🌐 Mailman 3 February 2021 - Tutor - python.org

Scripting Calligra sheets with Python
by Jim Byrnes Jan. 31, 2022

Jan. 31, 2022

I am trying to script Calligra Sheets (formerly KSpread) with python. I have gotten some of the included example scripts to run so I know python scripting is running. I found the following snippet on their website: import KSpread sheet = KSpread.view().sheet() # swap text of B5 and C6 t1 = sheet.text("B5") t2 = sheet.text(6,3) sheet.setText("B5", t2) sheet.setText(6, 3, t1) # swap value of D7 and E8 v1 = sheet.value("D7") v2 = sheet.value(8,5) sheet.setValue("D7", v2) sheet.setValue(8, 5, v1) Error: 'str' object has no attribute 'text' File "file:///usr/share/kde4/apps/sheets/scripts/extensions/myswap.py", line 4, in <module> This error message appeared in a dialog box. I've taught myself some Python but I don't understand what is happening here. Shouldn't t1 be a sheet object? What would cause it to be a str instead? Thanks, Jim

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elif statement doesn't work
by hongerlapjes Feb. 28, 2021

Feb. 28, 2021

For school I have to make a simple roulette game. If the user inputs "STOP", the game starts. But I cant seem to make the input STOP. ValueError: invalid literal for int() with base 10: 'STOP' What am I doing wrong? """ ROULETTESPEL """ import random invoer = ["STOP"] for x in range(37): invoer.append(x) getallen = [] def roulette (): print("Rien ne va plus.") winnendgetal=(random.randint(0,36)) print("De uitkomst is", winnendgetal) for winnendgetal in getallen: fiches +=35 print("Je hebt gewonnen en je hebt nu" , fiches, "fiches.") else: if fiches == 0: print("GAME OVER") def start (): fiches = 10 print("Je hebt", fiches, "fiches!") while fiches > 0: inzet=input("Op welk getal wil je inzetten?"'\n') print(inzet) if int(inzet) >= 0 and int(inzet) <= 36 : fiches -= 1 getallen.append(inzet) print(getallen) print("Je hebt nog", fiches, "fiches over.") elif str(inzet) == "STOP" : roulette() else: start() start () roulette ()

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How this list works
by Alan Gauld Feb. 28, 2021

Feb. 28, 2021

Over the last few months I've had several emails asking me to delete certain posts (including one from mid 2018!). The usual reason for this is that the sender has included personal details that they do not wish published. I thought I'd better make it clear how the list works and the limits of our (the moderators) abilities. When you post a mail to the list it comes to the server and possibly into the moderation queue. (Only a small proportion go to the queue, most go straight through. If it comes to the moderators there is a very narrow window in which you can catch or attention and prevent it being posted. If approved, (either automatically by the server or by a moderator) it goes to the mail relay server which then sends new email copies out to every subscriber (or adds it to the digest for those who use that feature) - that's currently over 800 people in total . Once those emails go out they *cannot be retrieved*. This is not Microsoft Exchange! Some of the addresses on the list are mail archive sites (there are at least 3 that I know of). When they receive the email they automatically publish it on their archive site, subject only to their local security policy. Those sites are independent of the mailing list (even the python.org archive) and we do not have any control or access to them. We do not even know how many there are for sure. These sites can be accessed by potentially tens of thousands (even millions!) of viewers - anyone on the internet in fact. So, in summary, when posting to the tutor list be mindful that it is a public message and do not include any personal information that you are not happy to share with the world. Once posted it cannot be retrieved. It is effectively a permanent record on public display. -- 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

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Re: [Tutor] Python problem
by Vakhtang Kvaratskhelia Feb. 27, 2021

Feb. 27, 2021

Thank you very much for the reply. i worked out other small problems but the main question was: can i use bisection search with the integer division? Every time i run a bisection search code through python tutor visualization the outcome is the same: the integer division turns into an infinite loop. What i mean is that if i write the code (for simplicity i have precalculated variables mentioned in the previous emails and narrowed the code down to the following ) : low=0 high=10000 epsilon=100 portion_saved=0 while abs(250,000 - portion_saved * 566,774) >= epsilon: if portion_saved * 566774 < 250,000 : low=portion_saved else: high=portion_saved portion_saved=(high+low) / 2 print(portion_saved) The result is the same as the answer in the book that had the problem. but if i run the code: low=0 high=10000 epsilon=100 portion_saved=0 while abs(250,000 - portion_saved * 566,774) >= epsilon: if portion_saved * 566,774 < 250,000 : low=portion_saved else: high=portion_saved portion_saved=(high+low) // 2 print(portion_saved) Where i use "//" instead of "/" i get infinite loop because the "//" gets stuck on one particular guess and doesn't continue the search as the "/" division would. So then how do i use the hint which was given to me in the book? the hint was the following: "Because we are searching for a value that is in principle a float, we are going to limit ourselves to two decimals of accuracy (i.e., we may want to save at 7.04% or 0.0704 in decimal – but we are not going to worry about the difference between 7.041% and 7.039%). This means we can search for an integer between 0 and 10000 (using integer division), and then convert it to a decimal percentage (using float division) to use when we are calculating the current_savings after 36 months. By using this range, there are only a finite number of numbers that we are searching over, as opposed to the infinite number of decimals between 0 and 1. This range will help prevent infinite loops. The reason we use 0 to 10000 is to account for two additional decimal places in the range 0% to 100%. Your code should print out a decimal (e.g. 0.0704 for 7.04%)." Is it possible that the hint doesn't mean "//" by the "integer division"? Thank you in advance. On Tue, Feb 23, 2021 at 10:42 PM Dennis Lee Bieber <wlfraed(a)ix.netcom.com> wrote: > On Sat, 20 Feb 2021 15:27:01 +0400, Vakhtang Kvaratskhelia > <va_kvaratskhelia(a)cu.edu.ge> declaimed the following: > > I'm going all the way back to your first post just to comment on a > number of inefficiencies seen in the provide code. > > Note: as you have NOT provided the code using // I can not comment > on > possible problems there... > > > > >However i am unable to prevent an infinite loop in case the annual salary > >is 10,000 > > > > Given the nature of the values provided in the code, I don't know > if > you posted scaled integers (for use in the // case?) or "real" values. I > would suspect that "annual salary" that low is a value that will not > converge to a solution -- that is, the income is too low to make payments > which will close out the cost in the time span specified. > > > >I want to use this hint : "Because we are searching for a value that is in > >principle a float, we are going to limit ourselves to two decimals of > >accuracy (i.e., we may want to save at 7.04% or 0.0704 in decimal – but we > >are not going to worry about the difference between 7.041% and 7.039%). > > This clause would appear to define the convergence/termination > criteria. It is unclear, however, if that criteria is to be abs(upper - > lower) < 0.01, or < 0.001, or something else like < 0.005. > > Now, comments on your posted code... > > > > >annual_salary=150000 > >total_cost=1000000 > >Months=36 > >semi_annual_raise=0.07 > >portion_down_payment=0.25 > >r=0.04 > > This is not used in your code -- I'm presuming it was meant to be > used > where your hard-coded 0.04 below. > > >portion_saved=0 > >epsilon=100 > >possible_savings=0 > >Steps=0 > >for i in range(0,Months): > > range(0, months) is the same as range(months) > > However, since you keep doing (i+1)%6 in subsequent code, it would > be > better to use range(1, months+1) and i%6 which