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Getting the name of a variable as a string
(32 answers)
Closed 4 months ago.
Is it possible to get the original variable name of a variable passed to a function? E.g.
foobar = "foo"
def func(var):
print var.origname
So that:
func(foobar)
Returns:
>>foobar
EDIT:
All I was trying to do was make a function like:
def log(soup):
f = open(varname+'.html', 'w')
print >>f, soup.prettify()
f.close()
.. and have the function generate the filename from the name of the variable passed to it.
I suppose if it's not possible I'll just have to pass the variable and the variable's name as a string each time.
EDIT: To make it clear, I don't recommend using this AT ALL, it will break, it's a mess, it won't help you in any way, but it's doable for entertainment/education purposes.
You can hack around with the inspect module, I don't recommend that, but you can do it...
import inspect
def foo(a, f, b):
frame = inspect.currentframe()
frame = inspect.getouterframes(frame)[1]
string = inspect.getframeinfo(frame[0]).code_context[0].strip()
args = string[string.find('(') + 1:-1].split(',')
names = []
for i in args:
if i.find('=') != -1:
names.append(i.split('=')[1].strip())
else:
names.append(i)
print names
def main():
e = 1
c = 2
foo(e, 1000, b = c)
main()
Output:
['e', '1000', 'c']
To add to Michael Mrozek's answer, you can extract the exact parameters versus the full code by:
import re
import traceback
def func(var):
stack = traceback.extract_stack()
filename, lineno, function_name, code = stack[-2]
vars_name = re.compile(r'\((.*?)\).*$').search(code).groups()[0]
print vars_name
return
foobar = "foo"
func(foobar)
# PRINTS: foobar
Looks like Ivo beat me to inspect, but here's another implementation:
import inspect
def varName(var):
lcls = inspect.stack()[2][0].f_locals
for name in lcls:
if id(var) == id(lcls[name]):
return name
return None
def foo(x=None):
lcl='not me'
return varName(x)
def bar():
lcl = 'hi'
return foo(lcl)
bar()
# 'lcl'
Of course, it can be fooled:
def baz():
lcl = 'hi'
x='hi'
return foo(lcl)
baz()
# 'x'
Moral: don't do it.
Another way you can try if you know what the calling code will look like is to use traceback:
def func(var):
stack = traceback.extract_stack()
filename, lineno, function_name, code = stack[-2]
code will contain the line of code that was used to call func (in your example, it would be the string func(foobar)). You can parse that to pull out the argument
You can't. It's evaluated before being passed to the function. All you can do is pass it as a string.
#Ivo Wetzel's answer works in the case of function call are made in one line, like
e = 1 + 7
c = 3
foo(e, 100, b=c)
In case that function call is not in one line, like:
e = 1 + 7
c = 3
foo(e,
1000,
b = c)
below code works:
import inspect, ast
def foo(a, f, b):
frame = inspect.currentframe()
frame = inspect.getouterframes(frame)[1]
string = inspect.findsource(frame[0])[0]
nodes = ast.parse(''.join(string))
i_expr = -1
for (i, node) in enumerate(nodes.body):
if hasattr(node, 'value') and isinstance(node.value, ast.Call)
and hasattr(node.value.func, 'id') and node.value.func.id == 'foo' # Here goes name of the function:
i_expr = i
break
i_expr_next = min(i_expr + 1, len(nodes.body)-1)
lineno_start = nodes.body[i_expr].lineno
lineno_end = nodes.body[i_expr_next].lineno if i_expr_next != i_expr else len(string)
str_func_call = ''.join([i.strip() for i in string[lineno_start - 1: lineno_end]])
params = str_func_call[str_func_call.find('(') + 1:-1].split(',')
print(params)
You will get:
[u'e', u'1000', u'b = c']
But still, this might break.
You can use python-varname package
from varname import nameof
s = 'Hey!'
print (nameof(s))
Output:
s
Package below:
https://github.com/pwwang/python-varname
For posterity, here's some code I wrote for this task, in general I think there is a missing module in Python to give everyone nice and robust inspection of the caller environment. Similar to what rlang eval framework provides for R.
import re, inspect, ast
#Convoluted frame stack walk and source scrape to get what the calling statement to a function looked like.
