[codecademy] 3.CONDITIONALS AND CONTROL FLOW

CONDITIONALS AND CONTROL FLOW

Conditionals & Control Flow 1/15

Go With the Flow

분기에 대해서 배울 것임

샘플코드를 실행하여 분기가 되는것을 확인해보아라

문제

Check out the code in the editor. You'll see the type of program you'll be able to write once you've mastered control flow. Click Save & Submit to see what happens!

def clinic(): print "You've just entered the clinic!" print "Do you take the door on the left or the right?" answer = raw_input("Type left or right and hit 'Enter'.").lower() if answer == "left" or answer == "l": print "This is the Verbal Abuse Room, you heap of parrot droppings!" elif answer == "right" or answer == "r": print "Of course this is the Argument Room, I've told you that already!" else: print "You didn't pick left or right! Try again." clinic() clinic()

실행하여 left, l, right, r을 입력하고 엔터키를 누르면 된다, 오타를 내서 분기가 되는것을 확인할 수있다

Conditionals & Control Flow 2/15

Compare Closely!

비교연산자 6가지

같다 ==

같지않다 !=

~보다 크다 <

~와 같거나 크다<=

~보다 작다 >

~보다 작거나 같다 >=

주의. ==와 =는 다르다, =는 변수를 세팅할때 사용

문제

Set each variable to True or False depending on what you think the result will be. For example, 1 < 2 will be True, because one is less than two.

Set bool_one equal to the result of 17 < 328

Set bool_two equal to the result of 100 == (2 * 50)

Set bool_three equal to the result of 19 <= 19

Set bool_four equal to the result of -22 >= -18

Set bool_five equal to the result of 99 != (98 + 1)

# Assign True or False as appropriate on the lines below! # Set this to True if 17 < 328 or to False if it is not. bool_one = True # We did this one for you! # Set this to True if 100 == (2 * 50) or to False otherwise. bool_two = # Set this to True if 19 <= 19 or to False if it is not. bool_three = # Set this to True if -22 >= -18 or to False if it is not. bool_four = # Set this to True if 99 != (98 + 1) or to False otherwise. bool_five =

문제에 나와있는 변수들의 값을 그대로 세팅하여 주면된다, 해당 변수에 들어갈 값을 말해주자면

bool_two = True

bool_three = True

bool_four = False

bool_five = False

Conditionals & Control Flow 3/15

Compare... Closelier!

비교연산자를 좀더 가깝게

앞장에서 해봤던것을 바탕으로 응용

문제

Let's run through the comparators again with more complex expressions. Set each variable to True or False depending on what you think the result will be.

Set bool_one to the result of (20 - 10) > 15

Set bool_two to the result of (10 + 17) == 3**16

Set bool_three to the result of 1**2 <= -1

Set bool_four to the result of 40 * 4 >= -4

Set bool_five to the result of 100 != 10**2

# Assign True or False as appropriate on the lines below! # (20 - 10) > 15 bool_one = False # We did this one for you! # (10 + 17) == 3**16 # Remember that ** can be read as 'to the power of'. 3**16 is about 43 million. bool_two = # 1**2 <= -1 bool_three = # 40 * 4 >= -4 bool_four = # 100 != 10**2 bool_five =

문제에 나와있는 변수들의 값을 그대로 세팅하여 주면된다, 해당 변수에 들어갈 값을 말해주자면

bool_two = False

bool_three = False

bool_four = True

bool_five = False

Conditionals & Control Flow 4/15

How the Tables Have Turned

이예제를 통하여 True와 False의 비교결과가 세팅되는것을 배운다

# Make me true! bool_one = 3 < 5

3 < 5라는 명제가 True이기 때문에 bool_one에는 True가 세팅된다

문제

For each boolean value in the editor, write an expression that evaluates to that value.

Remember, comparators are: ==, !=, >, >=, <, and <=.

Use at least three different ones!

Don't just use True and False! That's cheating!

