[Tutor] Need help
Hi, I am new in python. Could you guys please help me to short this code ? Want to write this iteration (Yellow one) using loop. r = 0 L1 = Label(bg = 'orange', text="Flat_No", relief=RIDGE,width=30) L1.grid(row=0,column=0) E1 = Entry(relief=SUNKEN,width=30) E1.grid(row=0,column=1) L2 = Label(bg = 'orange', text="Mains Unit", relief=RIDGE,width=30) L2.grid(row=1,column=0) E2 = Entry(relief=SUNKEN,width=30) E2.grid(row=1,column=1) L3 = Label(bg = 'orange', text="DG Unit", relief=RIDGE,width=30) L3.grid(row=2,column=0) E3 = Entry(relief=SUNKEN,width=30) E3.grid(row=2,column=1) L4 = Label(bg = 'orange', text="Month", relief=RIDGE,width=30) L4.grid(row=3,column=0) E4 = Entry(relief=SUNKEN,width=30) E4.grid(row=3,column=1) MyButton1 = Button(top, text="Submit", width=10, bg='red', command=lambda: database.data(E1.get(), E2.get(), E3.get(), E4.get())) MyButton1.grid(row=8, column=1) The output should be like this : [image: Inline image 1] Thanks Niraj -- Success occurs when opportunity and preparation meet ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Need help
Found the solution for this .
entry_option = ['Flat_No','Mains Unit','DG Unit','Month']
r = 0
entry = {}
label = {}
for item in entry_option:
lb = Label(bg = 'orange', text=item, relief=RIDGE,width=30)
lb.grid(row=r,column=0)
label[item] = lb
e = Entry(relief=SUNKEN,width=30)
e.grid(row=r,column=1)
entry[item] = e
r=r+1
But now how to pass these values as an argument for this function ?
command=lambda: database.data(E1.get(), E2.get(), E3.get(), E4.get())
Thanks
Niraj
On Wed, Sep 28, 2016 at 10:52 AM, niraj pandey
wrote:
> Hi,
>
> I am new in python. Could you guys please help me to short this code ?
>
> Want to write this iteration (Yellow one) using loop.
>
> r = 0
> L1 = Label(bg = 'orange', text="Flat_No", relief=RIDGE,width=30)
> L1.grid(row=0,column=0)
> E1 = Entry(relief=SUNKEN,width=30)
> E1.grid(row=0,column=1)
> L2 = Label(bg = 'orange', text="Mains Unit", relief=RIDGE,width=30)
> L2.grid(row=1,column=0)
> E2 = Entry(relief=SUNKEN,width=30)
> E2.grid(row=1,column=1)
> L3 = Label(bg = 'orange', text="DG Unit", relief=RIDGE,width=30)
> L3.grid(row=2,column=0)
> E3 = Entry(relief=SUNKEN,width=30)
> E3.grid(row=2,column=1)
> L4 = Label(bg = 'orange', text="Month", relief=RIDGE,width=30)
> L4.grid(row=3,column=0)
> E4 = Entry(relief=SUNKEN,width=30)
> E4.grid(row=3,column=1)
>
>
> MyButton1 = Button(top, text="Submit", width=10, bg='red', command=lambda:
> database.data(E1.get(), E2.get(), E3.get(), E4.get()))
> MyButton1.grid(row=8, column=1)
>
> The output should be like this :
>
> [image: Inline image 1]
>
> Thanks
> Niraj
> --
> Success occurs when opportunity and preparation meet
>
--
Success occurs when opportunity and preparation meet
___
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Re: [Tutor] differences between map and partial?
