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Check Names pointing to Python Object

How can I get a list of all the names that point to a Python object?
import my_function from example
a = my_function
b = my_function
get_names(my_function)
[a, b]
Edit: The goal is to help find how to monkey patch an object that is loaded in an unknown way.

Search the global namespace for objects matching via identity, and report the keys (names).
def my_func():
pass
a = my_func
b = my_func
def get_names(x):
for k, v in globals().items():
if v is x:
yield k
print(list(get_names(my_func))) #prints ['my_func', 'a', 'b']

See below (use globals and make sure you do not return the function itself)
from example import my_function
def get_names(func):
result = []
for k,v in globals().items():
if v == func and k not in str(v).split():
result.append(k)
return result
def foo():
pass
a = my_function
b = my_function
c = foo
print(get_names(my_function))
example.py
def my_function():
pass
output
['a','b']

How to find the name of a min (or max) attribute?

Given three or more variables, I want to find the name of the variable with the min value.
I can get the min value from the list, and I can get the index within the list of the min value. But I want the variable name.
I feel like there's another way to go about this that I'm just not thinking of.
a = 12
b = 9
c = 42
cab = [c,a,b]
# yields 9 (the min value)
min(cab)
# yields 2 (the index of the min value)
cab.index(min(cab))
What code would yield 'b'?

The magic of vars prevents you from having to make a dictionary up front if you want to have things in instance variables:
class Foo():
def __init__(self, a, b, c):
self.a = a
self.b = b
self.c = c
def min_name(self, names = None):
d = vars(self)
if not names:
names = d.keys()
key_min = min(names, key = (lambda k: d[k]))
return key_min
In action
>>> x = Foo(1,2,3)
>>> x.min_name()
'a'
>>> x.min_name(['b','c'])
'b'
>>> x = Foo(5,1,10)
>>> x.min_name()
'b'
Right now it'll crash if you pass an invalid variable name in the parameter list for min_name, but that's resolvable.
You can also update the dictionary and it's reflected in the source
def increment_min(self):
key = self.min_name()
vars(self)[key] += 1
Example:
>>> x = Foo(2,3,4)
>>> x.increment_min()
>>> x.a
3

You cannot get the name of the variable with the minimum/maximum value like this*, since as #jasonharper commented: cab is nothing more than a list containing three integers; there is absolutely no connection to the variables that those integers originally came from.
A simple workaround is to user pairs, like this:
>>> pairs = [("a", 12), ("b", 9), ("c", 42)]
>>> min(pairs)
('b', 9)
>>> min(pairs)[0]
'b'
See Green Cloak Guy's answer, but if you want to go for readability, I suggest following a similar approach to mine.

You'd have to get very creative for this to work, and the only solution I can think of is rather inefficient.
You can get the memory address of the data b refers to fairly easily:
>>> hex(id(b))
'0xaadd60'
>>> hex(id(cab[2]))
'0xaadd60'
To actually correspond that with a variable name, though, the only way to do that would be to look through the variables and find the one that points to the right place.
You can do this by using the globals() function:
# get a list of all the variable names in the current namespace that reference your desired value
referent_vars = [k for k,v in globals().items() if id(v) == id(cab[2])]
var_name = referent_vars[0]
There are two big problems with this solution:
Namespaces - you can't put this code in a function, because if you do that and then call it from another function, then it won't work.
Time - this requires searching through the entire global namespace.
The first problem could be alleviated by additionally passing the current namespace in as a variable:
def get_referent_vars(val, globals):
return [k for k,v in globals.items() if id(v) == id(val)]
def main():
a = 12
b = 9
c = 42
cab = [a, b, c]
var_name = get_referent_vars(
cab[cab.index(min(cab))],
globals()
)[0]
print(var_name)
# should print 'b'

convert an object to string in python [duplicate]

I would like to be able to get the name of a variable as a string but I don't know if Python has that much introspection capabilities. Something like:
>>> print(my_var.__name__)
'my_var'
I want to do that because I have a bunch of variables I'd like to turn into a dictionary like :
bar = True
foo = False
>>> my_dict = dict(bar=bar, foo=foo)
>>> print my_dict
{'foo': False, 'bar': True}
But I'd like something more automatic than that.
Python have locals() and vars(), so I guess there is a way.

