rose tinted glasses

[TIL] 파이썬 리스트 (del, sum, min, max, len, sort)

kyra — Sun, 20 Jul 2025 19:34:03 +0900

리스트 원소 삭제 (del)

fruits = ['apple', 'banana', 'orange', 'grapes']
del fruits[1]

print(fruits) # ['apple', 'orange', 'grapes']

----------------
fruits = ['apple', 'banana', 'orange', 'grape']

del fruits[1:3]
print(fruits) # ['apple', 'grapes']

리스트 원소 합(sum)

nums = [1, 2, 3, 4, 5]

print(sum(nums)) # 15 출력

리스트 원소 최댓값 최솟값(min(), max())

nums = [1, 2, 3, 4, 5, 6, 7]

print("max: ", max(nums))
print("min: ", min(nums))

리스트 원소 개수 (len)

cook = ["피자", "김밥", "만두", "양념치킨", "족발", "피자", "김치만두", "쫄면", "소시지", "라면", "팥빙수", "김치전"]

print(len(cook))

리스트 오름차순 정렬 (sort, sorted)

data = [2, 4, 3, 1, 5, 10, 9]
data.sort()

print(data)

----

data = [2, 4, 3, 1, 5, 10, 9]
data2 = sorted(data)

print(data2)

[TIL] 파이썬 문자열 (replace, split, rstrip)

kyra — Sun, 20 Jul 2025 18:28:13 +0900

문자열 슬라이싱 (거꾸로출력)

string = 'PYTHON'

print(string[::-1])

#[시작점:끝점:건너뛰기/거꾸로]

문자열 치환 (replace)

phone_numb = "010-1111-2222"
print(phone_numb.replace("-", " "))

# replace를 사용해 문자열 치환

string = 'abcdfe2a354a32a'

print(string.replace('a', 'A')) # Abcdfe2A354A32A 출력

문자열 분리 (split)

url = "http://sharebook.kr"
url_split = url.split(".")

print(url_split[-1]) # kr 출력

url_split # ['http://sharebook', 'kr'] 출력

오른쪽 공백 제거 (rstrip)

data = "039490     "
data = data.rstrip()

print(data) # 039490

[TIL] Append vs Extend vs Insert in python

kyra — Tue, 23 Apr 2024 11:05:57 +0900

Append

The append() method adds a single element to the end of a list.
If the element being added is a sequence (list, tuple, etc.), it will add that entire sequence as a single element to the list.

fruits = ['apple', 'banana']
fruits.append('orange')
print(fruits)  # ['apple', 'banana', 'orange']

nested_list = [1, 2]
fruits.append(nested_list)
print(fruits)  # ['apple', 'banana', 'orange', [1, 2]]

Extend

The extend() method adds all the elements of an iterable (sequence) to the end of a list.
If the iterable being added is a list, it will add each element of that list individually to the original list.

fruits = ['apple', 'banana']
fruits.extend(['orange', 'grape'])
print(fruits)  # ['apple', 'banana', 'orange', 'grape']

nested_list = [1, 2]
fruits.extend(nested_list)
print(fruits)  # ['apple', 'banana', 'orange', 'grape', 1, 2]

Insert

Insert() method enables to put value in a certain position in the list.

fruits = ['apple', 'banana', 'orange', 'grape']
fruits.insert(1, "blueberry")

print(fruits) # ['apple', 'blueberry', 'banana', 'orange', 'grape']

[Collocations] gain, win, earn, make, achieve, beat/defeat

kyra — Mon, 22 Apr 2024 08:42:22 +0900

Gain

1. 권력을 얻다.

2. 통제권을 얻다.

3. 접근할 수 있다.

4. 명성을 얻다.

5. 관심을 받다.

6. 인식을 가지다.

7. 장점을 갖다.

1. gain power

2. gain control

3. gain access

4. gain a reputation

5. gain publicity

6. gain recognition

7. gain an advantage

Win

1. 수상하다.

2. 상을 타다.

3. 메달을 따다.

4. 게임에서 이기다.

5. 전쟁에서 이기다.

6. 선거에서 이기다.

1. win an award

2. win a prize

3. win a medal

4. win a game

5. win a war

6. win an election

Make/ Earn, Achieve

1. 돈을 벌다.

2. 이윤을 내다.

