플로고리듬(Flowgorithm)은 프로그램의 구조를 시각적으로 나타내는 흐름도(flowchart)를 작성하는 도구이자, 그 자체로 프로그래밍 언어입니다. 플로고리듬으로 흐름도만 작성하면 다른 프로그래밍 언어로 코딩하는 과정 없이 곧바로 실행 결과를 볼 수 있으며, 의사코드(pseudocode) 및 각종 프로그래밍 언어 코드를 자동으로 생성할 수도 있습니다.
플로고리듬으로 아주 간단한 hello world 예제부터 시작해 점차 복잡한 알고리듬을 구현(!)해봄으로써, 순수한 재미도 느껴보고 교육이나 실무에 쓸모가 있을지도 알아보려고 합니다.
이 책을 따라 하다보면 다음을 자연스레 익힐 수 있을 것입니다.
흐름도 작성법
플로고리듬 사용법
프로그래밍의 기본 원리
의사코드
기초 알고리듬
흐름도를 직접 작성하지 않더라도, 여기에 실린 흐름도와 실행 영상을 참고해서 각자 사용하는 언어로 구현하는 것도 좋습니다.
hetechnical interviewis hard to master and can be a nerve-racking experience. Not only do you need to know what you are talking about, but you also have to prove it to the personinterviewingyou. Fortunately, most fears of failure in this regard are exaggerated, and often, the interview will boil down to only a few potentially difficult questions.
4 METHODS FOR SOLVING FIZZBUZZ IN PYTHON
Conditional statements.
String concatenation.
Itertools.
Lambda.
One very common problem that programmers are asked to solve in technical interviews and take-home assignments is theFizzBuzz problem. FizzBuzz is a word game designed for children to teach them about division. In the game, each number divisible by three will be returned with aFizzand any number divisible by four will return aBuzz. I was never a big fan of the test, but it can help weed out weaker applicants.
While the test is pretty easy to pass so long as you know the right operators, there are a variety of different ways to solve it. However, some solutions might prove to be more impressive than others, and I think this is something to keep in mind when working on this problem for a real interview. In addition todemonstratingthese alternative methods of solving FizzBuzz, we are going to time each solution and compare the respective results.
How to Solve FizzBuzz in Python
1. CONDITIONAL STATEMENTS
The most popular and well-known solution to this problem involves usingconditional statements. For every number in n, we are going to need to check if that number is divisible by four or three. If the number is divisible by three, it will printFizz; if the number is divisible by four, it will printBuzz. The key here is simply knowing what operators to use to check for divisibility. InPython, we can use the modulus operator,%.
In computing, the modulo operation is meant to return the signed remainder of division. If a number is divisible by another, the remainder will always be zero, so we can use that to our advantage whenever we make our FizzBuzz function. We will structure condition blocks like this, wherenumis the iteration in a list of numbers.
if num % 3 == 0:
print('Fizz')
We can now build an iterativeloopfollowing the same principle, except we’ll be addingFizzandBuzz:
for num in range(1,101):
string = ""
if num % 3 == 0:
string = string + "Fizz"
if num % 4 == 0:
string = string + "Buzz"
if num % 4 != 0 and num % 3 != 0:
string = string + str(num)
print(string)
Though incredibly similar to its regular conditional loop counterpart, the string concatenation method is another really great way to solve this problem. Of course, this method is also all but too similar to the conditional method. The significant difference here is that the conditionals are simply going to be affecting a small sequence of characters put into the string data-type.
for num in range(1,21):
string = “”
if num % 3 == 0:
string = string + “Fizz”
if num % 5 == 0:
string = string + “Buzz”
if num % 5 != 0 and num % 3 != 0:
string = string + str(num)
print(string)
3.ITERTOOLS
Another way we could approach this problem — as well as other iteration problems — is to use the standard library tool,itertools. This will create a loop with better performance than most other iteration methods. Itertools can be thought of as an iteration library that is built to mirror several other extremely performant libraries from other languages, except using pythonic methods for solving problems.
Itertools will need to be imported, however, it is in the standard library. This meanspipwon’t be necessary, but itertools is still considered a project dependency. We are going to utilize three different methods from this module:
cycle():Cycle is a function takes a basic data-type and creates an iterator out of it. This function is useful and makes building custom iterators incredibly easy in Python.
count():Count is another generator that iterates a range. This iterator is often called an “infinite iterator,” which basically means that the count() function could essentially loop on and on forever.
islice(): The islice function is short for “iteration slice.” We can use this iterator to cut out particular elements in a data structure and iterate them.