As unwind said, this isn't really something you do in Python - variables are actually name mappings to objects.
However, here's one way to try and do it:
>>> a = 1
>>> for k, v in list(locals().iteritems()):
if v is a:
a_as_str = k
>>> a_as_str
a
>>> type(a_as_str)
'str'

I've wanted to do this quite a lot. This hack is very similar to rlotun's suggestion, but it's a one-liner, which is important to me:
blah = 1
blah_name = [ k for k,v in locals().iteritems() if v is blah][0]
Python 3+
blah = 1
blah_name = [ k for k,v in locals().items() if v is blah][0]

Are you trying to do this?
dict( (name,eval(name)) for name in ['some','list','of','vars'] )
Example
>>> some= 1
>>> list= 2
>>> of= 3
>>> vars= 4
>>> dict( (name,eval(name)) for name in ['some','list','of','vars'] )
{'list': 2, 'some': 1, 'vars': 4, 'of': 3}

This is a hack. It will not work on all Python implementations distributions (in particular, those that do not have traceback.extract_stack.)
import traceback
def make_dict(*expr):
(filename,line_number,function_name,text)=traceback.extract_stack()[-2]
begin=text.find('make_dict(')+len('make_dict(')
end=text.find(')',begin)
text=[name.strip() for name in text[begin:end].split(',')]
return dict(zip(text,expr))
bar=True
foo=False
print(make_dict(bar,foo))
# {'foo': False, 'bar': True}
Note that this hack is fragile:
make_dict(bar,
foo)
(calling make_dict on 2 lines) will not work.
Instead of trying to generate the dict out of the values foo and bar,
it would be much more Pythonic to generate the dict out of the string variable names 'foo' and 'bar':
dict([(name,locals()[name]) for name in ('foo','bar')])

This is not possible in Python, which really doesn't have "variables". Python has names, and there can be more than one name for the same object.

I think my problem will help illustrate why this question is useful, and it may give a bit more insight into how to answer it. I wrote a small function to do a quick inline head check on various variables in my code. Basically, it lists the variable name, data type, size, and other attributes, so I can quickly catch any mistakes I've made. The code is simple:
def details(val):
vn = val.__name__ # If such a thing existed
vs = str(val)
print("The Value of "+ str(vn) + " is " + vs)
print("The data type of " + vn + " is " + str(type(val)))
So if you have some complicated dictionary / list / tuple situation, it would be quite helpful to have the interpreter return the variable name you assigned. For instance, here is a weird dictionary:
m = 'abracadabra'
mm=[]
for n in m:
mm.append(n)
mydic = {'first':(0,1,2,3,4,5,6),'second':mm,'third':np.arange(0.,10)}
details(mydic)
The Value of mydic is {'second': ['a', 'b', 'r', 'a', 'c', 'a', 'd', 'a', 'b', 'r', 'a'], 'third': array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9.]), 'first': [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]}
The data type of mydic is <type 'dict'>
details(mydic['first'])
The Value of mydic['first'] is (0, 1, 2, 3, 4, 5, 6)]
The data type of mydic['first'] is <type 'list'>
details(mydic.keys())
The Value of mydic.keys() is ['second', 'third', 'first']
The data type of mydic.keys() is <type 'tuple'>
details(mydic['second'][0])
The Value of mydic['second'][0] is a
The data type of mydic['second'][0] is <type 'str'>
I'm not sure if I put this in the right place, but I thought it might help. I hope it does.

