根据 if/else 条件对列表列表进行排序

Question

我需要根据条件对包含嵌套列表的列表进行排序。条件是：如果第一个索引为 0，则按索引 nr 7 排序。如果第二个索引为 0，则按索引 nr 8 排序。所以下面的列表（未排序）：

[[11, 37620, 'mat', 'xxx', 1, 28.5, 0, 11, 37620, 'Arp', 'xxx', 1, 28],
 [10, 24210, 'Skatt', 'xxx', 2, 40, 0, 0, 0, 0, 0, 0, 0],
 [10, 37010, 'test, a', 'xxx', 5, 36.75, 0, 10, 37010, '', 'xxx', 7, 50.75],
 [0, 0, 'mottagare', 'xxx', 3, 20.25, 0, 10, 21511, 0, 0, 0, 0],     
 [10, 40000, 'eld', 'xxx', 5, 30.5, 0, 10, 40000, 'Oly', 'xxx', 5, 30],
 [10, 17060, 'bok', 'xxx', 1, 28.5, 0, 0, 0, 0, 0, 0, 0]]

应该是这样的：

[[10, 17060, 'bok', 'xxx', 1, 28.5, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 'mottagare', 'xxx', 3, 20.25, 0, 10, 21511, 0, 0, 0, 0],
 [10, 24210, 'Skatt', 'xxx', 2, 40, 0, 0, 0, 0, 0, 0, 0],
 [10, 40000, 'eld', 'xxx', 5, 30.5, 0, 10, 40000, 'Oly', 'xxx', 5, 30],     
 [10, 37010, 'test, a', 'xxx', 5, 36.75, 0, 10, 37010, '', 'xxx', 7, 50.75],
 [11, 37620, 'mat', 'xxx', 1, 28.5, 0, 11, 37620, 'Arp', 'xxx', 1, 28]]

我试过 a_list.sort(key = itemgetter(0)) 和 lambda，但零总是先结束。我已经通过冒泡排序解决了这个问题，但是列表中有 6000 个元素已经非常慢了。我粘贴下面的冒泡排序以供参考。列表零值最初是 None，但为了在 Python 3 中排序，我已将它们转换为 int 0.

the_list_pos = 0
for key, value in kund_dict2.items():
    # Start sorting from the whole list, since the_list_pos is zero
    for passnum in range(len(komplett_lista)-1,the_list_pos,-1):
        # Loop the list, one by one
        for i in range(passnum):
            # If these helper variables are None at the end, nothing happens
            comp1 = None
            comp2 = None
            # Check that the variables to compare are not None, and set them to appropiate values
            if komplett_lista[i][0] is not None and komplett_lista[i][0] == key:
                comp1 = komplett_lista[i][1]
            elif komplett_lista[i][7] is not None and komplett_lista[i][7] == key:
                comp1 = komplett_lista[i][8]   
            if komplett_lista[i+1][0] is not None and komplett_lista[i+1][0] == key:
                comp2 = komplett_lista[i+1][1]
            elif komplett_lista[i+1][7] is not None and komplett_lista[i+1][7] == key:
                comp2 = komplett_lista[i+1][8]
            # Do the actual sorting
            if comp1 is not None and comp2 is not None:    
                if comp1 > comp2:
                    temp = komplett_lista[i]
                    komplett_lista[i] = komplett_lista[i+1]
                    komplett_lista[i+1] = temp
    # update the position so that we do not sort over the already sorted data
    the_list_pos += (value)

Answer 1

编写一个函数来包含查找要排序的列的逻辑：

def sort_col_value(row):
    # if first index is 0, sort on index nr 7. 
    if row[0] == 0:
        return row[7]
    # Then if second index is 0, sort on index nr 8
    if row[1] == 0:
        return row[8]
    return row[0]

然后使用这个函数作为key参数：

mylist.sort(key=sort_col_value)

根据 if/else 条件对列表列表进行排序

Sort a list of lists based on an if/else condition

python

sorting

list

python-3.5