Re: convert a string to a variable
On Friday, May 25, 2018 at 8:06:31 AM UTC-4, Ned Batchelder wrote: > On 5/24/18 6:54 PM, bruceg113...@gmail.com wrote: > > I am trying to convert a string to a variable. > > > > I got cases 1 & 2 to work, but not cases 3 & 4. > > > > The print statement in cases 3 & 4 reports the following: > > builtins.AttributeError: type object 'animal' has no attribute 'tiger' > > > > I am stuck on creating variables that can be accessed as follows. > >animal.tiger > >self.animal.tiger > > > > Any suggestions? > > > > > > Usually when people want to turn strings into variables, the best answer > is to not make variables, but instead have a dictionary. > > But I don't know what you are going to do with "animal.tiger", so I'm > not sure the best answer. Can you say more about the whole problem? > > --Ned. Hi Ned, I am writing a small interpreter just for fun. My program reads and processes text from a file. Good or bad, I prefer the variable syntax. I got animal.tiger syntax to work. (thanks Dieter) I am still trying to get self.animal.tiger syntax to work. Thanks, Bruce -- https://mail.python.org/mailman/listinfo/python-list
Re: convert a string to a variable
On Friday, May 25, 2018 at 1:56:14 AM UTC-4, dieter wrote:
> bruceg113...@gmail.com writes:
>
> > I am trying to convert a string to a variable.
> >
> > I got cases 1 & 2 to work, but not cases 3 & 4.
> >
> > The print statement in cases 3 & 4 reports the following:
> > builtins.AttributeError: type object 'animal' has no attribute 'tiger'
> >
> > I am stuck on creating variables that can be accessed as follows.
> > animal.tiger
> > self.animal.tiger
> >
> > Any suggestions?
> > ...
> > # Case 3: This does not work
> > indata = 'animal.tiger'
> > vars()[indata] = "Tigers, big and strong!"
> > print (animal.tiger)
>
> In the expression "animal.tiger", the "variable" is "animal",
> not "animal.tiger". It is evaluated as follows:
> determine the object bound to "animal", access its attribute "tiger".
> Your error message tells you that the first step (object bound
> to "animal") has been successful, but the result lacks the
> attribute "tiger".
>
> > #Case 4: This does not work
> > class animal():
> > def create (self, indata):
> > vars(self)[indata] = "Tigers, big and strong!"
>
> Here you want to define the attribute "tiger" (I think),
> not "animal.tiger". Note that the "." in a Python expression
> (not a string) separates two individual steps: determine
> an object corresponding to the leftside to the "."; access
> the attribute corresponding to the name following the ".".
> > print (self.animal.tiger)
> >
> > tmp = animal()
> > tmp.create('animal.tiger')
Hi Dieter,
After reading your response, I picked up on the 'attribute' word.
Updated case 3: (Success)
indata = 'tiger'
setattr (animal, indata, "Tigers, big and strong!!!")
print (animal.tiger)
animal.tiger = "Lions are bigger and stronger"
print (animal.tiger)
Output is:
Tigers, big and strong!!!
Lions are bigger and stronger
I am still working on case 4 using setattr.
Thanks,
Bruce
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Re: best way to remove leading zeros from a tuple like string
On Monday, May 21, 2018 at 1:05:52 PM UTC-4, Rodrigo Bistolfi wrote:
> >>> repr(tuple(int(i) for i in s[1:-1].split(',')))
> '(128, 20, 8, 255, -1203, 1, 0, -123)'
>
> 2018-05-21 4:26 GMT-03:00 Peter Otten <__pete...@web.de>:
>
> > bruceg113...@gmail.com wrote:
> >
> > > Looking over the responses, I modified my original code as follows:
> > >
> > s = "(128, 020, 008, 255, -1203,01,-000, -0123)"
> > ",".join([str(int(i)) for i in s[1:-1].split(",")])
> > > '128,20,8,255,-1203,1,0,-123'
> >
> > I think this looks better with a generator instead of the listcomp:
> >
> > >>> ",".join(str(int(i)) for i in s[1:-1].split(","))
> > '128,20,8,255,-1203,1,0,-123'
> >
> >
> > --
> > https://mail.python.org/mailman/listinfo/python-list
> >
I am not familiar with the repr statement.
Looking into your response, I like it.
There is no need to add-in the parentheses.
With the repr statment:
>>> bb = repr (tuple (int (i) for i in s [1: -1] .split (',')))
>>> bb
'(128, 20, 8, 255, -1203, 1, 0, -123)'
>>> type(bb)
>>>
Without the repr statement:
>>> aa = (tuple (int (i) for i in s [1: -1] .split (',')))
>>> aa
(128, 20, 8, 255, -1203, 1, 0, -123)
>>> type(aa)
>>>
Thanks,
Bruce
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Re: Is there a simpler way to modify all arguments in a function before using the arguments?
On Saturday, November 10, 2012 10:35:12 AM UTC-5, Aahz wrote:
> In article ,
>
> Peter Otten <__pete...@web.de> wrote:
>
> >Miki Tebeka wrote:
>
> >
>
> >>> Is there a simpler way to modify all arguments in a function before using
>
> >>> the arguments?
>
> >>
>
> >> You can use a decorator:
>
> >>
>
> >> from functools import wraps
>
> >>
>
> >> def fix_args(fn):
>
> >> @wraps(fn)
>
> >> def wrapper(*args):
>
> >> args = (arg.replace('_', '') for arg in args)
>
> >> return fn(*args)
>
> >>
>
> >> return wrapper
>
> >>
>
> >> @fix_args
>
> >> def foo(x, y):
>
> >> print(x)
>
> >> print(y)
>
> >
>
> >I was tempted to post that myself, but he said /simpler/ ;)
>
>
>
> From my POV, that *is* simpler. When you change the parameters for foo,
>
> you don't need to change the arg pre-processing. Also allows code reuse,
>
> probably any program needing this kind of processing once will need it
>
> again.
>
> --
>
> Aahz (a...@pythoncraft.com) <*> http://www.pythoncraft.com/
>
>
>
> "Normal is what cuts off your sixth finger and your tail..." --Siobhan
Using a decorator works when named arguments are not used. When named arguments
are used, unexpected keyword error is reported. Is there a simple fix?
Thanks to all,
Bruce
Code:
-
from functools import wraps
def fix_args(fn):
@wraps(fn)
def wrapper(*args):
args = (arg.replace('_', '') for arg in args)
return fn(*args)
return wrapper
@fix_args
def foo(a1="", a2="", b1="", b2=""):
print(a1)
print(a2)
print(b1)
print(b2)
foo ('a1a1_x', 'a2a2_x', 'b1b1_x', 'b2b2_x')
foo (a1='a1a1_x', a2='a2a2_x', b1='b1b1_x', b2='b2b2_x')
Results:
a1a1x
a2a2x
b1b1x
b2b2x
Traceback (most recent call last):
File "C:\WORK\masterDB_Update\argtest.py", line 19, in
foo (a1='a1a1_x', a2='a2a2_x', b1='b1b1_x', b2='b2b2_x')
TypeError: wrapper() got an unexpected keyword argument 'a1'
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