ValueError: invalid literal for int () with base 10
Question:
I wrote a program to solve y = a^x and then project it on a graph. The problem is that whenever a < 1 I get the error:
ValueError: invalid literal for int () with base 10.
Any suggestions?
Here’s the traceback:
Traceback (most recent call last):
File "C:UserskasutajaDesktopEksponentfunktsioonTEST - koopia.py", line 13, in <module>
if int(a) < 0:
ValueError: invalid literal for int() with base 10: '0.3'
The problem arises every time I put a number that is smaller than one, but larger than 0. For this example it was 0.3 .
This is my code:
# y = a^x
import time
import math
import sys
import os
import subprocess
import matplotlib.pyplot as plt
print ("y = a^x")
print ("")
a = input ("Enter 'a' ")
print ("")
if int(a) < 0:
print ("'a' is negative, no solution")
elif int(a) == 1:
print ("'a' is equal with 1, no solution")
else:
fig = plt.figure ()
x = [-2,-1.75,-1.5,-1.25,-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1,1.25,1.5,1.75,2]
y = [int(a)**(-2),int(a)**(-1.75),int(a)**(-1.5),int(a)**(-1.25),
int(a)**(-1),int(a)**(-0.75),int(a)**(-0.5),int(a)**(-0.25),
int(a)**(0),int(a)**(0.25),int(a)**(0.5),int(a)**(0.75),
int(a)**1,int(a)**(1.25),int(a)**(1.5),int(a)**(1.75), int(a)**(2)]
ax = fig.add_subplot(1,1,1)
ax.set_title('y = a**x')
ax.plot(x,y)
ax.spines['left'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_position('zero')
ax.spines['top'].set_color('none')
ax.spines['left'].set_smart_bounds(True)
ax.spines['bottom'].set_smart_bounds(True)
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
plt.savefig("graph.png")
subprocess.Popen('explorer "C:\Users\kasutaja\desktop\graph.png"')
def restart_program():
python = sys.executable
os.execl(python, python, * sys.argv)
if __name__ == "__main__":
answer = input("Restart program? ")
if answer.strip() in "YES yes Yes y Y".split():
restart_program()
else:
os.remove("C:\Users\kasutaja\desktop\graph.png")
Answers:
It might be better to validate a right when it is input.
try:
a = int(input("Enter 'a' "))
except ValueError:
print('PLease input a valid integer')
This either casts a to an int so you can be assured that it is an integer for all later uses or it handles the exception and alerts the user
Answer:
Your traceback is telling you that int() takes integers, you are trying to give a decimal, so you need to use float():
a = float(a)
This should work as expected:
>>> int(input("Type a number: "))
Type a number: 0.3
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '0.3'
>>> float(input("Type a number: "))
Type a number: 0.3
0.3
Computers store numbers in a variety of different ways. Python has two main ones. Integers, which store whole numbers (ℤ), and floating point numbers, which store real numbers (ℝ). You need to use the right one based on what you require.
(As a note, Python is pretty good at abstracting this away from you, most other language also have double precision floating point numbers, for instance, but you don’t need to worry about that. Since 3.0, Python will also automatically convert integers to floats if you divide them, so it’s actually very easy to work with.)
Previous guess at answer before we had the traceback:
Your problem is that whatever you are typing is can’t be converted into a number. This could be caused by a lot of things, for example:
>>> int(input("Type a number: "))
Type a number: -1
-1
>>> int(input("Type a number: "))
Type a number: - 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '- 1'
Adding a space between the - and 1 will cause the string not to be parsed correctly into a number. This is, of course, just an example, and you will have to tell us what input you are giving for us to be able to say for sure what the issue is.
Advice on code style:
y = [int(a)**(-2),int(a)**(-1.75),int(a)**(-1.5),int(a)**(-1.25),
int(a)**(-1),int(a)**(-0.75),int(a)**(-0.5),int(a)**(-0.25),
int(a)**(0),int(a)**(0.25),int(a)**(0.5),int(a)**(0.75),
int(a)**1,int(a)**(1.25),int(a)**(1.5),int(a)**(1.75), int(a)**(2)]
This is an example of a really bad coding habit. Where you are copying something again and again something is wrong. Firstly, you use int(a) a ton of times, wherever you do this, you should instead assign the value to a variable, and use that instead, avoiding typing (and forcing the computer to calculate) the value again and again:
a = int(a)
In this example I assign the value back to a, overwriting the old value with the new one we want to use.
y = [a**i for i in x]
This code produces the same result as the monster above, without the masses of writing out the same thing again and again. It’s a simple list comprehension. This also means that if you edit x, you don’t need to do anything to y, it will naturally update to suit.
Also note that PEP-8, the Python style guide, suggests strongly that you don’t leave spaces between an identifier and the brackets when making a function call.
As Lattyware said, there is a difference between Python2 & Python3 that leads to this error:
With Python2, int(str(5/2)) gives you 2.
With Python3, the same gives you: ValueError: invalid literal for int() with base 10: ‘2.5’
If you need to convert some string that could contain float instead of int, you should always use the following ugly formula:
int(float(myStr))
As float('3.0') and float('3') give you 3.0, but int('3.0') gives you the error.
int() casting can’t handle string numbers that have decimal points
– example –> int(‘13.5′) will give you error , but int(’13’) will convert the
string to integer
Why : This considered as explicit casting required by the user as it prevents you from losing information like 0.5 if you read dataset and don’t know it’s had floating-point numbers
Work Around >
int(Float("13.5"))
A real-world example I faced: where I wanted the numbers as int while int(I["mpg"]) directly didn’t work so I used float() then int()
sum([int(float(i["mpg"])) for i in file])//len(file)
I wrote a program to solve y = a^x and then project it on a graph. The problem is that whenever a < 1 I get the error:
ValueError: invalid literal for int () with base 10.
