month_name‘ “helpfully” pads the list with an empty string so that the list indices match up with the normal 1-based counting of the months. To work around this, I would slice the list and enumerate it with an explicit starting point of 1. for i, m in enumerate(calendar.month_name[1:], 1): Another, more cumbersome but possibly useful, … Read more
The main problem here is that you are calling functions as they were variables Here is your code with the functions being called correctly: import math print(‘input your height and weight to find BMI.’) num_a = int(input(“feet==”)) num_b = int(input(“inches==”)) num_c = int(input(“weight(lbs)==”)) def weight(): return int(num_c) * 2.2 def height(): return (int(num_a)) * 12 … Read more
I believe you want a dictionary with a single key and a list: l = [1,2,3,4,5] d = {l[0]:l[1:]} Output: {1: [2, 3, 4, 5]} 0 solved Turning a list into a dictionary?
If you use Python3 then input(question) returns string ‘1’, ‘2’, etc. You have to convert it into number int(input(question)). 6 solved Python Type Error Unsupported Operand [duplicate]
You might consider learning how to implement breadth-first searches and depth-first searches in order to accomplish your objective. The following example shows how both of these search strategies can be easily handled in one function. A modular approach should make the code simple to change. #! /usr/bin/env python3 from collections import deque from operator import … Read more
Your factorial() function returns 0 for any input because of how you defined your range. The range builtin starts at 0 unless otherwise defined so: for c in range(n): re = re * c # no semicolons in Python is doing: re = re * 0 on the first iteration so for all subsequent iterations: … Read more
Since this is homework, it would be a bad idea to write the code for you. So ideally, what you need to do, is get the words that fit into the criteria (that you mentioned) and store them in a list. Then take the length of the list. After that, you can divide the length … Read more
res = [] for item in s.split(‘ ‘): … try: … int(item) … res.append(item) … except: … res += list(item) 5 solved How to separate string and integers to a list [closed]
And yes , you can just pass with if/else and for loops : a = [[0,0,0,0,0], [0,1,1,0,0], [0,1,0,1,0], [0,0,1,0,0], [0,0,0,0,0]] to_replace = 1 replacement = 9 for i in range(len(a)): for x in range(len(a[i])): if a[i][x] == to_replace: for pos_x in range(i-1,i+2): for pos_y in range(x-1,x+2): try: if a[pos_x][pos_y] != to_replace: a[pos_x][pos_y] = replacement except … Read more
One easy way to do this is with variables. Create 4 variables, one for hundreds, one for tens, one for ones, and also a variable to store the actual number to be counted on. Set the values to 1 for the counters and the value to 111 for the number variable. Add 1 to the … Read more
The problem is the way you’re building your query string. String values need to be wrapped in single quotes, otherwise they will be interpreted as DB objects (table name, column name, …). strConsulta= “insert into tabla(nombre,autor,year) values (‘”+nombre1+”‘,'”+autor1+”‘,”+year1+”)” You can avoid this issue by using parameterized queries: params = [nombre1, autor1, year1] strConsulta= “insert into … Read more
Try this: def check(list_): last = None for element in list_: if element == last: return True else: last = element return False solved Check if there is a number near the same number in a list – python [duplicate]
Here: http://rubular.com/r/CCW7sAUMrs is an example regex that matches whole text within {{ }}. You can easily clean up Your string with it using re.sub. Note, that it will not work when You have text like {{test}} paragraph {{test}}, because it will match whole string. As larsmans said, it gets harder if You can nest braces. … Read more
You could simply use something like this: import re result = [x for x in numbers if !re.search(r”(.)\1″, str(x))] solved how to find element that has no duplicate number? [closed]
I just got a trick for you to achieving what you want. a = “hello world, hello there, hello python” a = a.replace(“hello”,”hi”,2).replace(“hi”,”hello”,1) print(a) So what exactly 2nd Line does it it is replacing first two occurrence of hello to hi and then replacing 1st occurrence of hi to hello. Thus getting the output you … Read more