파이썬(python) 스터디 – String

Programming Languages 에서 Google’s Python Class 의 String 을 스터디..

http://code.google.com/intl/ko-KR/edu/languages/google-python-class/strings.html

대충 읽고 나니 연습문제가!

지난번에 받은 google-python-exercises 에서 basic/string1.py 을 풀어보았다.

# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
    result = 'Number of donuts: ';
    if count >= 10: result = result + 'many'
    else: result = result + str(count)
    return result

# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
    if len(s) > 2: result = s[0:2] + s[-2:]
    else: result = ''
    return result

# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
    result = s.replace(s[0], '*')
    return s[0] + result[1:]

# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
#   'mix', pod' -> 'pox mid'
#   'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
    left = b[0:2] + a[2:]
    right = a[0:2] + b[2:]
    return left + " " + right

# A. donuts

# Given an int count of a number of donuts, return a string

# of the form 'Number of donuts: <count>', where <count> is the number

# passed in. However, if the count is 10 or more, then use the word 'many'

# instead of the actual count.

# So donuts(5) returns 'Number of donuts: 5'

# and donuts(23) returns 'Number of donuts: many'

def donuts(count):

result = 'Number of donuts: ';

if count >= 10: result = result + 'many'

else: result = result + str(count)

return result

# B. both_ends

# Given a string s, return a string made of the first 2

# and the last 2 chars of the original string,

# so 'spring' yields 'spng'. However, if the string length

# is less than 2, return instead the empty string.

def both_ends(s):

if len(s) > 2: result = s[0:2] + s[-2:]

else: result = ''

return result

# C. fix_start

# Given a string s, return a string

# where all occurences of its first char have

# been changed to '*', except do not change

# the first char itself.

# e.g. 'babble' yields 'ba**le'

# Assume that the string is length 1 or more.

# Hint: s.replace(stra, strb) returns a version of string s

# where all instances of stra have been replaced by strb.

def fix_start(s):

result = s.replace(s[0], '*')

return s[0] + result[1:]

# D. MixUp

# Given strings a and b, return a single string with a and b separated

# by a space '<a> <b>', except swap the first 2 chars of each string.

# e.g.

# 'mix', pod' -> 'pox mid'

# 'dog', 'dinner' -> 'dig donner'

# Assume a and b are length 2 or more.

def mix_up(a, b):

left = b[0:2] + a[2:]

right = a[0:2] + b[2:]

return left + " " + right

결과 :

basic apollo89$ python string1.py
donuts
OK  got: 'Number of donuts: 4' expected: 'Number of donuts: 4'
OK  got: 'Number of donuts: 9' expected: 'Number of donuts: 9'
OK  got: 'Number of donuts: many' expected: 'Number of donuts: many'
OK  got: 'Number of donuts: many' expected: 'Number of donuts: many'

both_ends
OK  got: 'spng' expected: 'spng'
OK  got: 'Helo' expected: 'Helo'
OK  got: '' expected: ''
OK  got: 'xyyz' expected: 'xyyz'

fix_start
OK  got: 'ba**le' expected: 'ba**le'
OK  got: 'a*rdv*rk' expected: 'a*rdv*rk'
OK  got: 'goo*le' expected: 'goo*le'
OK  got: 'donut' expected: 'donut'

mix_up
OK  got: 'pox mid' expected: 'pox mid'
OK  got: 'dig donner' expected: 'dig donner'
OK  got: 'spash gnort' expected: 'spash gnort'
OK  got: 'fizzy perm' expected: 'fizzy perm'
basic apollo89$

basic apollo89$ python string1.py

donuts

OK got: 'Number of donuts: 4' expected: 'Number of donuts: 4'

OK got: 'Number of donuts: 9' expected: 'Number of donuts: 9'

OK got: 'Number of donuts: many' expected: 'Number of donuts: many'

both_ends

OK got: 'spng' expected: 'spng'

OK got: 'Helo' expected: 'Helo'

OK got: '' expected: ''

OK got: 'xyyz' expected: 'xyyz'

fix_start

OK got: 'ba**le' expected: 'ba**le'

OK got: 'a*rdv*rk' expected: 'a*rdv*rk'

OK got: 'goo*le' expected: 'goo*le'

OK got: 'donut' expected: 'donut'

mix_up

OK got: 'pox mid' expected: 'pox mid'

OK got: 'dig donner' expected: 'dig donner'

