字典中的頂級值

Question

如何從字典中檢索前3列表？

>>> d 
{'a': 2, 'and': 23, 'this': 14, 'only.': 21, 'is': 2, 'work': 2, 'will': 2, 'as': 2, 'test': 4} 

預期結果：

and: 23 
only: 21 
this: 14 

Answer 1

使用collections.Counter：

>>> d = Counter({'a': 2, 'and': 23, 'this': 14, 'only.': 21, 'is': 2, 'work': 2, 'will': 2, 'as': 2, 'test': 4}) 
>>> d.most_common() 
[('and', 23), ('only.', 21), ('this', 14), ('test', 4), ('a', 2), ('is', 2), ('work', 2), ('will', 2), ('as', 2)] 
>>> for k, v in d.most_common(3): 
...  print '%s: %i' % (k, v) 
... 
and: 23 
only.: 21 
this: 14 

計數器對象提供各種其他優點，如使它幾乎微不足道收集計數擺在首位。

Answer 2

>>> d = {'a': 2, 'and': 23, 'this': 14, 'only.': 21, 'is': 2, 'work': 2, 'will': 2, 'as': 2, 'test': 4} 
>>> t = sorted(d.iteritems(), key=lambda x:-x[1])[:3] 

>>> for x in t: 
...  print "{0}: {1}".format(*x) 
... 
and: 23 
only.: 21 
this: 14 

Answer 3

你已經得到的答覆是正確的，但是我會創建我自己的關鍵函數來調用sort（）時使用。

d = {'a': 2, 'and': 23, 'this': 14, 'only.': 21, 'is': 2, 'work': 2, 'will': 2, 'as': 2, 'test': 4} 

# create a function which returns the value of a dictionary 
def keyfunction(k): 
    return d[k] 

# sort by dictionary by the values and print top 3 {key, value} pairs 
for key in sorted(d, key=keyfunction, reverse=True)[:3]: 
    print "%s: %i" % (key, d[key]) 

Answer 4

鑑於上面的解決方案：

def most_popular(L): 
    # using lambda 
    start = datetime.datetime.now() 
    res=dict(sorted([(k,v) for k, v in L.items()], key=lambda x: x[1])[-2:]) 
    delta=datetime.datetime.now()-start 
    print "Microtime (lambda:%d):" % len(L), str(delta.microseconds) 

    # using collections 
    start=datetime.datetime.now() 
    res=dict(collections.Counter(L).most_common()[:2]) 
    delta=datetime.datetime.now()-start 
    print "Microtime (collections:%d):" % len(L), str(delta.microseconds) 

# list of 10 
most_popular({el:0 for el in list(range(10))}) 

# list of 100 
most_popular({el:0 for el in list(range(100))}) 

# list of 1000 
most_popular({el:0 for el in list(range(1000))}) 

# list of 10000 
most_popular({el:0 for el in list(range(10000))}) 

# list of 100000 
most_popular({el:0 for el in list(range(100000))}) 

# list of 1000000 
most_popular({el:0 for el in list(range(1000000))}) 

從10^1至10^6字典對象的尺寸的數據集的dict工作像

print {el:0 for el in list(range(10))} 
{0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0} 

我們有以下的基準

Python 2.7.10 (default, Jul 14 2015, 19:46:27) 
[GCC 4.8.2] on linux 

Microtime (lambda:10): 24 
Microtime (collections:10): 106 
Microtime (lambda:100): 49 
Microtime (collections:100): 50 
Microtime (lambda:1000): 397 
Microtime (collections:1000): 178 
Microtime (lambda:10000): 4347 
Microtime (collections:10000): 2782 
Microtime (lambda:100000): 55738 
Microtime (collections:100000): 26546 
Microtime (lambda:1000000): 798612 
Microtime (collections:1000000): 361970 
=> None 

所以我們可以說小列表使用lambda，但對於巨大的列表，collections有更好的表現。

查看基準運行here。