3. 연봉을 받다.

4. 성공하다

5. 목표를 달성하다

1. make money/ earn money

2. make a profit

3. earn a salary

4. achieve success

5. achieve goals/ achieve your aims

[MachineLearning] K-means Clustering

kyra — Tue, 16 Apr 2024 13:36:53 +0900

https://youtu.be/4b5d3muPQmA?si=sA0n1YrxFFx8YhNo

1. Select the number of clusters(=K) you want to identify in your data.

2. Randomly select 3 distinct data points.

3. Measure the distance between the 1st point and the three initial clusters.

4. Assign the 1st point to the nearest cluster.

5. Do the same thing for the remaining points.

6. Assess the quality of clustering by adding up the variation within each cluster.

7. Calculates the mean of each cluster and then reclusters based on the new means.

8. This repeats until the clusters no longer change.

How to figure out what value to use for "K"?

1. try different values for K

2. find an "elbow plot" and pick "K"

KNN vs K-Means Clustering

The key differences between KNN (K-Nearest Neighbors) and K-Means Clustering are:

1. Purpose:
   - KNN is used for classification and regression problems.
   - K-Means is used for data clustering.

2. Input Data:
   - KNN requires a labeled dataset.
   - K-Means works with unlabeled, unsupervised data.

3. Working Mechanism:
   - KNN finds the K nearest neighbors of a new data point and classifies or regresses based on them.
   - K-Means partitions the data into K clusters and iteratively updates the cluster centroids until convergence.

4. Output:
   - KNN predicts the class label or regression value for a new data point.
   - K-Means outputs the cluster label for each data point.

5. Applications:
   - KNN is primarily used for classification and regression tasks.
   - K-Means is commonly used for customer segmentation, anomaly detection, data compression, etc.

In summary, KNN is a supervised learning algorithm that makes predictions on new data, while K-Means is an unsupervised learning algorithm that groups data into clusters. They serve different purposes and are used

in different types of machine learning problems.

[Collocations] Have, Take, Pay

kyra — Wed, 3 Apr 2024 10:58:52 +0900

Have

1. 사고가 나다

2. 말다툼을 하다/ 논쟁이 있다

3. 잠깐 쉬는시간을 갖다

4. 이야기하다

5. 어려움이 있다

6. 꿈/악몽을 꾸다

7. 경험이 있다

8. ~한 기분이 들다

9. 좋은 시간을 보내다

10. 살펴보다, 한번 보다

11. 파티를 하다

12. 문제가 있다, 어려움이 있다

13. 시도하다

1. have an accident

2. have an argument

3. have a break

4. have a chat/conversation

5. have difficulty

6. have a dream/nightmare

7. have an experience

8. have a feeling

9. have fun/ a good time

10. have a look

11. have a party

12. have a problem

13. have a try/go

Take

1. take a holiday

2. take a trip

3. take a bus

4. take a liking to

5. take a risk

6. take a photo

7. take a chance

8. taken a dislike

9. take advantage of

10. take action

11. take an interest in

1. 휴가를 보내다

2. 여행 가다

3. 버스를 타다

4. 좋아하게 되다, 마음에 들다

5. 위험을 감수하다

6. 사진 찍다

7. 기회를 잡다/ 이용하다

8. 싫어지다

9. 이용하다/ 장점을 활용하다

10. 행동하다

11. ~에 관심을 갖기 시작하다

[Collocations] Make, Do

kyra — Fri, 22 Mar 2024 14:06:59 +0900

Make

* connected with saying things

1. 준비를 하다

2. 바꾸다, 변화를 주다

3. 결정하다, 선택하다

4. 말하다, 의견을 제시하다

5. 기여하다

6. 결정하다

7. 노력하다

8. 변명하다

9. 친구를 사귀다

10. 개선하다

11. 실수를하다

12. 전화하다

13. 진전이 있다

1. make arrangements for

2. make a change

3. make a choice

4. make a comment

5. make a contribution to

6. make a decision

7. make an effort

8. make an excuse

9. make friends

10. make an improvement

11. make a mistake

12. make a phone call

13. make progress

Do

1. 최선을 다 하다

2. damage를 입히다.

