[ad_1] One approach using extended iterable unpacking and next: lst = [1, 7, 9, 8, 5, 5, 5, 5, 5, 5, 5, 5, 5] last, *front = reversed(lst) res = next((v for v in front if last != v), None) print(res) Output 8 The above approach is equivalent to the following for-loop: last, *front = … Read more
[ad_1] Add the “Fizz” “Buzz” “Bizz” if the division is possible, at the end if nothing has been added, it means that you have to print the number itself. n = 0 toTest = [ 3, 5, 7 ] outputCanBe = [ “Fizz”, “Buzz”, “Bizz” ] outputIndex = 0 iteration = (len(toTest)) while n <= … Read more
[ad_1] With the new example, you’re actually still dealing with tuples. If you print out what the right hand side is, you’ll get testVar = 1000, 1000, print(testVar) # result: (1000,1000) What is actually happening under the hood is that Python sees a tuple, then unpacks it into two values and assigns one to myvar … Read more
[ad_1] Given a non-empty list of strings of the same length, how do I create a list of lists of the letters at each position in the string without using zip? [closed] [ad_2] solved Given a non-empty list of strings of the same length, how do I create a list of lists of the letters … Read more
[ad_1] Use x[1][“number”] in your lambda n [40]: d Out[40]: {‘b’: {‘description’: ‘second letter’, ‘number’: 2}, ‘a’: {‘description’: ‘first letter’, ‘number’: 1}, ‘c’: {‘description’: ‘third letter’, ‘number’: 3}} In [41]: sorted(d.items(), key=lambda x:x[1][“number”]) Out[41]: [(‘a’, {‘description’: ‘first letter’, ‘number’: 1}), (‘b’, {‘description’: ‘second letter’, ‘number’: 2}), (‘c’, {‘description’: ‘third letter’, ‘number’: 3})] 1 [ad_2] solved … Read more
[ad_1] You could group by name, next count values and filter results which have count 3 (because you have 3 years) groups = df.groupby(‘name’).count() result = groups[ groups[‘date’] == 3 ].index.to_list() print(result) Or you could directly count names counts = df[‘name’].value_counts() result = counts[ counts == 3 ].index.to_list() print(‘result:’, result) Minimal working example: I use … Read more
[ad_1] I have the correct answer for that. When inviting a bot to any server, there is special 2Oauth link for inviting the bot. Now, this is something you need to worry about. There are two things you need to actually do to complete what you want to do. Assigning the Bot role proper permissions … Read more
[ad_1] You have to import bs4 first (assuming you have installed bs4), i.e. > import bs4 > soup=bs4.BeautifulSoup(r.Text, “xml”) 0 [ad_2] solved Why am i getting “NameError: name ‘bs4’ is not defined” error while trying to import BeautifulSoup? [closed]
[ad_1] Look at the index values of the list. alist=[[1],[2]] Here alist[1] is [2], so alist[1][0] is 2. 4%1 is 0. [ad_2] solved Why are lists giving different value?
[ad_1] The line number where you error occurs should be clearly noted. From my own call: while guess = password: ^ SyntaxError: invalid syntax In addition, it’s a logic error of sorts: you want the user to keep guessing while the guess is incorrect, that is guess != password. 1 [ad_2] solved Python code HW- … Read more
[ad_1] Use split(). Example. x = “Hello world” x.split( ) Please add any character you want to split at if you want to split on whitespace use like this x.split(” “) [ad_2] solved Convert list of characters into a list with strings [duplicate]
[ad_1] for is used to loop through the defined variable s, which in this case is a string containing letters, digits (numbers), and whitespace (spaces, tabs etc.). First time: ‘ Abs3 asdasd asd11 111 11ss’ ^ As we can see, the first character of the string is: “” (space). The code goes on to test … Read more
[ad_1] Happily, the atoms used by the format are similar enough to Python that we can use the tokenize and ast modules for an ad-hoc parser. It will probably break horribly on broken input, but works for your example data 🙂 import tokenize import token import ast import io import json def parse_valve_format(data): dest = … Read more
[ad_1] It will round to 0 if x is smaller then 0.005. x = 0.005 round(x, 2) > 0.01 x = 0.0049 round(x, 2) > 0.00 1 [ad_2] solved When I round x in python it returns 0.0 [closed]
[ad_1] If you convert the number in int then you should store in some variable. number = int(number) then do Oddoreven= number%2 And use = assignment sign not comparison == sign Because if u don’t store it in another variable it will not be casted to int and then it will be treated as string … Read more