min1 = score_arr should be changed to min1 = score_arr[i]. min1 = score_arr is setting min1 to the whole list, so the next iteration fails on the line if score_arr[i] < min1 since score_arr[i] is a float and min1 is now a list (as indicated by the error message). But, a better way to find … Read more
As @Aurora001 mentioned, you should not ask for completed code. However, I think addressing the method you might use could be a good answer to your question, and I hope others agree. This is how I would approach the problem: Prompt for input to get the coefficients a, b, and c. Write a method that … Read more
Or you can put your if statements between quotes: ex: if choice == “1” print(“nr 1”) elif choice == “2” so on .. solved Simple elif statement
You can build a list of lines skipping the digit ones this way : lines = [line for line in data.splitlines() if not line.strip().isdigit()] assuming data is your text file content. 1 solved How to extract a line from text file that is digits-only? [closed]
This should be fairly easy with something like the csv library, if you don’t want to get into learning dataframes. import os import csv new_data = [] for filename in os.listdir(‘./csv_dir’): if filename.endswith(‘.csv’): with open(‘./csv_dir/’ + filename, mode=”r”) as curr_file: reader = csv.reader(curr_file, delimiter=”,”) for row in reader: new_data.append(row[2]) # Or whichever column you need … Read more
Your code fragment doesn’t make a lot of sense without more context. To get a better understanding of the keyword look at this: Use of “global” keyword in Python As for your code it makes the variable only available in a global scope after it errors which is unusual. As the warning indicates you would/should … Read more
You’re not returning selection, you’re printing it instead. Replace print(selection) with return selection. 0 solved TypeError: ‘NoneType’ [duplicate]
Yes, you will have to overload the range function for i in range(0,n,24): print(i) Output: 0 24 48 72… 0 solved Python 3.4 i want creat list 0,24,48,…….6168 [closed]
Your dict is the wrong way round. The thing you want to look up (key) goes first in each pair pCode = {“T”: “Alberta”} a = input(“Enter a 6 character postal code: “) You can get the first character of the post code as a[0] print(pCode.get(a[0])) solved First character in a dictionary Python [closed]
This program should behave as you wanted: with open(“test.txt”,’r+’) as f: # Read the file lines=f.readlines() # Gets the word word=input(“Enter the word:”) print(word+” occured on line(s):”,end=’ ‘) # Puts a flag to see if the word occurs or not flag=False for i in range(0,len(lines)): # Creates a list of words which occured on the … Read more
If I am assuming right and you are using from bs4 import BeautifulSoup you need to understand that find_all is part of the bs4.element.Tag object findAll might not work obj = BeautifulSoup(html_text, ‘html.parser’) obj.find_all(“tr”,{“class”:”match”}) This should solve your problem. 3 solved when I write findAll it says: findAll is not defined [closed]
The easiest way to check if a number is a palindrom would probably to treat it as a string, flip it, and check if the flipped string is equal to the original. From there on, it’s just a loop that counts how many of these you’ve encounteres: begin = int(input(‘Enter begin: ‘)) end = int(input(‘Enter … Read more
Here’s a silly one-liner version: def print_table(seq, width=3): print(‘\n’.join([”.join( [(str(u[-i]) if len(u) >= i else ”).rjust(width) for u in seq]) for i in range(max(len(u) for u in seq), 0, -1)])) And here’s the same algorithm in a somewhat more readable form: def get_cell(u, i): return str(u[-i]) if len(u) >= i else ” def print_table(seq, width=3): … Read more
Simply use zip and list comprehension. z = [a+b for a,b in zip(x,y)] solved How to add integers from one list to another in ascending order? [closed]
#Updating code snippet by leplatrem result = {’44’: [‘2’, ‘4’], ’41’: [‘1’, ‘2’, ‘3’]} for (k,v) in result.viewitems(): temp_list = [str(i) for i in v] temp_str = “,”.join(temp_list) temp_str = “‘” + temp_str + “‘” print “UPDATE myTable SET myColumn={} where id={}”.format(temp_str,k) #Output UPDATE myTable SET myColumn=’2,4′ where id=44 UPDATE myTable SET myColumn=’1,2,3’ where id=41 … Read more