Python: How do I pass a string by reference? - python

From this link: How do I pass a variable by reference?, we know, Python will copy a string (an immutable type variable) when it is passed to a function as a parameter, but I think it will waste memory if the string is huge. In many cases, we need to use functions to wrap some operations for strings, so I want to know how to do it more effective?

Python does not make copies of objects (this includes strings) passed to functions:
>>> def foo(s):
... return id(s)
...
>>> x = 'blah'
>>> id(x) == foo(x)
True
If you need to "modify" a string in a function, return the new string and assign it back to the original name:
>>> def bar(s):
... return s + '!'
...
>>> x = 'blah'
>>> x = bar(x)
>>> x
'blah!'
Unfortunately, this can be very inefficient when making small changes to large strings because the large string gets copied. The pythonic way of dealing with this is to hold strings in an list and join them together once you have all the pieces.

Python does pass a string by reference. Notice that two strings with the same content are considered identical:
a = 'hello'
b = 'hello'
a is b # True
Since when b is assigned by a value, and the value already exists in memory, it uses the same reference of the string. Notice another fact, that if the string was dynamically created, meaning being created with string operations (i.e concatenation), the new variable will reference a new instance of the same string:
c = 'hello'
d = 'he'
d += 'llo'
c is d # False
That being said, creating a new string will allocate a new string in memory and returning a reference for the new string, but using a currently created string will reuse the same string instance. Therefore, passing a string as a function parameter will pass it by reference, or in other words, will pass the address in memory of the string.
And now to the point you were looking for- if you change the string inside the function, the string outside of the function will remain the same, and that stems from string immutability. Changing a string means allocating a new string in memory.
a = 'a'
b = a # b will hold a reference to string a
a += 'a'
a is b # False
Bottom line:
You cannot really change a string. The same as for maybe every other programming language (but don't quote me).
When you pass the string as an argument, you pass a reference. When you change it's value, you change the variable to point to another place in memory. But when you change a variable's reference, other variables that points to the same address will naturally keep the old value (reference) they held.
Wish the explanation was clear enough

In [7]: strs="abcd"
In [8]: id(strs)
Out[8]: 164698208
In [9]: def func(x):
print id(x)
x=x.lower() #perform some operation on string object, it returns a new object
print id(x)
...:
In [10]: func(strs)
164698208 # same as strs, i.e it actually passes the same object
164679776 # new object is returned if we perform an operation
# That's why they are called immutable
But operations on strings always return a new string object.

def modify_string( t ):
the_string = t[0]
# do stuff
modify_string( ["my very long string"] )

If you want to potentially change the value of something passed in, wrap it in a dict or a list:
This doesn't change s
def x(s):
s += 1
This does change s:
def x(s):
s[0] += 1
This is the only way to "pass by reference".

wrapping the string into a class will make it pass by reference:
class refstr:
"wrap string in object, so it is passed by reference rather than by value"
def __init__(self,s=""):
self.s=s
def __add__(self,s):
self.s+=s
return self
def __str__(self):
return self.s
def fn(s):
s+=" world"
s=refstr("hello")
fn(s) # s gets modified because objects are passed by reference
print(s) #returns 'hello world'

Just pass it in as you would any other parameter. The contents won't get copied, only the reference will.

Related

Passing list as an argument to a function . Can someone explain this behavior?

def test(lister):
lister.append('why am i being shown in output?')
def pass_the_list():
x = []
test(x)
print(x)
pass_the_list()
Output
['why am i being shown in output?']
Process finished with exit code 0
how to pass string as a reference without return?
The list x is passed as a reference not value, therefore changes made in the function will preserve also when the function finishes.
You can pass a string as a refence, but it doesnt matter, because string, unlike list, is immutable. So any change you do to the string will cause creation of a new object.
test_string = 'Test my immutability'
test_string[0]='B'
>> TypeError: 'str' object does not support item assignment
You can modify string like this though (by putting it into something mutable like list)
test_string = 'Test my immutability'
test_list = [test_string]
def reference_test(test_list):
test_list[0]=test_list[0].replace('Test','Fest')
reference_test(test_list)
print(test_list)
>> ['Rest my immutability']

Functions, methods, and how many arguments do I have to give them?

