Question

有一个元组列表。我想根据元组的第二个值进行排序并返回两个列表的元组。

代码如下：

def sort_artists(x):
    artist = []
    earnings = []
    z = (artist, earnings)
    for inner in x:
        artist.append(inner[0])
        earnings.append(inner[1])
    return z

artists = [("The Beatles", 270.8), ("Elvis Presley", 211.5), ("Michael Jackson", 183.9)]
print(sort_artists(artists))

期望的输出是

(['Elvis Presley', 'Michael Jackson', 'The Beatles'], [270.8, 211.5, 183.9])

我得到的输出是，

(['The Beatles', 'Elvis Presley', 'Michael Jackson'], [270.8, 211.5, 183.9])

Answer 1

您可以先对元组列表进行排序，然后再将它们转换为列表元组。只需将 x.sort(key = lambda x: x[1], reverse=False) 添加为函数的第一行，如下所示。

# your function
def sort_artists(x):
# sorts tuples by second value (x[1]), if reverse=True it will sort descending
    x.sort(key = lambda x: x[1], reverse=False)  
    artist = []
    earnings = []
    z = (artist, earnings)
    for inner in x:
        artist.append(inner[0])
        earnings.append(inner[1])
    return z

artists = [("The Beatles", 270.8), ("Elvis Presley", 211.5), ("Michael Jackson", 183.9)]

# call your function
list_artists = sort_artists(artists)

#print(list_artists)
'''
(['Michael Jackson', 'Elvis Presley', 'The Beatles'], [183.9, 211.5, 270.8])
'''

有关详细信息和其他选项，请查看 here。

Answer 2

试试这个，函数会完成工作。我更改了您的 var col_def = [title:"TheJobSource", field:"JobCount", formatter: "progress", formatterParams:{legend: <dataField_string>}} 列表的测试值以表明它有效。

artists

输出：

def sort_artists(x):
    artist = []
    earnings = []
    for inner in x:
        artist.append(inner[0])
        earnings.append(inner[1])
    index = sorted(range(len(earnings)), key=lambda k: earnings[k])
    artist = [artist[i] for i in index]
    earnings = [earnings[i] for i in index]
    z = (artist, earnings)
    return z

artists = [("The Beatles", 250), ("Elvis Presley", 530), ("Michael Jackson", 120)]
print(sort_artists(artists))

根据元组的第二个值对元组列表进行排序

2 个答案: