you can use something like that: sentence = input().split() ans = [] for i in range(len(sentence)): if i%2: ans.append(sentence[i]+’…’) else: ans.append(sentence[i]) print(‘ ‘.join(ans)) solved How to iterate odd indexes and add them back to original list
I haven’t closed the brackets on line 31 battleaction = str(input(‘[a]ttack [s]uper attack [i]tem or [r]un’) solved Error message when attempting to define a function [closed]
By restart the loop i’m assuming u mean call it again the easiest way to do is would be to make it a recursive function def fn(): for index, line in enumerate(lines): if line[:8] == Token.INCLUDE: system(“python3 {} {}”.format(argv[0], argv[1])) del lines[index] #make sure to add your conditions else it will go into an infinite … Read more
I’m not sure if this is what you are asking, but I interpreted this as list the files in a given directory, and then check if the input is in the file. import os string = input(“Please type the input “) directory = “c://files//python” for file in os.listdir(directory): if file.endswith(“.txt”): filecontent = open(file, “r”) if … Read more
Since you already have binned data, just use pyplot.errorbar or add yerr kwarg to bar plot from matplotlib import pyplot as plt plt.errorbar( [n/len(bins) for n, x in enumerate(bins)], [x[0][0] for x in bins], yerr = [x[0][1]**0.5 for x in bins], marker=”_”, fmt=”.”) plt.bar( [n/len(bins) for n, x in enumerate(bins)], [x[0][0] for x in bins], … Read more
That isn’t a concatenation: you’re not appending this to anything else. It’s only a value substitution. Thus, you should (1) use the variable name, rather than a string; (2) leave out the concatenation attempt. “device-id”: d1, This should substitute the desired value of varibale d1. solved Python String concatenation with in double quotes [closed]
def QuestionSet1(): print(“Challenge level 1 has being selected.”) print(“Can you translate these words into french?”) a=input(‘Q1. Hello! :’) score = 0 if ‘bonjour’ in a.lower(): score = score + 1 print(‘Correct!’) else: print(‘Wrong! ‘+’Its Bonjour’) print(‘You have finished and scored’, score, ‘out of 10’) You would need make a reference to score before you give … Read more
All you need to do is get rid of the first hangman() call: … for i in range(1, num_of_players + 1): print(players_dict[‘Player {}’.format(i)] if hangman() is False: results += False else: results += True If you need to keep hold of the value returned, assign it to a variable beforehand: … r = hangman() if … Read more
what about a = [1, 2, 3, 4, 5, 6, 7, 8] f = open(‘myFile.txt’, ‘a’) for n in a: f.write(‘%d\t’%n) f.write(‘\n’) f.close() and you get as file content after running it 4 times 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 … Read more
The text you are looking for is in the 2nd TR inside the table. So, print tbl.findAll(‘tr’)[2] There really isn’t any good structure inside that text, however, so you’re on your own about getting the contact name, etc. solved Why is my parser not working
I assume this is what you wanted to achive: def test(zen, zen2): if zen == 1 and zen2 == True: print (“hello”) zen2 = False else: print (“hello 2”) test(1, True) As some comments suggested, if one tries to run your code, it returns following (on linux GCC 4.8.2/Python 3.6.1): SyntaxError: name ‘zen2’ is used … Read more
print [a for a,b in zip(my_list[::2],my_list[1::2]) if a == b] you may need to sort my_list first … my_list = sorted(my_list) print [a for a,b in zip(my_list[::2],my_list[1::2]) if a == b] Im not at all sure this solves your issue. def do_this_weird_thing(a_list): from itertools import groupby for grp,item in groupby(sorted(a_list)): item = list(item) yield [grp] … Read more
Set your variable before the if : def Home(request):#django city = ” if request.method == ‘POST’: city = request.POST.get(‘search’) print(city) URL = f’http://api.openweathermap.org/data/2.5/weather?q={city} 0 solved I Want to access the variable of a if statement in the function .Python
Yes off course! It will work like that but it’s only for this condition in which you have only one condition so the compiler checks whether it’s true or not but what if there’s more than one conditions ?? Also not using elif will allow the program to output the other print statement, but using … Read more
If you have a mic on whatever device you are using, you can use that to read whatever sound is coming out of your computer. Then, you can compare those audio frames that you are recording to a sound file of what sound you are looking for Of course, this leaves it very vulnerable to … Read more