There are some problems with your code, but let’s look at the output first. You get that output, because you are not calling the function that you defined: print(“The lenght of the hypotenous is”,findhypot) So instead of just putting findhypot in your print you should call it and pass the two paramters that you just … Read more
Hope this works for you. code import time import datetime import pytz print(‘–‘*62) cont = 0 print(‘MENU’.center(115)) print(‘–‘*62) my_timezones = {} l = list() # MAIN LIST l.append(‘0’) my_timezones[‘United Arab Emirates’] = pytz.country_timezones[‘AE’][0] my_timezones[‘Canada’] = pytz.country_timezones[‘CA’][20] my_timezones[‘South Korea’] = pytz.country_timezones[‘KR’][0] my_timezones[‘United States’] = pytz.country_timezones[‘US’][17] my_timezones[‘New Zeland’] = pytz.country_timezones[‘NZ’][0] my_timezones[‘Norway’] = pytz.country_timezones[‘NO’][0] my_timezones[‘Ireland’] = pytz.country_timezones[‘IE’][0] my_timezones[‘Netherlands’] … Read more
Introduction If you are having trouble with an error and don’t know how to fix it, you are not alone. Many people have encountered similar issues and have found solutions. This introduction will provide an overview of how to troubleshoot and resolve the error you are experiencing. We will discuss the importance of understanding the … Read more
You can use range() instead. l = list(range(1,100)) print(l) k = [] for i in l: if i%10==8: k.append(i) print(k) solved I have an error and due to lack of knowledge i dont no how fix this: “Traceback (most recent call last): File ““, line 8, in [closed]
You can generally turn a nested for loop that builds a list into a list comprehension by using the same exact for … in statements in the same order: list3 = [k for l in lst1 for k in lst2 if l == k[0]] list4 = [k for l in lst1 for k in lst2 … Read more
try setx PATH “%PATH%;C:\Python34\Scripts” if it is not working install your python modules like this python -m pip install [packagename] solved Cannot install pip install numpy [duplicate]
I believe that you are trying for these : the_array = [] while True: sval = input(‘Enter a number: ‘) if sval == ‘done’: break try: ival = int(sval) the_array.append(ival) except: print(‘Invalid input’) continue def maximum(values): num1 = 0 for i in values: if i >= num1: num1 = i return num1 def minimum(values): num2 … Read more
You dont have on_delete. You have on_delet. Fix this and should be fine solved _int_() missing 1 required positional argument: on_delete [closed]
Assuming you want a list of dicts as the output of the json with each dict be the form in your question: The following one liner will put it into the data structure your looking for with each tuple generating it’s own complete structure: [{‘name’:i[0], ‘children’: [{‘name’: i[1], ‘value’: i[2]}]} for i in tuples] But … Read more
Try using: d = {i[‘service’]: i[‘price’] for i in a} print([{‘service’: i, ‘price’: d.get(i)} for i in b]) Output: [{‘service’: ‘basketball’, ‘price’: None}, {‘service’: ‘yoga’, ‘price’: 30}, {‘service’: ‘soccer’, ‘price’: None}, {‘service’: ‘golf’, ‘price’: 40}] 0 solved Compare in Python two lists and create a new one that is combinated [closed]
Files act like a long tape in a casette; you can read the file but by the time you are done you have passed the tape all the way to the end. Reading again won’t give you the data again. As such your second function tried to read data from a file that is already … Read more
It is hard to tell, but it looks like Python 3 is in use, and I suspect the book is requiring Python 2. If you can re-run the program with Python 2, you may find you have far fewer errors (if any). If you do have a new error, you can then ask a new … Read more
From your expected result, it appears that each “group” is based on contiguous id values. For this, you can use the compare-cumsum-groupby pattern, and then use agg to get the min and max values. # Sample data. df = pd.DataFrame( {‘id’: [1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, … Read more
You probably defined a function cercle that draws French circles. The function should look like: def cercle(canv, x, y, rad): return canv.create_oval(x-rad, y-rad, x+rad, y+rad, width=2) Note that the function must return the circle object. Otherwise self.ouvre is None and then you can not delete it as you do not have a reference to the … Read more
Short answer is: you can’t apply Trilinear Inerpolation. Let’s start with 2x2x2 blocks. Each block, is represented by it’s centre pixel ( 1,2,3,4 in ugly yellow on my sketch). Each pixel is located at the corner of a cell. A pixel (the red dot), will be shared by up to 4 blocks that overlap. With … Read more