OK got: 'spash gnort' expected: 'spash gnort'

OK got: 'fizzy perm' expected: 'fizzy perm'

basic apollo89$

내친김에 basic/string2.py 까지 고고~

# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
    if len(s) < 3: result = s
    else :
        if s[-3:] == 'ing': result = s + 'ly'
        if s[-3:] != 'ing': result = s + 'ing'
    return result

# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
    findNot = s.find('not')
    findBad = s.find('bad')
    if findBad > findNot: result = s[:findNot] + 'good' + s[findBad+3:]
    else: result = s
    return result

# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
#  a-front + b-front + a-back + b-back
def front_back(a, b):
    if len(a)%2 == 1:
        aFront = a[0:(len(a)/2)+1]
        aBack = a[(len(a)/2)+1:]
    else:
        aFront = a[0:(len(a)/2)]
        aBack = a[(len(a)/2):]

    if len(b)%2 == 1:
        bFront = b[0:(len(b)/2)+1]
        bBack = b[(len(b)/2)+1:]
    else:
        bFront = b[0:(len(b)/2)]
        bBack = b[(len(b)/2):]
    return aFront + bFront + aBack + bBack

# D. verbing

# Given a string, if its length is at least 3,

# add 'ing' to its end.

# Unless it already ends in 'ing', in which case

# add 'ly' instead.

# If the string length is less than 3, leave it unchanged.

# Return the resulting string.

def verbing(s):

if len(s) < 3: result = s

else :

if s[-3:] == 'ing': result = s + 'ly'

if s[-3:] != 'ing': result = s + 'ing'

return result

# E. not_bad

# Given a string, find the first appearance of the

# substring 'not' and 'bad'. If the 'bad' follows

# the 'not', replace the whole 'not'...'bad' substring

# with 'good'.

# Return the resulting string.

# So 'This dinner is not that bad!' yields:

# This dinner is good!

def not_bad(s):

findNot = s.find('not')

findBad = s.find('bad')

if findBad > findNot: result = s[:findNot] + 'good' + s[findBad+3:]

else: result = s

return result

# F. front_back

# Consider dividing a string into two halves.

# If the length is even, the front and back halves are the same length.

# If the length is odd, we'll say that the extra char goes in the front half.

# e.g. 'abcde', the front half is 'abc', the back half 'de'.

# Given 2 strings, a and b, return a string of the form

# a-front + b-front + a-back + b-back

def front_back(a, b):

if len(a)%2 == 1:

aFront = a[0:(len(a)/2)+1]

aBack = a[(len(a)/2)+1:]

else:

aFront = a[0:(len(a)/2)]

aBack = a[(len(a)/2):]

if len(b)%2 == 1:

bFront = b[0:(len(b)/2)+1]

bBack = b[(len(b)/2)+1:]

else:

bFront = b[0:(len(b)/2)]

bBack = b[(len(b)/2):]

return aFront + bFront + aBack + bBack

결과 :

basic apollo89$ python string2.py
verbing
OK  got: 'hailing' expected: 'hailing'
OK  got: 'swimingly' expected: 'swimingly'
OK  got: 'do' expected: 'do'

not_bad
OK  got: 'This movie is good' expected: 'This movie is good'
OK  got: 'This dinner is good!' expected: 'This dinner is good!'
OK  got: 'This tea is not hot' expected: 'This tea is not hot'
OK  got: "It's bad yet not" expected: "It's bad yet not"

front_back
OK  got: 'abxcdy' expected: 'abxcdy'
OK  got: 'abcxydez' expected: 'abcxydez'
OK  got: 'KitDontenut' expected: 'KitDontenut'
basic apollo89$

basic apollo89$ python string2.py

verbing

OK got: 'hailing' expected: 'hailing'

OK got: 'swimingly' expected: 'swimingly'

OK got: 'do' expected: 'do'

not_bad

OK got: 'This movie is good' expected: 'This movie is good'

OK got: 'This dinner is good!' expected: 'This dinner is good!'

OK got: 'This tea is not hot' expected: 'This tea is not hot'

OK got: "It's bad yet not" expected: "It's bad yet not"

front_back

OK got: 'abxcdy' expected: 'abxcdy'

OK got: 'abcxydez' expected: 'abcxydez'

OK got: 'KitDontenut' expected: 'KitDontenut'

basic apollo89$

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파이썬(python) 스터디 – String

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