No, there is no shorter way. Usually, you will even break it into two lines : important_airports = (airport for airport in airports if airport.is_important) for airport in important_airports: # do stuff This is more flexible, easier to read and still don’t consume much memory. 3 solved Single line for loop over iterator with an … Read more
Use re.findall. The regex: r’^(.*\S)\s+id:\s*(\d+)\s+type:\s*(.+)’ means: ^ : start of the string..* :any character, repeated 0 or more times.\S : non-whitespace character.\s+ : whitespace character, repeated 1 or more times.\d+ : any digit, repeated 1 or more times.(PATTERN) : capture the patterns and return it. Here we capture 3 patterns. import re string = “Wacom … Read more
With this function you can produce your underscore string based on your word def f(a): m = ” l = len(a) for i in range(l): m += ‘_’ return m 4 solved print a underscores on the basis of number of chrcters in a string python
The problem is with memo={}. When you just run one test, it will work locally (and also on LeetCode when you run it on one particular input), but when you would run multiple calls of the function with different nums inputs, you would get unexpected results too. memo is only assigned {} once, and is … Read more
You can do something like this: text=”Hello this is a test” unique_letters = set(text.lower()) unique_letters.remove(‘ ‘) print(unique_letters) print (‘odd’) if len(unique_letters) % 2 else print (‘even’) Result: {‘o’, ‘h’, ‘e’, ‘l’, ‘i’, ‘s’, ‘a’, ‘t’} even 0 solved How to check the distinct number of characters in a string [closed]
Is this what you’re looking for? In [2]: num_rows = 10 # should be divisible by 2 In [3]: np.repeat(np.eye(num_rows // 2), 2, axis=0) Out[3]: array([[1., 0., 0., 0., 0.], [1., 0., 0., 0., 0.], [0., 1., 0., 0., 0.], [0., 1., 0., 0., 0.], [0., 0., 1., 0., 0.], [0., 0., 1., 0., 0.], … Read more
Yes, the most common way to do this is with selenium and a webdriver manager. If you don’t need to open the whole webpage and just need the HTML, use beautifulsoup4 and requests. solved Is there a way to make python open a webpage?
Use lengths of the lists to terminate the recursion. In this example implementation, if length of either of the lists is 0, we have either mapped it all, or it is an empty list to begin with, thus we terminate the recursion. Also worth noting that l1[1:] is the slice of the list we haven’t … Read more
You can use the partition method for this. It’s part of the built-in str type. >>> help(str.partition) partition(self, sep, /) Partition the string into three parts using the given separator. This will search for the separator in the string. If the separator is found, returns a 3-tuple containing the part before the separator, the separator … Read more
import pandas as pd, numpy as np # filter results by showing records that meet condition # sample df d = {‘SBP’: [111, 100], ‘DBP’: [81, 40], ‘HEARTRATE’:[1,50]} df = pd.DataFrame(data=d) df # if an alert is found, prints alert, else normal if len(df[(df[ ‘SBP’ ] > 110) & (df[ ‘DBP’ ] > 80) & … Read more
A simple solution: def reversemat(mat): return [row[::-1] for row in mat] print(reversemat([[1,0],[1,0]])) # [[0, 1], [0, 1]] besides, you have returned nothing from your function, so you will get None from your second print. 3 solved Reverse of a list of list in Python [closed]
elemnts=[2,2,0,0,5,6] h1=9 h2=6 def func(elemts,a,b): list1=[] for i in range(0,len(elemts)): for j in range(i+1,len(elemts)): for k in range(0,len(elemts)): if(k not in [i,j]): temp=elemts[i]+elemts[j]+elemts[k] if(temp in [h1,h2]): list1.extend([elemts[i],elemts[j],elemts[k]]) return list1 list2=func(elemnts,h1,h2) @arthur_currry…just chnged one line in last if case.Its working fine solved Given a 6 blocks, of different height h1, h2 . Make 2 towers using … Read more
Is this what you want? def existingPlayers(name, players): ”’ Check if player name already exists ”’ for player in players: if player[0].lower() == name.lower(): return True return False def getPlayerNames(players): ”’ Get the names of the players. Check if the number of players and player names are valid ”’ while True: name = input(“Enter player … Read more
Here is the function that you were looking for. The Iterable and the corresponding import is just to provide type hints and can be removed if you like. from collections import Counter from typing import Iterable d = [1,1,2,3,4,5,5,5,6] def retainSingles(it: Iterable): counts = Counter(it) return [c for c in counts if counts[c] == 1] … Read more
From what I gather, you’re trying to get input from a user until that user types -1. If thats the case, please see my function below. public static void main (String[] args) { // Scanner is used for I/O Scanner input = new Scanner(System.in); // Prompt user to enter text System.out.println(“Enter something “); // Get … Read more