🌐 Mailman 3 March 2022 - Tutor - python.org

problem solving with lists
by marcus.luetolf＠bluewin.ch April 4, 2022

April 4, 2022

Hello Experts, I'am still biting my teeth out trying to put in to code the following pseudocode: 1.Set up a list of 16 letters and assign it to a variable : all_letters = list(abcdefghijklmnop) 2.Find an code to insert all 16 letters in 5 Lists in 4 sublists of 4 letters in each list: list1 = [['a', 'b', 'c', 'd'], ['e', 'f', 'g', 'k'].....['m', 'n', 'o', 'p']] list2= [['a', 'e', 'f','g'] ...............,['d', 'i', 'l',' p']] . . list5 =[[...],....................,[.........]] with constraint that a pair of letters, p.e. ['a', 'b', ..] can occur only once in all 20 sublists or if list1 contains sublist ['a', 'b',..], such a sublist cannot appear in list2 through list5. To write this code, as a first step I made a list (comb_list) containing all possible sublists and reducing their number to 80 by sorting them : ['a', 'b'..] equals ['b', 'a', ..] : with to code snippet below. >comb_list = [] >for dummy_i in range(4): >copy_all_letters = all_letters[:] >lst = (16*16)*copy_all_letters >f1 = [] >f2 = [] >f3 = [] >f4 = [] > f1.append(lst[0+n]) >f1.append(lst[2+n]) >f1.append(lst[4+n]) >f1.append(lst[6+n]) > f1.sort() >comb_list.append(f1) >f2.append(lst[8+n]) >f2.append(lst[10+n]) >f2.append(lst[12+n]) >f2.append(lst[14+n]) >f2.sort() >comb_list.append(f2) > f3.append(lst[1+n]) >f3.append(lst[3+n]) >f3.append(lst[5+n]) >f3.append(lst[7+n]) >f3.sort() >comb_list.append(f3) > f4.append(lst[9+n]) >f4.append(lst[11+n]) > f4.append(lst[13+n]) > f4.append(lst[15+n]) >f4.sort() > comb_list.append(f4) > print(f1, f2, f4, f4) > print(comb_list) > n += 3 The next step should be to extract out of the 80 sublists in comb_list 20 sublists fullfilling the above mentioned constraint. a) I don't know how to do that, b) I am not shure wether this is a way do it, nor c) if my problem can by solved by python at all. Marcus.

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Tkinter draw line segments
by Phil March 30, 2022

March 30, 2022

The segment table contains the start and end points of lines to be drawn on the canvas. seg = {0: (20, 20, 40, 20), 1: (44, 24, 44, 44), 2: (44, 48, 44, 68), 3: (20, 72, 40, 72), 4: (16, 48, 16, 68), 5: (16, 24, 16, 44), 6: (20, 46, 40, 46), } Say I want to draw segments 1 and 2. num = {1: (seg[1], seg[2])} The following will draw all of the segments correctly. for i in range(7): self.canvas.create_line(seg[i], width=4, fill="green") What I haven't sorted out is how to draw segments 1 and 2 without having two draw statements. The following does draw the two segments but without the break between them. self.canvas.create_line(num[1], width=4, fill="green") What I'm trying to come up with is a table entry that tells the draw routine how many draw statements there are, one for each segment. No doubt I'm making this more complicated than it needs to be. One way around this is to draw all of the segments in two colours. In the above case I would draw segments 1 and 2 green and the remainder black. That way I'd only need one draw statement in a for loop with a variable fill colour. This would work perfectly on a black background but it seems like an amateurish solution. -- Regards, Phil

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Re: [Tutor] problem solving with lists
by alan.gauld＠yahoo.co.uk March 26, 2022