3. 실험하다

4. 운동하다

5. 친절을 베풀다, 도와주다

6. 피해를 입히다

7. 머리하다

8. 숙제를 하다

1. do your best

2. do (some) damage

3. do an experiment

4. do exercises

5. do someone a favour

6. do harm

7. do your hair

8. do your homework

9. do the ironing/ shopping/ washing

10. do some work

[Udemy] 15 Days of SQL - Day 6 ① : JOINS(Inner join, Outer join, Left outer join, Right outer join

kyra — Tue, 12 Mar 2024 15:59:55 +0900

JOIN : combine information from multiple tables in one query

JOIN 5줄 요약

조인은 두 개의 테이블을 서로 묶어서 하나의 결과를 만들어 내는 것을 말한다.
INNER JOIN(내부 조인)은 두 테이블을 조인할 때, 두 테이블에 모두 지정한 열의 데이터가 있어야 한다.
OUTER JOIN(외부 조인)은 두 테이블을 조인할 때, 1개의 테이블에만 데이터가 있어도 결과가 나온다.
CROSS JOIN(상호 조인)은 한쪽 테이블의 모든 행과 다른 쪽 테이블의 모든 행을 조인하는 기능이다.
SELF JOIN(자체 조인)은 자신이 자신과 조인한다는 의미로, 1개의 테이블을 사용한다.

1. Inner join

Combin the two tables in one query
Always need a common column/reference - join column
Only rows where reference column value is in both tables
Order of tables (A and B/ B and A) does not matter
Repeated values in either table will also be repeated
Syntax

1)
SELECT * FROM TableA
INNER JOIN TableB
ON TableA.employee(column name) = TableB.employee(column name)

2)
SELECT * FROM TableA A
INNER JOIN TableB B
ON A.employee = B.employee

3)
SELECT A.employee FROM TableA A
INNER JOIN TableB B
ON A.employee = B.employee

-- Aliases help with writing & reading the code more easily

SELECT payment.*, first_name, last_name -- every column in payment table & 'first_name', 'last_name' column in customer table
FROM payment
INNER JOIN customer
ON payment.customer_id = customer.customer_id

- if the column is in both tables, specify where it belongs to

SELECT payment_id,
pa.customer_id,
amount,
first_name,
last_name
FROM payment pa
INNER JOIN customer cu
ON pa.customer_id = cu.customer_id

* Challenge

Write a query to know which category the airline company sells the most tikets. (category : Business, Economy, Comfort)

SELECT 
fare_conditions,
COUNT(*)
FROM boarding_passes b
INNER JOIN seats s
ON s.seat_no = b.seat_no
GROUP BY fare_conditions

Difficult to solve, need to check it over again

https://hongong.hanbit.co.kr/sql-%EA%B8%B0%EB%B3%B8-%EB%AC%B8%EB%B2%95-joininner-outer-cross-self-join/

SQL 기본 문법: JOIN(INNER, OUTER, CROSS, SELF JOIN)

조인은 두 개의 테이블을 서로 묶어서 하나의 결과를 만들어 내는 것을 말한다. INNER JOIN(내부 조인)은 두 테이블을 조인할 때, 두 테이블에 모두 지정한 열의 데이터가 있어야 한다.OUTER JOIN(외부

hongong.hanbit.co.kr

2. Full Outer Join

SELECT * FROM TableA
FULL OUTER JOIN TableB
ON TableA.employee(column name) = TableB.employee(column name)

SELECT * FROM boarding_passes b
FULL OUTER JOIN tickets t
ON b.ticket_no = t.ticket_no
WHERE b.tickets_no IS null

-- find data where there isn't ticket_no on both boarding_passes and tickets tables

SELECT COUNT(*) FROM boarding_passes b
FULL OUTER JOIN tickets t
ON b.ticket_no = t.ticket_no
WHERE b.tickets_no IS null

-- count how many tickets that don't have ticket numbers

3. LEFT OUTER JOIN

Find how many flights these aircrafts have been used

Left table : all aircrafts - outputput(default)

Right table : flights

SELECT * FROM aircrafts_data a
LEFT JOIN flights f
ON a.aircraft_code = f.aircraft_code

Find all aircrafts that have not been used in any flight.

SELECT * FROM aircrafts_data a
LEFT JOIN flights f
ON a.aircraft_code = f.aircraft_code
WHERE f.flight_id IS null

* Challenge

1. Write a query to find out which seats are the most popular - try to find out which seat has been chosen most frequently.