uses > only one > addition. > > > possible_savings+= annual_salary/12 + possible_savings*0.04/12 > > Similarly, you keep dividing annual salary by 12, and 0.04 by 12. > It > would be more efficient to precompute monthly salary and monthly interest > rate outside the loop (and recompute monthly salary when a raise is > applied). > > > if (i+1)%6 == 0: > > As mentioned, changing the range() call allows removing the +1 from > this. > > > annual_salary=annual_salary*(1+semi_annual_raise) > > And here, again, you have an addition that could have been done > outside > the loop. My solution keeps an addition, but changes where it takes place. > > >low=0 > >high=10000 > >while abs(total_cost*portion_down_payment - portion_saved * > >possible_savings) >= epsilon: > > Again, total cost * portion down payment never changes, precompute > it > outside the loop. > > I also note that "high" appears to be a scaled integer percentage, > yet > this code is not using the // operator. Low/high should be decimal fraction > percentage 0.0/1.0. As noted above, your convergence criteria is supposed > to be when the difference between low/high is < some limit. I don't > understand what your "while" statement is doing with some epsilon > (especially as I don't know if your "100" epsilon is supposed to represent > actual 100.0, or is a scaled integer again and represents 1.0) > > > if portion_saved * possible_savings < total_cost*portion_down_payment > : > > low=portion_saved > > else: > > high=portion_saved > > portion_saved=(high+low)/2 > > You never test for convergence -- should low/high end up with the > same > value, portion saved will not change. > > You also never showed us the result of running your code, so > knowing > what is a "right" result is difficult. My attempt at rewriting your logic > into a reusable format (your code requires editing to change any parameter) > resulted in... (watch out for line wraps) > > -=-=- > > def getInt(prompt): > while True: > sInt = input("Enter %s as an integer: " % prompt) > try: > iInt = int(sInt) > break > except: #yes, a bare except clause, this is not production > print("Invalid entry for integer: '%s'\n" % sInt) > return iInt > > def getFloat(prompt): > while True: > sFloat = input("Enter %s as a float: " % prompt) > try: > fFloat = float(sFloat) > break > except: #yes, a bare except clause, this is not production > print("Invalid entry for float: '%s'\n" % sFloat) > return fFloat > > def getPercent(prompt): > while True: > sFloat = input("Enter %s as a float percentage [nn.n, not 0.nnn]: " > % prompt) > try: > fFloat = float(sFloat) > if not (0.0 <= fFloat <= 100.0): > print("Invalid range for percent 0.0 - 100.0: %s\n" % > fFloat) > else: > break > except: #yes, a bare except clause, this is not production > print("Invalid entry for float: '%s'\n" % fFloat) > return fFloat / 100.0 #convert to decimal fraction 0.0-1.0 > > > annualSalary = getFloat("annual salary") > monthlySalary = annualSalary / 12.0 #algorithm works in months > > totalCost = getFloat("total cost") > > months = getInt("months") > > raisePercentRate = getPercent("rate of periodic raises") > raisePeriod = getInt("months between raises") > > downPaymentRate = getPercent("percent of cost used for down payment") > > annualInterestRate = getPercent("annual interest rate") > monthlyInterestRate = annualInterestRate / 12.0 > > convergenceCriteria = getPercent("convergence criteria") > > #compute maximum possible savings > possibleSavings = 0 > for month in range(1, months + 1): > possibleSavings += monthlySalary + possibleSavings * > monthlyInterestRate > if month % raisePeriod == 0: > annualSalary += annualSalary * raisePercentRate > monthlySalary = annualSalary / 12.0 > > #iterate over percentage (as decimal fraction) range > steps = 0 > low = 0.0 > high = 1.0 > downPayment = totalCost * downPaymentRate > while abs(high - low) > convergenceCriteria: > portionSaved = (high + low) / 2.0 > steps += 1 > if portionSaved * possibleSavings < downPayment: > low = portionSaved > else: > high = portionSaved > > print("Optimal portion saved is %7.2f %%" % (portionSaved * 100.0)) > print("Search required %s steps\n\n" % steps) > > -=-=- > > ... which