如何按键对字典进行排序?
示例输入:
{2:3, 1:89, 4:5, 3:0}
期望的输出:
{1:89, 2:3, 3:0, 4:5}
如何按键对字典进行排序?
示例输入:
{2:3, 1:89, 4:5, 3:0}
期望的输出:
{1:89, 2:3, 3:0, 4:5}
当前回答
l = dict.keys()
l2 = l
l2.append(0)
l3 = []
for repeater in range(0, len(l)):
smallnum = float("inf")
for listitem in l2:
if listitem < smallnum:
smallnum = listitem
l2.remove(smallnum)
l3.append(smallnum)
l3.remove(0)
l = l3
for listitem in l:
print(listitem)
其他回答
from operator import itemgetter
# if you would like to play with multiple dictionaries then here you go:
# Three dictionaries that are composed of first name and last name.
user = [
{'fname': 'Mo', 'lname': 'Mahjoub'},
{'fname': 'Abdo', 'lname': 'Al-hebashi'},
{'fname': 'Ali', 'lname': 'Muhammad'}
]
# This loop will sort by the first and the last names.
# notice that in a dictionary order doesn't matter. So it could put the first name first or the last name first.
for k in sorted (user, key=itemgetter ('fname', 'lname')):
print (k)
# This one will sort by the first name only.
for x in sorted (user, key=itemgetter ('fname')):
print (x)
如果你知道你所有的键都是相同的类型,或者有支持'< '(小于,python的__lt__)的类型,那么你可以使用dict(sorted(your_dict.items(), key=lambda _: _[0]))一个容易理解的一行程序
以下是建议解决方案的性能:
from collections import OrderedDict
from sortedcontainers import SortedDict
import json
keys = np.random.rand(100000)
vals = np.random.rand(100000)
d = dict(zip(keys, vals))
timeit SortedDict(d)
#45.8 ms ± 780 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit sorted(d.items())
#91.9 ms ± 707 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit OrderedDict(sorted(d.items(), key=lambda x: x[0]))
#93.7 ms ± 1.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit dict(sorted(dic.items()))
#113 ms ± 824 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit OrderedDict(sorted(dic.items()))
#122 ms ± 2.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit json.dumps(d, sort_keys=True)
#259 ms ± 9.42 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
如我们所见,格兰特·詹克斯的解决方案是目前为止最快的。
来自Python的集合库文档:
>>> from collections import OrderedDict
>>> # regular unsorted dictionary
>>> d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
>>> # dictionary sorted by key -- OrderedDict(sorted(d.items()) also works
>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
>>> # dictionary sorted by value
>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
>>> # dictionary sorted by length of the key string
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])
对于CPython/PyPy 3.6,以及任何Python 3.7或更高版本,这很容易做到:
>>> d = {2:3, 1:89, 4:5, 3:0}
>>> dict(sorted(d.items()))
{1: 89, 2: 3, 3: 0, 4: 5}