To check each digit you need to convert these number in string-object and then count each digit from other number which result 0 or some finite number and then using all we can check that all digit is present their or not. Check out this code: def same_digits(a: int, b: int)->bool: cond1 = all([str(b).count(i) for … Read more
You can use itertools.groupby to group everything into lists of adjacent identical values, then unpack the lists of length one out of the nested lists: from itertools import groupby def group_adjacent(iterable): lists = (list(g) for k, g in groupby(iterable)) return [subl[0] if len(subl) == 1 else subl for subl in lists] group_adjacent([1,2,4,5,5,7,6,6,6]) # [1, 2, … Read more
The condition is tested before each iteration of the loop, not after every statement inside the loop body. So even though queue.pop(0) empties the list, you still execute the next statement and print the message. Then it goes back to the beginning and tests queue again. This time the condition fails and the loop terminates. … Read more
This is not done with nested classes but with composition: class Tool: def __init__(self, target): self.target = target self.wifi = Wifi(self.target) class Wifi: def __init__(self, target): self.target = target def connect(self, id): print id self.target.write(“xxxxxxxxx”) def verify(self) pass if __name__ == ‘__main__’: T = Tool(target) T.wifi.connect(“dev”) 1 solved Python : How to write class in … Read more
Sounds like you are looking for the repr() function: >>> s=”hello” >>> print repr(s) ‘hello’ solved Printing a string with the parentheses attched [closed]
In Pig Latin, if the word starts with a consonant, then it moves to the end of the string, and you add “ay”, right? Let’s say my input is banana. Now, with slicing: banana[0] == ‘b’ banana[1:] == ‘anana’ Now try figure it out from here 🙂 1 solved Working on a project on codeacademy, … Read more
The question itself is easily answered, no there is no performance gain in using anonymous functions in Python. There is a good chance you are actually making it slower. A simple timeit tests on trivial functions show that there is no real difference between the two. We take these two functions def test(message): return message … Read more
Move the dictionaries in dicts.py and create a class in template.py to return required template.You can create as many templates as you want in templates.py. dicts.py teams = {…} nouns = {…} article_words = {…} template.py from grammer import getArticle import random class Template(): def __init__(self,**kwargs): self.__dict__.update(kwargs) # must define other variables like article which … Read more
You need to import redirect from django.shortcut with from django.shortcuts import redirect This is well documented in the Django’s shortcut documentation. solved undefined name in function ‘redirect’
Without looping: with open(file, ‘r’) as f: print(len(f.split(‘\n’)[34].split())) solved Python: number of words in line [duplicate]
You can count the length of the string using function len. str can contain all chars and letters. Try the following code: word = input(“Waiting for input: “) print(“Found “+str(len(word))+” letters!”) Testing it: Waiting for input: hello Found 5 letters! Testing it with numbers and other chars: Waiting for input: hello123!@# Found 11 letters! 2 … Read more
This is wrong with your code: dicts can only contain each key once. If you specify d1={1:1,-1:0} d2={1:2,1:1,1:0} You get dict d1 with contains keys 1 and -1 with its values and you get dict d2 wich contains only 1 as key (with value of 0 as this one was specified last). Test: d1={1:1,-1:0} d2={1:2,1:1,1:0} … Read more
Besides the error with the ( going through the code some modifications could make this a little more optimal. We can use a try, except block to eliminate the ERROR else statement and handle that directly when we receive the input if is not a valid int and we force a valid int between 0 … Read more
# Note that variable names cannot contain hyphens flag_list = [“–syn”,”–ack”,”–push”,”–fin”,”–urg”,”–rst”] # This clearly has to be in quotes user_input = raw_input(“Enter numbers separated by comma:” ) # Split the user input string at the commas, and turn each element into an integer. flag_nums = map(int, flag_num.split(‘,’)) # List indexes start at 0, so subtract … Read more
You should make sure your indentation is uniform (always 4 spaces). Starting with if Submit == “yes”:, your lines have one extra space. This will be easier to do if you use a better IDE than IDLE, which will automatically highlight and label problems like this. Some good alternatives are Spyder (which is free), and … Read more