Make sure all seats are included even if they have never been booked.

SELECT
s.seat_no,
COUNT(*)
FROM seats s
LEFT JOIN boarding_passes b 
ON s.seat_no = b.seat_no
GROUP BY s.seat_no -- doesn't matter 
ORDER BY COUNT(*) DESC
--WHERE b.seat_no is null

2. Are there seats that have never been booked?

3. Try to find which line (A, B, .., H) has been chosen most frequently.

SELECT
RIGHT(s.seat_no, 1) AS line,
COUNT(*)
FROM seats s
LEFT JOIN boarding_passes b
ON b.seat_no = s.seat_no
GROUP BY RIGHT(s.seat_no, 1)
ORDER BY COUNT(*) DESC

4. RIGHT OUTER JOIN

Rows that are only in the first table will result in the same.
Left outer join and Right outer join is interchangeable, so left outer join is more commonly used.
Syntax

1)
SELECT * FROM TableA
RIGHT OUTER JOIN TableB
ON TableA.employee(column name) = TableB.employee(column name)

2)
SELECT * FROM TableB
LEFT OUTER JOIN TableA
ON TableA.employee(column name) = TableB.employee(column name)

-- 1) and 2) will result the same

* Challenge

1. The company wants to run a phone call campaining on all customers in Texas(=district).

What are the customers (first_name, last_name, phone number and their district) from Texas?

SELECT
first_name,
last_name,
phone,
district
FROM customer c
LEFT JOIN address a
ON c.address_id = a.address_id
WHERE district = 'Texas'

2. Are there any (old) addresses that are not related to any customer?

SELECT * FROM address a
LEFT JOIN customer c
ON c.address_id = a.address_id
WHERE c.customer_id is null

-- this data is only seen in address table

[Udemy] 15 Days of SQL - Day 5 : CASE WHEN, CAST, COALESCE, REPLACE

kyra — Mon, 26 Feb 2024 21:50:16 +0900

1. Mathematical Functions and Operators

Syntax

SELECT
film_id,
rental_rate as old_rental_rate,
CEILING(rental_rate * 1.1) as new_rental_rate
FROM film

https://www.postgresql.org/docs/16/functions-math.html

9.3. Mathematical Functions and Operators

9.3. Mathematical Functions and Operators # Mathematical operators are provided for many PostgreSQL types. For types without standard mathematical conventions (e.g., …

www.postgresql.org

* Challenge

Write a query to create a list of the films including the relation of rental rate/ replacement cost where the rental rate is less than 4% of the replacement cost.

Create a list of that film_ids together with the percentage rounded to 2 decimal places.

SELECT
film_id,
ROUND(rental_rate / replacement_cost * 100, 2) AS percentage
FROM film
WHERE ROUND(rental_rate / replacement_cost * 100, 2) < 4
ORDER BY 2 ASC

2. CASE WHEN

Like If/Then statement, goes through a set of conditions and returns a value if a condition is met
Start with 'CASE' and end with 'END'
Syntax

CASE
WHEN condition1 THEN result1
WHEN condition2 THEN result2
WHEN condition3 THEN result3
ELSE result
END

SELECT
actual_departure-scheduled_departure,
CASE
WHEN actual_departure is null THEN 'no departure time',
WHEN actual_departure-scheduled_departure < '00:05' THEN 'On time'
WHEN actual_departure-scheduled_departure < '01:00' THEN 'A little late'
ELSE 'Late'
END
FROM flights

* Challenge

1) Write a query to find out how many tickets you have sold in the following categories:

- Low price ticket : total_amount < 20,000

- Mid price ticket : total_amount between 20,000 and 150,000

- High price ticket: total_amount >= 150,000

How many high price tickets has the company sold?

