在Python 3.5中基于if / else条件对列表进行排序

时间:2016-10-18 18:38:23

标签: python list sorting python-3.5

我需要根据条件对包含嵌套列表的列表进行排序。条件是:如果第一个索引是0,则在索引nr 7上排序。然后如果第二个索引是0,则在索引nr 8上排序。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]]

我尝试过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,但是在列表中有6000个元素已经非常慢了。我将下面的bubblesort粘贴以供参考。列表零值最初是None,但是对于Python 3中的排序,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)

非常感谢任何解决此问题的建议!

1 个答案:

答案 0 :(得分:2)

编写一个函数来包含用于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]

然后将此函数用作key参数:

mylist.sort(key=sort_col_value)