Python 3.3 PEP 418 example gives 'namespace' object is not iterable error

Question

The beta of Python 3.3 is out and wonderful.

The newly reworked time module has the get_clock_info method to get information about the platform's many logical clocks. PEP 418 describes the new time module.

When I try and run one of the example programs referenced in PEP 418, clock_resolution.py, I get TypeError: 'namespace' object is not iterable on line 54 below:

46 clocks = ['clock', 'perf_counter', 'process_time']
47 if hasattr(time, 'monotonic'):
48     clocks.append('monotonic')
49 clocks.append('time')
50 for name in clocks:
51     func = getattr(time, name)
52     test_clock("%s()" % name, func)
53     info = time.get_clock_info(name)
54     if 'precision' in info:
55         print("- announced precision: %s" % format_duration(info['precision']))
56     print("- implementation: %s" % info['implementation'])
57     print("- resolution: %s" % format_duration(info['resolution']))

'info' on line 53 contains:

>>> info
namespace(adjustable=True, implementation='gettimeofday()', monotonic=False, resolution=1e-06) 

So how does one iterate over a namespace object?

Answer 1

You don't want to iterate the object; you just want to test for presence of an attribute. Two ways:

# "easier to get forgiveness than permission" approach
try:
    print(info.precision)
except AttributeError:
    pass

# "look before you leap" approach
if hasattr(info, "precision"):
    print(info.precision)

The in test is used to check if something is in a dictionary, a list, a tuple, or some other sequence. In the general case, in will try to iterate something to find the value (dict and set are exceptions; Python special-cases them for efficiency). But info is an instance of a class that doesn't support iteration.

You could, if you wished, do this:

# alternate "look before you leap"
if "precision" in info.__dict__:
    print(info.precision)

Attributes actually are stored in a dict instance member variable named .__dict__.

EDIT: @DSM wrote a comment that showed an alternative to the above. The built-in function vars() will return the .__dict__ member variable, so this is equivalent to the above:

# nicer alternate "look before you leap"
if "precision" in vars(info):
    print(info.precision)