produced the following when entering what I think is your set of > parameters. I have no idea if the results are correct. > > -=-=- > C:\Users\Wulfraed\Documents\_Hg-Repositories\Python > Progs>bisectionSearch.py > Enter annual salary as a float: 150000.0 > Enter total cost as a float: 1000000.0 > Enter months as an integer: 36 > Enter rate of periodic raises as a float percentage [nn.n, not 0.nnn]: 7.0 > Enter months between raises as an integer: 6 > Enter percent of cost used for down payment as a float percentage [nn.n, > not 0.nnn]: 25.0 > Enter annual interest rate as a float percentage [nn.n, not 0.nnn]: 4.0 > Enter convergence criteria as a float percentage [nn.n, not 0.nnn]: 1.0 > Optimal portion saved is 44.53 % > Search required 7 steps > > > > C:\Users\Wulfraed\Documents\_Hg-Repositories\Python Progs> > -=-=- > > Out of curiosity, just what does "portion saved" really represent? > In > my tests it appears to be the percentage of salary required to make > payments, not the amount one saves. > > Also, there is no test for feasibility -- all three of these runs > came > with the same result, but that result likely means the salary is too small > to make the payments. > > -=-=- > C:\Users\Wulfraed\Documents\_Hg-Repositories\Python > Progs>bisectionSearch.py > Enter annual salary as a float: 15000.0 > Enter total cost as a float: 1000000.0 > Enter months as an integer: 36 > Enter rate of periodic raises as a float percentage [nn.n, not 0.nnn]: 7.0 > Enter months between raises as an integer: 6 > Enter percent of cost used for down payment as a float percentage [nn.n, > not 0.nnn]: 25.0 > Enter annual interest rate as a float percentage [nn.n, not 0.nnn]: 4.0 > Enter convergence criteria as a float percentage [nn.n, not 0.nnn]: 1.0 > Optimal portion saved is 99.22 % > Search required 7 steps > > > > C:\Users\Wulfraed\Documents\_Hg-Repositories\Python > Progs>bisectionSearch.py > Enter annual salary as a float: 10000.0 > Enter total cost as a float: 1000000.0 > Enter months as an integer: 36 > Enter rate of periodic raises as a float percentage [nn.n, not 0.nnn]: 7. > Enter months between raises as an integer: 6 > Enter percent of cost used for down payment as a float percentage [nn.n, > not 0.nnn]: 25 > Enter annual interest rate as a float percentage [nn.n, not 0.nnn]: 4 > Enter convergence criteria as a float percentage [nn.n, not 0.nnn]: 1 > Optimal portion saved is 99.22 % > Search required 7 steps > > > > C:\Users\Wulfraed\Documents\_Hg-Repositories\Python > Progs>bisectionSearch.py > Enter annual salary as a float: 9000 > Enter total cost as a float: 1000000 > Enter months as an integer: 36 > Enter rate of periodic raises as a float percentage [nn.n, not 0.nnn]: 7 > Enter months between raises as an integer: 6 > Enter percent of cost used for down payment as a float percentage [nn.n, > not 0.nnn]: 25 > Enter annual interest rate as a float percentage [nn.n, not 0.nnn]: 4 > Enter convergence criteria as a float percentage [nn.n, not 0.nnn]: 1 > Optimal portion saved is 99.22 % > Search required 7 steps > -=-=- > > In fact, annual salary needs to be around 70000.00 before one gets > something less than "99.22" > > -=-=- > C:\Users\Wulfraed\Documents\_Hg-Repositories\Python > Progs>bisectionSearch.py > Enter annual salary as a float: 70000 > Enter total cost as a float: 1000000 > Enter months as an integer: 36 > Enter rate of periodic raises as a float percentage [nn.n, not 0.nnn]: 7 > Enter months between raises as an integer: 6 > Enter percent of cost used for down payment as a float percentage [nn.n, > not 0.nnn]: 25 > Enter annual interest rate as a float percentage [nn.n, not 0.nnn]: 4 > Enter convergence criteria as a float percentage [nn.n, not 0.nnn]: 1 > Optimal portion saved is 94.53 % > Search required 7 steps > -=-=- > > A similar condition applies when the annual salary is larger than > total > cost. > > > -- > Wulfraed Dennis Lee Bieber AF6VN > wlfraed(a)ix.netcom.com > http://wlfraed.microdiversity.freeddns.org/ > > _______________________________________________ > Tutor maillist - Tutor(a)python.org > To unsubscribe or change subscription options: > https://mail.python.org/mailman/listinfo/tutor >

