Google python course basic exercise——list1

学习完string之后，现在来看看list的用法：

# -*- coding: cp936 -*-
#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.
def match_ends(words):
    count = 0
    for item in words:
        if len(item)>=2 and item[0] == item[-1]:
            count += 1
    return count

# 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']
def front_x(words):  #思路为将以'x'开头的和剩下的，分开作为两个list分别排序
    list_1 = []
    list_2 = []
    for item in words:
        if item[0] == 'x':
            list_1.append(item)
        else:
            list_2.append(item)
    list_1.sort()
    list_2.sort()
    return list_1+list_2  #此处常见错误为return list_1.extend(list_2)，返回值为None

# 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)]
def sort_last(tuples):  #采用关键字思想，先对最后一个元素排序，然后用匹配的方法把原tuples里的元素对应排序
    list_1 = []
    list_2 = []
    for tup in tuples:
       key = tup[-1]
       list_1.append(key)
    list_1.sort()
    for num in range(len(tuples)):
        for tup in tuples:
            if tup[-1] == list_1[num]:
                list_2.append(tup)
    return list_2
       
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got,expected):
    if got == expected:
        prefix = 'OK'
    else:
        prefix = 'X'
    print '%s got: %s expected: %s ' % (prefix, repr(got),repr(expected))

# Calls the above functions with interesting inputs.
def main():
    print 'match_ends'
    test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
    test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
    test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)

    print
    print 'front_x'
    test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
         ['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
    test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
         ['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
    test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
         ['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])

    print
    print 'sort_last'
    test(sort_last([(1, 3), (3, 2), (2, 1)]),
         [(2, 1), (3, 2), (1, 3)])
    test(sort_last([(2, 3), (1, 2), (3, 1)]),
         [(3, 1), (1, 2), (2, 3)])
    test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
         [(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
  main()

运行结果：

match_ends
OK got: 3 expected: 3 
OK got: 2 expected: 2 
OK got: 1 expected: 1 

front_x
OK got: ['xaa', 'xzz', 'axx', 'bbb', 'ccc'] expected: ['xaa', 'xzz', 'axx', 'bbb', 'ccc'] 
OK got: ['xaa', 'xcc', 'aaa', 'bbb', 'ccc'] expected: ['xaa', 'xcc', 'aaa', 'bbb', 'ccc'] 
OK got: ['xanadu', 'xyz', 'aardvark', 'apple', 'mix'] expected: ['xanadu', 'xyz', 'aardvark', 'apple', 'mix'] 

sort_last
OK got: [(2, 1), (3, 2), (1, 3)] expected: [(2, 1), (3, 2), (1, 3)] 
OK got: [(3, 1), (1, 2), (2, 3)] expected: [(3, 1), (1, 2), (2, 3)] 
OK got: [(2, 2), (1, 3), (3, 4, 5), (1, 7)] expected: [(2, 2), (1, 3), (3, 4, 5), (1, 7)] 

 

ps:

1. 在学完list和string后，容易发现他俩的异同之处：

　　　　list[0:2] = 'z'

　　　　string[0:2] = 'z'

　　　　前者能实现替换功能，而后者会报错。即list支持item assignment 但string不支持

　　2.对于list methods而言，调用函数不会返回新list，只会在原来的list上修改。

　　3.python中in语句，返回Ture/False，但像find之类的函数，当在搜索范围内未找到对象时，会抛出异常如ValueError。

补充：这两天无意中又翻了下sort的用法，顿时对C题有了新的认识。另一种代码：

def sort_last(tuples):  
    def get_last(tup):
        return tup[-1]
    return sorted(tuples,key=get_last)

不得不说key=功能太好用了！一般而言，sort()只针对list使用，但sorted()却没有这一限制，它适用于许多可列项形式的类型，所以推荐使用sorted()函数。