Why do the following lines give me the same result?
str.upper('hello')
and
'hello'.upper()
I tried to do the same with list.append but got a TypeError.
list.append([1])
Is the str type in Python overloaded? How can this be achieved by writing a class/function? I would appreciate an example.
list.append takes two arguments - the list to modify and the element to append. So you need to do it like this:
ls = [1]
list.append(ls, 2)
which is equivalent to the much more popular:
ls.append(2)
str.upper and list.append are both functions.
str.upper takes one argument.
>>> str.upper('test')
'TEST'
list.append takes two arguments.
>>> my_list = []
>>> list.append(my_list, 1)
>>> my_list
[1]
str.upper and list.append (like other functions) are also non-data-descriptors with a __get__ method which in this context has two implications:
When you access the function through the class via the dot notation (str.upper, list.append) the function's __get__ method (i.e. string.upper.__get__ and list.append.__get__) is called but it returns just the function itself.
When you access the function through an instance (my_string.upper, my_list.append) the function's __get__ method is called and it will return a new callable acting like the original function, but with whatever was "in front of the dot" automatically passed as the first argument. .
That's why you need to pass 1 - 1 = 0 arguments when calling my_string.upper() and 2 - 1 = 1 argument when calling my_list.append(1).
>>> 'my_string'.upper()
'MY_STRING'
>>>
>>> my_list = []
>>> my_list.append(1)
>>> my_list
[1]
You could even get these modified callables (methods) by explicitly calling __get__ and passing the argument to be bound (what has been before the dot) as its argument.
>>> my_string = 'my_string'
>>> upper_maker = str.upper.__get__(my_string)
>>> upper_maker()
'MY_STRING'
>>>
>>> my_list = []
>>> appender = list.append.__get__(my_list)
>>> appender(1)
>>> my_list
[1]
Finally, here's a short example demonstrating how descriptor instances can detect whether they are being accessed via their owner-class or via an instance.
class Descriptor:
def __get__(self, instance, owner_class):
if instance is None:
print('accessed through class')
# list.append.__get__ would return list.append here
else:
print('accessed through instance')
# list.append.__get__ would build a new callable here
# that takes one argument x and that internally calls
# list.append(instance, x)
class Class:
attribute = Descriptor()
Class.attribute # prints 'accessed through class'
instance = Class()
instance.attribute # prints 'accessed through instance'
Quoting Dave Kirbys answer from Relationship between string module and str:
There is some overlap between the string module and the str type,
mainly for historical reasons. In early versions of Python str objects
did not have methods, so all string manipulation was done with
functions from the string module. When methods were added to the str
type (in Python 1.5?) the functions were left in the string module for
compatibility, but now just forward to the equivalent str method.
However the string module also contains constants and functions that
are not methods on str, such as formatting, character translation etc.
There is nothing at all magical going on with str (except that we have a nice syntactic shortcut to creating one using ""). You can write a class that behaves like str and list to see more clearly what is happening here.
class MyClass():
def __init__(self, arg):
self.val=str(arg)
def do_thing(self):
self.val = "asdf"
def do_thing_with_arg(self, arg):
self.val = "asdf " + str(arg)
def __repr__(self):
return self.val
my_thing = MyClass("qwerty")
# this is like 'hello'.upper()
my_thing.do_thing()
print(my_thing)
# it prints 'asdf'
my_thing = MyClass("qwerty")
# this is like str.upper('hello')
MyClass.do_thing(my_thing)
print(my_thing)
# it prints 'asdf'
my_thing = MyClass("qwerty")
# this is like my_list.append('qwerty')
my_thing.do_thing_with_arg('zxcv')
print(my_thing)
# it prints 'asdf zxcv'
my_thing = MyClass("qwerty")
# this is like list.append(my_list, 'qwerty')
MyClass.do_thing_with_arg(my_thing, 'zxcv')
print(my_thing)
# it prints 'asdf zxcv'
The short version is, you're invoking what looks like an "instance method" on a class, but you are supplying the instance ('self') yourself as the first argument to the function call.

