try the following code. lst_val = [] class Sample: def func(any_list): global lst_val for value in any_list: lst_val.append(value) return lst_val Sample.func([1, 2, 3, 4]) print(lst_val) 1 solved Appending to the list outside function in python not changing the list [closed]
Here’s a way to do this using the .bit_length method of Python 3 integers. It also uses the string .format method to do the integer to binary string conversion. This function returns a string starting with ‘0’ for non-negative numbers so that they can be distinguished from negative numbers. def twos_complement(n): m = n + … Read more
in python, in order to access values in a dict, you have two ways: dict[‘key’] dict.get(‘key’) dict(‘key’) will not work. solved python3 programme raising TypeError: ‘dict’ object is not callable [closed]
If you want this to work for any arbitrary key(s) you can use a defaultdict of OrderedDicts.. from collections import defaultdict, OrderedDict result_dict = defaultdict(OrderedDict) data = [(‘Han Decane’,’12333′),(‘Can Decane’,’12333′),(‘AlRight’,’10110′)] for (v,k) in data: result_dict[k][v]=True >>> list(result_dict[‘12333’].keys()) [‘Han Decane’, ‘Can Decane’] And if you want all the results that had multiple values >>> [k for … Read more
Try: import random class player: def __init__(self, name): self.name = name self.score = random.randint(1,6) print(self.name, self.score) player_1 = player(“Richard”) player_2 = player(“Bob”) winner=”” if player_1.score > player_2.score: winner = player_1.score print(winner) elif player_2.score> player_1.score: winner = player_2.score print(winner) else: print(“Tie.”) Output: 7 solved Why does this code not output the winner variable?
Answering this specifically would be difficult without knowing the specifics of your node implementation, and as this is almost certainly a homework problem wouldn’t be sporting anyway. What you want to do is go through each element of the list, and compare it to the Min you have. If it’s higher, then just go to … Read more
.any(1) is the same as .any(axis=1), which means look row-wise instead of per column. With this sample dataframe: x1 x2 x3 0 1 1 0 1 0 0 0 2 1 0 0 See the different outcomes: import pandas as pd df = pd.read_csv(‘bool.csv’) print(df.any()) >>> x1 True x2 True x3 False dtype: bool So … Read more
Something like this? def decode (line): return ‘ ‘.join (token [:-3] if token.endswith (‘WAY’) else (lambda a, b: b [:-3] + a) (*token.split (‘X’) ) for token in line.split () ) Sample: >>> decode (‘elloXHWEY orldXwWEY isXthWEY isWAY eatXgrWEY’) ‘Hello world this is great’ Nota bene: Might fail with words conaining ‘X’. But it should … Read more
A simple example, define a list with five elements, print the max from the second until the third index. max is a built-in function. max: Return the largest item in an iterable or the largest of two or more arguments. l = [1,2,3,4,5] print (max(l[2:3])) solved Maximum value from contiguous part of a list
You could achieve a similar behaviour by using a dictionary and a for-loop iterating through it: #!/usr/bin/env python3 # coding: utf-8 # define a dict with sample data d = {‘fruit’: 13} # print keys and values of sample dict for key, val in d.items(): print(key, val) Output looks like: fruit 13 2 solved How … Read more
Oneliner: list(zip(*l)) # [(1, 3, 5), (2, 4, 6)] 1 solved Printing tuples in a list
import re inp = open(‘menu.bak’, “r”) outp = open(‘word.txt’,”w”) for line in inp: for word in line.split(): outp.write(str(word) + “,” + “1” + ‘\n’) inp.close() outp.close() solved How to print each word of a string printed as new record?
N1 is simply just a list of names (that are of type String). When you loop / iterate over your list (for names in N1), each loop iteration will then act on that index of the list. So for the first iteration of the loop, names is equal to Ryan and therefore since you are … Read more
If your object isn’t a dict but could be converted to a dict easily (e.g. list of pairs), you could use: def make_a_dict(some_object): if isinstance(some_object, dict): return some_object else: try: return dict(some_object) except: return {‘return’: some_object} print(make_a_dict({‘a’: ‘b’})) # {‘a’: ‘b’} print(make_a_dict([(1,2),(3,4)])) # {1: 2, 3: 4} print(make_a_dict(2)) # {‘return’: 2} solved how to make … Read more
You need may use an .get() to get the corresponding value from the dict, if it doesn’t exist you’ll a None order1 = input(“Welcome to the pizza store what size pizza can we get for you (small, medium, large, or x-large): ” ) size_price = pizzaSize.get(order1) if size_price is not None: price += size_price else: … Read more