For primitive types I can use the if in : boolean check. But if I use the in syntax to check for the existence of a class member I get a NameError exception. Is there a way in Python to check without an exception? Or is the only way to surround in try except block?
Here is my sample code.
class myclass:
i = 0
def __init__(self, num):
self.i = num
mylist = [1,2,3]
if 7 in mylist:
print "found it"
else:
print "7 not present" #prints 7 not present
x = myclass(3)
print x.i #prints 3
#below line NameError: name 'counter' is not defined
if counter in x:
print "counter in x"
else:
print "No counter in x"
You can use hasattr
if hasattr(x, 'counter'):
# whatever
The error you get is because you are using counter (a name) and not 'counter' (the string). However, even if you were to use 'counter' it would not do what you expect, you will get TypeError: argument of type 'a' is not iterable - that is you cannot iterate over your custom object.
Instead, use hasattr (thanks to Jon for the suggestion).
>>> x = A(3)
>>> x.i
3
>>> hasattr(x, 'counter')
False
>>> hasattr(x, 'i')
True
You can make a __contains__ function in your class, which reports back what attributes are in the class using the in operator.
class myclass:
def __init__(self, num):
self.i = num
def __contains__(self, attribute_name):
return hasattr(self, attribute_name)
Then (almost) the same as your code would work well.
x = myclass(3)
print x.i #prints 3
# prints 'No counter in x'
if 'counter' in x:
print "counter in x"
else:
print "No counter in x"
Note that you need to pass the string of the attribute name, rather than the attribute itself.
The correct answer to your question depends a bit on what you mean by a member existing in an object. If you mean, does an instance variable exist on a given object, use hasattr as other answers have explained.
However, if you're creating your own collection type and want to check for a specific value in its contents, then you'll want to give your class a __contains__ method. That magic method is called to implement the in operator. Here's a trivial example, where I simply wrap up a list in my own object.
class MyListWrapper(object):
def __init__(self, iterable=[]):
self.list = list(iterable)
def __contains__(self, value):
return value in self.list
Test session:
>>> m = MyListWrapper(["foo", "bar"])
>>> "foo" in m
True
Related
My question comes from this page, while I would like to create a pointer(-like thing) for an list element. The element is a primitive value (string) so I have to create a FooWrapper class as that page says.
I know that by setting __repr__ one can directly access this value.
class FooWrapper(object):
def __init__(self, value):
self.value = value
def __repr__(self):
return repr(self.value)
>>> bar=FooWrapper('ABC')
>>> bar
'ABC'
>>> bar=FooWrapper(3)
>>> bar
3
Now I can use it as an reference of string:
>>> L=[3,5,6,9]
>>> L[1]=FooWrapper('ABC')
>>> L
[3,'ABC',6,9]
>>> this=L[1]
>>> this.value='BCD'
>>> print(L)
[3,'BCD',6,9]
So now I have a pointer-like this for the list element L[1].
However it is still inconvenient since I must use this.value='BCD' to change its value. While there exists a __repr__ method to make this directly return this.value, is there any similar method to make this='BCD' to do this.value='BCD' ? I know this changes the rule of binding.. but anyway, is it possible?
I would also appreciate if there is a better solution for a list element pointer.
Thank you in advance:)
I'm not sure exactly what you are trying to do, but you could do something like:
class FooWrapper(object):
def __init__(self, value):
self.value = value
def __repr__(self):
return 'FooWrapper(' + repr(self.value) + ')'
def __str__(self):
return str(self.value)
def __call__(self,value):
self.value = value
Here I got rid of your idea of using __repr__ to hide FooWrapper since I think it a bad idea to hide from the programmer what is happening at the REPL. Instead -- I used __str__ so that when you print the object you will print the wrapped value. The __call__ functions as a default method, which doesn't change the meaning of = but is sort of what you want:
>>> vals = [1,2,3]
>>> vals[1] = FooWrapper("Bob")
>>> vals
[1, FooWrapper('Bob'), 3]
>>> for x in vals: print(x)
1
Bob
3
>>> this = vals[1]
>>> this(10)
>>> vals
[1, FooWrapper(10), 3]
However, I think it misleading to refer to this as a pointer. It is just a wrapper object, and is almost certain to make dealing with the wrapped object inconvenient.
