I’m using python 3.2.2 on windows 7 and I’m trying to create a program which accepts 7 numbers and then tells the user how many are positive, how many are negative and how many are zero. this is what I have got so far:
count=7
for i in count:
num = float(input("Type a number, any number:"))
if num == 0:
zero+=1
elif num > 0:
positive+=1
elif num < 0:
negative+=1
print (positive)
print (negative)
print (zero)
But when I run the code I get
TypeError: 'float' object is not iterable
If I replace float in line 3 with int I get the same problem except it says that the ‘int’ object is not iterable. I have also tried changing the value of count from 7 to 7.0
Now I took this challenge from a python tutorial book and they don’t have the answer, and from what I can tell I have done everything within the syntax they put forward.
The tutorial is here (PDF)
I’m trying to create a code that will solve for x using the quadratic formula. For the outputs I don’t want it to display the imaginary numbers…only real numbers. I set the value inside the square root to equal the variable called «root» to determine if this value would be pos/neg (if it’s neg, then the solution would be imaginary).
This is the code.
import math
print("Solve for x when ax^2 + bx + c = 0")
a = float(input("Enter a numerical value for a: "))
b = float(input("Enter a numerical value for b: "))
c = float(input("Enter a numerical value for c: "))
root = math.pow(b,2) - 4*a*c
root2 = ((-1*b) - math.sqrt(root)) / (2*a)
root1 = ((-1*b) + math.sqrt(root)) / (2*a)
for y in root1:
if root>=0:
print("x =", y)
elif root<0:
print('x is an imaginary number')
for z in root2:
if root>=0:
print("or x =", z)
elif root<0:
print('x is an imaginary number')
This is the error code:
File "/Users/e/Documents/Intro Python 2020/Project 1/Project 1 - P2.py", line 25, in <module>
for y in root1:
TypeError: 'float' object is not iterable
The error occurs at the line:
for y in root1:
How do I fix this error?
asked Apr 10, 2020 at 23:39
4
I understand you are using the quadratic equation here. An iterable is something like a list. A variable with more than 1 element. In your example
root1 is a single float value. root2 is also a single float value. For your purposes, you do not need either lines with the «for». Try removing the for y and for z lines and running your code.
To help you understand, a float value is simply a number that has decimals.
answered Apr 10, 2020 at 23:42
SriSri
2,2812 gold badges17 silver badges24 bronze badges
Well, the error is pretty self explanatory: you are trying to loop over root1 and root2 which are floats, not lists.
What I believe you wanted to do instead is simply use if/else blocks:
if root >= 0:
print("x =", root1)
print("x =", root2)
else:
print("x is an imaginary number")
answered Apr 10, 2020 at 23:45
Anis R.Anis R.
6,6862 gold badges15 silver badges37 bronze badges
I’m using python 3.2.2 on windows 7 and I’m trying to create a program which accepts 7 numbers and then tells the user how many are positive, how many are negative and how many are zero. this is what I have got so far:
count=7
for i in count:
num = float(input("Type a number, any number:"))
if num == 0:
zero+=1
elif num > 0:
positive+=1
elif num < 0:
negative+=1
print (positive)
print (negative)
print (zero)
But when I run the code I get
TypeError: 'float' object is not iterable
If I replace float in line 3 with int I get the same problem except it says that the ‘int’ object is not iterable. I have also tried changing the value of count from 7 to 7.0
Now I took this challenge from a python tutorial book and they don’t have the answer, and from what I can tell I have done everything within the syntax they put forward.
The tutorial is here (PDF)
One error that you might encounter when coding in Python is:
TypeError: 'float' object is not iterable
This error usually occurs when you pass a float object into a function or syntax where an iterable is expected. There are three common scenarios in which this error occurs:
- Using a float in a for loop
- Call list() with float objects
- Call the sum() function and pass float objects
This tutorial shows you examples that cause this error and how to fix it.
1. Using a float in a for loop
One common cause of this error is when you pass a float when attempting to iterate using a for loop.
Suppose you have a for loop code as follows:
num = 4.5
for i in num:
print(i)
The num variable is a float object, so it’s not iterable and causes the following error when used in a for loop:
Traceback (most recent call last):
File "main.py", line 3, in <module>
for i in num:
TypeError: 'float' object is not iterable
A for loop requires you to pass an iterable object, such as a list or a range object.
If you want to use a float, you need to convert it to an int using the int() function, then convert it to a range with range() as follows:
num = 4.5
for i in range(int(num)):
print(i)
Output:
Unlike the for loop in other programming languages where you can use an integer value, the for loop in Python requires you to pass an iterable. Both int and float objects are not iterable.
2. Call list() with float objects
This error also occurs when you call the list() function and pass float objects to it.
Suppose you want to create a list from float values as follows:
num = 4.5
my_list = list(num)
Python shows “TypeError: ‘float’ object is not iterable” because you can’t pass a float when creating a list.
To create a list from a float, you need to surround the argument in square brackets:
num = 4.5
my_list = list([num])
print(my_list) # [4.5]
The same error also occurs when you create a dictionary, tuple, or set using a float object. You need to use the appropriate syntax for each function:
num = 5.5
# Float to dictionary
val = dict(num=num)
print(val) # {'num': 5.5}
# Float to tuple
val = tuple([num])
print(val) # (5.5,)
# Float to set
val = set([num])
print(val) # {5.5}
For the dict() function, you need to pass arguments in the form of key=value for the key-value pair.
3. Call the sum() function and pass float objects
Another condition where this error might occur is when you call the sum() function and pass float objects to it.
Suppose you try to sum floating numbers as shown below:
x = 4.5
y = 6.8
z = sum(x, y)
You’ll get “float is not iterable” error because the sum() function expects a list.