# Create comparative statements as appropriate on the lines below! # Make me true! bool_one = 3 < 5 # We already did this one for you! # Make me false! bool_two = # Make me true! bool_three = # Make me false! bool_four = # Make me true! bool_five =

# Make me false!

bool_two = 3 <= 2

# Make me true!

bool_three = 3 > 2

# Make me false!

bool_four = 3 > 5

# Make me true!

bool_five = 2 > 1

Conditionals & Control Flow 5/15

To Be and/or Not to Be

Boolean 연산자 and, or, not 세개의 특징

and : 각각의 상태가 True인것을 체크

or : 어떤 하나라도 True인것을 체크

not : 현재상태의 반대로 전환 True->False, False->True

문제

Look at the truth table in the editor. Don't worry if you don't completely get it yet—you will by the end of this section!

Click Save & Submit to continue.

""" Boolean Operators ------------------------ True and True is True True and False is False False and True is False False and False is False True or True is True True or False is True False or True is True False or False is False Not True is False Not False is True """

노답

Conditionals & Control Flow 6/15

and

and연산자

A and b -> a,b 모두 True여야 True를 얻을 수 있다 그렇지 않을경우 False

문제

Let's practice with and. Assign each variable to the appropriate boolean value.

Set bool_one equal to the result of False and False

Set bool_two equal to the result of -(-(-(-2))) == -2 and 4 >= 16**0.5

Set bool_three equal to the result of 19 % 4 != 300 / 10 / 10 and False

Set bool_four equal to the result of -(1**2) < 2**0 and 10 % 10 <= 20 - 10 * 2

Set bool_five equal to the result of True and True

bool_one = bool_two = bool_three = bool_four = bool_five =

bool_one = False and False

bool_two = -(-(-(-2))) == -2 and 4 >= 16**0.5

bool_three = 19 % 4 != 300 / 10 / 10 and False

bool_four = -(1**2) < 2**0 and 10 % 10 <= 20 - 10 * 2

bool_five = True and True

Conditionals & Control Flow 7/15

or

or연산자

A or b -> a,b 둘중하나가 True여야 True를 얻을 수 있다, 둘다 False일 경우 False

문제

Time to practice with or!

Set bool_one equal to the result of 2**3 == 108 % 100 or 'Cleese' == 'King Arthur'

Set bool_two equal to the result of True or False

Set bool_three equal to the result of 100**0.5 >= 50 or False

Set bool_four equal to the result of True or True

Set bool_five equal to the result of 1**100 == 100**1 or 3 * 2 * 1 != 3 + 2 + 1

bool_one = bool_two = bool_three = bool_four = bool_five =

bool_one = 2**3 == 108 % 100 or 'Cleese' == 'King Arthur'

bool_two = True or False

bool_three = 100**0.5 >= 50 or False

bool_four = True or True

bool_five = 1**100 == 100**1 or 3 * 2 * 1 != 3 + 2 + 1

Conditionals & Control Flow 8/15

not

not연산자

현재 boolean상태를 반대로 전환

문제

Let's get some practice with not.

Set bool_one equal to the result of not True

Set bool_two equal to the result of not 3**4 < 4**3

Set bool_three equal to the result of not 10 % 3 <= 10 % 2

Set bool_four equal to the result of not 3**2 + 4**2 != 5**2

Set bool_five equal to the result of not not False

bool_one = bool_two = bool_three = bool_four = bool_five =

bool_one = not True

bool_two = not 3**4 < 4**3

bool_three = not 10 % 3 <= 10 % 2

bool_four = not 3**2 + 4**2 != 5**2

bool_five = not not False

Conditionals & Control Flow 9/15

This and That (or This, But Not That!)

이거저거 저거이거 다섞었을때

not은 첫번째에 평가된다

and는 다음번에 평가된다

or은 마지막에 평가된다

문제

Assign True or False as appropriate for bool_one through bool_five.

