回答編集履歴
1
heapq.nlargest についての説明を追加
answer
CHANGED
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@@ -19,4 +19,29 @@
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for i in range(N):
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new_dict[lists[i][0]] = lists[i][1]
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print(new_dict)
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```
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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。
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```python
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# coding: utf-8
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import heapq
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N = 3
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d = {'key1': 1,'key2': 14,'key3': 47,'key4': 2,'key5': 90}
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# keyでソートしたときの上位N個
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lists = heapq.nlargest(N, d.items(), key=lambda x: x[0])
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new_dict = {}
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for l in lists:
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new_dict[l[0]] = l[1]
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print(new_dict)
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# valueでソートしたときの上位N個
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lists = heapq.nlargest(N, d.items(), key=lambda x: x[1])
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new_dict = {}
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for l in lists:
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new_dict[l[0]] = l[1]
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print(new_dict)
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47
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```
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