I wrote a neat little useful function based on the answer to this question. I'm putting it here in case it's useful.
def what(obj, callingLocals=locals()):
"""
quick function to print name of input and value.
If not for the default-Valued callingLocals, the function would always
get the name as "obj", which is not what I want.
"""
for k, v in list(callingLocals.items()):
if v is obj:
name = k
print(name, "=", obj)
usage:
>> a = 4
>> what(a)
a = 4
>>|

I find that if you already have a specific list of values, that the way described by #S. Lotts is the best; however, the way described below works well to get all variables and Classes added throughout the code WITHOUT the need to provide variable name though you can specify them if you want. Code can be extend to exclude Classes.
import types
import math # mainly showing that you could import what you will before d
# Everything after this counts
d = dict(globals())
def kv_test(k,v):
return (k not in d and
k not in ['d','args'] and
type(v) is not types.FunctionType)
def magic_print(*args):
if len(args) == 0:
return {k:v for k,v in globals().iteritems() if kv_test(k,v)}
else:
return {k:v for k,v in magic_print().iteritems() if k in args}
if __name__ == '__main__':
foo = 1
bar = 2
baz = 3
print magic_print()
print magic_print('foo')
print magic_print('foo','bar')
Output:
{'baz': 3, 'foo': 1, 'bar': 2}
{'foo': 1}
{'foo': 1, 'bar': 2}

In python 3 this is easy
myVariable = 5
for v in locals():
if id(v) == id("myVariable"):
print(v, locals()[v])
this will print:
myVariable 5

Python3. Use inspect to capture the calling local namespace then use ideas presented here. Can return more than one answer as has been pointed out.
def varname(var):
import inspect
frame = inspect.currentframe()
var_id = id(var)
for name in frame.f_back.f_locals.keys():
try:
if id(eval(name)) == var_id:
return(name)
except:
pass

Here's the function I created to read the variable names. It's more general and can be used in different applications:
def get_variable_name(*variable):
'''gets string of variable name
inputs
variable (str)
returns
string
'''
if len(variable) != 1:
raise Exception('len of variables inputed must be 1')
try:
return [k for k, v in locals().items() if v is variable[0]][0]
except:
return [k for k, v in globals().items() if v is variable[0]][0]
To use it in the specified question:
>>> foo = False
>>> bar = True
>>> my_dict = {get_variable_name(foo):foo,
get_variable_name(bar):bar}
>>> my_dict
{'bar': True, 'foo': False}

In reading the thread, I saw an awful lot of friction. It's easy enough to give
a bad answer, then let someone give the correct answer. Anyway, here is what I found.
From: [effbot.org] (http://effbot.org/zone/python-objects.htm#names)
The names are a bit different — they’re not really properties of the object, and the object itself doesn't know what it’s called.
An object can have any number of names, or no name at all.
Names live in namespaces (such as a module namespace, an instance namespace, a function’s local namespace).
Note: that it says the object itself doesn’t know what it’s called, so that was the clue. Python objects are not self-referential. Then it says, Names live in namespaces. We have this in TCL/TK. So maybe my answer will help (but it did help me)
jj = 123
print eval("'" + str(id(jj)) + "'")
print dir()
166707048
['__builtins__', '__doc__', '__file__', '__name__', '__package__', 'jj']
So there is 'jj' at the end of the list.
Rewrite the code as:
jj = 123
print eval("'" + str(id(jj)) + "'")
for x in dir():
print id(eval(x))
161922920
['__builtins__', '__doc__', '__file__', '__name__', '__package__', 'jj']
3077447796
136515736
3077408320
3077656800
136515736
161922920
This nasty bit of code id's the name of variable/object/whatever-you-pedantics-call-it.
So, there it is. The memory address of 'jj' is the same when we look for it directly, as when we do the dictionary look up in global name space. I'm sure you can make a function to do this. Just remember which namespace your variable/object/wypci is in.
QED.

