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| from scipy import stats | |
| import math | |
| def confidence_interval(sd, mean, n, target_interval): | |
| """sd - выборочное среднее значение | |
| mean - среднее выборки | |
| n - размер выборки | |
| target interval - 0..1, например 0.95, 0.99 | |
| """ | |
| alpha = 1 - target_interval | |
| z = abs(stats.norm.ppf(alpha/2)) |
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| regexp: https://regex101.com/r/iqfDq3/5 | |
| ((eth|v)\d+|(en|sl|wl|ww|ib)(o|s\d+(f\d+)?d(\d+)?|(P\d+)?((p\d+)(s\d+)|(?<!s\d)(p\d+)|(s\d+)(?!p\d))((f\d+)?((u\d+){1,4})?(i\d)?|(d\d+)?)|x[0-9a-f]{12})) | |
| /* | |
| * Predictable network interface device names based on: | |
| * - firmware/bios-provided index numbers for on-board devices | |
| * - firmware-provided pci-express hotplug slot index number | |
| * - physical/geographical location of the hardware | |
| * - the interface's MAC address |
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| # https://www.codewars.com/kata/56a1c074f87bc2201200002e/ | |
| # modified merge sort counting left > right cases | |
| def countsort(arr, cntarr, cntdict): | |
| if len(arr) > 1: | |
| mid = len(arr) // 2 | |
| left = arr[:mid] | |
| leftcount = cntarr[:mid] | |
| right = arr[mid:] | |
| rightcount = cntarr[mid:] | |
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| from itertools import groupby | |
| def solution(args): | |
| # [-3,-2,-1,2,10,15,16,18,19,20] -> [[-3, -2, -1], [2], [10], [15, 16], [18, 19, 20]] | |
| grps = ([v[1] for v in g] for _,g in groupby(enumerate(args), lambda p: p[1]-p[0])) | |
| #[[-3, -2, -1], [2], [10], [15, 16], [18, 19, 20]] -> -3--1,2,10,15,16,18-20 | |
| return ','.join('{}{}{}'.format(g[0],'-'if len(g)>2 else',',g[-1]) | |
| if len(g)>1 else str(g[0]) for g in grps) |
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| import pandas as pd | |
| import requests | |
| from bs4 import BeautifulSoup | |
| def get_html(url): | |
| r = requests.get(url=url) | |
| return r.text | |
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| from itertools import permutations | |
| from functools import reduce | |
| import re | |
| import operator | |
| _DEBUG = False | |
| _DEBUG_PRINT_PER_TRIES = 1000 | |
| def infix_postfix(command): |
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| def estimate_timedelta(td): | |
| """returns estimate timedelta in format 'YY year MM mon DD day'. consider plural form or absence y/m/d""" | |
| if td.days > 0: | |
| y, rem = divmod(td.days, 365) | |
| m, d = divmod(rem, 31) | |
| dict_ = {'year': y, 'mon': m, 'day':d } | |
| return " ".join([f"{dict_[k]} {k}" + ("s" if dict_[k] > 1 else "") for k in dict_ if dict_[k] > 0 ]) | |
| else: | |
| return str(td) |
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| def day_of_week(date): | |
| """ get weekday from date (as instance 2022-12-31 iso format) """ | |
| # | |
| year = int(date[0:4]) | |
| month = int(date[5:7]) | |
| day = int(date[8:10]) | |
| offset = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334] | |
| week = ['Sunday', | |
| 'Monday', | |
| 'Tuesday', |