On Edit: The following is more of a pointer to a list. It allows you to create something like a pointer object with __call__ used to dereference the pointer (when no argument is passed to it) or to mutate the list (when a value is passed to __call__). It also implements a form of p++ called (pp) with wrap-around (though the wrap-around part could of course be dropped):
class ListPointer(object):
def __init__(self, myList,i=0):
self.myList = myList
self.i = i % len(self.myList)
def __repr__(self):
return 'ListPointer(' + repr(self.myList) + ',' + str(self.i) + ')'
def __str__(self):
return str(self.myList[self.i])
def __call__(self,*value):
if len(value) == 0:
return self.myList[self.i]
else:
self.myList[self.i] = value[0]
def pp(self):
self.i = (self.i + 1) % len(self.myList)
Used like this:
>>> vals = ['a','b','c']
>>> this = ListPointer(vals)
>>> this()
'a'
>>> this('d')
>>> vals
['d', 'b', 'c']
>>> this.pp()
>>> this()
'b'
>>> print(this)
b
I think that this is a more transparent way of getting something which acts like a list pointer. It doesn't require the thing pointed to to be wrapped in anything.
The __repr__ method can get a string however it wants to. Let's say it says return repr(self.value) + 'here'. If you say this = '4here', what should be affected? Should self.value be assigned to 4 or 4here? What if this had another attribute called key and __repr__ did return repr(self.key) + repr(self.value)? When you did this = '4here', would it assign self.key to the whole string, assign self.value to the whole string, or assign self.key to 4 and self.value to here? What if the string is completely made up by the method? If it says return 'here', what should this = '4here' do?
In short, you can't.
Self-Answer based on John Coleman's idea:
class ListPointer(object):
def __init__(self,list,index):
self.value=list[index]
list[index]=self
def __repr__(self):
return self.value
def __call__(self,value):
self.value=value
>>> foo=[2,3,4,5]
>>> this=ListPointer(foo,2)
>>> this
4
>>> foo
[2,3,4,5]
>>> this('ABC')
>>> foo
[2,3,'ABC',5]
>>> type(foo(2))
<class '__main__.ListPointer'>
ListPointer object accepts a list and an index, stores the list[index] in self.value and then substitutes the list element with self. Similarly a pp method can also be achieved, while the former element should be restored and the next element be substituted with self object. By directly referring foo[2] one also gets this object, which is what I want. (Perhaps this should be called as a reference rather than a pointer..)
Sorry for the title, I hope it reflects correctly my problem :
In the following code, I was expecting the result to be result 0 1 2 but instead I have 2 2 2. The code inside my_function seems to be interpreted with the last instance of obj. What is wrong ?
class Example:
def __init__(self, x):
self.x = x
def get(self):
return self.x
a_list = []
for index in range(3):
obj = Example(index)
def my_function(x):
#some stuff with x like obj.another_function(x)
return obj.get()
a_list.append(my_function)
for c in a_list:
print(c())
When you define this
def my_function():
return obj.get()
Python will understand that my_function should run the get() method of an object called obj and return the value. It won't know the value of obj and what the get() method does until you attempt to call it.
So, you are actually defining three different functions that will eventually do the same thing. And, in the end, running the same code thrice.
But why is the return 2 2 2?
Because after the last iteration, the value of obj is Example(2)* because you redefine its value at every iteration, and the last one remains.
*
because of this line obj = Example(index)
Understanding a few things about how python works will help you understand what's happening here. Here obj is a closure, closures are evaluated at call time, not when the function is defined so if I do this:
x = "hello"
def printX():
print x
x = "goodbye"
printX() # goodbye
I get "goodbye" because printX is referencing a global variable in my module, which changes after I create printX.