1) 

SELECT
COUNT(*),
CASE
WHEN total_amount < 20000 THEN 'low price ticket'
WHEN total_amount < 150000 THEN 'mid price ticket'
WHEN total_amount > 150000 THEN 'high price ticket'
END as ticket_price
FROM bookings
GROUP BY ticket_price



2) model answer

SELECT ticket_price, count(1)
FROM(SELECT
book_ref,
CASE
WHEN total_amount < 20000 THEN 'low price ticket'
WHEN total_amount < 150000 THEN 'mid price ticket'
ELSE 'high price ticket'
END as ticket_price
FROM bookings
) a
GROUP BY ticket_price;

* Tips

In terms of readability and simplicity, the second approach (directly categorizing data within the main query using CASE statements) is often preferred for straightforward categorizations. It is easier to understand and maintain, especially for simpler scenarios.
On the other hand, using subqueries (like in the first approach) can be advantageous when the categorization logic is complex or when the categorization needs to be reused in multiple queries.

2) Find out how many flights have departed in the following seasons:

- Winter : December, January, February

- Spring : March, April, May

- Summer : June, July, August

- Fall : September, October, November

SELECT
COUNT(scheduled_departure),
-- COUNT(*) is also available as we are using EXTRACT and specify 'scheduled_departure' in further lines.
CASE
WHEN EXTRACT(month from scheduled_departure) IN (12, 01, 02) THEN 'Winter'
WHEN EXTRACT(month from scheduled_departure) IN (03, 04, 05) THEN 'Spring'
WHEN EXTRACT(month from scheduled_departure) IN (06, 07, 08) THEN 'Summer'
WHEN EXTRACT(month from scheduled_departure) IN (09, 10, 11) THEN 'Fall'
END as season
FROM flights
GROUP BY season

3) Create a tier list in the following way:

1. Rating is 'PG' or 'PG-13' or length is more than 210 min: 'Great rating or long(tier 1)'

2. Description contains 'Drama' and length is more than 90 min: 'Long drama(tier 2)'

3. Description contains 'Drama' and length is not more than 9 min: 'Short drama(tier 3)'

4. Rental_rate less than $1: 'Very cheap(tier 4)'

If one movie can be in multiple categories it gets the higher tier assigned.

How can you filter to only those movies that appear in one of these 4 tiers?

SELECT
title,
CASE
WHEN rating IN ('PG','PG-13') OR length > 210 THEN 'Great rating or long (tier 1)'
WHEN description LIKE '%Drama%' AND length>90 THEN 'Long drama (tier 2)'
WHEN description LIKE '%Drama%' THEN 'Short drama (tier 3)'
WHEN rental_rate<1 THEN 'Very cheap (tier 4)'
END as tier_list
FROM film
WHERE 
CASE
WHEN rating IN ('PG','PG-13') OR length > 210 THEN 'Great rating or long (tier 1)'
WHEN description LIKE '%Drama%' AND length>90 THEN 'Long drama (tier 2)'
WHEN description LIKE '%Drama%' THEN 'Short drama (tier 3)'
WHEN rental_rate<1 THEN 'Very cheap (tier 4)'
END is not null

--WHERE case is not null cannot be used here as AS doesn't affect WHERE clause
-- try using NULLIF / NVL / COALESCE functions

3. CASE WHEN & SUM

Like If/Then statement, goes through a set of conditions and returns a value if a condition is met
Start with 'CASE' and end with 'END'

SELECT
SUM(CASE
WHEN rating in ('PG', 'G') THEN 1
ELSE 0
END) AS no_of_ratings_with_g_or_pg
FROM film

SELECT
SUM(CASE WHEN rating in ('G') THEN 1 ELSE 0 END) AS "G",
SUM(CASE WHEN rating in ('R') THEN 1 ELSE 0 END) AS "R",
SUM(CASE WHEN rating in ('NC-17') THEN 1 ELSE 0 END) AS "NC-17",
SUM(CASE WHEN rating in ('PG-13') THEN 1 ELSE 0 END) AS "PG-13",
SUM(CASE WHEN rating in ('PG') THEN 1 ELSE 0 END) AS "PG"
FROM film

4. COALESCE

Returns first value of list of values which is not null
Syntax

COALESCE(actual_arrival, scheduled_arrival)

COALESCE(actual_arrival, '1970-01-01 0:00') -- particular values can also be used, if both data types are the same.

5. CAST

Changes the data type of a value
Syntax

CAST(scheduled_arrival AS data type)

CAST(scheduled_arrival AS VARCHAR) -- change the data type to text

CAST(schedued_arrival AS DATE) -- timestamp to date

SELECT
COALESCE(CAST(actual_arrival-scheduled_arrival AS VARCHAR), 'not arrived')
FROM flights

null values have changed to 'not arrived'

6. REPLACE

Replaces text from a string in a column with another text
can be used to remove unnecessary blanks or letters
Syntax

REPLACE(flight_no, 'PG', '')

* delete blank space

SELECT
passenger_id,
REPLACE(passenger_id, ' ', ''), -- delete blanks
CAST(REPLACE(passenger_id, ' ', '') AS BIGINT)
FROM tickets

* remove letters

SELECT
flight_no,
CAST(REPLACE(flight_no, 'PG', '') AS INT)
FROM flights

[Udemy] 15 Days of SQL - Day 4 ② : EXTRACT, TO_CHAR, Timestamp

kyra — Tue, 20 Feb 2024 14:13:57 +0900

6. EXTRACT

Used to extract parts of timestamp/date
Syntax

EXTRACT (field from date/time/interval)

SELECT
EXTRACT(day from rental_date),
COUNT(*)
FROM rental
GROUP BY EXTRACT(day from rental_date)
ORDER BY COUNT(*) DESC

* Challenge

Write a query to find out the following:

1) What's the month with the highest total payment amount?

2) What's the day of week with the highest total payment amount?(0 is Sunday)

3) What's the highest amount one customer has spent in a week?

--#1
SELECT
EXTRACT(month from payment_date) AS month,
SUM(amount) AS total_payment_amount
FROM payment
GROUP BY EXTRACT(month from payment_date)
-- GROUP BY month can also be fine as we named it as 'month' in line 3
ORDER BY SUM(amount) DESC

--#2
SELECT
EXTRACT(week from payment_date) as day_of_week,
SUM(amount) as total_payment_amount, 
customer_id
FROM payment
GROUP BY EXTRACT(week from payment_date), customer_id
-- GROUPY BY day_of_week, customer_id
ORDER BY SUM(amount) DESC

-- #3
SELECT
EXTRACT(dow from payment_date),
SUM(amount)
FROM payment
GROUP BY EXTRACT(dow from payment_date)
ORDER BY SUM(amount) DESC

7. TO_CHAR

Used to get custome formats timestamp/date/numbers
Syntax

TO_CHAR(date/time/interval, format)
TO_CHAR(rental_date, 'MM-YYYY')

SELECT
SUM(amount),
TO_CHAR(payment_date, 'Dy, Month YYYY')
FROM payment
GROUP BY TO_CHAR(payment_date, 'Dy, Month YYYY')

reference document : https://www.postgresql.org/docs/current/functions-formatting.html

* Challenge

Write a query to sum payments and group in the following formats

SELECT
SUM(amount) AS total_amount,
TO_CHAR(payment_date, 'Dy, DD/MM/YYYY') AS day
FROM payment
GROUP BY TO_CHAR(payment_date, 'Dy, DD/MM/YYYY')
ORDER BY SUM(amount) ASC

SELECT
SUM(amount) AS total_amount,
TO_CHAR(payment_date, 'Mon,YYYY') AS day
FROM payment
GROUP BY TO_CHAR(payment_date, 'Mon,YYYY')
ORDER BY SUM(amount) ASC

SELECT
SUM(amount) AS total_amount,
TO_CHAR(payment_date, 'Dy,HH:MI') AS day
FROM payment
GROUP BY TO_CHAR(payment_date, 'Dy,HH:MI')
ORDER BY SUM(amount) DESC

8. Timestamp and Interval

Used to get a current timestamp

SELECT CURRENT_TIMESTAMP

Used to get an intervals between dates

SELECT 
CURRENT_TIMESTAMP,
rental_date,
CURRENT_TIMESTAMP - rental_date
FROM rental

Used with EXTRACT to extract particular numbers

SELECT 
CURRENT_TIMESTAMP,
rental_date,
EXTRACT(day from CURRENT_TIMESTAMP - rental_date)
FROM rental

* Challenge

Write a query to create a list for the support team of all rental durations of customer with customer_id 35.

Find out which customer has the longest average rental duration.

SELECT
customer_id,
return_date - rental_date AS rental_durations
FROM rental
WHERE customer_id = 35

SELECT
customer_id,
AVG(return_date - rental_date) AS avg
FROM rental
GROUP BY customer_id
ORDER BY avg DESC