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Re: [Tutor] Python problem
by Mark Lawrence Feb. 27, 2021

Feb. 27, 2021

On 27/02/2021 13:32, Dennis Lee Bieber wrote: > On Sat, 27 Feb 2021 11:23:19 +0400, Vakhtang Kvaratskhelia > <va_kvaratskhelia(a)cu.edu.ge> declaimed the following: > > >> low=0 >> high=10000 >> epsilon=100 >> portion_saved=0 >> while abs(250,000 - portion_saved * 566,774) >= epsilon: >> if portion_saved * 566774 < 250,000 : >> low=portion_saved >> else: >> high=portion_saved >> portion_saved=(high+low) / 2 >> print(portion_saved) >> >> The result is the same as the answer in the book that had the problem. >> >> but if i run the code: >> >> low=0 >> high=10000 >> epsilon=100 >> portion_saved=0 >> while abs(250,000 - portion_saved * 566,774) >= epsilon: >> if portion_saved * 566,774 < 250,000 : >> low=portion_saved >> else: >> high=portion_saved >> portion_saved=(high+low) // 2 >> print(portion_saved) >> >> Where i use "//" instead of "/" i get infinite loop because the "//" gets >> stuck on one particular guess and doesn't continue the search as the "/" >> division would. >> > > In both versions, you NEVER TEST for high/low converging on a value. > Your WHILE loop is testing on the monetary (?) computation, and I don't > know what units epsilon is supposed to be -- scaled integer for "1.00", or > an actual difference of 100(currency units). > > Reread your "hint"... > >> "Because we are searching for a value that is in principle a float, we are >> going to limit ourselves to two decimals of accuracy (i.e., we may want to >> save at 7.04% or 0.0704 in decimal – but we are not going to worry about >> the difference between 7.041% and 7.039%). This means we can search for an >> integer between 0 and 10000 (using integer division), and then convert it >> to a decimal percentage (using float division) to use when we are > > The convergence is on the scaled %age, NOT on the monetary value you > compute. The binary search is for %, not money result. > > You have to determine how close high and low can be to each other to > count as "same value". Note that 7.039..7.041 spans a range of 0.002. So: > abs(high - low) < 0.002 would indicate you are done with the loop. Now > scale that to your integer version (which is going to be a problem -- using > the suggested scale means the above would be 703.9..704.1, but you don't > have the .9/.1 -- your smallest epsilon is 1 [704 - 703] corresponding to > 0.01%, not the 0.002% of the example) > > Note: the number of steps taken will be consistent, and based solely > upon how many significant digits you are carrying: 0..100% with epsilon 1% > will always take 7 steps. 0..100% with epsilon 0.1% will take 10 steps. > Binary search, 2^7 spans 0..127; 2^10 spans 0..1024. This is why I state > that using scaled integers and integer division is futile -- the search > ends when the difference between high and low is less than your selected > epsilon value, so "infinite number of floating values between 0.0 .. 100.0" > just doesn't matter -- if epsilon is 1.0% (100 as scaled integer) the loop > will take 7 steps; if 0.1% (10 as scaled integer) it will take 10 steps. > > The only justification I can see for using scaled integers is that one > is running on a microcontroller that does not have floating point > operations at all. > I do not see the purpose of all this analysis when the code given does not even run:- cat mytest.py low = 0 high = 10000 epsilon = 100 portion_saved = 0 while abs(250, 000 - portion_saved * 566, 774) >= epsilon: if portion_saved * 566774 < 250, 000: low = portion_saved else: high = portion_saved portion_saved = (high+low) / 2 print(portion_saved) mark@mark-HP-15-Notebook-PC:~/mypython$ python3 ./mytest.py File "./mytest.py", line 6 if portion_saved * 566774 < 250, 000: ^ SyntaxError: invalid syntax The call to abs gets away with it as there are three parameters listed, which is wrong anyway as abs only takes one argument. In fact check the difference between 566, 774 in the call to abs and 566774 in the line below. Correct all the problems and the output I get is 0.4410743713378906. Is that correct? I've no idea and I'm not bothered enough to check. -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence

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(no subject)
by ੧ Feb. 27, 2021

Feb. 27, 2021

Sir, For the question - Define a function called myfunc that takes in a string, and returns a matching string where every even letter is uppercase, and every odd letter is lowercase. Assume that the incoming string only contains letters, and don't worry about numbers, spaces or punctuation. The output string can start with either an uppercase or lowercase letter, so long as letters alternate throughout the string. I tried - def mufunc(**kwargs): finaloutput = '' " i = 0 while i < len(kwargs): if i%2 == 0: finaloutput = finaloutput + kwargs[i].upper() else: finaloutput = finaloutput + kwargs[i].lower() i = i + 1 return finaloutput myfunc('Anthropomorphism') It's not working. Please guide me. Thanking you Yours sincerely Tarandeep Singh

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Re: [Tutor] Python problem
by dn Feb. 23, 2021

Feb. 23, 2021

On 23/02/2021 12.56, Dennis Lee Bieber wrote: > On Tue, 23 Feb 2021 12:03:48 +1300, dn via Tutor <tutor(a)python.org> > declaimed the following: > > >> The second is the list *index* (or "pointer") which *must* be an >> integer. [will assume "list" for convenience = any collection] >> > > My apologies, but I think you went off the rails at this point... As I > recall the original post, there is no list/index involved. It is all just > an iterative numeric computation -- compute a candidate split point (in the > scaled integer range 0..10000 => 0.00%..100.00%), determine if the optimal > value is above or below that split, determine a new split point within > 0..Split or Split..10000 (depending upon which side one determined is > optimal), taking into account some epsilon used to determine if a solution > has been found. > > It is not a binary search of a collection object... > > OTOH: the code that was in the original post contains floating point > division operators, so one has not seen the supposedly failing code. You're correct. The algorithm can be used against previously-computed data, ie sitting in a list or a table; and at other times computing ever more-precise results 'dynamically' and by 'splitting the difference'. Sigh! Not seeing the entire assignment-question, nor the actual code... This is a Python list. What's all this talk about "Rails", Mizz Ruby? -- Regards, =dn

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Questions about Creating Executables and Backward Compatibility
by Shaun Keyes-McClements Feb. 22, 2021

Feb. 22, 2021

Hi Team, I am a hobbyist programmer with the following questions: Since Python is a scripting language and does not need to be compiled, how would I make executables of programs written in either Python 3.1 or 3.9? PIP seems to support specific versions of Python, such as 3.4 and 3.7, respectively. What I would like, is to share text adventures I've written with those who don't have Python installed on their systems. Is there a way to do this, I mean, do I need to create a library and import it into another language such as C++, to be compiled? Next question, with the exception of updates made, are Python 3x versions typically backward compatible? For example, have the Input and Print Functions remained the same in terms of syntax from let's say 3.1 onward? Most of my experience with Python is with 3.1; I recently installed 3.9 and don't want to reinvent the wheel by rewriting all of my programs from scratch, though I am not opposed to revising for efficiency and improved User Experience. Thank you for your time and attention. Shaun

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Re: [Tutor] Python problem
by Vakhtang Kvaratskhelia Feb. 22, 2021