What you want to do is create a function with a closure that references a specific object. The functional way to do this is to create a function that returns another function:
x = "hello"
def makePrintX(a):
def printX():
# We print a, the object passed to `makePrintX`
print a
return printX
# x is evaluated here when it is still "hello"
myPrintX = makePrintX(x)
x = "goodbye"
myPrintX() # "hello"
If you're having trouble understanding the above example I would recommend reading up on python's scoping rules. For your example, you could do something like this:
class Example:
def __init__(self, x):
self.x = x
def get(self):
return self.x
def makeObjFunction(obj):
def objFunction(x):
return obj.get()
return objFunction
a_list = []
for index in range(3):
obj = Example(index)
my_function = makeObjFunction(obj)
a_list.append(my_function)
for c in a_list:
print(c("some value"))
You are appending three my_functions to the a_list which are all closures over the same Example object. Try:
def my_function():
return obj
<__main__.Example object at 0x0054EDF0>
<__main__.Example object at 0x0054EDF0>
<__main__.Example object at 0x0054EDF0>
You can see they have the same id so calling get() on each should give the same answer.
If you just append the obj.get function (and drop the my_function) it'll work fine.
a_list.append(obj.get)
....
0
1
2
Edit: You've updated your question so to let you do more stuff in my_function(). It's still basically a scoping problem.
def my_func_factory(p_obj):
def my_function(x):
#some stuff with x like obj.another_function(x)
return p_obj.get()
return my_function
for index in range(3):
obj = Example(index)
a_list.append(my_func_factory(obj))
Since my_function can't see obj being reassigned, each instance doesn't pick up the change.
I think append() during the for just append the function address in a_list[]. After for iteration, the a_list is really given the number. Then it discovers the address of my_function, and they get the number in my_function, this is, 2. That's why you get [2,2,2].
Or maybe, in my_function, function give the method of "obj". But for iteration change the "obj" memory address each time, so the symbol "obj" always aim to the newest object Example. Due to my_function always get "obj", you get the same number from the last object.
I have seen source code where more than one methods are called on an object eg x.y().z() Can someone please explain this to me, does this mean that z() is inside y() or what?
This calls the method y() on object x, then the method z() is called on the result of y() and that entire line is the result of method z().
For example
friendsFavePizzaToping = person.getBestFriend().getFavoritePizzaTopping()
This would result in friendsFavePizzaTopping would be the person's best friend's favorite pizza topping.
Important to note: getBestFriend() must return an object that has the method getFavoritePizzaTopping(). If it does not, an AttributeError will be thrown.
Each method is evaluated in turn, left to right. Consider:
>>> s='HELLO'
>>> s.lower()
'hello'
>>> s='HELLO '
>>> s.lower()
'hello '
>>> s.lower().strip()
'hello'
>>> s.lower().strip().upper()
'HELLO'
>>> s.lower().strip().upper().replace('H', 'h')
'hELLO'
The requirement is that the object to the left in the chain has to have availability of the method on the right. Often that means that the objects are similar types -- or at least share compatible methods or an understood cast.
As an example, consider this class:
class Foo:
def __init__(self, name):
self.name=name
def m1(self):
return Foo(self.name+'=>m1')
def m2(self):
return Foo(self.name+'=>m2')
def __repr__(self):
return '{}: {}'.format(id(self), self.name)
def m3(self):
return .25 # return is no longer a Foo
Notice that as a type of immutable, each return from Foo is a new object (either a new Foo for m1, m2 or a new float). Now try those methods:
>>> foo
4463545376: init
>>> foo.m1()
4463545304: init=>m1
^^^^ different object id
>>> foo
4463545376: init
^^^^ foo still the same because you need to assign it to change
Now assign:
>>> foo=foo.m1().m2()
>>> foo
4464102576: init=>m1=>m2
Now use m3() and it will be a float; not a Foo anymore:
>>> foo=foo.m1().m2().m3()
>>> foo
.25
Now a float -- can't use foo methods anymore:
>>> foo.m1()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'float' object has no attribute 'm1'
But you can use float methods:
>>> foo.as_integer_ratio()
(1, 4)
In the case of:
x.y().z()
You're almost always looking at immutable objects. Mutable objects don't return anything that would HAVE a function like that (for the most part, but I'm simplifying). For instance...
class x:
def __init__(self):
self.y_done = False
self.z_done = False
def y(self):
new_x = x()
new_x.y_done = True
return new_x
def z(self):
new_x = x()
new_x.z_done = True
return new_x
You can see that each of x.y and x.z returns an x object. That object is used to make the consecutive call, e.g. in x.y().z(), x.z is not called on x, but on x.y().