Difference between 'array[:] = something' and 'array = something'

I know they practically do the same thing, but if you were to lets say do something like...
curpop = this_other_ndarray
i = 0;
while i<20:
curpop[:] = select(curpop, parameter, parameter1)
stuff
more stuff
curpop[:] = some_stuff_I_did
i += 1;
So the above code is just saying, before I enter a generational loop I am going to take an initial generation of populations from 'this other ndarray'.
Then I am planning on changing that array over and over and everytime I restart the loop I will only select some from itself but I will declare that as what it is equal to now. Is this okay to do in Python3?
Is the declaration of
'array[:] = some of it self'
versus
'array = some of itself'
different at all?
These are two totally different things.
The first is simple assignment.
foo = bar
This assignment statement merely says that the name on the left-hand side now refers to the same object as the name on the right-hand side. These statements do not modify either object.
Objects are neither created nor necessarily destroyed. If you lose the last name of an object, however, you will have lost the object. The CPython runtime uses reference counting as a memory management strategy, and will automatically reclaim objects that have a zero reference count.
In Python, variables act simply like object names that you can create, destroy, and change what they reference. Think of them like name-tags.
Now, a statement like:
foo[:] = bar
Is actually a method call. It can be translated to:
foo.__setitem__(slice(None, None, None), bar)
Observe:
>>> class Foo:
... def __setitem__(self, key, value):
... print("Key:", key, "Value:", value)
...
>>> class Bar: pass
...
>>> foo = Foo()
>>> bar = Bar()
>>> foo[:] = bar
Key: slice(None, None, None) Value: <__main__.Bar object at 0x104aa5c50>
So, really, the type of the objects control the ultimate effects of this statement. In the case of numpy.ndarray objects, slice-based assignment works similarly to list slice based assignment in that it mutates the array object in-place, with a few more caveats like broadcasting to throw into the mix. See the relevant docs:
https://docs.scipy.org/doc/numpy-1.13.0/user/basics.indexing.html#assigning-values-to-indexed-arrays
In many cases
curpop[:]= iterable_value_as_tuple_string_dictionary_and_list_etc
do the same thing as
curpop=iterable_value_as_tuple_string_dictionary_and_list_etc
of course assign a string first or at any step will remove the ability in the next steps to use [:] to assign something again
note that
curpop[:]= notiterable_value != curpop=notiterable_value
as the first assign notiterable_value to each element of curpop and the second assign the value notiterable_value to curpop itself

Python: make a copy of object when equal old object to new

I've created new class based on default str class. I've also changed default methods like __add__, __mul__, __repr__ etc. But I want to change default behaviour when user equal new variable to old one. Look what I have now:
a = stream('new stream')
b = a
b += ' was modified'
a == b
>>> True
print a
>>> stream('new stream was modified')
print b
>>> stream('new stream was modified')
So as you see each time I modify second variable Python also changes original variable. As I understand Python simply sends adress of variable a to variable b. Is it possible to make a copy of variable on creation like in usual str? As I think I need smth like new in C++.
a = 'new string'
b = a
b += ' was modified'
a == b
>>> False
P.S. Creation of the object begins in self.new() method. Creation is made like this:
def __new__(self, string):
return(str.__new__(self, string))
It is more complicated, because it takes care of unicode and QString type, first getting str object from them, but I think it's not neccessary.
I don't believe you can change the behavior of the assignment operator, but there are explicit ways to create a copy of an object rather than just using a reference. For a complex object, take a look at the copy module. For a basic sequence type (like str), the following works assuming you're implementing slice properly:
Code
a = str('abc')
#A slice creates a copy of a sequence object.
#[:] creates a copy of the entire thing.
b = a[:]
#Since b is a full copy of a, this will not modify a
b += ' was modified'
#Check the various values
print('a == b' + str(a == b))
print(a)
print(b)
Output
False
abc
abc was modified

Convert Variable Name to String?

