对具有固定位置的特定值的对象列表进行排序

Question

我有一个对象列表。我只想对满足特定条件的值进行排序。考虑这个列表：

class Destination:
    def __init__(self, name, tag):
        self.name = name
        self.tag = tag

dest = [
Destination(name="language", tag=['lang']), 
Destination(name="operationId", tag=['order_02']), 
Destination(name="eventTime", tag=['order_01']),
Destination(name="countryCode", tag=[]),
Destination(name="eventDate", tag=['order_00']),
Destination(name="offerId", tag=['primary_key'])
]

我想对其中包含

tag

的

order

值进行排序。结果应该如下所示：

output = [
Destination(name="language", tag=['lang']), 
Destination(name="eventDate", tag=['order_00']),
Destination(name="eventTime", tag=['order_01']),
Destination(name="operationId", tag=['order_02']), 
Destination(name="countryCode", tag=[]),
Destination(name="offerId", tag=['primary_key'])
]

目前，我可以使用

cols = sorted(dest, key=lambda x: x.tag)

对所有标签进行排序，但我需要一些帮助来重写它，以便仅对包含特定值的

tags

进行排序，同时保持其他值不变。如有任何帮助，我们将不胜感激！

Answer 1

假设您想要所有无序项，然后满足第一个顺序时所有已排序的顺序，然后是所有其他对象，您可以使用循环和列表来收集中间体：

def is_order(x):
    return x.tag and x.tag[0].startswith('order')

orders = []
others = []

out = []
flag = True              # directly add to "out" if True
for x in dest:
    if is_order(x):
        flag = False     # start collecting in lists
        orders.append(x)
    elif flag:
        out.append(x)    # directly add to "out"
    else:
        others.append(x) # add to "others" temporarily

# add sorted "orders"
out.extend(sorted(orders, key=lambda x: x.tag))
# add others
out.extend(others)

输出：

[Destination(name=language, tag=['lang']),
 Destination(name=eventDate, tag=['order_00']),
 Destination(name=eventTime, tag=['order_01']),
 Destination(name=operationId, tag=['order_02']),
 Destination(name=countryCode, tag=[]),
 Destination(name=offerId, tag=['primary_key'])]