python - 获取具有任何所需值的组

Question

gr = []
for i in range(12000): gr.extend([i] * 2)
np.random.seed(0)
df = pd.DataFrame({'gr': gr,
                   'col1': np.random.choice(200, 24000)})
anyOfThese = np.array([50, 60]) #randomly chosen
t = time()
out = df[df.groupby('gr')['col1'].transform(lambda x: np.any(np.in1d(np.array(x), anyOfThese))).astype(bool)].gr.unique()
print(round(time() - t,2))
>>> 1.87

我需要获取在 col1 中具有任何两个所需值的所有组。

有没有办法更快地实现这一点？我需要重复相同的过程 ~100k 次。

Answer 1

使用boolean indexing按 isin 过滤:

out = df.loc[df['col1'].isin(anyOfThese), 'gr'].unique()

或通过 numpy.in1d 查看成员(member)资格:

out = df.loc[np.in1d(df['col1'], anyOfThese), 'gr'].unique()

时间:

np.random.seed(218)

gr = []
for i in range(12000): 
    gr.extend([i] * 2)
np.random.seed(0)
df = pd.DataFrame({'gr': gr,
                   'col1': np.random.choice(200, 24000)})
anyOfThese = np.array([50, 60]) #randomly chosen

a = df[df.groupby('gr')['col1'].transform(lambda x: np.any(np.in1d(np.array(x), anyOfThese))).astype(bool)].gr.unique()
out = df.loc[df['col1'].isin(anyOfThese), 'gr'].unique()
print ((a == out).all())
True

---

In [314]: %timeit df[df.groupby('gr')['col1'].transform(lambda x: np.any(np.in1d(np.array(x), anyOfThese))).astype(bool)].gr.unique()
2.9 s ± 79.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [315]: %timeit df.loc[df['col1'].isin(anyOfThese), 'gr'].unique()
746 µs ± 32.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [316]: %timeit df.loc[np.in1d(df['col1'], anyOfThese), 'gr'].unique()
325 µs ± 14.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)