python字典值排序

Question

我有2个字典,dict1并且dict2其中包含相同的密钥,但是不同的值的密钥.我想要做的是为每个字典,从最大到最小的值排序,然后给每个值一个等级1-N,1是最大的值.从这里开始,我希望得到相同键的每个字典中值的等级差异.例如:

dict1 = {a:0.6, b:0.3, c:0.9, d:1.2, e:0.2}
dict2 = {a:1.4, b:7.7, c:9.0, d:2.5, e:2.0}

# sorting by values would look like this:
dict1 = {d:1.2, c:0.9, a:0.6, b:0.3, e:0.2}
dict2 = {c:9.0, b:7.7, d:2.5, e:2.0, a:1.4}

#ranking the values would produce this:
dict1 = {d:1, c:2, a:3, b:4, e:5}
dict2 = {c:1, b:2, d:3, e:4, a:5}

#computing the difference between ranks would be something like this:
diffs = {}
for x in dict1.keys():
    diffs[x] = (dict1[x] - dict2[x])

#diffs would look like this:
diffs[a] = -2
diffs[b] = 2
diffs[c] = 1
diffs[d] = -2
diffs[e] = 1

我知道词典是随机的而不是可排序的,但也许有一种方法可以将键和值放入列表中？我面临的主要挑战是获取按值(从最大到最小)排序的键和值,然后将值更改为排序列表中的相应排名.

Answer 1

一个简单的小解决方案是

dict1 = {"a":0.6, "b":0.3, "c":0.9, "d":1.2, "e":0.2}
dict2 = {"a":1.4, "b":7.7, "c":9.0, "d":2.5, "e":2.0}
k1 = sorted(dict1, key=dict1.get)
k2 = sorted(dict2, key=dict2.get)
diffs = dict((k, k2.index(k) - k1.index(k)) for k in dict1)

对于较大的dicts,更高效,更不易阅读的版本:

ranks1 = dict(map(reversed, enumerate(sorted(dict1, key=dict1.get))))
ranks2 = dict(map(reversed, enumerate(sorted(dict2, key=dict2.get))))
diffs = dict((k, ranks2[k] - ranks1[k]) for k in dict1)

Answer 2

您可能对collections.OrderedDict感兴趣

下面是一个示例,我最初的吼声,你也在寻找词典与值排序键,东西od1和od2有.

d1 = {"a":0.6, "b":0.3, "c":0.9, "d":1.2, "e":0.2}
d2 = {"a":1.4, "b":7.7, "c":9.0, "d":2.5, "e":2.0}

od1 = OrderedDict(sorted(d1.items(), key=lambda t: t[1]))
od2 = OrderedDict(sorted(d2.items(), key=lambda t: t[1]))

k1 = od1.keys()
k2 = od2.keys()

diff = dict((k, n - k2.index(k)) for n, k in enumerate(k1))

如果你不需要它们,那么Sven解决方案可能更快.

编辑:老实说没那么快......(sven.py是他的第二个,效率更高的版本):

$ cat /tmp/mine.py | time python -m timeit
10000000 loops, best of 3: 0.0842 usec per loop
real    0m 3.69s
user    0m 3.38s
sys 0m 0.03s
$ cat /tmp/sven.py | time python -m timeit
10000000 loops, best of 3: 0.085 usec per loop
real    0m 3.86s
user    0m 3.42s
sys 0m 0.03s

如果有人想发布格式更大的dicts,我也会测试它们.