Python通过多个键对字典列表进行排序
我有一个列表：
b = [{u'TOT_PTS_Misc': u'Utley, Alex', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Russo, Brandon', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Chappell, Justin', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Foster, Toney', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Lawson, Roman', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Lempke, Sam', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Gnezda, Alex', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Kirks, Damien', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Worden, Tom', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Korecz, Mike', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Swartz, Brian', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Burgess, Randy', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Smugala, Ryan', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Harmon, Gary', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Blasinsky, Scott', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Carter III, Laymon', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Coleman, Johnathan', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Venditti, Nick', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Blackwell, Devon', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Kovach, Alex', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Bolden, Antonio', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Smith, Ryan', u'Total_Points': 60.0}]
我需要使用Total_Points反转的多键排序，然后不会被TOT_PTS_Misc反转。
这可以在命令提示符下完成，如下所示：
a = sorted(b, key=lambda d: (-d['Total_Points'], d['TOT_PTS_Misc']))
但是我必须通过一个函数来运行它，在那里我传入列表和排序键。 例如， def multikeysort(dict_list, sortkeys):
如何使用lambda行来对列表进行排序，对于传递给multikeysort函数的任意数量的键，并考虑到sortkeys可以具有任意数量的键并且需要反向排序的那些键将被识别与之前的' - '？
这个答案适用于字典中的任何一种字段 - 否定字段不一定是数字。
def multikeysort(items, columns):
from operator import itemgetter
comparers = [((itemgetter(col[1:].strip()), -1) if col.startswith('-') else
(itemgetter(col.strip()), 1)) for col in columns]
def comparer(left, right):
for fn, mult in comparers:
result = cmp(fn(left), fn(right))
if result:
return mult * result
else:
return 0
return sorted(items, cmp=comparer)
你可以这样称呼它：
b = [{u'TOT_PTS_Misc': u'Utley, Alex', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Russo, Brandon', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Chappell, Justin', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Foster, Toney', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Lawson, Roman', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Lempke, Sam', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Gnezda, Alex', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Kirks, Damien', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Worden, Tom', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Korecz, Mike', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Swartz, Brian', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Burgess, Randy', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Smugala, Ryan', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Harmon, Gary', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Blasinsky, Scott', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Carter III, Laymon', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Coleman, Johnathan', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Venditti, Nick', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Blackwell, Devon', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Kovach, Alex', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Bolden, Antonio', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Smith, Ryan', u'Total_Points': 60.0}]
a = multikeysort(b, ['-Total_Points', 'TOT_PTS_Misc'])
for item in a:
print item
尝试任一列否定。 你会看到排序顺序相反。
下一步：改变它，所以它不使用额外的类....
2016年1月17日
从这个答案中得到我的灵感从一个可迭代匹配条件中获得第一个条目的最佳方式是什么？我缩短了代码：
from operator import itemgetter as i
def multikeysort(items, columns):
comparers = [
((i(col[1:].strip()), -1) if col.startswith('-') else (i(col.strip()), 1))
for col in columns
]
def comparer(left, right):
comparer_iter = (
cmp(fn(left), fn(right)) * mult
for fn, mult in comparers
)
return next((result for result in comparer_iter if result), 0)
return sorted(items, cmp=comparer)
如果你喜欢你的代码简洁。
之后2016-01-17
这与python3（它消除了cmp参数sort ）一起工作：
from operator import itemgetter as i
from functools import cmp_to_key
def multikeysort(items, columns):
comparers = [
((i(col[1:].strip()), -1) if col.startswith('-') else (i(col.strip()), 1))
for col in columns
]
def comparer(left, right):
comparer_iter = (
cmp(fn(left), fn(right)) * mult
for fn, mult in comparers
)
return next((result for result in comparer_iter if result), 0)
return sorted(items, key=cmp_to_key(comparer))
受此答案的启发我应该如何在Python 3中进行自定义排序？
def sortkeypicker(keynames):
negate = set()
for i, k in enumerate(keynames):
if k[:1] == '-':
keynames[i] = k[1:]
negate.add(k[1:])
def getit(adict):
composite = [adict[k] for k in keynames]
for i, (k, v) in enumerate(zip(keynames, composite)):
if k in negate:
composite[i] = -v
return composite
return getit
a = sorted(b, key=sortkeypicker(['-Total_Points', 'TOT_PTS_Misc']))
http://stygianvision.net/updates/python-sort-list-object-dictionary-multiple-key/有很好的概述了各种技术。 如果您的要求比“全双向多键”更简单，请看一看。 很明显，接受的答案和我刚才提到的博客文章以某种方式影响了彼此，但我不知道该订单。
如果链接在这里死去，那么上面没有提到的例子就是一个非常快速的概要：
mylist = sorted(mylist, key=itemgetter('name', 'age'))
mylist = sorted(mylist, key=lambda k: (k['name'].lower(), k['age']))
mylist = sorted(mylist, key=lambda k: (k['name'].lower(), -k['age']))