Pandas3 Solutions

calculate-special-bonus.py

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+"""
+
+Table: Employees
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| employee_id | int     |
+| name        | varchar |
+| salary      | int     |
++-------------+---------+
+employee_id is the primary key (column with unique values) for this table.
+Each row of this table indicates the employee ID, employee name, and salary.
+
+
+Write a solution to calculate the bonus of each employee. The bonus of an employee is 100% of their salary if the ID of the employee is an odd number and the employee's name does not start with the character 'M'. The bonus of an employee is 0 otherwise.
+
+Return the result table ordered by employee_id.
+
+The result format is in the following example.
+
+
+Example 1:
+
+Input: 
+Employees table:
++-------------+---------+--------+
+| employee_id | name    | salary |
++-------------+---------+--------+
+| 2           | Meir    | 3000   |
+| 3           | Michael | 3800   |
+| 7           | Addilyn | 7400   |
+| 8           | Juan    | 6100   |
+| 9           | Kannon  | 7700   |
++-------------+---------+--------+
+Output: 
++-------------+-------+
+| employee_id | bonus |
++-------------+-------+
+| 2           | 0     |
+| 3           | 0     |
+| 7           | 7400  |
+| 8           | 0     |
+| 9           | 7700  |
++-------------+-------+
+Explanation: 
+The employees with IDs 2 and 8 get 0 bonus because they have an even employee_id.
+The employee with ID 3 gets 0 bonus because their name starts with 'M'.
+The rest of the employees get a 100% bonus.
+
+"""
+
+# The function calculates a special bonus for employees based on given conditions. 
+# Employees receive a bonus equal to their salary if their employee ID is odd and their name does not start with 'M'. 
+# The result is a DataFrame containing employee IDs and their corresponding bonuses, sorted by employee ID.
+
+import pandas as pd
+
+def calculate_special_bonus(employees: pd.DataFrame) -> pd.DataFrame:
+
+    employees['bonus'] = employees.apply(
+        lambda x: x['salary'] if x['employee_id'] % 2 and not x['name'].startswith('M') else 0, 
+        axis=1
+    )
+
+    df = employees[['employee_id', 'bonus']].sort_values('employee_id')
+    return df
+

fix-names-in-a-table.py

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+"""
+
+Table: Users
+
++----------------+---------+
+| Column Name    | Type    |
++----------------+---------+
+| user_id        | int     |
+| name           | varchar |
++----------------+---------+
+user_id is the primary key (column with unique values) for this table.
+This table contains the ID and the name of the user. The name consists of only lowercase and uppercase characters.
+
+
+Write a solution to fix the names so that only the first character is uppercase and the rest are lowercase.
+
+Return the result table ordered by user_id.
+
+The result format is in the following example.
+
+
+
+Example 1:
+
+Input: 
+Users table:
++---------+-------+
+| user_id | name  |
++---------+-------+
+| 1       | aLice |
+| 2       | bOB   |
++---------+-------+
+Output: 
++---------+-------+
+| user_id | name  |
++---------+-------+
+| 1       | Alice |
+| 2       | Bob   |
++---------+-------+
+
+"""
+
+# This function capitalizes the first letter of each user's name while keeping the rest lowercase.  
+# It then sorts the DataFrame based on the 'user_id' column in ascending order.  
+# The modified DataFrame is returned with the changes applied.  
+
+import pandas as pd
+
+def fix_names(users: pd.DataFrame) -> pd.DataFrame:
+    users['name'] = users['name'].str.capitalize()
+    return users.sort_values(by=['user_id'])
+

patients-with-a-condition.py

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+"""
+
+Table: Patients
+
++--------------+---------+
+| Column Name  | Type    |
++--------------+---------+
+| patient_id   | int     |
+| patient_name | varchar |
+| conditions   | varchar |
++--------------+---------+
+patient_id is the primary key (column with unique values) for this table.
+'conditions' contains 0 or more code separated by spaces. 
+This table contains information of the patients in the hospital.
+
+
+Write a solution to find the patient_id, patient_name, and conditions of the patients who have Type I Diabetes. Type I Diabetes always starts with DIAB1 prefix.
+
+Return the result table in any order.
+
+The result format is in the following example.
+
+
+
+Example 1:
+
+Input: 
+Patients table:
++------------+--------------+--------------+
+| patient_id | patient_name | conditions   |
++------------+--------------+--------------+
+| 1          | Daniel       | YFEV COUGH   |
+| 2          | Alice        |              |
+| 3          | Bob          | DIAB100 MYOP |
+| 4          | George       | ACNE DIAB100 |
+| 5          | Alain        | DIAB201      |
++------------+--------------+--------------+
+Output: 
++------------+--------------+--------------+
+| patient_id | patient_name | conditions   |
++------------+--------------+--------------+
+| 3          | Bob          | DIAB100 MYOP |
+| 4          | George       | ACNE DIAB100 | 
++------------+--------------+--------------+
+Explanation: Bob and George both have a condition that starts with DIAB1.
+
+"""
+
+# This function filters patients with conditions containing "DIAB1".  
+# It selects rows where the "conditions" column starts with "DIAB1" or contains " DIAB1" to avoid partial matches within other words.  
+# The `regex=False` ensures an exact substring match without using regular expressions.  
+
+import pandas as pd
+def find_patients(patients: pd.DataFrame) -> pd.DataFrame:
+    return patients[patients["conditions"].str.startswith("DIAB1") | patients["conditions"].str.contains(" DIAB1", regex=False)]
+