回答編集履歴
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@@ -1,4 +1,8 @@
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## シンプルな解答
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```python
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df.loc[:, (df != 'NaN').sum() > 5]
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@@ -25,3 +29,81 @@
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# 9 10 NaN NaN
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```
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`FutureWarning`が出ますが気にしなくて大丈夫です。エラーが気になる場合は`.isin()`を使った、より速いコードがあります。
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```python
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%%timeit
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df.loc[:, (df != 'NaN').sum() > 5]
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# 1.23 ms ± 35.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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%%timeit
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df.loc[:, (~df.isin({'NaN'})).sum() > 5]
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# 1.06 ms ± 7.88 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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```
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## @can110氏のコード
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```python
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%%timeit
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df.drop(columns=[c for c in df.columns
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if sum(~np.isnan(df.replace('NaN', np.nan)[c])) < 5])
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# 5.56 ms ± 30.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
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```
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*今回の場合、`df.replace('NaN', np.nan)`は`df.astype(float)`で代用可能です。
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---
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```python
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%%timeit
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df.loc[:, (~np.isnan(df.astype(float).to_numpy())).sum(0) > 5]
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# 779 µs ± 3.16 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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%%timeit
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df.loc[:, np.isfinite(df.astype(float).to_numpy()).sum(0) > 5]
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# 767 µs ± 3.87 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
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```
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さらに速くなりました。
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