回答編集履歴
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修正
answer
CHANGED
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@@ -9,7 +9,7 @@
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9
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n: 取り出す個数（int）
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10
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"""
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11
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vln = len(v)
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12
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-
base = arr.max() ** np.arange(vln)
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12
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base = (arr.max()+1) ** np.arange(vln)
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search_idx, = np.nonzero(np.convolve(arr, base) == np.sum(base*v[::-1]))
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idx = search_idx + np.arange(1, n+1)[:, None]
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return arr[idx]
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@@ -31,7 +31,11 @@
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[93, 16, 45, 68, 55, 23, 1, 25, 35, 21]])
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# @aokikenichi さんの作成した配列に対して
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In [12]: l = [0, 1, 2, 3,
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35
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...: 13, 10, 5, 4, 3, 2, 1, 0,
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...: 13, 10, 1, 2, 3, 4, 5, 6, 7,
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...: 13, 10, 99, 98, 97, 96, 95, 94,
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-
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...: 13, 10, 100, 101, 102, 103, 104, 105]
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...: extract(np.array(l), [13, 10], 3)
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Out[12]:
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array([[ 5, 1, 99, 100],
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@@ -39,10 +43,36 @@
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[ 3, 3, 97, 102]])
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# 条件を変えることもできます
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In [
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In [13]: extract(np.array(l), [1, 2, 3], 4)
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Out[
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Out[13]:
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array([[13, 4],
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[10, 5],
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[ 5, 6],
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[ 4, 7]])
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```
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---
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@aokikenichi さんの回答は一般化すると以下のような関数になりますが、
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```python
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def extract_for(l, v, n):
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vln = len(v)
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out = []
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for i in range(len(l) - 1):
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if all(l[i+j] == vi for j, vi in enumerate(v)):
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out.append(l[i+vln:i+vln+n])
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return np.array(out).T
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```
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速度を比較するとNumpyの関数を用いたほうが効率が良いことがわかります。
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```python
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# arrは長さ100000の配列
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In [21]: %timeit extract(arr, [13, 10], 8)
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698 µs ± 15.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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In [22]: l = arr.tolist()
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...: %timeit extract_for(l, [13, 10], 8)
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78 ms ± 1.32 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
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```
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