回答編集履歴
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追記
test
CHANGED
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@@ -17,3 +17,39 @@
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[1, 2]
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```
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----
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一番多い要素なら、`most_common`で先頭の要素を取りますか。
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```python
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>>> from collections import Counter
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>>> a = [1,1,2,3]
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>>> a_count = Counter(a)
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>>> max_a_count = a_count.most_common()[0][1]
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>>> [k for k, v in a_count.most_common() if v == max_a_count]
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[1]
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>>> b = [1,1,2,2,3]
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>>> b_count = Counter(b)
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>>> max_b_count = b_count.most_common()[0][1]
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>>> [k for k, v in b_count.most_common() if v == max_b_count]
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[1, 2]
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```
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要素がたくさんあるなら探索を打ち切った方がいいですが、tell_k さんが挙げているようなループで十分ですね。
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