Isinstance() doesn't work

Question

For some reason I can't get isinstance() to work on Python 2.7.2

def print_lol(the_list, indent=False, level=0):
    for item in the_list:
        if isinstance(item, list):
            print_lol(item, indent, level+1)
        else:
            print(item)

And when I compile and run it:

>>> list = ["q", "w", ["D", ["E", "I"]]]
>>> print_lol(list)

I get the error message:

if isinstance(item, list):
TypeError: isinstance() arg 2 must be a class, type, or tuple of classes and types

What am I mising?

Everybody makes a list called list in almost every python problem on SO... — jamylak
– jamylak, Commented Apr 14, 2012 at 6:10

agf · Accepted Answer · 2012-04-14 06:14:25Z

8

You've named your variable list:

>>> list = ["q", "w", ["D", ["E", "I"]]]

Which is hiding the built-in list.

Once you've renamed your variable, restart Python (or your IDE). Since your list is a global variable, it will stick around otherwise.

edited Apr 14, 2012 at 6:14

answered Apr 14, 2012 at 6:09

agf

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3 Comments

torek Over a year ago

You can also re-bind list manually: import __builtin__; list = __builtin__.list will do the trick in this case.

Chris Morgan Over a year ago

@torek: that's not the ideal way; the best way is del list. Then list will go back to being the built-in type.

torek Over a year ago

@ChrisMorgan: interesting, never knew that before. Doesn't work with other (non-builtins) overwrites that you want to recover from in the interpreter, of course. :-)

Abhijit · Accepted Answer · 2012-04-14 06:15:01Z

2

This is where you have issue

>>> list = ["q", "w", ["D", ["E", "I"]]]

overwriting python named types like list binds the variable name with a new object instance. So later when you tried isinstance with list it failed.

When ever you create new variable, please refrain from naming it differently from built-in and not to conflict with the namespace.

In this example using the following would work like a breeze

>>> mylist = [w for w in mylist if len(w) >= 3 and diff(w)]
>>> isinstance(mylist,list)
True

Please note, if you have polluted the namespace and you are running in IDLE, restarting IDLE is a good alternative.

edited Apr 14, 2012 at 6:15

answered Apr 14, 2012 at 6:09

Abhijit

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3 Comments

agf Over a year ago

It doesn't "bind a new behavior to an existing type". The type still functions exactly the same, the name just doesn't point to the type any more.

Abhijit Over a year ago

@agf: Ahh typo I mean bind to a variable

agf Over a year ago

It doesn't bind anything to something existing. It creates a name, list, in the global namespace, which hides the list name in the builtin namespace.

Mark Byers · Accepted Answer · 2012-04-14 06:09:10Z

1

The problem is you have shadowed the built-in list by assigning to a variable called list:

list = ["q", "w", ["D", ["E", "I"]]]

You should try to avoid using any of the built-in names for variables, because this will often result in confusing errors. The solution is to use a different name for your variable.

answered Apr 14, 2012 at 6:09

Mark Byers

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Comments

Shep · Accepted Answer · 2012-04-14 06:23:09Z

1

You are overwriting list, try print list before isinstance to diagnose this sort of problem.

edited Apr 14, 2012 at 6:23

answered Apr 14, 2012 at 6:10

Shep

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