I wrote the package sorcery to do this kind of magic robustly. You can write:
from sorcery import dict_of
my_dict = dict_of(foo, bar)

Maybe I'm overthinking this but..
str_l = next((k for k,v in locals().items() if id(l) == id(v)))
>>> bar = True
>>> foo = False
>>> my_dict=dict(bar=bar, foo=foo)
>>> next((k for k,v in locals().items() if id(bar) == id(v)))
'bar'
>>> next((k for k,v in locals().items() if id(foo) == id(v)))
'foo'
>>> next((k for k,v in locals().items() if id(my_dict) == id(v)))
'my_dict'

import re
import traceback
pattren = re.compile(r'[\W+\w+]*get_variable_name\((\w+)\)')
def get_variable_name(x):
return pattren.match( traceback.extract_stack(limit=2)[0][3]) .group(1)
a = 1
b = a
c = b
print get_variable_name(a)
print get_variable_name(b)
print get_variable_name(c)

I uploaded a solution to pypi. It's a module defining an equivalent of C#'s nameof function.
It iterates through bytecode instructions for the frame its called in, getting the names of variables/attributes passed to it. The names are found in the .argrepr of LOAD instructions following the function's name.

Most objects don't have a __name__ attribute. (Classes, functions, and modules do; any more builtin types that have one?)
What else would you expect for print(my_var.__name__) other than print("my_var")? Can you simply use the string directly?
You could "slice" a dict:
def dict_slice(D, keys, default=None):
return dict((k, D.get(k, default)) for k in keys)
print dict_slice(locals(), ["foo", "bar"])
# or use set literal syntax if you have a recent enough version:
print dict_slice(locals(), {"foo", "bar"})
Alternatively:
throw = object() # sentinel
def dict_slice(D, keys, default=throw):
def get(k):
v = D.get(k, throw)
if v is not throw:
return v
if default is throw:
raise KeyError(k)
return default
return dict((k, get(k)) for k in keys)

Well, I encountered the very same need a few days ago and had to get a variable's name which was pointing to the object itself.
And why was it so necessary?
In short I was building a plug-in for Maya. The core plug-in was built using C++ but the GUI is drawn through Python(as its not processor intensive). Since I, as yet, don't know how to return multiple values from the plug-in except the default MStatus, therefore to update a dictionary in Python I had to pass the the name of the variable, pointing to the object implementing the GUI and which contained the dictionary itself, to the plug-in and then use the MGlobal::executePythonCommand() to update the dictionary from the global scope of Maya.
To do that what I did was something like:
import time
class foo(bar):
def __init__(self):
super(foo, self).__init__()
self.time = time.time() #almost guaranteed to be unique on a single computer
def name(self):
g = globals()
for x in g:
if isinstance(g[x], type(self)):
if g[x].time == self.time:
return x
#or you could:
#return filter(None,[x if g[x].time == self.time else None for x in g if isinstance(g[x], type(self))])
#and return all keys pointing to object itself
I know that it is not the perfect solution in in the globals many keys could be pointing to the same object e.g.:
a = foo()
b = a
b.name()
>>>b
or
>>>a
and that the approach isn't thread-safe. Correct me if I am wrong.
At least this approach solved my problem by getting the name of any variable in the global scope which pointed to the object itself and pass it over to the plug-in, as argument, for it use internally.
I tried this on int (the primitive integer class) but the problem is that these primitive classes don't get bypassed (please correct the technical terminology used if its wrong). You could re-implement int and then do int = foo but a = 3 will never be an object of foo but of the primitive. To overcome that you have to a = foo(3) to get a.name() to work.

