I have declared global variable in function a
def a():
global my_var
my_var = 3
If I want to read this variable from function main, everything works
if __name__ == "__main__":
main()
def main():
print(my_var)
Output:
3
However I have another function b, where the var is not defined
def b():
try:
random_numbers = (random.sample(range(1, my_var), 2))
except Exception as e:
print(e)
Output:
NameError: name 'my_var' is not defined
Could someone explain me, why in function main I can access my_var, while in another it is not defined? How can I solve this problem?
Related
I am working on accessing a variable outside of a function. Here is part of my code:
def main():
trigger_gmm() = 0
log = []
def spatter_tracking_cb(ts, clusters):
global trigger_gmm
for cluster in clusters:
log.append([ts, cluster['id'], int(cluster['x']), int(cluster['y']), int(cluster['width']),
int(cluster['height'])])
if cluster['width'] >= 200:
trigger_gmm = 1
else:
trigger_gmm = 0
print(trigger_gmm)
while True:
print(trigger_gmm)
if trigger_gmm == 1:
print("print something")
if __name__ == "__main__":
main()
I get the output like this:
NameError: name 'trigger_gmm' is not defined
Any ideas would be much appreciated!
You have three issues in that code:
trigger_gmm() = 0 - You need to remove the parenthesis
You need to move the global variable definition up to the beginning of the main function
The if __name__ == "__main__": is not reached as it is after the while loop, you need to move it up.
EDIT:
I added a global declaration to the main module (above the main function) and inside the spatter_tracking_cb function. This is because you need to indicate that the variable trigger_gmm is a global variable whenever you use it.
This code seems to work for me:
global trigger_gmm
def main():
global trigger_gmm
trigger_gmm = 0
log = []
def spatter_tracking_cb(ts, clusters):
global trigger_gmm
for cluster in clusters:
log.append([ts, cluster['id'], int(cluster['x']), int(cluster['y']), int(cluster['width']),
int(cluster['height'])])
if cluster['width'] >= 200:
trigger_gmm = 1
else:
trigger_gmm = 0
print(trigger_gmm)
if __name__ == "__main__":
main()
while True:
print(trigger_gmm)
if trigger_gmm == 1:
print("print something")
trigger_gmm = 0
Remove parenthesis.
You also don't need global trigger_gmm as this variable will be available within scope of main function.
https://realpython.com/python-scope-legb-rule/#nested-functions-the-enclosing-scope
I am no Python user, but from it looks like you are calling variables that are not in a "global" scope.
Every variable defined in a function or loop, is not accessible by another function unless stated so.
Try defining your variable outside of the function, or make it global
As stated before.
W3Schools
def user(choose):
if (choose == "1"):
play = game()
elif (choose == "2"):
return stats(play)
else:
return quit()
I want to take the value from function game() and use it in stats(), but I get an error saying that play is not defined. How do I declare game() and use it in another function?
You could "postpone" execution of func1:
def func1():
return 'abc'
def something_else(callable):
callable()
def main():
hello = None
def f():
"""set result of func1 to variable hello"""
nonlocal hello
hello = func1()
# over here func1 is not executed yet, hello have not got its value
# you could pass function f to some other code and when it is executed,
# it would set result for hello
print(str(hello)) # would print "None"
call_something_else(f)
print(str(hello)) # would print "abc"
main()
After question has changed...
Right now, your local variable play is out of scope for stats.
Also, looks like you expect that function would be called twice.
You need to save play in global content
play = None # let's set it to some default value
def user(choose):
global play # let python know, that it is not local variable
if choose == "1": # no need for extra brackets
play = game()
if choose == "2" and play: # double check that play is set
return stats(play)
return quit()
I have tried to search this but I don't quite understand. I am coming across this error so I formed a quick easy example.
def test():
global a
a = 0
a+=1
def test2():
a+=1
print (a)
inp = input('a?')
if inp == 'a':
test()
test2()
When I input a. I expected the code to output 2. However, I get this error UnboundLocalError: local variable 'a' referenced before assignment. When I searched around about this, I found that you need to use global, but I already am using it.
So I don't understand. Can someone briefly explain what I'm doing wrong?
Thanks.
