[Solved] Printing every item on a new line in Python [duplicate]
Should be just: print ‘\n’.join([str(factorial(x)) for x in range(0, 6)]) 1 solved Printing every item on a new line in Python [duplicate]
Should be just: print ‘\n’.join([str(factorial(x)) for x in range(0, 6)]) 1 solved Printing every item on a new line in Python [duplicate]
The for loops are nested, from left to right. You can write it out as regular loops like this: words = [] for line in open(‘words.txt’, ‘r’): for word in line.split(): words.append(word) So the expression before the for loops is the final value added to the produced list, and all the for loops (and any … Read more
You can use reduce (functools.reduce in Python3) if the integer is non-zero >>> x = [False, False, False, 10, False, False] >>> reduce(lambda i,j:i or j, x) 10 You can use a generator expression also here >>> x = [False, False, False, 10, False, False] >>> (i for i in x if i!=False).next() 10 Also … Read more
your first question: [i for i in range(n) for j in range(5)] your second question: [j for i in range(k) for j in range(1,6)] where k is the number of repeats 5 solved How to create the following lists with list comprehensions? [closed]
def f(x): if x == 1: return [] return [f(x-1)] print(f(5)) Output: [[[[[[]]]]]] 1 solved Python. List comprehension. List in the list and so on [closed]
as pointed out by @Delgan, it can be done directly via d1 = dict(keyval.split(“, “) for keyval in a_list) without the inner nesting 🙂 older approach which were not really correct :- d = [a.split(‘,’) for a in a_list] d1 = {key: val for key,val in d} or d1 = {key: val for key,val in … Read more
It works like a foreach loop, so foreach (x in [1..3]) { foreach (y in [x .. x * 2]) { yield y; } } First x is 1, so y in [1 .. 2] Then x is 2, so y in [2 .. 4] Then x is 3, so y in [3 .. 6] … Read more
Assuming that sorted_indices contains only integers: tmaxima = [t[i] for i in sorted_indices] ag_maxima = [a_g[i] for i in sorted_indices] solved how to turn this python lines code into a list comprehension?
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
Not beautiful solution, but working one values = [{‘id’: 1, ‘x’: 2}, {‘id’: 1, ‘y’: 4}, {‘id’: 1, ‘z’: 6}, {‘id’: 2, ‘j’: 5}, {‘id’: 2, ‘k’: 10}, {‘id’: 3, ‘w’: 1}, {‘id’: 3, ‘x’: 3}, {‘id’: 3, ‘y’: 5}, {‘id’: 3, ‘z’: 7}] # get all unique possible keys unique_keys = set((x[‘id’] for x … Read more
I believe the goal in this case to make your code as concise and efficient as possible. At times it can seem convoluted, but the computer looping through multiple lines as opposed to a single line adds processing time, which in large applications and across many iterations can cause some delays. Additionally, although it seems … Read more
Introduction List comprehension is a powerful tool in Python that allows developers to create lists in a concise and efficient manner. It is a popular feature of the language and is used extensively by developers to create lists quickly and easily. In this article, we will discuss why list comprehension is so prevalent in Python … Read more
Order of elements The PEP 202 is not very… comprehensive, but you can found some information in the Python Language Reference: The comprehension consists of a single expression followed by at least one for clause and zero or more for or if clauses. In this case, the elements of the new container are those that … Read more
It appears to me that within your list comprehension you just have one minor issue! Instead of: item for item in textString within your list comprehension, I would suggest: item for item in listText as currently you are iterating through each char of the whole text, rather than each element in the list of the … Read more
Could you have been thinking of something like this? def isPrime(n): if n < 2: return None factors = [ i for i in range(2, n) if n%i == 0 ] return True if not factors else False print( [ i for i in range(1, 100) if isPrime(i) ] ) #PRINTS: [2, 3, 5, 7, … Read more