I would like to convert a python variable name into the string equivalent as shown. Any ideas how?
var = {}
print ??? # Would like to see 'var'
something_else = 3
print ??? # Would print 'something_else'
TL;DR: Not possible. See 'conclusion' at the end.
There is an usage scenario where you might need this. I'm not implying there are not better ways or achieving the same functionality.
This would be useful in order to 'dump' an arbitrary list of dictionaries in case of error, in debug modes and other similar situations.
What would be needed, is the reverse of the eval() function:
get_indentifier_name_missing_function()
which would take an identifier name ('variable','dictionary',etc) as an argument, and return a
string containing the identifier’s name.
Consider the following current state of affairs:
random_function(argument_data)
If one is passing an identifier name ('function','variable','dictionary',etc) argument_data to a random_function() (another identifier name), one actually passes an identifier (e.g.: <argument_data object at 0xb1ce10>) to another identifier (e.g.: <function random_function at 0xafff78>):
<function random_function at 0xafff78>(<argument_data object at 0xb1ce10>)
From my understanding, only the memory address is passed to the function:
<function at 0xafff78>(<object at 0xb1ce10>)
Therefore, one would need to pass a string as an argument to random_function() in order for that function to have the argument's identifier name:
random_function('argument_data')
Inside the random_function()
def random_function(first_argument):
, one would use the already supplied string 'argument_data' to:
serve as an 'identifier name' (to display, log, string split/concat, whatever)
feed the eval() function in order to get a reference to the actual identifier, and therefore, a reference to the real data:
print("Currently working on", first_argument)
some_internal_var = eval(first_argument)
print("here comes the data: " + str(some_internal_var))
Unfortunately, this doesn't work in all cases. It only works if the random_function() can resolve the 'argument_data' string to an actual identifier. I.e. If argument_data identifier name is available in the random_function()'s namespace.
This isn't always the case:
# main1.py
import some_module1
argument_data = 'my data'
some_module1.random_function('argument_data')
# some_module1.py
def random_function(first_argument):
print("Currently working on", first_argument)
some_internal_var = eval(first_argument)
print("here comes the data: " + str(some_internal_var))
######
Expected results would be:
Currently working on: argument_data
here comes the data: my data
Because argument_data identifier name is not available in the random_function()'s namespace, this would yield instead:
Currently working on argument_data
Traceback (most recent call last):
File "~/main1.py", line 6, in <module>
some_module1.random_function('argument_data')
File "~/some_module1.py", line 4, in random_function
some_internal_var = eval(first_argument)
File "<string>", line 1, in <module>
NameError: name 'argument_data' is not defined
Now, consider the hypotetical usage of a get_indentifier_name_missing_function() which would behave as described above.
Here's a dummy Python 3.0 code: .
# main2.py
import some_module2
some_dictionary_1 = { 'definition_1':'text_1',
'definition_2':'text_2',
'etc':'etc.' }
some_other_dictionary_2 = { 'key_3':'value_3',
'key_4':'value_4',
'etc':'etc.' }
#
# more such stuff
#
some_other_dictionary_n = { 'random_n':'random_n',
'etc':'etc.' }
for each_one_of_my_dictionaries in ( some_dictionary_1,
some_other_dictionary_2,
...,
some_other_dictionary_n ):
some_module2.some_function(each_one_of_my_dictionaries)
# some_module2.py
def some_function(a_dictionary_object):
for _key, _value in a_dictionary_object.items():
print( get_indentifier_name_missing_function(a_dictionary_object) +
" " +
str(_key) +
" = " +
str(_value) )
######
Expected results would be:
some_dictionary_1 definition_1 = text_1
some_dictionary_1 definition_2 = text_2
some_dictionary_1 etc = etc.
some_other_dictionary_2 key_3 = value_3
some_other_dictionary_2 key_4 = value_4
some_other_dictionary_2 etc = etc.
......
......
......
some_other_dictionary_n random_n = random_n
some_other_dictionary_n etc = etc.
Unfortunately, get_indentifier_name_missing_function() would not see the 'original' identifier names (some_dictionary_,some_other_dictionary_2,some_other_dictionary_n). It would only see the a_dictionary_object identifier name.
