123
TypeError: cannot unpack non-iterable int object

If you would like to fix this problem change the right hand side to have three values.

a,b,c = 1,2,3
print(a)
print(b)
print(c)
print(a,b,c)

Output:
1
2
3
1 2 3

Process finished with exit code 0

In essence, what is going on is that an evaluation checking that both sides have the same amount of values.

It is important to remember, the code above we used to show the error is an integer, which cannot be unpacked.

So if you take 123 for an example, which we used here it cannot be split into say 100 and 10 and 13.

In this case, even though when they are added up to 123, integers cannot be unpacked.

For this reason in the code for our solution, the difference is that the values used are tuples as follows:

a,b,c = 1,2,3
print(a)
print(b)
print(c)
print(a,b,c)
print(type((a,b,c)))

Or 
a,b,c = (1,2,3)
print(a)
print(b)
print(c)
print(a,b,c)
print(type((a,b,c)))

yield the same result:

1
2
3
1 2 3
<class 'tuple'>

Process finished with exit code 0

So in summary:

When unpacking there are a number of things to remember:

  • Integers on their own cannot be unpacked.
  • You need to make sure that if you have a number of variables, that you have the same number of integers if they the values.
    • This will make it a tuple and unpacking can then happen.

TypeError object of type ‘int’ has no len()

I have seen this data type error come up numerous times while working on my data analytics projects, and recently decided to investigate further. On initial inspection, it can seem a bit of a funny one, but in actual fact, it is quite straight forward.

Lets break it down and see what is going on

So in the below code, there are a number of things:

On line 1 we have a variable that is an integer. If we think about this logically, something that is a single numeric number cannot have a length.

An integer by design is purely to count up a number of apples or no of people, it cannot be viewed as having a length as it is descriptive of the number of occurrences of an object.

data = 100
print(type(data))
print(len(data))

Output Error:
<class 'int'>
Traceback (most recent call last):
  File "object of type int.py", line 3, in <module>
    print(len(data))
TypeError: object of type 'int' has no len()

So for it to in anyway allow a length to be calculated, the object needs to be one of the following data types:

  • List
  • String
  • Tuple
  • Dictionary

Opposite to an integer, these are datatypes that have values that would be more appropriate to having values that a length can be calculated on.

data = "100"
print(type(data))
print("Length of string is: ", len(data))

data = [100,200,300]
print(type(data))
print("Length of list is: ", len(data))

data = (100,200,300)
print(type(data))
print("Length of tuple is: ", len(data))

data = {"Age": 1, "Name": 2}
print(type(data))
print("Length of dictionary is: ", len(data))

And the output is:
<class 'str'>
Length of string is:  3
<class 'list'>
Length of list is:  3
<class 'tuple'>
Length of tuple is:  3
<class 'dict'>
Length of dictionary is:  2

In summary, to understand this error and fix it:

An integer describes the number of things that exist for an object, they are actually not the actual object in existence.

Anything that can have a length method applied to it actually exists and can be counted. In the above four examples, they are actually values that you could describe as existing as you can count each one of them.

The explanation here hopefully clears up the matter, if you have any questions leave a comment and I will answer for you!

TypeError: type object is not subscriptable

Estimated reading time: 2 minutes

I was recently working on our last blog post how to reverse a string in python and I came across this error.

The thought passed me what does it mean and how can I fix it?

So what does the error actually mean?

Essentially it means that , you are trying to access an object type, that has a property of “type”.

What are typical property of types? Well they can be:

  • int()
  • str()
  • tuple()
  • dict()

The above allows you to change your data to these data types, so the data contained within them can be further manipulated.

In essence you are trying to call type in the wrong way and in the wrong place in your code.

By calling it , it will throw this error, and they should be avoided, as they are a built in function.

Lets take an example of how we can replicate this error and fix it

name1 = "joe" # These have index values of [0,1,2]
emptylist =[]
strlength = len(name1) # Returns length of three
while strlength > 0:
    emptylist += str[strlength - 1] #This is the last index value of the variable "name1"
    strlength = strlength - 1
print(emptylist)

In the above code all appears well, but in line 5 the “str” before the [ is the problem. The code automatically looks to call this function.

The simple answer to fixing this is to rename it to name1 as follows:

name1 = "joe" # These have index values of [0,1,2]
emptylist =[]
strlength = len(name1) # Returns length of three
while strlength > 0:
    emptylist += name1[strlength - 1] #This is the last index value of the variable "name1"
    strlength = strlength - 1
print(emptylist)

which gives you the following error free output:

Result with no error: ['e', 'o', 'j']

In summary and what not to do

So it is clear that referencing types as a string variable should be avoided, and keep your code clean from this perspective.

This would also apply to any reserved words in Python as well.

Have you seen type error list object is not an iterator?

Also you may have come across type error float object is not callable

How do I fix TypeError: unhashable type: ‘list’ Error?

Estimated reading time: 3 minutes

When programming in python you will come across this error quite often, in this case is quite easily fixed once understood.

The problem usually arises when you try to loop through a dictionary with key-value pairs. If you are unsure what a dictionary looks like see W3 Schools

Lets examine those loops that don’t throw the error.

Using a list, produces the following output with no error:

list = [['a'],['b'],['c'],['d'],['e'],['f']]
print(type(list))
for i in list:
    print(i)
<class 'list'>
['a']
['b']
['c']
['d']
['e']
['f']

If there is a need for a tuple, then the following outputs with no error:

list = (['a'],['b'],['c'],['d'],['e'],['f'])
print(type(list))
for i in list:
    print(i)
<class 'tuple'>
['a']
['b']
['c']
['d']
['e']
['f']

Using a dictionary, it gives the error you are looking to resolve, but why?

list = {['a'],['b'],['c'],['d'],['e'],['f']}
print(type(list))
for i in list:
    print(i)
list = {['a'],['b'],['c'],['d'],['e'],['f']}
TypeError: unhashable type: 'list'

To understand the error it is important to step back and figure out what is going on in each scenario:

(A) Looping through the list, it looks at the values on their own, thus the loop completes with no problem.

(B) As with lists, Tuples are immutable ( cannot be modified), more importantly, they can be looped through with no error.

In this case the lists have single values, the dictionary above has only one value, it expects two, hence the error.

How do we fix this error going forward?

The simplest way is to loop through a list of single items with the iterable code below:

fixlist = [['a'],['b'],['c'],['d'],['e'],['f'],['f'],['c']]
# Converts fixlist from a list of lists to a flat list, and removes duplicates with set
fixlist  = list(set(list(itertools.chain.from_iterable(fixlist))))
print(fixlist)
Result : ['d', 'f', 'c', 'b', 'a', 'e']

Now your code is only looking to loop through some single values within your list, compared to dictionary key-value pairs.

Approaching solving this problem through an iteration line by line helped to pinpoint the problem.

Consequently the steps I went through to fix the problem involved:

(A) print(type(variable)) – Use this on passing data to see what the data types are, clarifies if this is the problem.

(B) Consequently once the line of code that was throwing the error was found, removing the dictionary fixed the problem.

Or

If a dictionary is required to be looped through, it needs the proper key, value pairs setup.

Conclusion

In conclusion, in order to remove this error it is important to identify the line and or lines, that have a dictionary and covert them to a list

or

if a dictionary is needed ensure that the lists are converted to a dictionary with key, value pairs.

If you would like to see a very good video explanation of this error head over to Brandon Jacobson’s YouTube channel , and make sure to subscribe.

His explantion is below: