|
| 1 | +# Problem 585: Investments in 2016 |
| 2 | +# Difficulty: Medium |
| 3 | + |
| 4 | +# Table: Insurance |
| 5 | + |
| 6 | +# +-------------+-------+ |
| 7 | +# | Column Name | Type | |
| 8 | +# +-------------+-------+ |
| 9 | +# | pid | int | |
| 10 | +# | tiv_2015 | float | |
| 11 | +# | tiv_2016 | float | |
| 12 | +# | lat | float | |
| 13 | +# | lon | float | |
| 14 | +# +-------------+-------+ |
| 15 | +# pid is the primary key (column with unique values) for this table. |
| 16 | +# Each row of this table contains information about one policy where: |
| 17 | +# pid is the policyholder's policy ID. |
| 18 | +# tiv_2015 is the total investment value in 2015 and tiv_2016 is the total investment value in 2016. |
| 19 | +# lat is the latitude of the policy holder's city. It's guaranteed that lat is not NULL. |
| 20 | +# lon is the longitude of the policy holder's city. It's guaranteed that lon is not NULL. |
| 21 | + |
| 22 | +# Problem Statement: |
| 23 | +# Write a function to report the sum of all total investment values in 2016 tiv_2016, |
| 24 | +# for all policyholders who: |
| 25 | +# - have the same tiv_2015 value as one or more other policyholders, and |
| 26 | +# - are not located in the same city as any other policyholder (i.e., the (lat, lon) attribute pairs must be unique). |
| 27 | +# Round tiv_2016 to two decimal places. |
| 28 | + |
| 29 | +# Solution |
| 30 | + |
| 31 | +import pandas as pd |
| 32 | + |
| 33 | +def find_investments(insurance: pd.DataFrame) -> pd.DataFrame: |
| 34 | + # I create a column counting how many times each tiv_2015 value appears |
| 35 | + insurance['tiv_2015_cnt'] = insurance.groupby('tiv_2015')['tiv_2015'].transform('count') |
| 36 | + |
| 37 | + # I create another column counting how many times each (lat, lon) pair appears |
| 38 | + insurance['latlon_pair_cnt'] = insurance.groupby(['lat', 'lon'])['pid'].transform('count') |
| 39 | + |
| 40 | + # I filter to records where tiv_2015 is shared by at least one other policyholder |
| 41 | + # and (lat, lon) is unique |
| 42 | + temp_df = insurance.loc[(insurance['tiv_2015_cnt'] > 1) & (insurance['latlon_pair_cnt'] == 1), ] |
| 43 | + |
| 44 | + # I compute and return the sum of tiv_2016 rounded to 2 decimal places |
| 45 | + result_df = pd.DataFrame({'tiv_2016': [round(temp_df['tiv_2016'].sum(), 2)]}) |
| 46 | + return result_df |
| 47 | + |
| 48 | +# Intuition: |
| 49 | +# I need to sum the 2016 investment values (tiv_2016) for only those policyholders |
| 50 | +# who share their 2015 investment value with others but live in a unique (lat, lon) location. |
| 51 | + |
| 52 | +# Explanation: |
| 53 | +# First, I use `groupby().transform('count')` to calculate how many times each tiv_2015 value occurs. |
| 54 | +# Then, I check for uniqueness of (lat, lon) by counting how many policyholders are in each city. |
| 55 | +# The condition is to keep only those with tiv_2015 repeated more than once and (lat, lon) seen exactly once. |
| 56 | +# After filtering, I compute the sum of tiv_2016 for these filtered rows and round the result to two decimal places. |
| 57 | +# Problem 585: Investments in 2016 |
| 58 | +# Difficulty: Medium |
| 59 | + |
| 60 | +# Table: Insurance |
| 61 | + |
| 62 | +# +-------------+-------+ |
| 63 | +# | Column Name | Type | |
| 64 | +# +-------------+-------+ |
| 65 | +# | pid | int | |
| 66 | +# | tiv_2015 | float | |
| 67 | +# | tiv_2016 | float | |
| 68 | +# | lat | float | |
| 69 | +# | lon | float | |
| 70 | +# +-------------+-------+ |
| 71 | +# pid is the primary key (column with unique values) for this table. |
| 72 | +# Each row of this table contains information about one policy where: |
| 73 | +# pid is the policyholder's policy ID. |
| 74 | +# tiv_2015 is the total investment value in 2015 and tiv_2016 is the total investment value in 2016. |
| 75 | +# lat is the latitude of the policy holder's city. It's guaranteed that lat is not NULL. |
| 76 | +# lon is the longitude of the policy holder's city. It's guaranteed that lon is not NULL. |
| 77 | + |
| 78 | +# Problem Statement: |
| 79 | +# Write a function to report the sum of all total investment values in 2016 tiv_2016, |
| 80 | +# for all policyholders who: |
| 81 | +# - have the same tiv_2015 value as one or more other policyholders, and |
| 82 | +# - are not located in the same city as any other policyholder (i.e., the (lat, lon) attribute pairs must be unique). |
| 83 | +# Round tiv_2016 to two decimal places. |
| 84 | + |
| 85 | +# Solution |
| 86 | + |
| 87 | +import pandas as pd |
| 88 | + |
| 89 | +def find_investments(insurance: pd.DataFrame) -> pd.DataFrame: |
| 90 | + # I create a column counting how many times each tiv_2015 value appears |
| 91 | + insurance['tiv_2015_cnt'] = insurance.groupby('tiv_2015')['tiv_2015'].transform('count') |
| 92 | + |
| 93 | + # I create another column counting how many times each (lat, lon) pair appears |
| 94 | + insurance['latlon_pair_cnt'] = insurance.groupby(['lat', 'lon'])['pid'].transform('count') |
| 95 | + |
| 96 | + # I filter to records where tiv_2015 is shared by at least one other policyholder |
| 97 | + # and (lat, lon) is unique |
| 98 | + temp_df = insurance.loc[(insurance['tiv_2015_cnt'] > 1) & (insurance['latlon_pair_cnt'] == 1), ] |
| 99 | + |
| 100 | + # I compute and return the sum of tiv_2016 rounded to 2 decimal places |
| 101 | + result_df = pd.DataFrame({'tiv_2016': [round(temp_df['tiv_2016'].sum(), 2)]}) |
| 102 | + return result_df |
| 103 | + |
| 104 | +# Intuition: |
| 105 | +# I need to sum the 2016 investment values (tiv_2016) for only those policyholders |
| 106 | +# who share their 2015 investment value with others but live in a unique (lat, lon) location. |
| 107 | + |
| 108 | +# Explanation: |
| 109 | +# First, I use `groupby().transform('count')` to calculate how many times each tiv_2015 value occurs. |
| 110 | +# Then, I check for uniqueness of (lat, lon) by counting how many policyholders are in each city. |
| 111 | +# The condition is to keep only those with tiv_2015 repeated more than once and (lat, lon) seen exactly once. |
| 112 | +# After filtering, I compute the sum of tiv_2016 for these filtered rows and round the result to two decimal places. |
0 commit comments