Hello when I try the next code with sympy I can not cath the exeption of SyntaxError
from sympy.abc import x, y
try:
e =x + y + x)
# except Sympifyerror
except exception:
# except SyntaxError as error:
# except SyntaxError as ex:
raise SyntaxError('Error during evaluation of sympy expression: '
+ str(ex))
print ("errores")Thanks
You cannot catch
SyntaxError, since it is raised prior code execution (i.e. before try and except are executed); If you use
eval, you can be able to catch it, e.g.
from sympy.abc import x, y
code = """
e = x + y + x)
"""
try:
eval(code)
except SyntaxError:
print("Hello")
But when I try I get:
Error:
psm@psm:~/Documentos$ python scriptsympy.py
Traceback (most recent call last):
File "scriptsympy.py", line 18, in <module>
main()
File "scriptsympy.py", line 8, in main
eval(e =x + y + x)
TypeError: eval() takes no keyword arguments
what's that mean?
eval() cannot evaluate a statement with an equal sign in it. The interpreter is understanding "e = ..." as assigning the keyword argument "e" to the value of "x + y + x".
from sympy.abc import x, y
code = """
x + y + x
"""
try:
e = eval(code)
except SyntaxError:
print("Hello")
Hello again, new problems, when I run my code:
#!/usr/bin/python3.6
def main():
from sympy.abc import x, y
code = """
x + y + x
"""
try:
e=eval(code)
except SyntaxError:
print ("errores")
print (e)
if __name__ == "__main__":
main()I've found the following mistake:
Error:
UnboundLocalError: local variable 'e' referenced before assignment
What I'm doing wrong?
Thanks
Ok, exec would be more appropriate in this case.
As far as I understand, we want to raise SyntaxError intentionally, so why I left extra closing parenthesis.
So, variable assignment e =... or extra ) both causes SyntaxError, that is good.
But what about the error message:
UnboundLocalError
Since your code snippet is unformatted, I can just suppose what exactly code you ran.
def main():
from sympy.abc import x, y
code = """
x + y + x
"""
try:
e = eval(code)
except SyntaxError:
print ("errores")
print(e)
if __name__ == "__main__":
main()Error:
UnboundLocalError: local variable 'e' referenced before assignment
This is because the expression to be evaluated (the
code string) includes spaces that causes
IndentationError.
IndentationError, in turn, is subclass of
SyntaxError, so,
print(errors) is executed and variable
e is not being assigned. This is why
UnboundLocalError raised.
Hello, I take out the sapaces:
def main():
from sympy.abc import x, y
code = """
x+y+x
"""
try:
e=eval(code)
except SyntaxError:
print ("errores")
print (e)
if __name__ == "__main__":
main()And I have the same problem, how can I resolve my problem?
Error:
UnboundLocalError: local variable 'e' referenced before assignment
def main():
from sympy.abc import x, y
code = """x+y+x"""
e = None
try:
e = eval(code)
except SyntaxError:
print ("errores")
if e is not None:
print(e)
if __name__ == "__main__":
main()