Any suggestions?
Here’s the traceback:
Traceback (most recent call last):
File "C:UserskasutajaDesktopEksponentfunktsioonTEST - koopia.py", line 13, in <module>
if int(a) < 0:
ValueError: invalid literal for int() with base 10: '0.3'
The problem arises every time I put a number that is smaller than one, but larger than 0. For this example it was 0.3 .
This is my code:
# y = a^x
import time
import math
import sys
import os
import subprocess
import matplotlib.pyplot as plt
print ("y = a^x")
print ("")
a = input ("Enter 'a' ")
print ("")
if int(a) < 0:
print ("'a' is negative, no solution")
elif int(a) == 1:
print ("'a' is equal with 1, no solution")
else:
fig = plt.figure ()
x = [-2,-1.75,-1.5,-1.25,-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1,1.25,1.5,1.75,2]
y = [int(a)**(-2),int(a)**(-1.75),int(a)**(-1.5),int(a)**(-1.25),
int(a)**(-1),int(a)**(-0.75),int(a)**(-0.5),int(a)**(-0.25),
int(a)**(0),int(a)**(0.25),int(a)**(0.5),int(a)**(0.75),
int(a)**1,int(a)**(1.25),int(a)**(1.5),int(a)**(1.75), int(a)**(2)]
ax = fig.add_subplot(1,1,1)
ax.set_title('y = a**x')
ax.plot(x,y)
ax.spines['left'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_position('zero')
ax.spines['top'].set_color('none')
ax.spines['left'].set_smart_bounds(True)
ax.spines['bottom'].set_smart_bounds(True)
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
plt.savefig("graph.png")
subprocess.Popen('explorer "C:\Users\kasutaja\desktop\graph.png"')
def restart_program():
python = sys.executable
os.execl(python, python, * sys.argv)
if __name__ == "__main__":
answer = input("Restart program? ")
if answer.strip() in "YES yes Yes y Y".split():
restart_program()
else:
os.remove("C:\Users\kasutaja\desktop\graph.png")
It might be better to validate a right when it is input.
try:
a = int(input("Enter 'a' "))
except ValueError:
print('PLease input a valid integer')
This either casts a to an int so you can be assured that it is an integer for all later uses or it handles the exception and alerts the user
Answer:
Your traceback is telling you that int() takes integers, you are trying to give a decimal, so you need to use float():
a = float(a)
This should work as expected:
>>> int(input("Type a number: "))
Type a number: 0.3
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '0.3'
>>> float(input("Type a number: "))
Type a number: 0.3
0.3
Computers store numbers in a variety of different ways. Python has two main ones. Integers, which store whole numbers (ℤ), and floating point numbers, which store real numbers (ℝ). You need to use the right one based on what you require.
(As a note, Python is pretty good at abstracting this away from you, most other language also have double precision floating point numbers, for instance, but you don’t need to worry about that. Since 3.0, Python will also automatically convert integers to floats if you divide them, so it’s actually very easy to work with.)
Previous guess at answer before we had the traceback:
Your problem is that whatever you are typing is can’t be converted into a number. This could be caused by a lot of things, for example:
>>> int(input("Type a number: "))
Type a number: -1
-1
>>> int(input("Type a number: "))
Type a number: - 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '- 1'
Adding a space between the - and 1 will cause the string not to be parsed correctly into a number. This is, of course, just an example, and you will have to tell us what input you are giving for us to be able to say for sure what the issue is.
Advice on code style:
y = [int(a)**(-2),int(a)**(-1.75),int(a)**(-1.5),int(a)**(-1.25),
int(a)**(-1),int(a)**(-0.75),int(a)**(-0.5),int(a)**(-0.25),
int(a)**(0),int(a)**(0.25),int(a)**(0.5),int(a)**(0.75),
int(a)**1,int(a)**(1.25),int(a)**(1.5),int(a)**(1.75), int(a)**(2)]
This is an example of a really bad coding habit. Where you are copying something again and again something is wrong. Firstly, you use int(a) a ton of times, wherever you do this, you should instead assign the value to a variable, and use that instead, avoiding typing (and forcing the computer to calculate) the value again and again:
a = int(a)
In this example I assign the value back to a, overwriting the old value with the new one we want to use.
y = [a**i for i in x]
This code produces the same result as the monster above, without the masses of writing out the same thing again and again. It’s a simple list comprehension. This also means that if you edit x, you don’t need to do anything to y, it will naturally update to suit.
Also note that PEP-8, the Python style guide, suggests strongly that you don’t leave spaces between an identifier and the brackets when making a function call.
As Lattyware said, there is a difference between Python2 & Python3 that leads to this error:
With Python2, int(str(5/2)) gives you 2.
With Python3, the same gives you: ValueError: invalid literal for int() with base 10: ‘2.5’
If you need to convert some string that could contain float instead of int, you should always use the following ugly formula:
int(float(myStr))
As float('3.0') and float('3') give you 3.0, but int('3.0') gives you the error.
int() casting can’t handle string numbers that have decimal points
– example –> int(‘13.5′) will give you error , but int(’13’) will convert the
string to integer
Why : This considered as explicit casting required by the user as it prevents you from losing information like 0.5 if you read dataset and don’t know it’s had floating-point numbers
Work Around >
int(Float("13.5"))
A real-world example I faced: where I wanted the numbers as int while int(I["mpg"]) directly didn’t work so I used float() then int()
sum([int(float(i["mpg"])) for i in file])//len(file)