Hi, Danny, Thank you for your reply I checked the mail, the attachment is attached, i don't know why you can't see it ( I change the extension of .tar.gz to .tar.gz_src, as gmail can't pass the security check of the .tar.gz) I would like to inline the code as well as attached again( the are all simple files ) -- accouting -log1 -log2 logtool -__init__.py -tools.py run.py log1,log2 ( write anything you like, it pretend to be log file ) __init__.py :blank --tools.py #!/bin/env python def identy(string): print string; def unity(string): return 1 def line_op(string,op=identy): return apply(op,(string,)) def file_op(filename,lineop): f=open(filename) res=None; try: res=map(lambda x:line_op(x,lineop),f) finally: f.close() return res run.py-- import os #log_location="/home/source/python/pg/accounting" log_location='accounting' from multiprocessing import Pool from functools import partial if __name__=="__main__": from logtool.tools import * p=Pool(processes=4); for root,dir,file in os.walk(log_location): input=map(lambda x: os.path.join(root,x),file) print p.map(partial(file_op,lineop=unity),input) #print p.map(lambda x:file_op(x,unity), input) == THAT'S ALL OF THE CODE Thank you On Wed, Sep 28, 2016 at 3:26 AM, Danny Yoo wrote: > On Tue, Sep 27, 2016 at 2:23 AM, source liu wrote: >> Hi, List >> >> the test code as attached > > > Unfortunately, it didn't attach. If you can inline the content of the > test code, that would be helpful. > > >> this one works print p.map(partial(file_op,lineop=unity),input) >> this one doesn't work print p.map(lambda x:file_op(x,unity), input) > > > I don't know what the function signature of file_op is. This is an > important detail! > > > > I would expect: > > p.map(partial(file_op,lineop=unity),input) > > and: > > p.map(lambda x:file_op(x,lineop=unity), input) > > to be functionality equivalent. > > > > However, I do not know, without seeing file_op, whether: > > p.map(lambda x:file_op(x, lineop=unity), input) > > versus: > > p.map(lambda x:file_op(x, unity), input) > > Depending on method signature, this might not be equivalent, > especially if file_op takes in multiple keyword arguments. > > > > Also, I don't know what you mean by "works" vs. "doesn't work": those > are human expressions that are ambiguous enough that I only know > something is wrong, but I don't know what. :) Can you be more > specific? If you see error messages or stack traces, please post > them, as they are additional "signal" that may be helpful in figuring > out what's happening. -- Liu An Institution of modern physics, Shanghai, China ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] differences between map and partial?
source liu wrote: > Hi, Danny, > > > Thank you for your reply > > I checked the mail, the attachment is attached, i don't know why you > can't see it ( I change the extension of .tar.gz to .tar.gz_src, as > gmail can't pass the security check of the .tar.gz) They are stripped off by the mailing list software, so don't bother trying. > from multiprocessing import Pool > p=Pool(processes=4); > print p.map(partial(file_op,lineop=unity),input) > #print p.map(lambda x:file_op(x,unity), input) The missing part of information was that you are using multiprocessing (The traceback would also have shown the problem). Multiprocessing uses pickle to pass data around between processes, and anonymous functions (aka lambdas) cannot be pickled. Instead you have to use an ordinary function defined on the module level def whatever(x): return file_op(x, unity) p.map(whatever, input) For these pickle need not pass the code, it just remembers the module and function name which are then used to look up the function during unpickling. As you found out functools.partial() can be pickled, too, and thus works when all of its arguments can be pickled (in particular its first argument has to be a global function rather than a local one or a lambda). ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Need help
niraj pandey wrote:
> Found the solution for this.
You can further simplifiy it with enumerate()
> entry_option = ['Flat_No','Mains Unit','DG Unit','Month']
> entry = {}
> label = {}
for r, item in enumerate(entry_option):
> lb = Label(bg = 'orange', text=item, relief=RIDGE,width=30)
> lb.grid(row=r,column=0)
> label[item] = lb
> e = Entry(relief=SUNKEN,width=30)
> e.grid(row=r,column=1)
> entry[item] = e
>
> But now how to pass these values as an argument for this function ?
>
> command=lambda: database.data(E1.get(), E2.get(), E3.get(), E4.get())
Well, you saved the Entry instances in a dict, so you can retrieve them:
command=lambda: database.data(*[entry[k].get() for k in entry_option])
If you use a list instead of or in addition to the dict
entries = []
for ...: # your loop from above
...
entries.append(e)
the lambda becomes
command=lambda: database.data(*[e.get() for e in entries])
If you have not come across it before: the * operator unpacks the list, so
args = ["a", "b"]
f(*args)
is equivalent to calling f with all items in the list,
f(args[0], args[1])
in the above example.
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Re: [Tutor] Google is not your friend
On 28/09/16 01:25, Ben Finney wrote: > Alan Gauld via Tutor writes: > >> I'm not sure if they are up to 3.5 yet or not but the easiest way is >> just to grab a full distro such as Anaconda or Canopy. Google is your >> friend. > > I do wish that meme would die. Google is many things, but it is *not* > your friend. Corporations are not friends of anyone. Valid point, I actually use duck-duck-go personally, but I was intending to imply google in the generic sense of a search engine. So I should just have said search... -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Need help
Thanks Peter for the help.
Best Regards
Niraj
On Wed, Sep 28, 2016 at 2:47 PM, Peter Otten <__pete...@web.de> wrote:
> niraj pandey wrote:
>
> > Found the solution for this.