Therefore the real result would rather be:
a_dictionary_object definition_1 = text_1
a_dictionary_object definition_2 = text_2
a_dictionary_object etc = etc.
a_dictionary_object key_3 = value_3
a_dictionary_object key_4 = value_4
a_dictionary_object etc = etc.
......
......
......
a_dictionary_object random_n = random_n
a_dictionary_object etc = etc.
So, the reverse of the eval() function won't be that useful in this case.
Currently, one would need to do this:
# main2.py same as above, except:
for each_one_of_my_dictionaries_names in ( 'some_dictionary_1',
'some_other_dictionary_2',
'...',
'some_other_dictionary_n' ):
some_module2.some_function( { each_one_of_my_dictionaries_names :
eval(each_one_of_my_dictionaries_names) } )
# some_module2.py
def some_function(a_dictionary_name_object_container):
for _dictionary_name, _dictionary_object in a_dictionary_name_object_container.items():
for _key, _value in _dictionary_object.items():
print( str(_dictionary_name) +
" " +
str(_key) +
" = " +
str(_value) )
######
In conclusion:
Python passes only memory addresses as arguments to functions.
Strings representing the name of an identifier, can only be referenced back to the actual identifier by the eval() function if the name identifier is available in the current namespace.
A hypothetical reverse of the eval() function, would not be useful in cases where the identifier name is not 'seen' directly by the calling code. E.g. inside any called function.
Currently one needs to pass to a function:
the string representing the identifier name
the actual identifier (memory address)
This can be achieved by passing both the 'string' and eval('string') to the called function at the same time. I think this is the most 'general' way of solving this egg-chicken problem across arbitrary functions, modules, namespaces, without using corner-case solutions. The only downside is the use of the eval() function which may easily lead to unsecured code. Care must be taken to not feed the eval() function with just about anything, especially unfiltered external-input data.
Totally possible with the python-varname package (python3):
from varname import nameof
s = 'Hey!'
print (nameof(s))
Output:
s
Install:
pip3 install varname
Or get the package here:
https://github.com/pwwang/python-varname
I searched for this question because I wanted a Python program to print assignment statements for some of the variables in the program. For example, it might print "foo = 3, bar = 21, baz = 432". The print function would need the variable names in string form. I could have provided my code with the strings "foo","bar", and "baz", but that felt like repeating myself. After reading the previous answers, I developed the solution below.
The globals() function behaves like a dict with variable names (in the form of strings) as keys. I wanted to retrieve from globals() the key corresponding to the value of each variable. The method globals().items() returns a list of tuples; in each tuple the first item is the variable name (as a string) and the second is the variable value. My variablename() function searches through that list to find the variable name(s) that corresponds to the value of the variable whose name I need in string form.
The function itertools.ifilter() does the search by testing each tuple in the globals().items() list with the function lambda x: var is globals()[x[0]]. In that function x is the tuple being tested; x[0] is the variable name (as a string) and x[1] is the value. The lambda function tests whether the value of the tested variable is the same as the value of the variable passed to variablename(). In fact, by using the is operator, the lambda function tests whether the name of the tested variable is bound to the exact same object as the variable passed to variablename(). If so, the tuple passes the test and is returned by ifilter().
The itertools.ifilter() function actually returns an iterator which doesn't return any results until it is called properly. To get it called properly, I put it inside a list comprehension [tpl[0] for tpl ... globals().items())]. The list comprehension saves only the variable name tpl[0], ignoring the variable value. The list that is created contains one or more names (as strings) that are bound to the value of the variable passed to variablename().
