As zipa pointed out, you never break your while loop in the else statement on line 146 unless you run out of tries. A one line “fix” is to simply put an exit() after line 98 within the traverse() function. 3 solved Udacity python code a quiz [closed]
This is one way to achieve what you are looking to do: number = “12345678910” print( ‘number id {0}.{1}.{2}-{3}’.format( *[number[i:i+3] for i in range(0,len(number), 3)] ) ) #number id 123.456.789-10. There are a few problems with your code. First the first number in the “{0…}” refers to the positional argument in the format() function. In … Read more
First, use split() to split the input into a list. Then, you need to test isnumeric() on every element of the list, not the whole string. Call int() to convert the list elements to integers. Finally, add the sum to counter, not counter_item. my_list = input(“Input your list.(Numbers)\n”).split() while not all(item.isnumeric() for item in my_list): … Read more
A = [s for s in A if s.strip() != ”] or A = [s for s in A if len(s.strip()) > 0] 1 solved How to remove str. Of spaces from list [closed]
According to: https://en.wikipedia.org/wiki/Wikipedia:Special:Random The endpoint for a categorized random subject is: https://en.wikipedia.org/wiki/Special:RandomInCategory/ And not: https://en.wikipedia.org/wiki/Special:Random/ So you should change your url to: url = requests.get(f”https://en.wikipedia.org/wiki/Special:RandomInCategory/{topic}”) 1 solved A Random Wikipedia Article Generator
If you want to check if a string i contains string j, you can do it using i in j. Then comes the part how do you want the output actually. What my approach was to go through the list1 and for each of them go though the list2 and check if element of list2 … Read more
Given a list of triples, return a list with the second element of each triple (If an item is not a triple, return None for that element ) solved Given a list of triples, return a list with the second element of each triple (If an item is not a triple, return None for that … Read more
list=[] for i in columns: a=df_train[i].isnull().sum() list.append(a) print(list) 2 solved how to get only last line
import itertools import pandas as pd dfs_in_list = [df1, df2, df3, df4, df5] combinations = [] for length in range(2, len(dfs_in_list)): combinations.extend(list(itertools.combinations(dfs_in_list, length))) for c in combinations: pd.concat(c, axis=1) 0 solved Combinations of DataFrames from list
Yes, you can do exactly what you suggest: class D(C, B): a = A.a 5 solved Python multiple inheritance name clashes [closed]
For DataFrames, the value to be passed for axis is columns. frame.sort_index(axis=”columns”) 1 solved No axis named 1 for object type
You should be using = to assign values and not : You can do like this. import datetime example_timestamp = int(‘1629617204525776950’) timestamp_to_date_time = datetime.datetime.fromtimestamp(example_timestamp/1000000000).strftime(‘%Y-%m-%d %H:%M:%S,%f’) print(timestamp_to_date_time) 2021-08-22 07:26:44,525777 solved Convert Timestamp(Nanoseconds format) to datetime [closed]
Here’s how to do what you say you want. Your description of the desired “table” output is somewhat vague, so I made my best guess. def get_rows(data): rows = [] for line in data.splitlines(): fields = line.split(‘,’) if not any(field == ‘NULL’ for field in fields): # Not defective row. rows.append(fields) return rows csv_string = … Read more
ids = [int(row[‘id’]) for row in b[‘cricket’][‘Arts & Entertainment’]] will give you a list of the numbers, [6003760631113, 6003350271605, 6003106646578, 6003252371486, 6003370637135] And for your new edits; ids = [int(row[‘id’]) for row in b[‘cricket’][‘Arts & Entertainment’] + b[‘News, Media & Publications’]] 3 solved Parse a nested dictionary [closed]
You can’t fill a list in this way. Because you start with an empty list, Jewels = [], your attempt to assign to Jewels[0] will result in an IndexError. You can instead use the list.append method jewels = [] for i in range(total_jewels): jewels.append(raw_input(‘Please Enter approx price for Jewel #{}: ‘.format(i))) or a list comprehension … Read more