Abuton

import re

with open("company (2).txt") as file:
    com = file.read()
com_split = com.split("-------")
com_list = com_split[3].replace('--', '').split("\n")[1:-3]
a = [x.split('(') for x in com_list]
dict_prod = {}
for i in a:
    dict_prod[i[0]] = i[1].replace(')', '')

Abuton

CREATE TABLE IF NOT EXISTS `Cages`
(
    cageId INT NOT NULL AUTO_INCREMENT,
    noOfBirds INT DEFAULT 0,
    PRIMARY KEY (`cageId`)
)
ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE utf8mb4_unicode_ci;

CREATE TABLE IF NOT EXISTS `Customer`
(

Abuton

EntityName = {attribute1 datatype, attribute2 datatype, …..} 

Cages = {cageId INT, noOfBirds INT}
Employees = {employeeId INT, firstName TEXT, lastName TEXT, dateJoined DATE, salary DOUBLE, position VARCHAR(25)}
Sales = {id INT, date DATE, customerId INT, employeeId INT, price DOUBLE, noOfEggSold INT}

Abuton

import pandas as pd
import os
import shutil


def extract(path: str = "s3://my_bucket_name/file0.parquet") -> pd.DataFrame:
  df = pd.read_parquet(path)
  return df

Abuton

def replace_word(roots: list, sentence: str) -> str:
  output = []
  sentence_list = sentence.split()
  for word in sentence_list:
    output.append(word)
    for r in roots:
      if word.startswith(r):
        output[-1] = r

  return " ".join(output)

Abuton

def find_bigrams(S: str) -> list:
  # 1. create a list of words from the string
  new_s = S.lower().split()
  # 2. create an empty list to append tuples to
  bigrams = []
  # 3. loop through this list k-1 times say k is the length of list
  for i in range(len(new_s)-1):
    # 4. append the current word and the next to the empty list
    bigrams.append((new_s[i], new_s[i+1]))
  # 5. return the list

Abuton

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
import collections
def solution(S):
    # write your code in Python 3.6
    # create an empty dictionary to store the values of S
    f_dict = collections.defaultdict(int)
    # loop S
    for i in S:
        # increase number of times each character exists in S

Abuton

-- write your code in PostgreSQL 9.4
SELECT r.task_id, t.name as task_name, 
    case when avg(r.score) <= 20 then 'Hard'
        when avg(r.score) > 20 and avg(score) <= 60 then 'Medium'
        when avg(r.score) > 60 then 'Easy' End As difficulty

    from reports r left join tasks t on r.task_id = t.id 
    group by r.task_id, task_name order by r.task_id

Abuton

def solution(U, L, C):
    # write your code in Python 3.6
    if U + L != sum(C):
        return "IMPOSSIBLE"
    res = [[0 for j in range(len(C))] for i in range(2)]
    for i, s in enumerate(C):
        if s == 2:
            if U == 0 or L == 0:
                return "IMPOSSIBLE"

Abuton

With amounts as (
    select t1.id, t1.position, t1.salary, sum(t2.salary) as running_salary
    from candidates t1 join candidates t2 on t1.id >= t2.id
    and t1.position = t2.position
    where t1.salary <= 50000
    group by t1.id, t1.position, t1.salary
    order by t1.id
),
snrs as (
    select id, position, salary, running_salary from amounts