March 26, 2022

On 19 Mar 2022 10:57, marcus.luetolf(a)bluewin.ch wrote: Many thanks again. Using your first advice my code got even clearer : >lst = [['a', 'b', 'c'], ['d', 'e', 'f'], ['a', 'b', 'g'], ['b', 'c', 'h']] >sub_lst = [] >for p in lst: > pair1 = p[0:2] > pair2 = list(p[0]+ p[2]) > pair3 = p[1:] > if pair1 not in sub_lst or pair2 not in sub_lst or pair3 not in sub_lst: > sub_lst.append(p) >print(sub_lst) But your second advice : >if pair1 not in sub_lst or pair2 not in sub_lst or pair3 not in sub_lst: Still yields all four sublists instead of eliminating the last 2 sublists , ['a', 'b', 'g'], ['b', 'c', 'h'], since ‘a’, ‘b’ and ‘b’, ’c’ were already in the first 2 sublists. I think python does not allow to check if an item is in a list with only slices of the item to be checked ?? Python does what you ask. If you get the wrong answer then you are asking the wrong question. But I still don't really understand what you are trying to do. Have you looked at Dennis replies, his approach is more likely to get you what you want.... Alan g Von: alan.gauld(a)yahoo.co.uk <alan.gauld(a)yahoo.co.uk> Gesendet: Freitag, 18. März 2022 19:16 An: marcus.luetolf(a)bluewin.ch Cc: tutor(a)python.org Betreff: Re: AW: AW: [Tutor] problem solving with lists On 18 Mar 2022 15:12, [1]marcus.luetolf(a)bluewin.ch wrote: …many thanks for your quick replay. When I used a for loop : >lst = [['a', 'b', 'c'], ['d', 'e', 'f'], ['a', 'b', 'g'], ['b', 'c', 'h']] >sub_lst = [] >for p in lst: > pair1 = lst[p][0:2] P is not an index its the actual sublist. So you just need Pair1 = p[0:2] > > ected. But I don’t understand what you mean with If pair 1 not empty Or pair2 not empty Or pair3 not in newlist What is the difference between pair not beeing empty or not beeing in new_list ? Pair not empty means pair contains anything other than the empty list. Pair not in newlist means pair is not empty and it's contents are not I. Newlist. Very different! How should I define the if condition properly ? You need, I assume, If Pair1 not in newlist Or pair2 not in newlist Or pair3 not in newlist Alan g. References Visible links 1. mailto:marcus.luetolf@bluewin.ch

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Re: [Tutor] problem solving with lists
by dn March 21, 2022

March 21, 2022

On 21/03/2022 21.51, marcus.luetolf(a)bluewin.ch wrote: > ...yes, the weekly Draws below are exactly what I have in mind !!! > And I realised earlier that "I asked the wrong questions". But with your > Cooments about the solution I think I can find a solution myself. > It was never my intention to let tutor do my "homework", I had learned nothing. > Thanks, Marcus. Excellent news! Keep going! (here) Follow earlier suggestions of defining meaningful terminology by first writing (specifying) the problem, as completely as you can. All those detailed clauses, as we say in English: "the if-s, the but-s, and the maybe-s" are the details which cause trips and stumbles. A lot of programmers rush to start coding. The above is well-worth spending the time - and if the spec[ification] has come from someone else (as it would in a commercial environment), this is where we find, as early as possible, that the client hasn't given us the full-story! As you say, the key to it all is to ask the 'right' questions and document the requirements and 'rules'. Once the problem is clearly understood, we can start looking for solutions. Often there is more than one, and certainly more than one way to approach building a solution. These are the major steps of "program[me] design". Again, it is worth checking and re-checking, because errors caught early in the development process are cheaper (time and/or $) to fix, than those not realised until later. Thus, the completed work described earlier was a second, and longer, version of the first - so a complete re-write became necessary because I did not understand what was needed! Now you understand why there was evidence of frustration in the way people were replying to you! Next try to break-up the chosen solution into stages or phases. In this case, I think of a League (or competition) which will take (number_of_weeks) five weeks before all 'unique' combinations of players can be played. Each week requires its own "draw", which will consist of (number_of_groupings) four groups of players. Each group of players will consist of (players_per_grouping) four players - who have never played with each-other before (in this league/competition). You can see that the overall solution consists of sub-solutions to sub-problems. It's a bit like those inter-locking Russian dolls, or (?Swiss) chocolates with a center and multiple layers/flavors inside a hard chocolate coating. We also refer to it as like an onion. This approach to problem-solving is called "stepwise decomposition" (or "- refinement). It is the process of dividing a large problem into smaller problems, and those into smaller... until the small problem is one that is easy to solve. The beauty of this is that if you have a correct and sensible design (easily said, not so easily done!), it should be possible to write and test (or 'prove') each 'unit' or sub-solution of the code, by itself - which is a good deal easier than trying to test the whole 'onion' and not being sure where some failure or error is caused within the mass of code! What happens when you have to cut-up a (real) onion? I tend to code each 'layer' as a function. Indeed after choosing the name of the function (and writing the def ... line), I'll add a docstring which is basically that section of the 'design' (from earlier). I'll repeat this until the whole problem has been covered. Only then, do I sit down to start writing code for the smallest of the sub-solutions (or, the inner-most layer of the onion/chocolate). Now, to help you on your way, here are a few other data-structures which I used: weekly_draws = list() - to hold the [five] weeks' draws - perhaps only necessary if you're going to do something with this data (I was only printing it out) player_history = { player:{ player } for player in players } - to hold the set of players who this player has previously played - rather than complicating loop-controls (or maybe adding lots of one-time if-conditions) within the code, it seemed easier to initialise each player's set with him-/her-self - after all, under the 'rules' you cannot play someone you've already played, nor can you play yourself. (two 'rules' for the price of one!) - this 'one-liner' code-construction is called a "set-comprehension" and I very rarely see/use one - whereas their close-relatives, "list-comprehensions", are very common. If you'd like to post your sub-solutions (text-design, plus Python-code), we'll be happy to help you along... -- Regards, =dn

