Google Python Tutorial - Basic Python Exercises #2 List

Google Python Tutorial -  Basic Python Exercises #2

Basic Python Exercises

There are 3 exercises that go with the first sections of Google's Python class. They are located in the "basic" directory within the google-python-exercises directory. Download the google-python-exercises.zip if you have not already (see the Set-Up page for details).

• string1.py -- complete the string functions in string1.py, based on the material in the Python Strings section (additional exercises available in string2.py)
• list1.py -- complete the list functions in list1.py, based on the material in the Python Lists and Python Sorting sections (additional exercises available in list2.py)
• wordcount.py -- this larger, summary exercise in wordcount.py combines all the basic Python material in the above sections plus Python Dicts and Files (a second exercise is available in mimic.py)

  With all the exercises, you can take a look at our solution code inside the solution subdirectory.

*참고:  google-python-exercises.zip 의 경우, python 2.x로 작성되어 있다.

나의 경우 python 3.x로 작업하고 있기때문에 2to3를 이용해 우선 3.x로 변경하는 사전 작업후 코드를 작성하였다.

Google Python Tutirial의 Basic Python Exercises에는 총 3개의 예제가 존재가 있다.

그 중 두번째 예제는 List 섹션의 내용을 학습하고 list1.py의 function들을 완성하는 예제이다.

예제는 우선 직접 코드를 작성하였고,  solution폴더에 있는 답과 비교해보는 방식을 취하였다.

직접 작성하여 답과는 다를 수 있다.

#1 list1.py

# Basic list exercises

# Fill in the code for the functions below. main() is already set up

# to call the functions with a few different inputs,

# printing 'OK' when each function is correct.

# The starter code for each function includes a 'return'

# which is just a placeholder for your code.

# It's ok if you do not complete all the functions, and there

# are some additional functions to try in list2.py.

list1.py에는 총 3개의 문제가 있다.

# 1-1 인덱스, len()

# A. match_ends

# Given a list of strings, return the count of the number of

# strings where the string length is 2 or more and the first

# and last chars of the string are the same.

# Note: python does not have a ++ operator, but += works.

def match_ends(words):

    # +++your code here+++

    cnt = 0

    for word in words:

        if len(word) >= 2:

            if word[0] == word[-1]:

                cnt += 1

    return cnt

# 1-2 sorted(), 인덱스, append(), startswith()

# B. front_x

# Given a list of strings, return a list with the strings

# in sorted order, except group all the strings that begin with 'x' first.

# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields

# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']

# Hint: this can be done by making 2 lists and sorting each of them

# before combining them.

def front_x(words):

    # +++your code here+++

    xlist = []

    otherlist = []

    words = sorted(words)

    

    for word in words:

        ''' 시작 문자 x를 찾는 다른방법 

        if word.startswith('x'):

            xlist.append(word)

        '''

        if word[0] == 'x': 

            xlist.append(word)

        else:

            otherlist.append(word)

    

    xlist = sorted(xlist);

    return xlist + otherlist

# 1-3 sorted(), custom key sort, 인덱스

# C. sort_last

# Given a list of non-empty tuples, return a list sorted in increasing

# order by the last element in each tuple.

# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields

# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]

# Hint: use a custom key= function to extract the last element form each tuple.

def sort_last(tuples):

    # +++your code here+++

    tuples = sorted(tuples, key=myfnc)

    return tuples

def myfnc(s):

    return s[-1]

---

#2 list2.py

# Additional basic list exercises

list2.py에는 총 2개의 문제가 있다.

# 2-1 인덱스, len(), append()

# D. Given a list of numbers, return a list where

# all adjacent == elements have been reduced to a single element,

# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or

# modify the passed in list.

def remove_adjacent(nums):

    # +++your code here+++

    outputlist = []

    for num in nums:

        cnt = len(outputlist)

        if cnt == 0 or outputlist[-1] != num:

            outputlist.append(num)

    return outputlist

# 2-2 len(), pop(0), pop(-1), append(), extend()

# E. Given two lists sorted in increasing order, create and return a merged

# list of all the elements in sorted order. You may modify the passed in lists.

# Ideally, the solution should work in "linear" time, making a single

# pass of both lists.

def linear_merge(list1, list2):

    # +++your code here+++

    outputlist = []

    while len(list1) != 0 and len(list2) != 0:

        if list1[0] < list2[0]:

            outputlist.append(list1.pop(0))

        else:

            outputlist.append(list2.pop(0))

    

    outputlist.extend(list1)

    outputlist.extend(list2)

    

    ''' pop(-1), reverse()를 이용한 방법 

    while len(list1) != 0 and len(list2) != 0:

        if list1[-1] < list2[-1]:

            outputlist.append(list2.pop(-1))

        else:

            outputlist.append(list1.pop(-1))

    outputlist.extend(list1)

    outputlist.extend(list2)

    outputlist.reverse()

    '''

    return outputlist

        

# Note: the solution above is kind of cute, but unforunately list.pop(0)

# is not constant time with the standard python list implementation, so

# the above is not strictly linear time.

# An alternate approach uses pop(-1) to remove the endmost elements

# from each list, building a solution list which is backwards.

# Then use reversed() to put the result back in the correct order. That

# solution works in linear time, but is more ugly.

+ 오름차순 정렬된 리스트 두개를 인자로 받아서 머지한 정렬된 리스트를 반환하는 함수

선형시간안에 머지된 리스트를 만들도록 구현..

나온 방법대로 했지만..note내용을 정확히 이해는 못했음....ㅠㅠ