With python 2.7 and newer there is also dictionary comprehension which makes it a bit shorter. If possible I would use getattr instead eval (eval is evil) like in the top answer. Self can be any object which has the context your a looking at. It can be an object or locals=locals() etc.
{name: getattr(self, name) for name in ['some', 'vars', 'here]}

I was working on a similar problem. #S.Lott said "If you have the list of variables, what's the point of "discovering" their names?" And my answer is just to see if it could be done and if for some reason you want to sort your variables by type into lists. So anyways, in my research I came came across this thread and my solution is a bit expanded and is based on #rlotun solution. One other thing, #unutbu said, "This idea has merit, but note that if two variable names reference the same value (e.g. True), then an unintended variable name might be returned." In this exercise that was true so I dealt with it by using a list comprehension similar to this for each possibility: isClass = [i for i in isClass if i != 'item']. Without it "item" would show up in each list.
__metaclass__ = type
from types import *
class Class_1: pass
class Class_2: pass
list_1 = [1, 2, 3]
list_2 = ['dog', 'cat', 'bird']
tuple_1 = ('one', 'two', 'three')
tuple_2 = (1000, 2000, 3000)
dict_1 = {'one': 1, 'two': 2, 'three': 3}
dict_2 = {'dog': 'collie', 'cat': 'calico', 'bird': 'robin'}
x = 23
y = 29
pie = 3.14159
eee = 2.71828
house = 'single story'
cabin = 'cozy'
isClass = []; isList = []; isTuple = []; isDict = []; isInt = []; isFloat = []; isString = []; other = []
mixedDataTypes = [Class_1, list_1, tuple_1, dict_1, x, pie, house, Class_2, list_2, tuple_2, dict_2, y, eee, cabin]
print '\nMIXED_DATA_TYPES total count:', len(mixedDataTypes)
for item in mixedDataTypes:
try:
# if isinstance(item, ClassType): # use this for old class types (before 3.0)
if isinstance(item, type):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isClass.append(mapping_as_str)
isClass = [i for i in isClass if i != 'item']
elif isinstance(item, ListType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isList.append(mapping_as_str)
isList = [i for i in isList if i != 'item']
elif isinstance(item, TupleType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isTuple.append(mapping_as_str)
isTuple = [i for i in isTuple if i != 'item']
elif isinstance(item, DictType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isDict.append(mapping_as_str)
isDict = [i for i in isDict if i != 'item']
elif isinstance(item, IntType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isInt.append(mapping_as_str)
isInt = [i for i in isInt if i != 'item']
elif isinstance(item, FloatType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isFloat.append(mapping_as_str)
isFloat = [i for i in isFloat if i != 'item']
elif isinstance(item, StringType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isString.append(mapping_as_str)
isString = [i for i in isString if i != 'item']
else:
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
other.append(mapping_as_str)
other = [i for i in other if i != 'item']
except (TypeError, AttributeError), e:
print e
print '\n isClass:', len(isClass), isClass
print ' isList:', len(isList), isList
print ' isTuple:', len(isTuple), isTuple
print ' isDict:', len(isDict), isDict
print ' isInt:', len(isInt), isInt
print ' isFloat:', len(isFloat), isFloat
print 'isString:', len(isString), isString
print ' other:', len(other), other
# my output and the output I wanted
'''
MIXED_DATA_TYPES total count: 14
isClass: 2 ['Class_1', 'Class_2']
isList: 2 ['list_1', 'list_2']
isTuple: 2 ['tuple_1', 'tuple_2']
isDict: 2 ['dict_1', 'dict_2']
isInt: 2 ['x', 'y']
isFloat: 2 ['pie', 'eee']
isString: 2 ['house', 'cabin']
other: 0 []
'''

you can use easydict
>>> from easydict import EasyDict as edict
>>> d = edict({'foo':3, 'bar':{'x':1, 'y':2}})
>>> d.foo
3
>>> d.bar.x
1
>>> d = edict(foo=3)
>>> d.foo
3
another example:
>>> d = EasyDict(log=False)
>>> d.debug = True
>>> d.items()
[('debug', True), ('log', False)]

On python3, this function will get the outer most name in the stack:
import inspect
def retrieve_name(var):
"""
Gets the name of var. Does it from the out most frame inner-wards.
:param var: variable to get name from.
:return: string
"""
for fi in reversed(inspect.stack()):
names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
if len(names) > 0:
return names[0]
It is useful anywhere on the code. Traverses the reversed stack looking for the first match.