In the uses of variablename() shown below, the desired string is returned as an element in a list. In many cases, it will be the only item in the list. If another variable name is assigned the same value, however, the list will be longer.
>>> def variablename(var):
... import itertools
... return [tpl[0] for tpl in
... itertools.ifilter(lambda x: var is x[1], globals().items())]
...
>>> var = {}
>>> variablename(var)
['var']
>>> something_else = 3
>>> variablename(something_else)
['something_else']
>>> yet_another = 3
>>> variablename(something_else)
['yet_another', 'something_else']
as long as it's a variable and not a second class, this here works for me:
def print_var_name(variable):
for name in globals():
if eval(name) == variable:
print name
foo = 123
print_var_name(foo)
>>>foo
this happens for class members:
class xyz:
def __init__(self):
pass
member = xyz()
print_var_name(member)
>>>member
ans this for classes (as example):
abc = xyz
print_var_name(abc)
>>>abc
>>>xyz
So for classes it gives you the name AND the properteries
This is not possible.
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 = 1
baz = 1
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.
In Python, a name is a way to access an object, so there is no way to work with names directly. There might be some clever way to hack the Python bytecodes or something to get the value of the name, but that is at best a parlor trick.
If you know you want print foo to print "foo", you might as well just execute print "foo" in the first place.
EDIT: I have changed the wording slightly to make this more clear. Also, here is an even better example:
foo = 1
bar = foo
baz = foo
In practice, Python reuses the same object for integers with common values like 0 or 1, so the first example should bind the same object to all three names. But this example is crystal clear: the same object is bound to foo, bar, and baz.
Technically the information is available to you, but as others have asked, how would you make use of it in a sensible way?
>>> x = 52
>>> globals()
{'__builtins__': <module '__builtin__' (built-in)>, '__name__': '__main__',
'x': 52, '__doc__': None, '__package__': None}
This shows that the variable name is present as a string in the globals() dictionary.
>>> globals().keys()[2]
'x'
In this case it happens to be the third key, but there's no reliable way to know where a given variable name will end up
>>> for k in globals().keys():
... if not k.startswith("_"):
... print k
...
x
>>>
You could filter out system variables like this, but you're still going to get all of your own items. Just running that code above created another variable "k" that changed the position of "x" in the dict.
But maybe this is a useful start for you. If you tell us what you want this capability for, more helpful information could possibly be given.
By using the the unpacking operator:
>>> def tostr(**kwargs):
return kwargs
>>> var = {}
>>> something_else = 3
>>> tostr(var = var,something_else=something_else)
{'var' = {},'something_else'=3}
You somehow have to refer to the variable you want to print the name of. So it would look like:
print varname(something_else)
There is no such function, but if there were it would be kind of pointless. You have to type out something_else, so you can as well just type quotes to the left and right of it to print the name as a string:
print "something_else"
What are you trying to achieve? There is absolutely no reason to ever do what you describe, and there is likely a much better solution to the problem you're trying to solve..
The most obvious alternative to what you request is a dictionary. For example:
>>> my_data = {'var': 'something'}
>>> my_data['something_else'] = 'something'
>>> print my_data.keys()
['var', 'something_else']
>>> print my_data['var']
something
Mostly as a.. challenge, I implemented your desired output. Do not use this code, please!
#!/usr/bin/env python2.6
class NewLocals:
"""Please don't ever use this code.."""
def __init__(self, initial_locals):
self.prev_locals = list(initial_locals.keys())
def show_new(self, new_locals):
output = ", ".join(list(set(new_locals) - set(self.prev_locals)))
self.prev_locals = list(new_locals.keys())
return output
# Set up
eww = None
eww = NewLocals(locals())
# "Working" requested code
var = {}
print eww.show_new(locals()) # Outputs: var
something_else = 3
print eww.show_new(locals()) # Outputs: something_else
# Further testing
another_variable = 4
and_a_final_one = 5
