PYTHON 用空值、字符串和数值排序列表

Question

我想对短列表进行排序，例如：

# we can have only 3 types of value: any string numeric value like '555', 'not found' and '' (can have any variation with these options)
row = ['not found', '', '555']

至

# numeric values first, 'not found' less prioritize and '' in the end
['555', 'not found', ''] 

我尝试使用

row.sort(key=lambda x: str(x).isnumeric() and not bool(x))

但它不工作

我该如何排序？ （数值在前，“未找到”优先级较低，最后是“”）

Answer 1

def custom_sort(list):
    L1 = []
    L2 = []
    L3 = []
    for element in list:
        if element.isnumeric():
            L1.append(element)
        if element == 'Not found':
            L2.append(element)
        else : L3.append(element)
    L1.sort()
    L1.append(L2).append(L3)
    return L1

Answer 2

编辑：按要求对非数值进行排序

ar = [i for i in row if not i.isnumeric()]
ar.sort(reverse=True)
row = [i for i in row if i.isnumeric()] + ar

Answer 3

这将对您的列表进行排序并给予'not found'比''更高的优先级：

l = [int(a) for a in row if a.isnumeric()] # Or float(a)
l.sort()
row = [str(a) for a in l] +\
    ['not found'] * row.count('not found') +\
    [''] * row.count('')

Answer 4

这也可以解决问题：

row = ['not found', '', 555, 1, '5' , 444]
print(row)

def func(x):
    if str(x).isnumeric():
        return  1/-int(x) # ordering numerics event if they are strings 
    elif str(x) == 'not found':
        return  2
    elif str(x) == '':
        return  3

row2 = row.sort(key=func)

print(row)

结果：

['not found', '', 555, 1, '5', 444]
[1, '5', 444, 555, 'not found', '']