Set bool_one equal to the result of False or not True and True

Set bool_two equal to the result of False and not True or True

Set bool_three equal to the result of True and not (False or False)

Set bool_four equal to the result of not not True or False and not True

Set bool_five equal to the result of False or not (True and True)

bool_one = bool_two = bool_three = bool_four = bool_five =

bool_one = False or not True and True

bool_two = False and not True or True

bool_three = True and not (False or False)

bool_four = not not True or False and not True

bool_five = False or not (True and True)

Conditionals & Control Flow 10/15

Mix 'n' Match

섞어 매치-> 연산자를 섞어서 연습

# Make me false bool_one = (2 <= 2) and "Alpha" == "Bravo"

one에는 True, False값이 들어간다 이유는 boolean 연산을 했기 때문에 

문제

This time we'll give the expected result, and you'll use some combination of boolean operators to achieve that result.

Remember, the boolean operators are and, or, and not. Use each one at least once!

# Use boolean expressions as appropriate on the lines below! # Make me false! bool_one = (2 <= 2) and "Alpha" == "Bravo" # We did this one for you! # Make me true! bool_two = # Make me false! bool_three = # Make me true! bool_four = # Make me true! bool_five =

bool_two = (2 <= 2) or "Alpha" == "Bravo"

bool_three = not (2 <= 2) and "Alpha" == "Bravo"

bool_four = (2 == 2) and not ("Alpha" == "Bravo")

bool_five = (2 <= 2) and "Alpha" != "Bravo"

Conditionals & Control Flow 11/15

Conditional Statement Syntax

조건문

if문으로 분기하는것을 배울 수 있다

if 8 < 9: print "Eight is less than nine!"

if문 옆에 조건이 True일경우 아래 print문이 출력된다

문제

If you think the print statement will print to the console, set response equal to 'Y'; otherwise, set response equal to 'N'.

response = answer = "Left" if answer == "Left": print "This is the Verbal Abuse Room, you heap of parrot droppings!" # Will the above print statement print to the console? # Set response to 'Y' if you think so, and 'N' if you think not.

response = 'Y'

answer = "Left"

if answer == "Left":

    print "This is the Verbal Abuse Room, you heap of parrot droppings!"

    

# Will the above print statement print to the console?

# Set response to 'Y' if you think so, and 'N' if you think not.

Conditionals & Control Flow 12/15

If You're Having...

조건문과 함수

if문은 함수의 return값에 따라도 분기된다

물론 함수안에서도 조건문으로 분기가능

문제

In the editor you'll see two functions. Don't worry about anything unfamiliar. We'll explain soon enough.

Replace the underline on line 2 with an expression that returns True.

Replace the underline on line 6 with an expression that returns True.

If you do it successfully, then both "Success #1" and "Success #2" are printed.

def using_control_once(): if _______: return "Success #1" def using_control_again(): if _______: return "Success #2" print using_control_once() print using_control_again()

def using_control_once():

    if True:

        return "Success #1"

def using_control_again():

    if True:

        return "Success #2"

print using_control_once()

print using_control_again()

Conditionals & Control Flow 13/15

Else Problems, I Feel Bad for You, Son...

else문

else문은 if문과 짝을 이루어 사용되는 문이다, 단일로 사용불가능

if 8 > 9: print "I don't printed!" else: print "I get printed!"

if문에서 분기되지 못했을때 무조건 else문이 호출되는 분기 문법

문제

Complete the else statements to the right. Note the indentation for each line!

answer = "'Tis but a scratch!" def black_knight(): if answer == "'Tis but a scratch!": return True else: return # Make sure this returns False def french_soldier(): if answer == "Go away, or I shall taunt you a second time!": return True else: return # Make sure this returns False

answer = "'Tis but a scratch!"

def black_knight():

    if answer == "'Tis but a scratch!":

        return True

    else:             

        return False       # Make sure this returns False

def french_soldier():

    if answer == "Go away, or I shall taunt you a second time!":

        return True

    else:             

        return False       # Make sure this returns False

Conditionals & Control Flow 14/15

I Got 99 Problems, But a Switch Ain't One

elif문

각각의 조건으로 여러가지 분기구문을 만들어 낼 수 있다

if 8 > 9: print "I don't get printed!" elif 8 < 9: print "I get printed!" else: print "I also don't get printed!"