#Specifically return the name of the variable passed as parameter found at position pos in the parameter list.
def _caller_param_name(pos):
#The parameter name to return
param = None
#Get the frame object for this function call
thisframe = inspect.currentframe()
try:
#Get the parent calling frames details
frames = inspect.getouterframes(thisframe)
#Function this function was just called from that we wish to find the calling parameter name for
function = frames[1][3]
#Get all the details of where the calling statement was
frame,filename,line_number,function_name,source,source_index = frames[2]
#Read in the source file in the parent calling frame upto where the call was made
with open(filename) as source_file:
head=[source_file.next() for x in xrange(line_number)]
source_file.close()
#Build all lines of the calling statement, this deals with when a function is called with parameters listed on each line
lines = []
#Compile a regex for matching the start of the function being called
regex = re.compile(r'\.?\s*%s\s*\(' % (function))
#Work backwards from the parent calling frame line number until we see the start of the calling statement (usually the same line!!!)
for line in reversed(head):
lines.append(line.strip())
if re.search(regex, line):
break
#Put the lines we have groked back into sourcefile order rather than reverse order
lines.reverse()
#Join all the lines that were part of the calling statement
call = "".join(lines)
#Grab the parameter list from the calling statement for the function we were called from
match = re.search('\.?\s*%s\s*\((.*)\)' % (function), call)
paramlist = match.group(1)
#If the function was called with no parameters raise an exception
if paramlist == "":
raise LookupError("Function called with no parameters.")
#Use the Python abstract syntax tree parser to create a parsed form of the function parameter list 'Name' nodes are variable names
parameter = ast.parse(paramlist).body[0].value
#If there were multiple parameters get the positional requested
if type(parameter).__name__ == 'Tuple':
#If we asked for a parameter outside of what was passed complain
if pos >= len(parameter.elts):
raise LookupError("The function call did not have a parameter at postion %s" % pos)
parameter = parameter.elts[pos]
#If there was only a single parameter and another was requested raise an exception
elif pos != 0:
raise LookupError("There was only a single calling parameter found. Parameter indices start at 0.")
#If the parameter was the name of a variable we can use it otherwise pass back None
if type(parameter).__name__ == 'Name':
param = parameter.id
finally:
#Remove the frame reference to prevent cyclic references screwing the garbage collector
del thisframe
#Return the parameter name we found
return param
If you want a Key Value Pair relationship, maybe using a Dictionary would be better?
...or if you're trying to create some auto-documentation from your code, perhaps something like Doxygen (http://www.doxygen.nl/) could do the job for you?
I wondered how IceCream solves this problem. So I looked into the source code and came up with the following (slightly simplified) solution. It might not be 100% bullet-proof (e.g. I dropped get_text_with_indentation and I assume exactly one function argument), but it works well for different test cases. It does not need to parse source code itself, so it should be more robust and simpler than previous solutions.
#!/usr/bin/env python3
import inspect
from executing import Source
def func(var):
callFrame = inspect.currentframe().f_back
callNode = Source.executing(callFrame).node
source = Source.for_frame(callFrame)
expression = source.asttokens().get_text(callNode.args[0])
print(expression, '=', var)
i = 1
f = 2.0
dct = {'key': 'value'}
obj = type('', (), {'value': 42})
func(i)
func(f)
func(s)
func(dct['key'])
func(obj.value)
Output:
i = 1
f = 2.0
s = string
dct['key'] = value
obj.value = 42
Update: If you want to move the "magic" into a separate function, you simply have to go one frame further back with an additional f_back.