x.y().z() =>
tmp = x.y()
result = tmp.z()
In #dawg's excellent example, he's using strings (which are immutable in Python) whose methods return strings.
string = 'hello'
string.upper() # returns a NEW string with value "HELLO"
string.upper().replace("E","O") # returns a NEW string that's based off "HELLO"
string.upper().replace("E","O") + "W"
# "HOLLOW"
The . "operator" is Python syntax for attribute access. x.y is (nearly) identical to
getattr(x, 'y')
so x.y() is (nearly) identical to
getattr(x, 'y')()
(I say "nearly identical" because it's possible to customize attribute access for a user-defined class. From here on out, I'll assume no such customization is done, and you can assume that x.y is in fact identical to getattr(x, 'y').)
If the thing that x.y() returns has an attribute z such that
foo = getattr(x, 'y')
bar = getattr(foo(), 'z')
is legal, then you can chain the calls together without needing the name foo in the middle:
bar = getattr(getattr(x, 'y')(), 'z')
Converting back to dot notation gives you
bar = getattr(x.y(), 'z')
or simply
bar = x.y().z()
x.y().z() means that the x object has the method y() and the result of x.y() object has the method z() . Now if you first want to apply the method y() on x and then on the result want to apply the z() method, you will write x.y().z(). This is like,
val = x.y()
result = val.z()
Example:
my_dict = {'key':'value'}
my_dict is a dict type object. my_dict.get('key') returns 'value' which is a str type object. now I can apply any method of str type object on it. which will be like,
my_dict.get('key').upper()
This will return 'VALUE'.
That is (sometimes a sign of) bad code.
It violates The law of Demeter. Here is a quote from Wikipedia explaining what is meant:
Each unit should have only limited knowledge about other units: only units "closely" related to the current unit.
Each unit should only talk to its friends; don't talk to strangers.
Only talk to your immediate friends.
Suppose you have a car, which itself has an engine:
class Car:
def __init__(self):
self._engine=None
#property
def engine(self):
return self._engine
#engine.setter
def engine(self, value):
self._engine = value
class Porsche_engine:
def start(self):
print("starting")
So if you make a new car and set the engine to Porsche you could do the following:
>>> from car import *
>>> c=Car()
>>> e=Porsche_engine()
>>> c.engine=e
>>> c.engine.start()
starting
If you are maing this call from an Object, it has not only knowledge of a Car object, but has too knowledge of Engine, which is bad design.
Additionally: if you do not know whether a Car has an engine, calling directly start
>>> c=Car()
>>> c.engine.start()
May result in an Error
AttributeError: 'NoneType' object has no attribute 'start'
Edit:
To avoid (further) misunterstandings and misreadings, from what I am saying.
There are two usages:
1) as I pointed out, Objects calling methods on other objects, returned from a third object is a violation of LoD. This is one way to read the question.
2) an exception to that is method chaining, which is not bad design.
And a better design would be, if the Car itself had a start()-Method which delegates to the engine.
All the id() in the following python program prints the same value. I had thought different member functions are different objects therefore id() should return different values. Why does id() return the same values?
class A():
def x(self):
print 'x'
def __y(self):
print 'y'
def y(self):
self.__y()
class B(A):
def x(self):
self.jj = 20
print 'bx'
def __y(self):
self.sj = 20
print 'by'
print 'by'
ta = A()
tb = B()
ta.x()
ta.y()
ta._A__y()
tb.x()
tb._A__y()
tb._B__y()
tb.y()
print id(ta.x)
print id(ta.y)
print id(ta._A__y)
print id(tb.x)
print id(tb._A__y)
print id(tb._B__y)
print id(tb.y)
The next question (actually that was the original question I had) would be, how to examine if tb._A__y and tb.y really invokes the same function. I had though using id() will tell me the answer if id() returns the address of a function. But now I udnerstand it doesn't work that way in Python.
I have a variable x in python. How can i find the string 'x' from the variable. Here is my attempt:
def var(v,c):
for key in c.keys():
if c[key] == v:
return key
def f():
x = '321'
print 'Local var %s = %s'%(var(x,locals()),x)
x = '123'
print 'Global var %s = %s'%(var(x,locals()),x)
f()
The results are:
Global var x = 123
Local var x = 321
The above recipe seems a bit un-pythonesque. Is there a better/shorter way to achieve the same result?