>
> You can further simplifiy it with enumerate()
> > entry_option = ['Flat_No','Mains Unit','DG Unit','Month']
>
> > entry = {}
> > label = {}
> for r, item in enumerate(entry_option):
> > lb = Label(bg = 'orange', text=item, relief=RIDGE,width=30)
> > lb.grid(row=r,column=0)
> > label[item] = lb
> > e = Entry(relief=SUNKEN,width=30)
> > e.grid(row=r,column=1)
> > entry[item] = e
> >
> > But now how to pass these values as an argument for this function ?
> >
> > command=lambda: database.data(E1.get(), E2.get(), E3.get(), E4.get())
>
> Well, you saved the Entry instances in a dict, so you can retrieve them:
>
> command=lambda: database.data(*[entry[k].get() for k in entry_option])
>
> If you use a list instead of or in addition to the dict
>
> entries = []
> for ...: # your loop from above
>...
>entries.append(e)
>
> the lambda becomes
>
> command=lambda: database.data(*[e.get() for e in entries])
>
> If you have not come across it before: the * operator unpacks the list, so
>
> args = ["a", "b"]
> f(*args)
>
> is equivalent to calling f with all items in the list,
>
> f(args[0], args[1])
>
> in the above example.
>
> ___
> Tutor maillist - Tutor@python.org
> To unsubscribe or change subscription options:
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>
--
Success occurs when opportunity and preparation meet
___
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Re: [Tutor] Need help
Ignore this I have done it.
Thanks
Niraj
On Wed, Sep 28, 2016 at 12:30 PM, niraj pandey
wrote:
> Found the solution for this .
>
> entry_option = ['Flat_No','Mains Unit','DG Unit','Month']
>
> r = 0
> entry = {}
> label = {}
> for item in entry_option:
> lb = Label(bg = 'orange', text=item, relief=RIDGE,width=30)
> lb.grid(row=r,column=0)
> label[item] = lb
> e = Entry(relief=SUNKEN,width=30)
> e.grid(row=r,column=1)
> entry[item] = e
> r=r+1
>
> But now how to pass these values as an argument for this function ?
>
> command=lambda: database.data(E1.get(), E2.get(), E3.get(), E4.get())
>
> Thanks
> Niraj
>
> On Wed, Sep 28, 2016 at 10:52 AM, niraj pandey > wrote:
>
>> Hi,
>>
>> I am new in python. Could you guys please help me to short this code ?
>>
>> Want to write this iteration (Yellow one) using loop.
>>
>> r = 0
>> L1 = Label(bg = 'orange', text="Flat_No", relief=RIDGE,width=30)
>> L1.grid(row=0,column=0)
>> E1 = Entry(relief=SUNKEN,width=30)
>> E1.grid(row=0,column=1)
>> L2 = Label(bg = 'orange', text="Mains Unit", relief=RIDGE,width=30)
>> L2.grid(row=1,column=0)
>> E2 = Entry(relief=SUNKEN,width=30)
>> E2.grid(row=1,column=1)
>> L3 = Label(bg = 'orange', text="DG Unit", relief=RIDGE,width=30)
>> L3.grid(row=2,column=0)
>> E3 = Entry(relief=SUNKEN,width=30)
>> E3.grid(row=2,column=1)
>> L4 = Label(bg = 'orange', text="Month", relief=RIDGE,width=30)
>> L4.grid(row=3,column=0)
>> E4 = Entry(relief=SUNKEN,width=30)
>> E4.grid(row=3,column=1)
>>
>>
>> MyButton1 = Button(top, text="Submit", width=10, bg='red',
>> command=lambda: database.data(E1.get(), E2.get(), E3.get(), E4.get()))
>> MyButton1.grid(row=8, column=1)
>>
>> The output should be like this :
>>
>> [image: Inline image 1]
>>
>> Thanks
>> Niraj
>> --
>> Success occurs when opportunity and preparation meet
>>
>
>
>
> --
> Success occurs when opportunity and preparation meet
>
--
Success occurs when opportunity and preparation meet
___
Tutor maillist - Tutor@python.org
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Re: [Tutor] NumPy and SciPy
On Tue, Sep 27, 2016 at 03:23:23PM +, Floeck, Thomas wrote: > Hi there, > > you have an idea where I can find NumPy and SciPy windows *.exe-files > for Python 3.5? Why does it have to be an .exe file? http://www.scipy.org/install.html -- Steve ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
