如何用for循环计算嵌套字典列表中length总和以找出最长专辑

我需要实现一个函数，返回总时长最长的专辑名称。目前手动累加专辑歌曲列表中各字典的length值的方法可正常运行，但代码冗余，希望改用for循环优化。尝试编写的循环代码无法正确累加时长，album1和album2的总和始终为0，求正确的for循环写法。

可行但冗余的代码 #

from cisc108 import assert_equal
album1={
    'name':'Album1',
    'songs':[{'title':'rock','explicit':True,'length':2.5},
       {'title':'starlight','explicit':False,'length':5.2},
       {'title':'smile','explicit':False,'length':1.75}]
    }

album2={
    'name':'Album2',
    'songs':[{'title':'stone','explicit':False,'length':4.75},
       {'title':'moonlight','explicit':False,'length':3.0},
       {'title':'happy','explicit':True,'length':2.8}]}

def longest_album(A1,A2):
    x1 = A1['songs'][0]['length']
    x2 = A1['songs'][1]['length']
    x3 = A1['songs'][2]['length']
    album_1_len= x1 + x2 + x3
    y1 = A2['songs'][0]['length']
    y2 = A2['songs'][1]['length']
    y3 = A2['songs'][2]['length']
    album_2_len = y1 + y2 + y3
    if album_2_len > album_1_len:
        return A2['name']
    else:
        return A1['name']

assert_equal(longest_album(album1,album2), "Album2")

尝试的错误循环代码 #

def longest_album(A1,A2):
    album1 = 0
    album2 = 0
    for i in A1['songs']:
        if i == 'length':
            album1+= i
    for i in A2['songs']:
        if i == 'length':
            album2+= i
    if album2 > album1:
        return A2['name']
    else:
        return A1['name']

错误原因分析 #

遍历A1['songs']时，变量i拿到的是每首歌的完整字典对象，不是字典的键。你用i == 'length'做判断，永远不会成立，所以累加逻辑从未执行，总和始终为0。

正确的for循环实现 #

from cisc108 import assert_equal

album1={
    'name':'Album1',
    'songs':[{'title':'rock','explicit':True,'length':2.5},
       {'title':'starlight','explicit':False,'length':5.2},
       {'title':'smile','explicit':False,'length':1.75}]
    }

album2={
    'name':'Album2',
    'songs':[{'title':'stone','explicit':False,'length':4.75},
       {'title':'moonlight','explicit':False,'length':3.0},
       {'title':'happy','explicit':True,'length':2.8}]}

def longest_album(A1,A2):
    album1_total = 0
    # 遍历专辑1的每首歌字典，直接取length值累加
    for song in A1['songs']:
        album1_total += song['length']
    
    album2_total = 0
    # 遍历专辑2的每首歌字典，直接取length值累加
    for song in A2['songs']:
        album2_total += song['length']
    
    return A2['name'] if album2_total > album1_total else A1['name']

assert_equal(longest_album(album1,album2), "Album2")

进一步优化：抽离时长计算函数 #

为避免重复代码，可将单张专辑的时长计算逻辑抽成独立函数，让代码更易维护：

from cisc108 import assert_equal

album1={
    'name':'Album1',
    'songs':[{'title':'rock','explicit':True,'length':2.5},
       {'title':'starlight','explicit':False,'length':5.2},
       {'title':'smile','explicit':False,'length':1.75}]
    }

album2={
    'name':'Album2',
    'songs':[{'title':'stone','explicit':False,'length':4.75},
       {'title':'moonlight','explicit':False,'length':3.0},
       {'title':'happy','explicit':True,'length':2.8}]}

def calculate_album_length(album):
    total = 0
    for song in album['songs']:
        total += song['length']
    return total

def longest_album(A1,A2):
    len1 = calculate_album_length(A1)
    len2 = calculate_album_length(A2)
    return A2['name'] if len2 > len1 else A1['name']

assert_equal(longest_album(album1,album2), "Album2")

极简写法：用生成器表达式求和 #

Python的sum()函数配合生成器表达式，可一行完成时长计算，代码更简洁：

from cisc108 import assert_equal

album1={
    'name':'Album1',
    'songs':[{'title':'rock','explicit':True,'length':2.5},
       {'title':'starlight','explicit':False,'length':5.2},
       {'title':'smile','explicit':False,'length':1.75}]
    }

album2={
    'name':'Album2',
    'songs':[{'title':'stone','explicit':False,'length':4.75},
       {'title':'moonlight','explicit':False,'length':3.0},
       {'title':'happy','explicit':True,'length':2.8}]}

def longest_album(A1,A2):
    len1 = sum(song['length'] for song in A1['songs'])
    len2 = sum(song['length'] for song in A2['songs'])
    return A2['name'] if len2 > len1 else A1['name']

assert_equal(longest_album(album1,album2), "Album2")

内容的提问来源于stack exchange，提问作者baylee