回答編集履歴
1
heapq.nlargest についての説明を追加
test
CHANGED
|
@@ -41,3 +41,53 @@
|
|
|
41
41
|
print(new_dict)
|
|
42
42
|
|
|
43
43
|
```
|
|
44
|
+
|
|
45
|
+
|
|
46
|
+
|
|
47
|
+
Nが小さいならば、[heapq.nlargest モジュール](https://docs.python.jp/3/library/heapq.html#heapq.nlargest)を使ったほうが高速です。
|
|
48
|
+
|
|
49
|
+
|
|
50
|
+
|
|
51
|
+
|
|
52
|
+
|
|
53
|
+
```python
|
|
54
|
+
|
|
55
|
+
# coding: utf-8
|
|
56
|
+
|
|
57
|
+
import heapq
|
|
58
|
+
|
|
59
|
+
|
|
60
|
+
|
|
61
|
+
N = 3
|
|
62
|
+
|
|
63
|
+
d = {'key1': 1,'key2': 14,'key3': 47,'key4': 2,'key5': 90}
|
|
64
|
+
|
|
65
|
+
|
|
66
|
+
|
|
67
|
+
# keyでソートしたときの上位N個
|
|
68
|
+
|
|
69
|
+
lists = heapq.nlargest(N, d.items(), key=lambda x: x[0])
|
|
70
|
+
|
|
71
|
+
new_dict = {}
|
|
72
|
+
|
|
73
|
+
for l in lists:
|
|
74
|
+
|
|
75
|
+
new_dict[l[0]] = l[1]
|
|
76
|
+
|
|
77
|
+
print(new_dict)
|
|
78
|
+
|
|
79
|
+
|
|
80
|
+
|
|
81
|
+
# valueでソートしたときの上位N個
|
|
82
|
+
|
|
83
|
+
lists = heapq.nlargest(N, d.items(), key=lambda x: x[1])
|
|
84
|
+
|
|
85
|
+
new_dict = {}
|
|
86
|
+
|
|
87
|
+
for l in lists:
|
|
88
|
+
|
|
89
|
+
new_dict[l[0]] = l[1]
|
|
90
|
+
|
|
91
|
+
print(new_dict)
|
|
92
|
+
|
|
93
|
+
```
|