Your code is using Python’s list comprehension to create a list from another list based on the condition. See more about list comprehensions here: https://www.programiz.com/python-programming/list-comprehension The [1] is accessing the item at index 1 (2nd item) of your sorted list. As for .join, each line in the sorted list is being printed out with a … Read more
You can use a dictionary to store the values: x=[‘aze’,’qsd’,’frz’,…] vars = {} for i in x: vars[“MAX” + i] = [] Or in order for them to become real variables you can add them to globals: x=[‘aze’,’qsd’,’frz’,…] for i in x: globals()[“MAX” + i] = [] You can now use: MAXaze = [1,2,3] 3 … Read more
Many thanks for all. I used this code: count = 3 urls = {} def test(count): a = 1 while a <= count: urls[a] = ‘test’ a += 1 test(count) print(urls) solved Python3 – create global variable(list) from function
Finally print that list to the console >>> cubes_by_four = [x**3 for x in range(1,11) if x**3 % 4 == 0] >>> print(cubes_by_four) [8, 64, 216, 512, 1000] It says print the list, not print each item in the list to console 0 solved Print cubes of the numbers 1 through 10 only if the … Read more
You can do something like this : input_num = 123 list1 = [0, 1, 2, 3, 4] # if list1 is not dervied from input number list2 = [“0”, “hello”, “my”, “name”, “is”, “Daniel”] print(‘ ‘.join([list2[list1.index(int(ind))] for ind in str(input_num)])) This will result in : hello my name Also, i would suggest you to look … Read more
Sui_Q1<- list() is written more than once. it cause the list() to override. The solution is only to put Sui_Q1<- list() and Sui_Q2<-list() before looping. solved R : assigning value in looping and else if
The part I am getting hung up on is how I can loop through the values for each key and get to the individual data in each value. I can do the rest after that So if I understand you correctly then you want to iterate through a dictionary which has lists as its values? … Read more
Your best way of doing this, is probably going to be a list comprehension. user_input = raw_input(“Please enter five numbers separated by a single space only: “) input_numbers = [int(i) for i in user_input.split(‘ ‘) if i.isdigit()] This will split the user’s input at the spaces, and create an integer list. 1 solved User input … Read more
The problem here is in your conditional while len(lists) is True: is checks for identity, not equality. [1, 2, 3] == [1, 2, 3] # True [1, 2, 3] is [1, 2, 3] # False, they are two distinct (but equivalent) lists. However even equality would be incorrect here, since 42 == True # False … Read more
This should work- for word in file: if isinstance(word, int): … elif isinstance(word, list): … elif isinstance(word, tuple): … elif isinstance(word, str): … 1 solved how to check str int list and tuple? [closed]
>>> X = range(4, 7) # List of number from 4 to 6 >>> Y = range(2) # List of number from 0 to 1 >>> X [4, 5, 6] >>> Y [0, 1] >>> X[2] = Y # Stored ‘Y’ at X[2] in place of ‘6’ # X[2] is referencing Y >>> X [4, … Read more
The problem you’re experiencing is due to a misunderstanding of lists, list-references, and possibly the concept of mutability. In your case, you are binding the name user_data to a list object. In other words, user_data acts as a list-reference to the actual list object in memory. When you say user_list.append(user_data), all you’re doing is appending … Read more
You have a constant for what the scoring is scoring :: [Int] scoring = [25, 18, 15, 12, 10, 8, 6, 4, 2, 1] Then you need a way for pairing up a driver with the score they got. Whenever you’re pairing two things in Haskell, the canonical choice is to use a tuple. The … Read more
By the suggestions in the comments, partialy @Charles_May ‘s: Simply looping: List<string> Source = {}; //source list List<string> Females = new List<string>(), Males = new List<string>(); foreach (string value in Source) { if (value.ToUpper().Contains(“,F,”)) //Female { Females.Add(value); } else { Males.Add(value); } }//result: both lists are with values That’s it. if you’s like to make … Read more
The type of a variable and the type of the object the variable refers to do not need to be the same, so long as the type of the object is compatible with the type of the variable. So what does it mean to be compatible? It means that the object’s type has an “is … Read more