문제

01. On line 2, fill in the if statement to check if answer is greater than 5.

02. On line 4, fill in the elif so that the function outputs -1 if answer is less than 5.

def greater_less_equal_5(answer): if ________: return 1 elif ________: return -1 else: return 0 print greater_less_equal_5(4) print greater_less_equal_5(5) print greater_less_equal_5(6)

def greater_less_equal_5(answer):

    if answer > 5:

        return 1

    elif answer < 5:          

        return -1

    else:

        return 0

        

print greater_less_equal_5(4)

print greater_less_equal_5(5)

print greater_less_equal_5(6)

Conditionals & Control Flow 15/15

The Big If

마무으리 , 복습

비교연산자

3 < 4 5 >= 5 10 == 10 12 != 13

boolean연산자

True or False (3 < 4) and (5 >= 5) this() and not that()

분기문법

if this_might_be_true(): print "This really is true." elif that_might_be_true(): print "That is true." else: print "None of the above."

문제

Write an if statement in the_flying_circus(). It must include:

01. if, elif, and else statements;

02. At least one of and, or, or not;

03. A comparator (==, !=, <, <=, >, or >=);

04. Finally, the_flying_circus() must return True when evaluated.

Don't forget to include a : after your if statements!

# Make sure that the_flying_circus() returns True def the_flying_circus(): if ________: # Start coding here! # Don't forget to indent # the code inside this block! elif ________: # Keep going here. # You'll want to add the else statement, too!

# Make sure that the_flying_circus() returns True

def the_flying_circus():

    if 3 < 5:    # Start coding here!

        # Don't forget to indent

        # the code inside this block!

        return True

    elif 3 > 2:

        # Keep going here.

        # You'll want to add the else statement, too!

        return True

PygLatin 1/11

Break It Down

알파벳(String) 가지고 놀꺼임 

문제

When you're ready to get coding, click Save and Submit. Since we took the time to write out the steps for our solution, you'll know what's coming next!

노답

PygLatin 2/11

Ahoy! (or Should I Say Ahoyay!)

기본적인 message 출력

문제

Please print the phrase "Pig Latin".

print "Pig Latin"

PygLatin 3/11

Input!

입력방법

name = raw_input("What's your name?") print name

실행화면

[#콘솔] What's your name? >

문제

01. On line 4, use raw_input("Enter a word:") to ask the user to enter a word. Save the results of raw_input() in a variable called original.

02. Click Save & Submit Code

03. Type a word in the console window and press Enter (or Return).

print 'Welcome to the Pig Latin Translator!' # Start coding here!

print 'Welcome to the Pig Latin Translator!'

# Start coding here!

original = raw_input("Enter a word:")

PygLatin 4/11

Check Yourself!

문자열에 대한 체크방법

empty_string = "" if len(empty_string) > 0: # Run this block. # Maybe print something? else: # That string must have been empty.

문제

Write an if statement that verifies that the string has characters.

01. Add an if statement that checks that len(original) is greater than zero. Don't forget the : at the end of the if statement!

02. If the string actually has some characters in it, print the user's word.

03. Otherwise (i.e. an else: statement), please print "empty".

You'll want to run your code multiple times, testing an empty string and a string with characters. When you're confident your code works, continue to the next exercise.

print 'Welcome to the Pig Latin Translator!' # Start coding here! original = raw_input("Enter a word:")

print 'Welcome to the Pig Latin Translator!'

# Start coding here!

original = raw_input("Enter a word:")

if len(original) > 0:

    print original

else:

    print "empty"

PygLatin 5/11

Check Yourself... Some More

문자열체크 더더더더

x = "J123" x.isalpha() # False

isalpha()는 알파벳인지 확인하는 함수 알파벳이면 상수 그렇지 않을경우 0

문제

Use and to add a second condition to your if statement. In addition to your existing check that the string contains characters, you should also use .isalpha() to make sure that it only contains letters.