Feb. 22, 2021

Thank you very much for the reply. I understand those comments, however what i don't understand is how can you use integer division "//" with the bisection search code? The "/" division works just fine but the "//" doesn't work with the bisection search code. Is it possible that the hint is wrong? Or maybe integer division doesn't mean "//" On Sat, Feb 20, 2021 at 9:31 PM Dennis Lee Bieber <wlfraed(a)ix.netcom.com> wrote: > On Sat, 20 Feb 2021 15:27:01 +0400, Vakhtang Kvaratskhelia > <va_kvaratskhelia(a)cu.edu.ge> declaimed the following: > > > > >I want to use this hint : "Because we are searching for a value that is in > >principle a float, we are going to limit ourselves to two decimals of > >accuracy (i.e., we may want to save at 7.04% or 0.0704 in decimal – but we > >are not going to worry about the difference between 7.041% and 7.039%). > >This means we can search for an integer between 0 and 10000 (using integer > >division), and then convert it to a decimal percentage (using float > >division) to use when we are calculating the current_savings after 36 > >months. By using this range, there are only a finite number of numbers > that > >we are searching over, as opposed to the infinite number of decimals > >between 0 and 1. This range will help prevent infinite loops. The reason > we > >use 0 to 10000 is to account for two additional decimal places in the > range > >0% to 100%. Your code should print out a decimal (e.g. 0.0704 for 7.04%)." > >but i am not able to. > > Bottom up. Python floats are double precision -- approximately 15 > significant digits -- ALWAYS. > > If you need specific formatting for output, you have to provide the > correct formatting string when outputting the result. Your code is just > outputting a sequence of a string and a numeric with no formatting codes -- > so naturally Python is going to display as many significant decimals as it > can... > > >>> import math > >>> math.pi > 3.141592653589793 > >>> "%7.4f" % math.pi > ' 3.1416' > >>> > > I'm not so certain about that justification for using integers and > then > converting to float for output... While real world may have an infinite > range between 0.0 and 1.0, computers do not. Furthermore, if you are using > an epsilon to define when two values are "close enough" you will never be > computing in increments of less than that epsilon. > > abs(a-b)<0.00005 > > Of course, going further from the track... If you are discussing > financial values, using the Decimal module configured for the desired > resolution is probably recommended. > > > -- > Wulfraed Dennis Lee Bieber AF6VN > wlfraed(a)ix.netcom.com > http://wlfraed.microdiversity.freeddns.org/ > > _______________________________________________ > Tutor maillist - Tutor(a)python.org > To unsubscribe or change subscription options: > https://mail.python.org/mailman/listinfo/tutor >

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Python problem
by Vakhtang Kvaratskhelia Feb. 20, 2021

Feb. 20, 2021

Hello, i have written the following code that uses the bisection search to find the optimal percentage to save. However i am unable to prevent an infinite loop in case the annual salary is 10,000 also i do not get how i can use only integer division while using the bisection search? when i substitute // division instead of / division the loop runs infinitely. I want to use this hint : "Because we are searching for a value that is in principle a float, we are going to limit ourselves to two decimals of accuracy (i.e., we may want to save at 7.04% or 0.0704 in decimal – but we are not going to worry about the difference between 7.041% and 7.039%). This means we can search for an integer between 0 and 10000 (using integer division), and then convert it to a decimal percentage (using float division) to use when we are calculating the current_savings after 36 months. By using this range, there are only a finite number of numbers that we are searching over, as opposed to the infinite number of decimals between 0 and 1. This range will help prevent infinite loops. The reason we use 0 to 10000 is to account for two additional decimal places in the range 0% to 100%. Your code should print out a decimal (e.g. 0.0704 for 7.04%)." but i am not able to. this is the code i have written and it gives the correct amount as a solution except it is not rounded to 4 decimals as it would have been if i was using integer division: annual_salary=150000 total_cost=1000000 Months=36 semi_annual_raise=0.07 portion_down_payment=0.25 r=0.04 portion_saved=0 epsilon=100 possible_savings=0 Steps=0 for i in range(0,Months): possible_savings+= annual_salary/12 + possible_savings*0.04/12 if (i+1)%6 == 0: annual_salary=annual_salary*(1+semi_annual_raise) low=0 high=10000 while abs(total_cost*portion_down_payment - portion_saved * possible_savings) >= epsilon: if portion_saved * possible_savings < total_cost*portion_down_payment : low=portion_saved else: high=portion_saved portion_saved=(high+low)/2 Steps+=1 print("Optimal portion saved is = ",portion_saved) print("Steps in bisection search: ",Steps) Can you help with this? Thank you in advance.

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