Whats wrong with this code:
n = 10
((n/3)).is_integer()
I do not understand why I cannot set n = any number and check if it is an integer or not.
Thanks for your help!
python 2.7.4
error:
Traceback (most recent call last):
File "/home/userh/Arbeitsfläche/übung.py", line 2, in <module>
print ((n/3)).is_integer()
AttributeError: 'int' object has no attribute 'is_integer'
The reason you get this error is because you divide the integer 10 by 3 using integer division, getting the integral number 3 in the form of an int instance as a result. You then try to call the method is_integer() on that result but that method is in the float class and not in the int class, just as the error message says.
A quick fix would be to change your code and divide by 3.0 instead of 3 which would result in floating point division and give you a float instance on which you can call the is_integer() method like you are trying to. Do this:
n = 10
((n/3.0)).is_integer()
You are using Python 2.7. Unless you use from __future__ import division, dividing two integers will return you and integer. is_integer exists only in float, hence your error.
the other answers say this but aren't very clear (imho).
in python 2, the / sign means "integer division" when the arguments are integers. that gives you just the integer part of the result:
>>> 10/3
3
which means that in (10/3).is_integer() you are calling is_integer() on 3, which is an integer. and that doesn't work:
>>> (3.0).is_integer()
True
>>> (3).is_integer()
AttributeError: 'int' object has no attribute 'is_integer'
what you probably want is to change one of the numbers to a float:
>>> (10/3.0).is_integer()
False
this is fixed in python 3, by the way (which is the future, and a nicer language in many small ways).
You can use isdigit, this is a good function provided by Python itself
You can refer documentation here https://docs.python.org/2/library/stdtypes.html#str.isdigit
if token.isdigit():
return int(token)
...
When I wrote this answer there was no information about language.
But in python2 you can use the following to check if it's an integer or not
isinstance( <var>, ( int, long ) )
Related
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
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
Stop stomping on round somewhere else by binding an int to it.
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.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
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
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 )
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.
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.
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
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
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
Stop stomping on round somewhere else by binding an int to it.
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.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
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
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 )
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.
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.
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
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
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
Stop stomping on round somewhere else by binding an int to it.
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.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
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
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 )
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.
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.
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
I am learning about decimal type in python when I came under this doubt
Passing str(1/2) to decimal.Decimal() returns Decimal('0.5')
>>>import decimal
>>>decimal.Decimal(str(1/2))
>>>Decimal('0.5')
But when I pass '1/2' as argument it returns error:
>>>import decimal
>>> decimal.Decimal('1/2')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
decimal.InvalidOperation: [<class 'decimal.ConversionSyntax'>]
Can anyone please explain the reason behind this ?
Thanks in advance
When you make str(1/2) is being evaluated to str(0.5) and then to '0.5'. On your second example, passing the string '1/2' returns an error because the evaluation of expressions is not supported on initializing an instance of the Decimal class.
str(1/2) is evaluated to a float (0.5) during compilation, long before it is passed to the Decimal() constructor (which expects a float or a string representation of a float). I doesn't try to evaluate a string that is passed to it.
The point is, when you pass 1/2 to str(), it gets calculated. So str(1/2) makes absolutely no difference from str(0.5).
So the question turns to decimal.Decimal('0.5') and decimal.Decimal('1/2'). When initializing instances of Decimal, the expression is supposed to be convertible to float. Surely the string 1/2 is not the case: float('1/2') gives ValueError.
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
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
Stop stomping on round somewhere else by binding an int to it.
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.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
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
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 )
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.
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.
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