回答編集履歴
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test
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@@ -20,7 +20,7 @@
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idx = cartesian([np.arange(length) for length in lens])
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offset =
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offset = lens.cumsum() - lens
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追記
test
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コードを整理すればもっといい方法があるかもしれませんが、いかがでしょうか。`itertools.product`より速いと思われます（特に各要素が数値なら）。
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コードを整理すればもっといい方法があるかもしれませんが、以下の方法でいかがでしょうか。`itertools.product`より速いと思われます（特に各要素が数値なら）。
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動作確認
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## 動作確認
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# (後略)
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```
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## 速度比較
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```python
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In [21]: def df_product_itertools(df_list):
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: lsts = [df.to_numpy().tolist() for df in df_list]
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: new_list = [reduce(add, e) for e in itertools.product(*lsts)]
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: return pd.DataFrame(new_list, columns=np.concatenate(
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: [df.columns.to_numpy() for df in df_list]))
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In [22]: l_1 = [[1, 2], [3, 4]]
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: l_2 = [[11, 22], [33, 44]]
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: l_3 = [[111, 222], [333, 444]]
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: l_4 = [[1111, 2222], [3333, 4444]]
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: l_5 = [[11111, 22222], [33333, 44444]]
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:
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: df1 = pd.DataFrame(l_1, columns=['A', 'B'])
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: df2 = pd.DataFrame(l_2, columns=['C', 'D'])
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: df3 = pd.DataFrame(l_3, columns=['E', 'F'])
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: df4 = pd.DataFrame(l_4, columns=['G', 'H'])
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: df5 = pd.DataFrame(l_5, columns=['I', 'J'])
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In [23]: %timeit df_product([df1, df2, df3, df4])
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: %timeit df_product_itertools([df1, df2, df3, df4])
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355 µs ± 18.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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928 µs ± 16.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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In [24]: %timeit df_product([df1, df2, df3, df4, df5])
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: %timeit df_product_itertools([df1, df2, df3, df4, df5])
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378 µs ± 12.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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1.12 ms ± 17.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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```
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