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identifying if a path is heading into a zip file - windows
by Nathan Smith March 20, 2022

March 20, 2022

Take a path, for instance: path="c:\my folder" os.path.isdir(path) = True Take a second path: path= 'c:\my folder\a file.zip\a folder in the zip' Is there a way for python to know, without me physically breaking it down, that that path is in a zip file? For instance, a way for it to know it's dealing with a zip archive file, and not just a folder with .zip in the name? thanks -- Best Wishes, Nathan Smith, BSC My Website: https://nathantech.net

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Re: [Tutor] Anonymous function
by Devanshu Nalavade March 17, 2022

March 17, 2022

Respected Tutor, I am very thankful that you took the time to clear my doubts and explain this concept to me in a manner I can understand. -- Thanks and Regards, Devanshu Nalavade

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Anonymous function
by Devanshu Nalavade March 17, 2022

March 17, 2022

Respected Tutor, I have written a code that finds the sum of numbers passed as arguments in a user-defined function. This function has an anonymous function that takes the same arguments and does the summation. However for some reason the code is not giving me the desired output, there are no syntax errors so I'm not understanding what I'm doing wrong. Your help is much appreciated, the program I wrote is as follows: >>>def sum(arg1,arg2,arg3=10): ... m = lambda arg1,arg2,arg3:arg1+arg2+arg3 >>>print(sum(5,6)) The output comes out as 'None' Thanks and Regards, Devanshu Nalavade

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finding sum of digits in a string
by Manprit Singh March 15, 2022

March 15, 2022

Dear Sir, Take an example of finding the sum of digits in a string : st = "ab23YH6UJ7ik83LO" To find the sum of digits in the string st, i can do it in below given manners : 1) sum(int(ele) for ele in st if ele in "123456789") 2) sum(int(ele) for ele in st if ele.isdigit()) Although the 2nd one is more compact, using isdigit() method , what bothers me is - isdigit() method. It should be used to check a substring, not a single character. Here I have to check all characters of the string one by one, why not use in operator as done in the 1st way. Any guidelines regarding this ? Regards Manprit Singh

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Example of good use of enumerate()
by Manprit Singh March 13, 2022

March 13, 2022

Dear Sir, Consider an example of printing all elements of a principal diagonal of a rectangular matrix. lx = [[2, 5, 9, 0], [3, 8, 1, 5], [4, 8, 1, 2], [6, 1, 2, 4], [3, 1, 6, 7]] for i, nlist in enumerate(lx): for j, num in enumerate(nlist): if i==j: print(num) The above use of enumerate() works well. Another solution is also given below: for i in range(len(lx)): for j in range(len(lx[0])): if i == j: print(lx[i][j]) also works, but lots of indexing and other operations don't look good and effective to me . Is there any other better solution other than the first one using enumerate() ? Is this a valid and preferred use case of using enumerate() function ? Regards Manprit Singh

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I need help
by Amer Saeed March 13, 2022

March 13, 2022

I need help about how to represent these equations in python Task which is index start with 1 to m Tasks ti=[t1, t2, t3, t4,…,tm] Each t1 is assigned to either Iot device, sy,z, when y=o, z=n Fog cloud or s0,n then sy,z, when y=1, z=1,..f Cloud or s0,n then s1,z then sy,z, when y=2, z=1,…c Xn,i= 0 when it is first case above Xn,i=1 when it is second case above Xn,i=2 when it is third case above Xn,i= 0 depicts that the ith task is assigned to the nth IoT device (s^0,n) for local execution, and Xn,i =1 and xn;i = 2 denote that the ith task is assigned to one of fog servers and cloud servers, respectively, for the remote execution. Moreover, the f and c show the number of available

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