Unladen Swallow
Dec-27-2019, 05:23 AM https://docs.python.org/3/faq/programming.html#why-am-i-getting-an-unboundlocalerror-when-the-variable-has-a-value
I used both global , nonlocal but both throws different error
throws this error
nonlocal test_var ^
SyntaxError: invalid syntax
throws this error ^
NameError: global name 'test_var' is not defined
Any suggestions ? I have gone through top 10 search results , all suggest to use nonlocal or global...
Dec-27-2019, 09:08 AM
I think that this boils down to this:
You should also understand difference between reference and assignment:
When you reference a variable in an expression, the Python interpreter will traverse the scope to resolve the reference in following order:
- current function’s scope
- any enclosing scopes (like containing functions)
- scope of the module that contains the code (global scope)
- built-in scope (that contains functions like int and abs)
If Python doesn't find defined variable with the referenced name, then a NameError exception is raised.
Assigning a value to a variable works differently. If the variable is already defined in the current scope, then it will just take on the new value. If the variable doesn’t exist in the current scope, then Python treats the assignment as a variable definition. The scope of the newly defined variable is the function that contains the assignment. Bucky Katt, Get Fuzzy
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UnboundLocalError: local variable referenced before assignment
I have following simple function to get percent values for different cover types from a raster. It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment
which isn't clear to me. Any suggestions?
- arcgis-10.1
- unboundlocalerror
- 1 Because if row.getValue("Value") == 1 might be false and so a never gets assigned. – Nathan W Commented May 20, 2013 at 2:39
- It has value and do gets assigned. I checked it in arcmap interactive python window but can't get it to work in a stand alone script. – Ibe Commented May 20, 2013 at 2:44
- 1 your loop will also only give you the values of the last loop iteration as you are returning out of the loop and not doing anything with each value. – Nathan W Commented May 20, 2013 at 2:49
- You could use 3 x elif and an else to see if any values other than 1-4 are encountered. – PolyGeo ♦ Commented May 20, 2013 at 3:44
- I tried that way as well but still hung up with error. – Ibe Commented May 20, 2013 at 4:15
This error is pretty much explained here and it helped me to get assignments and return values for all variables.
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having a problem with error code: UnboundLocalError: local variable 'player' referenced before assignment
hi hoping someone here can actually help me. I asked this question on stack overflow and they closed it and redirected to similar previously asked questions. none of which actually helped me to understand how to correct the error code.
I am following a video tutorial series for learning to code with python/pygame. and i have checked my coding aginst the source material and they are the same, because i typed it as was told to in the video.
here's the full tracebackback including the error:
File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 160, in <module>
File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 152, in main
GameInterface(num_player=1, screen=screen)
File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 82, in GameInterface
if player.cooling_time > 0:
UnboundLocalError: local variable 'player' referenced before assignment.
here's the actual code:
if player.cooling_time > 0: player.cooling_time -= 1
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Help with Python UnboundLocalError: local variable referenced before assignment
I have post the similar question before,however,I think I may have misinterpreted my question,so may I just post my origin code here,and looking for someone can help me,I am really stuck now..thanks alot.
when i call force python gives:
can anyone give me some tips?thanks......
- 2 The use of "from numpy import *" is a bad practice. It pollutes the global namespace. "import numpy as np" is better. If you have specific functions you use a lot, and you are tired of writing np.sin(), np.cos, etc, you should import those specifically ("from numpy import sin"). cheers. – Ofri Raviv Commented Nov 14, 2009 at 19:51
5 Answers 5
where do you call force in this code?
In any event, the problem is in update_r. You reference vs in the first line of update_r even though vs is not defined in this function. Python is not looking at the vs defined above. Try adding
as the first line of update_r or adding vs to the parameter list for update_r
In the first line of update_r , you have vs,ve=update_v(n,vs,ve,ms,me,dt,fs,fe) . Look at the function that you are calling. You are calling update_v with a bunch of parameters. One of these parameters is vs . However, that is the first time in that function that vs appears. The variable vs does not have a value associated with it yet. Try initializing it first, and your error should disappear
- 1 there is a global definition of vs. normally, it would have been used. the only reason that it's not, is the vs in the left hand side of the assignment. that defines an unintialized local variable with the same name. – Ofri Raviv Commented Nov 14, 2009 at 19:43
- I did intend to convey that, but perhaps I should have worded my answer better. Sorry for the miscommunicaton – inspectorG4dget Commented Nov 14, 2009 at 19:48
- thanks for replay,however,in my understanding(newb's),is that vs on the left hand side doesnt not really matter if it has the sam name with one of the global variables, it is local.please help me with understanding this – user211037 Commented Nov 14, 2009 at 23:43
- Okay, so you're assigning a local variable called vs to the return value of some computation on the global variable vs. While you and I understand this distinction, the Python interpreter does not. When you create the local vs (which is being assigned), then the next call to vs will be to the local vs. So what happens is that the Python interpreter is trying to perform the computation on the local vs (which has no value yet). I would fix it by doing this: temp_vs,ve=update_v(n,vs,ve,ms,me,dt,fs,fe) vs = temp_vs This should fix your problem. Also, do what foosion says (before the vs init) – inspectorG4dget Commented Nov 15, 2009 at 0:36
Put an additional global statement containing all your globals after each def statement. Otherwise, all globals are transformed into locals within your def without it.