Combining these methods will allow us to create a new function where we can solve the FizzBuzz problem without using the typical iteration methods in Python that we might be used to.
import itertools as its
def fizz_buzz(n):
fizzes = its.cycle([""] * 2 + ["Fizz"])
buzzes = its.cycle([""] * 4 + ["Buzz"])
fizzes_buzzes = (fizz + buzz for fizz, buzz in zip(fizzes, buzzes))
result = (word or n for word, n in zip(fizzes_buzzes, its.count(1)))
for i in its.islice(result, 100):
print(i)
The benefits of using this methodology is that the itertools library’s methods of iteration are typically going to be a lot faster than the pythonic methods of iteration. While itertools is still pythonic, it is likely that the speed of iterative looping is going to improve when using this library over the typical for loop in Python. Needless to say, creating a faster algorithm than any other applicant could certainly put you on the map for getting the job. This is a valuable module and application of the module for programmers who are still searching for employment.
A tutorial on how to solve FizzBuzz in Python. | Video: Programming with Mosh
4.LAMBDA
Another method we could use to solve this problem is even more Pythonic of a solution, and it makes use of Python’s bridge to scientific computing,lambda. There are a lot of standard functions that can be used with these lambda expressions, and they certainly come in handy. One of the most frequently used methods in this regard is themap()method. This method is going to take an expression that we can create using lambda as well as an iterative data structure.
print(*map(lambda i: 'Fizz'*(not i%3)+'Buzz'*(not i%5) or i, range(1,101)),sep='\n')
For this example, I used the range generator, and the “not” keywords in order to reverse the polarity of the modulus operators usage.
With all of these new ways to solve the problem, you might be wondering which one you should use. Of course, there are going to be trade-offs between the solutions, but in order to really make a great impression, we could narrow our decision down to using either the lambda method or the itertools method.
The lambda method has the advantage of being incredibly concise. However, depending on what code the map() method uses for iteration, it might trail behind the itertools method in terms of speed due to its less efficient iteration. The only way to figure out whether or not this is the case is to run some tests and compare our interpreter return times. So, that is going to be the mission between comparing these two heaps of code. In order to facilitate this comparison, I am going to be using theIPython magicin-line command,%timeit. Let’s start by trying it out on the itertools method. Since I wrote this as a function earlier, I can simply time the function call:
%timeit fizz_buzz(101)
Timed results for FizzBuzz using itertools. | Image: Emmett Boudreau
We will do the same with the lambda method:
%timeit print(*map(lambda i: 'Fizz'*(not i%3)+'Buzz'*(not i%5) or i, range(1,101)),sep='\n')
Timed results for FizzBuzz using the lambda method. | Image: Emmett Boudreau
Just as I predicted, the itertools method came in just a little faster, while the lambda method lagged slightly behind losing less than a millisecond off of the overall interpretation time. The answer here is somewhat of a mixed bag because the concise nature of the lambda expression andmap()function in tandem make the lambda method appear to be a lot more impressive. But the compile time of the itertools method is most certainly impressive because of its speed.
As is often the case in programming, there are multiple ways to do one thing, and as is also often the case, some ways are significantly better than others. There are certainly some trade-offs depending on what methodology you select, but this is what defines your own style as a programmer. I believe regardless of the decision that is made, using these faster methods will almost certainly make any aspiring programmer look a lot more proficient in their take home assignment. Furthermore, any aspiring programmer could certainly learn a lot more about programming and the language they are programming in by trying out different methods of doing the same thing.
from calendar import*
year = int(input('EnterYear: '))
print(calendar(year, 2, 1, 8, 3))
#2 = 2 characters for days (Mo, Tu, etc)
#1 = 1 line (row) for each week
#8 = 8 rows for each month
#3 = 3 columns for all months of the year.
PyWhatKit은다양한 유용한 기능을 갖춘 Python 라이브러리입니다.사용하기 쉽고 추가 설정이 필요하지 않습니다.현재 WhatsApp 및 YouTube 자동화를 위한 가장 인기 있는 라이브러리 중 하나입니다.새로운 기능과 버그 수정이 포함된 새로운 업데이트가 자주 출시됩니다.
PyWhatKitis a Python library with various helpful features. It's easy-to-use and does not require you to do any additional setup. Currently, it is one of the most popular library for WhatsApp and YouTube automation. New updates are released frequently with new features and bug fixes.
import pywhatkit as kit
import datetime
import time
# Replace these values with your own
phone_number = "+1234567890" # Include the country code without '+' or '0'
message = "Hello, this is a test message!"
# Set the time to send the message (24-hour format)
hour = 12
minute = 0
# Get the current time
now = datetime.datetime.now()
current_hour = now.hour
current_minute = now.minute
# Calculate the delay in seconds until the specified time
delay_seconds = ((hour - current_hour) * 60 + (minute - current_minute)) * 60
# Wait until the specified time
if delay_seconds > 0:
print(f"Waiting for {delay_seconds} seconds until {hour}:{minute}")
time.sleep(delay_seconds)
# Send the WhatsApp message
kit.sendwhatmsg(phone_number, message, now.hour, now.minute + 1) # Adding 1 minute to the current time
print("Message sent successfully!")