Don't forget to keep the colon at the end of the if statement!

print 'Welcome to the Pig Latin Translator!' # Start coding here! original = raw_input("Enter a word:") if len(original) > 0: print original else: print "empty"

print 'Welcome to the Pig Latin Translator!'

# Start coding here!

original = raw_input("Enter a word:")

if original.isalpha()> 0:

    print original

else:

    print "empty"

PygLatin 6/11

Pop Quiz!

만든코드를 테스트해보시오

문제

Take some time to test your current code. Try some inputs that should pass and some that should fail. Enter some strings that contain non-alphabetical characters and an empty string.

When you're convinced your code is ready to go, click Save & Submit to move forward!

print 'Welcome to the Pig Latin Translator!' # Start coding here! original = raw_input("Enter a word:") if original.isalpha()> 0: print original else: print "empty"

노답

PygLatin 7/11

Ay B C

이제부터 문자열을 가지고 바꿀 장난질을 시작할꺼임

문제

Create a variable named pyg and set it equal to the suffix 'ay'.

pyg = 'ay'

PygLatin 8/11

Word Up

글자 끌어올리기

글자 소문자로 변환하는 함수

the_string = "Hello" the_string = the_string.lower()

문자열에 대한 index 위치

주의. 항상 0부터 시작한다

first_letter = the_string[0] second_letter = the_string[1] third_letter = the_string[2]

문제

Inside your if statement:

01. Create a new variable called word that holds the .lower()-case conversion of original.

02. Create a new variable called first that holds word[0], the first letter of word.

pyg = 'ay' original = raw_input('Enter a word:') if len(original) > 0 and original.isalpha(): print original else: print 'empty'

pyg = 'ay'

original = raw_input('Enter a word:')

if len(original) > 0 and original.isalpha():

    print original

    word = original.lower()

    first = word[0]

else:

    print 'empty'

PygLatin 9/11

Move it on Back

문자열 연산

greeting = "Hello " name = "D. Y." welcome = greeting + name

문자열은 위와 같이 변수 더하듯(주의. 숫자는 문자열로 형변환하여 더해야함) 더하여 출력 및 저장이 가능하다

문제

On a new line after where you created the first variable:

Create a new variable called new_word and set it equal to the concatenation of word, first, and pyg.

pyg = 'ay' original = raw_input('Enter a word:') if len(original) > 0 and original.isalpha(): print original word = original.lower() print word first = word[0] else: print 'empty'

pyg = 'ay'

original = raw_input('Enter a word:')

if len(original) > 0 and original.isalpha():

    print original

    word = original.lower()

    first = word[0]

    new_word = word+first + pyg

else:

    print 'empty'

PygLatin 10/11

Ending Up

엔딩끌어올리기

s = "Charlie" print s[0] # will print "C" print s[1:4] # will print "har"

s[0]은 문자열중 문자하나 [0]의 위치를 사용하겠다는 의미이다.

s[1:4]는 해당 문자열의 s의 index인 1~4까지 사용하겠다는 의미이다. print를 사용하였으니 출력하겠다는의미

문제

Set new_word equal to the slice from the 1st index all the way to the end of new_word. Use [1:len(new_word)] to do this.

pyg = 'ay' original = raw_input('Enter a word:') if len(original) > 0 and original.isalpha(): print original word = original.lower() first = word[0] new_word = word+first + pyg else: print 'empty'

pyg = 'ay'

original = raw_input('Enter a word:')

if len(original) > 0 and original.isalpha():

    print original

    word = original.lower()

    first = word[0]

    new_word = word+first + pyg

    new_word = new_word[1:len(new_word)]

else:

    print 'empty'

PygLatin 11/11

Testing, Testing, is This Thing On?

테스트하고 마무으리

문제

When you're sure your translator is working just the way you want it, click Save & Submit Code to finish this project.

pyg = 'ay' original = raw_input('Enter a word:') if len(original) > 0 and original.isalpha(): print original word = original.lower() first = word[0] new_word = word+first + pyg new_word = new_word[1:len(new_word)] else: print 'empty'

노답

참고 : codecademy