On line 39 you do
while you are inside a function. Since you defined a global variable called vs , you would expect this to work. It would have worked if you had:
because then the interpreter knows vs in the function arguments is the global one. But since you had vs in the left hand side, you created an uninitialized local variable.
But dude, you have a much bigger problem in your code : update_r calls update_v, update_v calls force, and force calls update_r - you will get a stack overflow :)
- thanks for replay..the thing is how can I solve this bigger problem,that is actually the reason I post this,it does really make me feel bad,please help me.. – user211037 Commented Nov 14, 2009 at 23:39
I got that error when my class name was assigned to a variable that is called exactly like its name. example ClassName = ClassName. You may do this if you come from .Net
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File "weird.py", line 5, in main. print f(3) UnboundLocalError: local variable 'f' referenced before assignment. Python sees the f is used as a local variable in [f for f in [1, 2, 3]], and decides that it is also a local variable in f(3). You could add a global f statement: def f(x): return x. def main():
Output. Hangup (SIGHUP) Traceback (most recent call last): File "Solution.py", line 7, in <module> example_function() File "Solution.py", line 4, in example_function x += 1 # Trying to modify global variable 'x' without declaring it as global UnboundLocalError: local variable 'x' referenced before assignment Solution for Local variable Referenced Before Assignment in Python
Actually, every situation with this bug is the same: you are defining a variable in a global context, referencing it in a local context, and then modifying it later in that context. When Python interprets your function, it identifies all variables you modify in the function and creates local versions of them.
Avoid Reassignment of Global Variables. Below, code calculates a new value (local_var) based on the global variable and then prints both the local and global variables separately.It demonstrates that the global variable is accessed directly without being reassigned within the function.
The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable. To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function. I hope this tutorial is useful.
The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function. To solve the error, mark the variable as global in the function definition, e.g. global my_var .
UnboundLocalError: local variable referenced before assignment. Example #1: Accessing a Local Variable. Solution #1: Passing Parameters to the Function. Solution #2: Use Global Keyword. Example #2: Function with if-elif statements. Solution #1: Include else statement. Solution #2: Use global keyword. Summary.
2 Solutions for the Problem. 2.1 Method 1: Initializing the Variable. 2.2 Method 2: Using Global Variables. 2.3 Method 3: Using Nonlocal Variables.
Trying to assign a value to a variable that does not have local scope can result in this error: UnboundLocalError: local variable referenced before assignment. Python has a simple rule to determine the scope of a variable. If a variable is assigned in a function, that variable is local. This is because it is assumed that when you define a ...
Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable declaration have been explained, and multiple solutions regarding the issue have been provided.
What is UnboundLocalError: local variable referenced before assignment? Trying to assign a value to a variable that does not have local scope can result in this error: 1 UnboundLocalError: local variable referenced before assignment. Python has a simple rule to determine the scope of a variable.
value = value + 1 print (value) increment() If you run this code, you'll get. BASH. UnboundLocalError: local variable 'value' referenced before assignment. The issue is that in this line: PYTHON. value = value + 1. We are defining a local variable called value and then trying to use it before it has been assigned a value, instead of using the ...
To fix this, you can either move the assignment of the variable x before the print statement, or give it an initial value before the print statement. def example (): x = 5 print (x) example()
Assigning a value to a variable works differently. If the variable is already defined in the current scope, then it will just take on the new value. If the variable doesn't exist in the current scope, then Python treats the assignment as a variable definition. The scope of the newly defined variable is the function that contains the assignment.
The Unboundlocalerror: local variable referenced before assignment is raised when you try to use a variable before it has been assigned in the local context. Python doesn't have variable declarations , so it has to figure out the scope of variables itself. It does so by a simple rule: If there is an assignment to a variable inside a function ...
This is a piece of code i made for a project to mimic an ATM like machine, this is just the start of it (inputting the pin and accessing balance and other things) . and im wondering why it all works until i input the wrong integer for the pin, thats when it gives me 'UnboundLocalError: local variable 'attempts' referenced before assignment' any ideas?
I have following simple function to get percent values for different cover types from a raster. It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment whic...
UnboundLocalError: local variable 'player' referenced before assignment. here's the actual code: if player.cooling_time > 0: player.cooling_time -= 1 Share Sort by: Best. Open comment sort options ... UnboundLocalError: local variable 'player' referenced before assignment Reply reply
Okay, so you're assigning a local variable called vs to the return value of some computation on the global variable vs. While you and I understand this distinction, the Python interpreter does not. When you create the local vs (which is being assigned), then the next call to vs will be to the local vs.