To fix this, surround the arguments you passed to sum() with square brackets as follows:
x = 4.5
y = 6.8
z = sum([x, y])
print(z) # 11.3
Notice how the sum function works without any error this time.
Conclusion
The Python error “‘float’ object is not iterable” occurs when you pass a float object when an iterable is expected.
To fix this error, you need to correct the assignments in your code so that you don’t pass a float in place of an iterable, such as a list or a range.
If you’re using a float in a for loop, you can use the range() and int() functions to convert that float into a range. You can also convert a float into a list by adding square brackets [] around the float objects.
I hope this tutorial is helpful. Happy coding! 👍
Floats and iterables are distinct objects In Python. A float is any decimal point number, and an iterable is an object capable of returning elements one at a time, for example, a list. A float is a single value and does not contain other values. If you try to iterate over a float, you will raise the error “TypeError: ‘float’ object is not iterable“.
To solve this error, ensure you use the range() method, for example,
for number in range(floating_point_number)
to iterate over a range of numbers up to the specified floating_point_number.
This tutorial will go through the error in detail. We will go through two example scenarios and learn how to solve them.
Table of contents
- TypeError: ‘float’ object not iterable
- What is a TypeError?
- Difference Between a Float and an Iterable
- Example #1: Iterate Over a Floating Point Number
- Solution #1: Convert float to string Using the str() Method
- Solution #2: Iterate Using the range() Method
- Example #2: Determine if a Number is Prime
- Solution
- Summary
TypeError: ‘float’ object not iterable
What is a TypeError?
A TypeError occurs when we try to perform an operation on the wrong type of object. For example, if we try to calculate the square root of a list instead of an integer or a floating-point number, then a TypeError will be generated by the Python interpreter.
Difference Between a Float and an Iterable
Iterables are containers that can store multiple values and return them one by one. Iterables can store any number of values, and the values can either be the same type or different types. You can go to the next item in an iterable object using the next() method.
A floating-point number is any number with a decimal point. You can define a floating-point number in Python by defining a variable and assigning a decimal point number to it.
x = 4.2 print(type(x))
class 'float'
Floating-point numbers do not store multiple values like a list or a dictionary. If you try to iterate over a float, you will raise the error “TypeError: ‘float’ object is not iterable” because float does not support iteration.
You will get a similar error if you try to iterate over an integer or a NoneType object.
Example #1: Iterate Over a Floating Point Number
Let’s look at an example where we initialize a floating-point number and iterate over it.
# Define floating point number
float_num = 10.0
# Iterate over the floating point number
for num in float_num:
print(num)
Let’s see what happens when we run the code:
TypeError Traceback (most recent call last)
1 for num in float_num:
2 print(num)
3
TypeError: 'float' object is not iterable
We raise the error because Python does not support iteration on a floating-point number.
Solution #1: Convert float to string Using the str() Method
The first solution involves converting the float_num object to a string using the str() method and iterating over every digit. We can do iteration over a string because the string type is iterable. However, the for loop will return each character of the float_num string.
# Define floating point number
float_num = 10.0
# Iterate over the floating point number
for digit in str(float_num):
print(digit)
Solution #2: Iterate Using the range() Method
To print the range of numbers, we can use the int() method to convert the number to an integer and then use the integer as input to the range() method. The range() method only accepts integer numbers, in which case we have to convert any floating-point number that we want to iterate over to an integer. The range() method returns an iterable object which we can iterate over using a for loop.
# Define floating point number
float_num = 30.0
# Convert the float number to an integer
int_num = int(float_num)
# Iterate over the floating point number
for num in range(int_num):
print(num)
Let’s see what happens when we run the revised code:
0 1 2 3 4 5 6 7 8 9
Example #2: Determine if a Number is Prime
Let’s look at another example where we write a program to check if a number entered by the user is prime or not. Prime numbers are only divisible by themselves, and 1. To start, we will ask the user to insert the number to determine whether it is prime or not.
number = float(input("Enter a number: "))
Then we define a function that determines whether the number is prime by using the modulo operator %. If the remained of x % y equals zero, then y is a factor of x. Prime numbers have two factors, one and itself.
# Function to determine if number is prime or not
def prime_number_calc(num):
for i in num:
# Ensure we do not divide the number by zero or one
if i > 1:
if num % i == 0:
print(f'{num} is not prime')
break
# If the loop runs completely the number is prime
else:
print(f'{num} is a prime number')
prime_number_calc(number)
Let’s run the code to see what happens:
TypeError Traceback (most recent call last)
1 def prime_number_calc(num):
2 for i in num:
3 if i > 1:
4 if num % i == 0:
5 print(f'{num} is not prime')
TypeError: 'float' object is not iterable
The Python interpreter throws the TypeError because a floating-point number is not a suitable type to iterate over. A for loop requires an iterable object to loop over.
Solution
To solve this problem, we need to convert the input number to an integer using int() and pass the integer the range() method instead of the float. Let’s look at the revised code:
def prime_number_calc(num):
for i in range(int(num)):
if i > 1:
if num % i == 0:
print(f'{num} is not prime')
break
else:
print(f'{num} is a prime number')
number = float(input("Enter a number: "))
prime_number_calc(number)
Let’s run the code to see what happens:
Enter a number: 17.0 17.0 is a prime number
Our code successfully prints the result that 17.0 is a prime number.
Summary
Congratulations on reading to the end of this tutorial. The error “TypeError: ‘float’ object is not iterable” occurs when you try to iterate over a floating-point number as if it were an iterable object like a list.
You must use the range() method to iterate over a collection of numbers. However, you must convert the float to an integer beforehand passing it to the range() method. If you want to iterate over the digits of the float, you can convert the float to a string and use the range() function.
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