A global declaration only applies within that function. So the declaration in test() means that uses of the variable a in that function will refer to the global variable. It doesn't have any effect on other functions, so if test2 also wants to access the global variable, you need the same declaration there as well.
def test2():
global a
a += 1
print(a)
1) You can return the modified value like:
def test():
a = 0
a+=1
return a
def test2(a):
a+=1
print (a)
inp = input('a?')
if inp == 'a':
a = test()
test2(a)
2) Or you can use a class:
class TestClass:
a = 0
def test(self):
self.a = 0
self.a+=1
def test2(self):
self.a+=1
print (self.a)
Usage of option 2:
>>> example = TestClass()
>>> example.test()
>>> example.test2()
2
If I try to run the following code:
def func():
a = 5
print 'done'
return a
temp = raw_input('')
if temp == '':
func()
print func()
Say temp is '' and the function is run. It prints done and returns variable a. How can I print the returned variable without running the function once more, so done isn't printed again?
You should assign the returned value to a variable (e.g. a).
Update: you could either print inside the function (version1) or use global variable (version2)
def func():
a = 5
print 'done'
return a
# version 1: if print doesn't have to be outside of the function
def main():
temp = raw_input('')
if temp == '':
local_a = func()
else:
# use else to avoid UnboundLocalError: local variable 'a' referenced
# before assignment
local_a = None
print local_a
if __name__ == "__main__":
main()
# # version 2: if print have to be outside of the function, then I can only
# # think of using global variable, but it's bad.
# global_a = None
# def main():
# temp = raw_input('')
# if temp == '':
# global global_a
# global_a = func()
# if __name__ == "__main__":
# main()
# print global_a
You could use #zyxue's answer above and store the return value to a variable or you could also just not return anything from the function and just assign you final value in a function to a global variable if you have need for that.
I should warn you that it isn't good practice to overuse global variables unnecessarily or overly. See: https://stackoverflow.com/a/19158418/4671205
I have some python 3.4 code that works fine:
def run():
m = 0
while m != 1:
p = input('Please choose p: ')
p = makeInt(p)
#Some other code
print(p)
m = makeInt(input('Enter 1 if you would like to quit: '))
def makeInt(i):
try:
i = int(i)
except ValueError:
i = input('Incorrect input! Enter your answer: ')
i = makeInt(i)
return i
#Some other functions
if __name__ == '__main__':
run()
I want to put all this code in a class (Except possibly if __name__ ==...) When I put all the code including if __name__ ==... in a class like so:
class Foo(object):
def run(self):
m = 0
while m != 1:
p1 = input('Please choose p: ')
p1 = self.makeInt(p1)
#Some other code
print(p1)
m = self.makeInt(input('Enter 1 if you would like to quit: '))
def makeInt(self, i):
try:
i = int(i)
except ValueError:
i = input('Incorrect input! Enter your answer: ')
i = self.makeInt(i)
return i
#Some other functions and stuff
if __name__ == '__main__':
run()
I get the following error: TypeError: run() missing 1 required positional argument: 'self'. When I remove the self argument from run() it runs until makeInt() is called and then I get: NameError: name 'makeInt' is not defined. I get the same error if I take the if statement out of the class and call Foo.run(). I have some other code earlier in this program that works when I call functions of a class from another function in that same class. I realize I don't have to put all my code in a class, but in this case I want to. Why am I getting these errors and what can I do to put my working code in a class?
As others mentioned, by putting your functions in a class, you've made them methods, that means they need an instance of this class as first argument. So you can indeed call your run method using Foo().run() as Foo() will create an instance of Foo.
Another way (e.g. if you don't need the class for anything else than encapsulation) is to make them static, using the staticmethod decorator:
class Foo(object):
#staticmethod
def run():
...
#staticmethod
def makeInt(i):
...
if __name__ == '__main__':
Foo.run() # don't need an instance as run is static
In Python, a method can be static, i.e. no need for any special argument, a class method, i.e. first argument is the class itself, or a standard method, i.e. the first argument is an instance of the class.
Since you wrap your code within a class, your run() is a method now. You should remove your main from your class by unindenting it and initialize an instance of your class:
if __name__ == '__main__':
Foo().run()
It thinks the guard is a part of your class due to the indentation: you have your guard indented to the same level as the other class members. Unindent the
if __name__ == '__main__'
Also change it to be
if __name__ == '__main__':
main()
and then instantiate a new object of type Foo in your newly created main() function
def main():
newFoo = Foo()
newFoo.run()