def get_name_of_argument():
callFrame = inspect.currentframe().f_back.f_back
callNode = Source.executing(callFrame).node
source = Source.for_frame(callFrame)
return source.asttokens().get_text(callNode.args[0])
def func(var):
print(get_name_of_argument(), '=', var)
If you want to get the caller params as in #Matt Oates answer answer without using the source file (ie from Jupyter Notebook), this code (combined from #Aeon answer) will do the trick (at least in some simple cases):
def get_caller_params():
# get the frame object for this function call
thisframe = inspect.currentframe()
# get the parent calling frames details
frames = inspect.getouterframes(thisframe)
# frame 0 is the frame of this function
# frame 1 is the frame of the caller function (the one we want to inspect)
# frame 2 is the frame of the code that calls the caller
caller_function_name = frames[1][3]
code_that_calls_caller = inspect.findsource(frames[2][0])[0]
# parse code to get nodes of abstract syntact tree of the call
nodes = ast.parse(''.join(code_that_calls_caller))
# find the node that calls the function
i_expr = -1
for (i, node) in enumerate(nodes.body):
if _node_is_our_function_call(node, caller_function_name):
i_expr = i
break
# line with the call start
idx_start = nodes.body[i_expr].lineno - 1
# line with the end of the call
if i_expr < len(nodes.body) - 1:
# next expression marks the end of the call
idx_end = nodes.body[i_expr + 1].lineno - 1
else:
# end of the source marks the end of the call
idx_end = len(code_that_calls_caller)
call_lines = code_that_calls_caller[idx_start:idx_end]
str_func_call = ''.join([line.strip() for line in call_lines])
str_call_params = str_func_call[str_func_call.find('(') + 1:-1]
params = [p.strip() for p in str_call_params.split(',')]
return params
def _node_is_our_function_call(node, our_function_name):
node_is_call = hasattr(node, 'value') and isinstance(node.value, ast.Call)
if not node_is_call:
return False
function_name_correct = hasattr(node.value.func, 'id') and node.value.func.id == our_function_name
return function_name_correct
You can then run it as this:
def test(*par_values):
par_names = get_caller_params()
for name, val in zip(par_names, par_values):
print(name, val)
a = 1
b = 2
string = 'text'
test(a, b,
string
)
to get the desired output:
a 1
b 2
string text
Since you can have multiple variables with the same content, instead of passing the variable (content), it might be safer (and will be simpler) to pass it's name in a string and get the variable content from the locals dictionary in the callers stack frame. :
def displayvar(name):
import sys
return name+" = "+repr(sys._getframe(1).f_locals[name])
If it just so happens that the variable is a callable (function), it will have a __name__ property.
E.g. a wrapper to log the execution time of a function:
def time_it(func, *args, **kwargs):
start = perf_counter()
result = func(*args, **kwargs)
duration = perf_counter() - start
print(f'{func.__name__} ran in {duration * 1000}ms')
return result
I am trying to write a class with 2 functions in Python, like this:
class Arithmetic(Enum):
add = 1
addi = 2
sub = 3
lui = 4
auipc = 5
def parse(inst, *argv):
if inst == "add":
return calculate_add(argv)
elif inst == "addi":
pass
elif inst == "sub":
pass
elif inst == "lui":
pass
elif inst == "auipc":
pass
def calculate_add(*argv):
return argv[0] + argv[1]
and I keep getting this error:
NameError: name 'calculate_add' is not defined. What am I doing wrong?
In your code, you have defined calculate_add after the parse function. So, when you run calculate_add in your parse function, Python does not recognize it, since it hasn't been defined yet.
To fix it, define calculate_add before you define parse.
I believe you must declare the function before trying to access it. 🙂
(It's not cool like Java)
this funstion works well :
def create_layer(layers_list,position=0,layer_obj=None):
if not layer_obj:
layer_obj,created = Layer.objects.get_or_create(name=layers_list[0])
return create_layer(layers_list,position,layer_obj)
else:
position +=1
if position < len(layers_list) :
obj,created = Layer.objects.get_or_create(name=layers_list[position],parent=layer_obj)
return create_layer(layers_list,position,obj)
else:
return layer_obj
But when I put it into class, I got error :
create_layer() takes at most 4 arguments (5 given)
class Test():
def test1(self):
self.create_layer(layers_list)
def create_layer(self,layers_list,position=0,layer_obj=None):
if not layer_obj:
layer_obj,created = Layer.objects.get_or_create(name=layers_list[0])
return self.create_layer(self,layers_list,position,layer_obj)
else:
position +=1
if position < len(layers_list) :
obj,created = Layer.objects.get_or_create(name=layers_list[position],parent=layer_obj)
return self.create_layer(self,layers_list,position,obj)
else:
return layer_obj
What happened??
