From my point of view, the best you can do is to read the whole documentation https://docs.djangoproject.com/en/3.0/ it answers most of the questions you might have and it shows what is possible to do with Django. Its very well written and I really enjoyed reading it. You should be able to write your Django app … Read more
if you but braces end of a list (x.favourite_channels is list), then you get such error. if you want i’th element of list x.favourite_channels you should write x.favourite_channels[i-1] instead. solved python object oriented programming type:list object is not callable
You can try something like this : input_st = “Field1:’abc’,Field2:’HH:MM:SS’,Field3:’d,b,c'” output_st = {item.split(“:”)[0]:”:”.join(item.split(“:”)[1:]).replace(“‘”, “”) for item in input_st.split(“‘,”)} outputs : {‘Field1’: ‘abc’, ‘Field2’: ‘HH:MM:SS’, ‘Field3’: ‘d,b,c’} Kind of ugly but it does the job. 4 solved Convert string to Ordered Dictionary [closed]
The two main concept that you see in the last line are list comprehension and nested loops. Have a look. for understanding better what is going on, we’re going to split that line in simplier part: TENS for tens in “twenty thirty forty fifty sixty seventy eighty ninety”.split(): print(tens) OUTPUT: twenty thirty forty fifty sixty … Read more
python 3 simple calculator , need assistance understanding subject and completing task, new to python and coding in general [closed] solved python 3 simple calculator , need assistance understanding subject and completing task, new to python and coding in general [closed]
You are trying to call an unbound function; zxLolBoT is a class, not an instance. As such, the self parameter is not being passed in for you, and the method only received 2 arguments instead of 3 (with self being that 3rd argument). self is already that instance, access the method on that to get … Read more
Supporting Python 3 means supporting Python 3 up to the current point release. Python point releases are backward compatible. What works on 3.0 should generally work on 3.4. See PEP 387 for the general guidelines on this policy. Python 3.4 added some deprecations but none of these will affect packages that were once only written … Read more
how about this def display(x): for i in x: for j in i: j = chr(j) print (j, end = ‘ ‘) print() if each i represent a line, you need add a extra print to start over in the next line 1 solved Printing a smile face 🙂
Convert each string into a list, then chain the lists: from itertools import chain list(chain.from_iterable(map(list, character))) #[‘#’, ‘$’, ‘@’, ‘#’, ‘%’, ‘%’, ‘#’, ‘@’, ‘#’, ‘%’] 3 solved Adding spaces within the (string) elements of an array [closed]
In python 3, print is a function. See here: https://docs.python.org/3/whatsnew/3.0.html#print-is-a-function Try: with open(‘list.txt’) as f: print(” “.join(line.strip() for line in f)) solved What is wrong with the syntax of this code? [duplicate]
Welcome to python. In python everything is easy. Firstly you need to understand your problem on a slightly deeper level though. The problem here is just realizing that you’re taking two slightly different sums. A good way to go about this would be to represent them both in a common form (minutes) then do all … Read more
You can try something like this without importing any module : s1=’appleandgold’ s2=’appleblue’ track=[] for k in range(len(s1)): if k!=0: for ka in range(0,len(s1),k): if s1[ka:ka+k] in s2: track.append((len(s1[ka:ka+k]),s1[ka:ka+k])) print(max(track)[1]) output: apple solved String intersection in Python
list = [1,2,3,4,5,6] for x in range(len(list)): if list[x] % 3 == 0: list[x] = list[x]**2 print(list) solved How to divide specific elements in a list and then square the values? [closed]
Firstly, you can simply write while True: Instead of a = True while a: Secondly, you don’t use total or t anywhere. You can delete them. Thirdly, if x has to belong to [a, b], you can write if a <= x <= b: Fourthly, you can use else statement instead of negation of if. … Read more
>>> from __future__ import division … … result = [] … text = “20 7/8 16 1/4” … for num in text.split(): … try: … numerator, denominator = (int(a) for a in num.split(“https://stackoverflow.com/”)) … result.append(numerator / denominator) … except ValueError: … result.append(int(num)) … >>> result [20, 0.875, 16, 0.25] 2 solved How to convert a … Read more