While this is probably an awful idea, it is along the same lines as rlotun's answer but it'll return the correct result more often.
import inspect
def getVarName(getvar):
frame = inspect.currentframe()
callerLocals = frame.f_back.f_locals
for k, v in list(callerLocals.items()):
if v is getvar():
callerLocals.pop(k)
try:
getvar()
callerLocals[k] = v
except NameError:
callerLocals[k] = v
del frame
return k
del frame
You call it like this:
bar = True
foo = False
bean = False
fooName = getVarName(lambda: foo)
print(fooName) # prints "foo"

should get list then return
def get_var_name(**kwargs):
"""get variable name
get_var_name(var = var)
Returns:
[str] -- var name
"""
return list(kwargs.keys())[0]

It will not return the name of variable but you can create dictionary from global variable easily.
class CustomDict(dict):
def __add__(self, other):
return CustomDict({**self, **other})
class GlobalBase(type):
def __getattr__(cls, key):
return CustomDict({key: globals()[key]})
def __getitem__(cls, keys):
return CustomDict({key: globals()[key] for key in keys})
class G(metaclass=GlobalBase):
pass
x, y, z = 0, 1, 2
print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}

With python-varname you can easily do it:
pip install python-varname
from varname import Wrapper
foo = Wrapper(True)
bar = Wrapper(False)
your_dict = {val.name: val.value for val in (foo, bar)}
print(your_dict)
# {'foo': True, 'bar': False}
Disclaimer: I'm the author of that python-varname library.

>>> a = 1
>>> b = 1
>>> id(a)
34120408
>>> id(b)
34120408
>>> a is b
True
>>> id(a) == id(b)
True
this way get varname for a maybe 'a' or 'b'.

Is it possible to assign variables within the assignment of another variable?

I have two questions:
1) I want to write a statement like: superVar = [subVar1 = 'a', subVar2 = 'b']
After this statement, I want:
superVar = ['a', 'b']
subVar1 = 'a'
subVar2 = 'b'
Is this somehow achievable in python?
2) I want to do the following inside a class's init function:
self.superVar = [v1 = 'a', v2 = 'b']
After this statement, I want:
self.superVar = ['a', 'b']
self.superVar.v1 = 'a'
self.superVar.v2 = 'b'
i.e. have 'superVar' be part of the name of the other variables.
As you might have guessed, what I really want to do is (2) above. (1) is just the first step and I would be happy even with that, if (2) is not possible.
Thanks a lot!
Gaurav

>>> from collections import namedtuple
>>> T = namedtuple('supervar', ['v1', 'v2'])
>>> superVar = T('a', 'b')
>>> superVar.v1
'a'
>>> superVar.v2
'b'

Do you mean a dictionary?
superVar = {'subVar1': 'a', 'subVar2': 'b'}
superVar['subVar1']
# 'a'
superVar['subVar2']
# 'b'
In a class:
class Foo(object):
def __init__(self, a, b):
self.superVar = {'subVar1' : a, 'subVar2' : b}
bar = Foo('one', 'two')
bar.superVar.values()
# ['one', 'two']
If you want to keep order, you can use an ordered dictionary from the collections module.

Why not create a dictionary for your subVars and make superVar a list of values from the subVar dictionary?
d = {'subVar1': 'value1', 'subVar2': 'value2'}
superVar = d.values()

In Python, assignments are statements, not expressions. You can't chain together statements. So (1) is impossible.
However, you can use keyword arguments in the __init__ method of a class to get the behavior in (2):
>>> class Test:
def __init__(self, **kwargs):
print "the keyword arguments are", kwargs #kwargs is a dictionary
for x in kwargs:
setattr(self, x, kwargs[x]) #adds an attribute from kwargs to self
>>> t = Test(var1=1234,var2='hello',var3=45.234)
the keyword arguments are {'var1': 1234, 'var3': 45.234, 'var2': 'hello'}
>>> t.var1
1234
>>> t.var2
'hello'
>>> t.var3
45.234
>>>