print eww.show_new(locals()) # Outputs: another_variable, and_a_final_one
Does Django not do this when generating field names?
http://docs.djangoproject.com/en/dev//topics/db/models/#verbose-field-names
Seems reasonable to me.
I think this is a cool solution and I suppose the best you can get. But do you see any way to handle the ambigious results, your function may return?
As "is" operator behaves unexpectedly with integers shows, low integers and strings of the same value get cached by python so that your variablename-function might priovide ambigous results with a high probability.
In my case, I would like to create a decorator, that adds a new variable to a class by the varialbename i pass it:
def inject(klass, dependency):
klass.__dict__["__"+variablename(dependency)]=dependency
But if your method returns ambigous results, how can I know the name of the variable I added?
var any_var="myvarcontent"
var myvar="myvarcontent"
#inject(myvar)
class myclasss():
def myclass_method(self):
print self.__myvar #I can not be sure, that this variable will be set...
Maybe if I will also check the local list I could at least remove the "dependency"-Variable from the list, but this will not be a reliable result.
Here is a succinct variation that lets you specify any directory.
The issue with using directories to find anything is that multiple variables can have the same value. So this code returns a list of possible variables.
def varname( var, dir=locals()):
return [ key for key, val in dir.items() if id( val) == id( var)]
I don't know it's right or not, but it worked for me
def varname(variable):
for name in list(globals().keys()):
expression = f'id({name})'
if id(variable) == eval(expression):
return name
it is possible to a limited extent. the answer is similar to the solution by #tamtam .
The given example assumes the following assumptions -
You are searching for a variable by its value
The variable has a distinct value
The value is in the global namespace
Example:
testVar = "unique value"
varNameAsString = [k for k,v in globals().items() if v == "unique value"]
#
# the variable "varNameAsString" will contain all the variable name that matches
# the value "unique value"
# for this example, it will be a list of a single entry "testVar"
#
print(varNameAsString)
Output : ['testVar']
You can extend this example for any other variable/data type
I'd like to point out a use case for this that is not an anti-pattern, and there is no better way to do it.
This seems to be a missing feature in python.
There are a number of functions, like patch.object, that take the name of a method or property to be patched or accessed.
Consider this:
patch.object(obj, "method_name", new_reg)
This can potentially start "false succeeding" when you change the name of a method. IE: you can ship a bug, you thought you were testing.... simply because of a bad method name refactor.
Now consider: varname. This could be an efficient, built-in function. But for now it can work by iterating an object or the caller's frame:
Now your call can be:
patch.member(obj, obj.method_name, new_reg)
And the patch function can call:
varname(var, obj=obj)
This would: assert that the var is bound to the obj and return the name of the member. Or if the obj is not specified, use the callers stack frame to derive it, etc.
Could be made an efficient built in at some point, but here's a definition that works. I deliberately didn't support builtins, easy to add tho:
Feel free to stick this in a package called varname.py, and use it in your patch.object calls:
patch.object(obj, varname(obj, obj.method_name), new_reg)
Note: this was written for python 3.
import inspect
def _varname_dict(var, dct):
key_name = None
for key, val in dct.items():
if val is var:
if key_name is not None:
raise NotImplementedError("Duplicate names not supported %s, %s" % (key_name, key))
key_name = key
return key_name
def _varname_obj(var, obj):
key_name = None
for key in dir(obj):
val = getattr(obj, key)
equal = val is var
if equal:
if key_name is not None:
raise NotImplementedError("Duplicate names not supported %s, %s" % (key_name, key))
key_name = key
return key_name
def varname(var, obj=None):
if obj is None:
if hasattr(var, "__self__"):
return var.__name__
caller_frame = inspect.currentframe().f_back
try:
ret = _varname_dict(var, caller_frame.f_locals)
except NameError:
ret = _varname_dict(var, caller_frame.f_globals)
else:
ret = _varname_obj(var, obj)
if ret is None:
raise NameError("Name not found. (Note: builtins not supported)")
return ret
This will work for simnple data types (str, int, float, list etc.)
>>> def my_print(var_str) :
print var_str+':', globals()[var_str]
>>> a = 5
>>> b = ['hello', ',world!']
>>> my_print('a')
a: 5
>>> my_print('b')
b: ['hello', ',world!']
It's not very Pythonesque but I was curious and found this solution. You need to duplicate the globals dictionary since its size will change as soon as you define a new variable.
def var_to_name(var):
# noinspection PyTypeChecker
dict_vars = dict(globals().items())
var_string = None
for name in dict_vars.keys():
if dict_vars[name] is var:
var_string = name
break
return var_string
if __name__ == "__main__":
test = 3
print(f"test = {test}")
print(f"variable name: {var_to_name(test)}")
which returns:
test = 3
variable name: test
To get the variable name of var as a string:
var = 1000
var_name = [k for k,v in locals().items() if v == var][0]
print(var_name) # ---> outputs 'var'
Thanks #restrepo, this was exactly what I needed to create a standard save_df_to_file() function. For this, I made some small changes to your tostr() function. Hope this will help someone else:
def variabletostr(**df):
variablename = list(df.keys())[0]
return variablename
variabletostr(df=0)
The original question is pretty old, but I found an almost solution with Python 3. (I say almost because I think you can get close to a solution but I do not believe there is a solution concrete enough to satisfy the exact request).
First, you might want to consider the following:
objects are a core concept in Python, and they may be assigned a variable, but the variable itself is a bound name (think pointer or reference) not the object itself
var is just a variable name bound to an object and that object could have more than one reference (in your example it does not seem to)
in this case, var appears to be in the global namespace so you can use the global builtin conveniently named global
different name references to the same object will all share the same id which can be checked by running the id builtin id like so: id(var)
This function grabs the global variables and filters out the ones matching the content of your variable.
def get_bound_names(target_variable):
'''Returns a list of bound object names.'''
return [k for k, v in globals().items() if v is target_variable]
The real challenge here is that you are not guaranteed to get back the variable name by itself. It will be a list, but that list will contain the variable name you are looking for. If your target variable (bound to an object) is really the only bound name, you could access it this way:
bound_names = get_variable_names(target_variable)
var_string = bound_names[0]
Possible for Python >= 3.8 (with f'{var=}' string )
Not sure if this could be used in production code, but in Python 3.8(and up) you can use f' string debugging specifier. Add = at the end of an expression, and it will print both the expression and its value:
my_salary_variable = 5000
print(f'{my_salary_variable = }')
Output:
my_salary_variable = 5000
To uncover this magic here is another example:
param_list = f'{my_salary_variable=}'.split('=')
print(param_list)
Output:
['my_salary_variable', '5000']
Explanation: when you put '=' after your var in f'string, it returns a string with variable name, '=' and its value. Split it with .split('=') and get a List of 2 strings, [0] - your_variable_name, and [1] - actual object of variable.
Pick up [0] element of the list if you need variable name only.
my_salary_variable = 5000
param_list = f'{my_salary_variable=}'.split('=')
print(param_list[0])
Output:
my_salary_variable
or, in one line
my_salary_variable = 5000
print(f'{my_salary_variable=}'.split('=')[0])
Output:
my_salary_variable
Works with functions too:
def my_super_calc_foo(number):
return number**3
print(f'{my_super_calc_foo(5) = }')
print(f'{my_super_calc_foo(5)=}'.split('='))
Output:
my_super_calc_foo(5) = 125
['my_super_calc_foo(5)', '125']
Process finished with exit code 0
This module works for converting variables names to a string:
https://pypi.org/project/varname/
Use it like this:
from varname import nameof
variable=0
name=nameof(variable)
print(name)
//output: variable
Install it by:
pip install varname
print "var"
print "something_else"
Or did you mean something_else?

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