Q: I have a variable x in python. How can i find the string 'x' from the variable.
A: If I am understanding your question properly, you want to go from the value of a variable to its name. This is not really possible in Python.
In Python, there really isn't any such thing as a "variable". What Python really has are "names" which can have objects bound to them. It makes no difference to the object what names, if any, it might be bound to. It might be bound to dozens of different names, or none.
Consider this example:
foo = 1
bar = foo
baz = foo
Now, suppose you have the integer object with value 1, and you want to work backwards and find its name. What would you print? Three different names have that object bound to them, and all are equally valid.
print(bar is foo) # prints True
print(baz is foo) # prints True
In Python, a name is a way to access an object, so there is no way to work with names directly. You might be able to search through locals() to find the value and recover a name, but that is at best a parlor trick. And in my above example, which of foo, bar, and baz is the "correct" answer? They all refer to exactly the same object.
P.S. The above is a somewhat edited version of an answer I wrote before. I think I did a better job of wording things this time.
I believe the general form of what you want is repr() or the __repr__() method of an object.
with regards to __repr__():
Called by the repr() built-in function
and by string conversions (reverse
quotes) to compute the “official”
string representation of an object.
See the docs here: object.repr(self)
stevenha has a great answer to this question. But, if you actually do want to poke around in the namespace dictionaries anyway, you can get all the names for a given value in a particular scope / namespace like this:
def foo1():
x = 5
y = 4
z = x
print names_of1(x, locals())
def names_of1(var, callers_namespace):
return [name for (name, value) in callers_namespace.iteritems() if var is value]
foo1() # prints ['x', 'z']
If you're working with a Python that has stack frame support (most do, CPython does), it isn't required that you pass the locals dict into the names_of function; the function can retrieve that dictionary from its caller's frame itself:
def foo2():
xx = object()
yy = object()
zz = xx
print names_of2(xx)
def names_of2(var):
import inspect
callers_namespace = inspect.currentframe().f_back.f_locals
return [name for (name, value) in callers_namespace.iteritems() if var is value]
foo2() # ['xx', 'zz']
If you're working with a value type that you can assign a name attribute to, you can give it a name, and then use that:
class SomeClass(object):
pass
obj = SomeClass()
obj.name = 'obj'
class NamedInt(int):
__slots__ = ['name']
x = NamedInt(321)
x.name = 'x'
Finally, if you're working with class attributes and you want them to know their names (descriptors are the obvious use case), you can do cool tricks with metaclass programming like they do in the Django ORM and SQLAlchemy declarative-style table definitions:
class AutonamingType(type):
def __init__(cls, name, bases, attrs):
for (attrname, attrvalue) in attrs.iteritems():
if getattr(attrvalue, '__autoname__', False):
attrvalue.name = attrname
super(AutonamingType,cls).__init__(name, bases, attrs)
class NamedDescriptor(object):
__autoname__ = True
name = None
def __get__(self, instance, instance_type):
return self.name
class Foo(object):
__metaclass__ = AutonamingType
bar = NamedDescriptor()
baaz = NamedDescriptor()
lilfoo = Foo()
print lilfoo.bar # prints 'bar'
print lilfoo.baaz # prints 'baaz'
There are three ways to get "the" string representation of an object in python:
1: str()
>>> foo={"a":"z","b":"y"}
>>> str(foo)
"{'a': 'z', 'b': 'y'}"
2: repr()
>>> foo={"a":"z","b":"y"}
>>> repr(foo)
"{'a': 'z', 'b': 'y'}"
3: string interpolation:
>>> foo={"a":"z","b":"y"}
>>> "%s" % (foo,)
"{'a': 'z', 'b': 'y'}"
In this case all three methods generated the same output, the difference is that str() calls dict.__str__(), while repr() calls dict.__repr__(). str() is used on string interpolation, while repr() is used by Python internally on each object in a list or dict when you print the list or dict.
As Tendayi Mawushe mentiones above, string produced by repr isn't necessarily human-readable.
Also, the default implementation of .__str__() is to call .__repr__(), so if the class does not have it's own overrides to .__str__(), the value returned from .__repr__() is used.