How can I fix it ??
Don't pass self to self.create_layer when calling it recursively. This is done for you by Python when it calls a method of a class instance. What happens here is that Python adds another self to the front of your argument list, thereby making it one argument too many.
in fucntion getLink(urls), I have return (cloud,parent,children)
in main function, I have (cloud,parent,children) = getLink(urls) and I got error of this line: TypeError: 'NoneType' object is not iterable
parent and children are all list of http links. since, it is not able to paste them here, parent is a list contains about 30 links; children is a list contains about 30 items, each item is about 10-100 links which is divide by ",".
cloud is a list contain about 100 words, like that: ['official store', 'Java Applets Centre', 'About Google', 'Web History'.....]
I didnot know why I get an error. Is there anything wrong in passing parameter? Or because the list take too much space?
#crawler url: read webpage and return a list of url and a list of its name
def crawler(url):
try:
m = urllib.request.urlopen(url)
msg = m.read()
....
return (list(set(list(links))),list(set(list(titles))) )
except Exception:
print("url wrong!")
#this is the function has gone wrong: it throw an exception here, also the error I mentioned, also it will end while before len(parent) reach 100.
def getLink(urls):
try:
newUrl=[]
parent = []
children =[]
cloud =[]
i=0
while len(parent)<=100:
url = urls[i]
if url in parent:
i += 1
continue
(links, titles) = crawler(url)
parent.append(url)
children.append(",".join(links))
cloud = cloud + titles
newUrl= newUrl+links
print ("links: ",links)
i += 1
if i == len(urls):
urls = list(set(newUrl))
newUrl = []
i = 0
return (cloud,parent,children)
except Exception:
print("can not get links")
def readfile(file):
#not related, this function will return a list of url
def main():
file='sampleinput.txt'
urls=readfile(file)
(cloud,parent,children) = getLink(urls)
if __name__=='__main__':
main()
There might be a way that your function ends without reaching the explicit return statement.
Look at the following example code.
def get_values(x):
if x:
return 'foo', 'bar'
x, y = get_values(1)
x, y = get_values(0)
When the function is called with 0 as parameter the return is skipped and the function will return None.
You could add an explicit return as the last line of your function. In the example given in this answer it would look like this.
def get_values(x):
if x:
return 'foo', 'bar'
return None, None
Update after seing the code
When the exception is triggered in get_link you just print something and return from the function. You have no return statement, so Python will return None. The calling function now tries to expand None into three values and that fails.
Change your exception handling to return a tuple with three values like you do it when everything is fine. Using None for each value is a good idea for it shows you, that something went wrong. Additionally I wouldn't print anything in the function. Don't mix business logic and input/output.
except Exception:
return None, None, None
Then in your main function use the following:
cloud, parent, children = getLink(urls)
if cloud is None:
print("can not get links")
else:
# do some more work
I'm trying to get a sort by genre function to work with some python code, but I just can't seem to get it working. Currently, my sort function gets a TypeError stating that it takes exactly 2 arguments and there is only 1 given. I am getting the data from a json file.
Here is the beginning of my class:
import json
import operator
from operator import itemgetter
class Movielist():
dataFile = "mymoviedata.json"
def __init__(self):
print "Loading Program..."
self.welcome()
def loadData(self, dataFile):
movieObject = []
json_data=open(dataFile)
movieObject = json.load(json_data)
json_data.close()
return movieObject
Here is where I call the function
if option == "4":
print self.sortByGenre()
Here is the function:
def sortByGenre(self, FoundMedia):
sortedgenrelist = []
sortedgenrelist = sorted(FoundMedia, key = itemgetter('genre'))
return sortedgenrelist
The problem is that:
print self.sortByGenre()
Only passes 1 argument (self). The function requires 2:
def sortByGenre(self, FoundMedia):
...
You should be passing another argument -- Some argument for FoundMedia.