Thanks for the response and I can certainly see that this old code can be improved, but I
respectfully disagree on the utility of this function. The flexibility of passing any
number of arguments to the function and returning a dictionary is much easier than
writing out dict(x=x, y=y, ...n=n). I also believe that "makeDict" makes my
code very readable.
My question is around finding the names of the variables passed to a function
from within the function. I have spent many hours looking on the web looking
for where in the frames and stacks this information is stored. Does anyone
have any insight where/how I can find the variables associated with the
arguments in
dctA = makeDict(strA, intA)
from within the function
def makeDict(*args):
.... #need to find the variable names "strA" and "intA" in this
context
---Andrew
On Apr 26, 2014, at 05:10 AM, Ned Batchelder <n...@nedbatchelder.com> wrote:
On 4/26/14 1:50 AM, Andrew Konstantaras wrote:
> I wrote the following code that works in Python 2.7 that takes the
> variables passed to the function into a dictionary. The following call:
>
> strA = 'a'
> intA = 1
> dctA = makeDict(strA, intA)
>
> produces the following dictionary:
>
> {'strA':'a', 'intA':1}
>
> To access the names passed into the function, I had to find the
> information by parsing through the stack. The code that used to work
is:
>
> from traceback import extract_stack
>
> def makeDict(*args):
> strAllStack = str(extract_stack())
> intNumLevels = len(extract_stack())
> intLevel = 0
> blnFinished = False
> while not blnFinished:
> strStack = str(extract_stack()[intLevel])
> if strStack.find("makeDict(") >0:
> blnFinished = True
> intLevel += 1
> if intLevel >= intNumLevels:
> blnFinished = True
> strStartText = "= makeDict("
> intLen = len(strStartText)
> intOpenParenLoc = strStack.find(strStartText)
> intCloseParenLoc = strStack.find(")", intOpenParenLoc)
> strArgs = strStack[ intOpenParenLoc+intLen : intCloseParenLoc ].strip()
> lstVarNames = strArgs.split(",")
> lstVarNames = [ s.strip() for s in lstVarNames ]
> if len(lstVarNames) == len(args):
> tplArgs = map(None, lstVarNames, args)
> newDict = dict(tplArgs)
> return newDict
> else:
> return "Error, argument name-value mismatch in function 'makeDict'.
> lstVarNames: " + str(lstVarNames) + "\n args: " + str(args),
strAllStack
>
> The same code does not work in Python 3.3.4. I have tried parsing
> through the stack information and frames and I can't find any reference
> to the names of the arguments passed to the function. I have tried
> inspecting the function and other functions in the standard modules,
but
> I can't seem to find anything that will provide this information.
1) This is a very strange function. Instead of going to all this
trouble, why not use:
dctA = dict(strA=strA, intA=intA)
Yes, you have to repeat the names, but you'd be done by now, it works on
both versions of Python, and people reading your code would understand
what it does.
2) Why is your code examining the entire stack? The frame you want is
the one just above you. Why are you stringifying the tuple produced by
extract_stack when the source line you want is the fourth element? Why
are you using a while loop and a counter to iterate over elements of a list?
3) In the error case you return a tuple when the caller is expecting a
dict? Why not raise an exception?
4) Your code only works if makeDict is assigned to a name. When I first
tried it, I used "print(makeDict(...))", and it failed completely.
5) You haven't mentioned what goes wrong on Python 3, but when I tried
it, I got:
Traceback (most recent call last):
File "foo3.py", line 34, in <module >
d = makeDict(a, b)
File "foo3.py", line 27, in makeDict
newDict = dict(list(tplArgs))
TypeError: 'NoneType' object is not callable
Looking at your code, I see:
tplArgs = map(None, lstVarNames, args)
I didn't realize map accepted a callable of None (TIL!), but it no
longer does in Python 3. You'll have to do this a different way.
But seriously: just use dict() and be done with it. There's a lot here
that can be much simpler if you learn to use Python as it was meant to
be used. You are swimming upstream, there are easier ways.
>
> Can anyone point me in the direction to find this information? Any help
> is appreciated.
>
> ---Andrew
>
>
>
--
Ned Batchelder, http://nedbatchelder.com
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