Dynamically Load Classes in Python

Question

I'm trying to define a class dynamically and then instantiate it . For this what i'm doing is , i have the class definition as a string and i'm feeding it to the eval function to define it during runtime .

f="""class f:
    def pr(self):
        print "asdfg"
    """
>>> eval(f)

Traceback (most recent call last):
  File "<pyshell#5>", line 1, in <module>
    eval(f)
  File "<string>", line 1
     class f:
         ^
SyntaxError: invalid syntax

Why im doing this is because there are lots of classes that can be used but rarely are they all needed together. Plus i allow user's to add new classes to the application.

Thanks in advance .

Edit -------------------

This is python 2.5 output

data = """class f:
    def pr(self):
        print "asdfg"
    """
>>> exec(data)

Traceback (most recent call last):
  File "<pyshell#14>", line 1, in <module>
    exec(data)
  File "<string>", line 4

   ^
SyntaxError: invalid syntax

#What i think the problem is "\n"
>>> data
'class f:\n    def pr(self):\n        print "asdfg"\n    '

im not sure, but that is where the error is pointing.

the reason for using python 2.5 is that , scapy is a part of the program and it has support for only python 2.5.

If users of your program need to create new classes, you're doing something wrong. — jterrace
– jterrace, Commented Jan 11, 2014 at 3:29
Are you trying to desing some sort of plug-in system, where f would be filled with the content of a Python source file? — ixe013
– ixe013, Commented Jan 11, 2014 at 3:47
@ixe013 : yup , i think the below given solution will work. i actually forgot about it " exec() ". — thecreator232
– thecreator232, Commented Jan 11, 2014 at 4:38
@thecreator232 Hm. Maybe try altering data so that the closing triple quotes """ appear on the end of the third line (so like print "asdfg" """? I don't really know why this is happening, though. — senshin
– senshin, Commented Jan 11, 2014 at 5:12

senshin · Accepted Answer · 2014-01-11 05:15:36Z

4

Use exec, not eval. exec will execute arbitrary code, while eval will only evaluate expressions (e.g. eval('2+3') == 5).

>>> data = """class f:
    def pr(self):
        print "asdfg" """
>>> exec data
>>> instance = f()
>>> instance.pr()
asdfg

There is probably a better way to achieve what you want to do, though. Having users of your program create new classes seems like a suboptimal design choice.

edited Jan 11, 2014 at 5:15

answered Jan 11, 2014 at 3:42

senshin

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5 Comments

thecreator232 Over a year ago

Thanx ! i forgot about it "exec()".

thecreator232 Over a year ago

i do know that it isn't the best way to do it, but its the only way i currently know . Pls share what are the other ways of doing it .

thecreator232 Over a year ago

does the above given method work in Python 2.5 also ?

senshin Over a year ago

@thecreator232 I believe that the functionality of exec is essentially unchanged from Python 2.4 forward (so I think it's the same in Python 2.5), but I don't have an install of Python 2.5 available to test this.

senshin Over a year ago

@thecreator232 Please edit that into your question so that it can be properly formatted as code.

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