This is basically the same as Jeremy Bentham's answer,
class T:
def __init__(self):
self.SuperVar = {'v1':'a', 'v2':'b'}
print self.SuperVar.v1
print self.SuperVar.v2
def __setattr__(self, k, v):
class SuperVar:
def __setattr__(self, k, v):
self.__dict__[k] = v
if k == "SuperVar":
self.__dict__["SuperVar"] = SuperVar()
for k in v:
self.__dict__["SuperVar"].__setattr__(k, v[k])
else:
self.__dict__[k] = v
t = T()

Difference between dict.clear() and assigning {} in Python

In python, is there a difference between calling clear() and assigning {} to a dictionary? If yes, what is it?
Example:d = {"stuff":"things"}
d.clear() #this way
d = {} #vs this way

If you have another variable also referring to the same dictionary, there is a big difference:
>>> d = {"stuff": "things"}
>>> d2 = d
>>> d = {}
>>> d2
{'stuff': 'things'}
>>> d = {"stuff": "things"}
>>> d2 = d
>>> d.clear()
>>> d2
{}
This is because assigning d = {} creates a new, empty dictionary and assigns it to the d variable. This leaves d2 pointing at the old dictionary with items still in it. However, d.clear() clears the same dictionary that d and d2 both point at.

d = {} will create a new instance for d but all other references will still point to the old contents.
d.clear() will reset the contents, but all references to the same instance will still be correct.

In addition to the differences mentioned in other answers, there also is a speed difference. d = {} is over twice as fast:
python -m timeit -s "d = {}" "for i in xrange(500000): d.clear()"
10 loops, best of 3: 127 msec per loop
python -m timeit -s "d = {}" "for i in xrange(500000): d = {}"
10 loops, best of 3: 53.6 msec per loop

As an illustration for the things already mentioned before:
>>> a = {1:2}
>>> id(a)
3073677212L
>>> a.clear()
>>> id(a)
3073677212L
>>> a = {}
>>> id(a)
3073675716L

In addition to #odano 's answer, it seems using d.clear() is faster if you would like to clear the dict for many times.
import timeit
p1 = '''
d = {}
for i in xrange(1000):
d[i] = i * i
for j in xrange(100):
d = {}
for i in xrange(1000):
d[i] = i * i
'''
p2 = '''
d = {}
for i in xrange(1000):
d[i] = i * i
for j in xrange(100):
d.clear()
for i in xrange(1000):
d[i] = i * i
'''
print timeit.timeit(p1, number=1000)
print timeit.timeit(p2, number=1000)
The result is:
20.0367929935
19.6444659233

Mutating methods are always useful if the original object is not in scope:
def fun(d):
d.clear()
d["b"] = 2
d={"a": 2}
fun(d)
d # {'b': 2}
Re-assigning the dictionary would create a new object and wouldn't modify the original one.

One thing not mentioned is scoping issues. Not a great example, but here's the case where I ran into the problem:
def conf_decorator(dec):
"""Enables behavior like this:
#threaded
def f(): ...
or
#threaded(thread=KThread)
def f(): ...
(assuming threaded is wrapped with this function.)
Sends any accumulated kwargs to threaded.
"""
c_kwargs = {}
#wraps(dec)
def wrapped(f=None, **kwargs):
if f:
r = dec(f, **c_kwargs)
c_kwargs = {}
return r
else:
c_kwargs.update(kwargs) #<- UnboundLocalError: local variable 'c_kwargs' referenced before assignment
return wrapped
return wrapped
The solution is to replace c_kwargs = {} with c_kwargs.clear()
If someone thinks up a more practical example, feel free to edit this post.

In addition, sometimes the dict instance might be a subclass of dict (defaultdict for example). In that case, using clear is preferred, as we don't have to remember the exact type of the dict, and also avoid duplicate code (coupling the clearing line with the initialization line).
x = defaultdict(list)
x[1].append(2)
...
x.clear() # instead of the longer x = defaultdict(list)