熊猫系列到2d阵列

Question

因此，我使用了将2d数组放入Pandas系列的答案， 将 2D numpy数组放入pandas系列。 简而言之，它是

a = np.zeros((5,2))
s = pd.Series(list(a))

现在，将熊猫系列转换回2D阵列最便宜的方法是什么？ 如果我尝试s.values ，我会获得带有object s.values的数组。

到目前为止，我尝试了np.vstack(s.values)但它当然会复制数据。

Answer 1

我相信你需要：

a = np.array(s.values.tolist())
print (a)
[[ 0.  0.]
 [ 0.  0.]
 [ 0.  0.]
 [ 0.  0.]
 [ 0.  0.]]

---

a = np.zeros((50000,2))
s = pd.Series(list(a))

In [131]: %timeit (np.vstack(s.values))
10 loops, best of 3: 107 ms per loop

In [132]: %timeit (np.array(s.values.tolist()))
10 loops, best of 3: 19.7 ms per loop

In [133]: %timeit (np.array(s.tolist()))
100 loops, best of 3: 19.6 ms per loop

但如果转置差异很小（但缓存 ）：

a = np.zeros((2,50000))
s = pd.Series(list(a))
#print (s)

In [159]: %timeit (np.vstack(s.values))
The slowest run took 23.31 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 55.7 µs per loop

In [160]: %timeit (np.array(s.values.tolist()))
The slowest run took 7.20 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 49.8 µs per loop

In [161]: %timeit (np.array(s.tolist()))
The slowest run took 7.31 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 62.6 µs per loop