合并/压缩两个系列到ndarray的ndarray

Question

我有两个相同长度的pandas series，如下所示：

S1 = 
0      -0.483415
1      -0.514082
2      -0.515724
3      -0.519375
4      -0.505685
...

S2 = 
1      -0.961871
2      -0.964762
3      -0.963798
4      -0.962112
5      -0.962028
...

我想将它们压缩成一个numpy ndarray of ndarray，使它看起来像这样：

<class 'numpy.ndarray'>
[[-0.483415 -0.961871]
 [-0.514082 -0.964762]
 [-0.515724 -0.963798]
 ...
]

如果我想要list的tuple，我可以这样说：

v = list(zip(S1, S2))

这给了我：

<class 'list'>
[(-0.48341467662344273, -0.961871075696243), 
 (-0.5140815458448855, -0.9647615371349125),
  ...
]

我如何做同样的“拉链”但回到ndarray of ndarray？我不想要循环。

Answer 1

Zip在这里没有必要，为了更好的性能使用numpy或pandas：

arr = np.hstack((S1.values[:, None], S2.values[:, None]))

要么：

arr = np.vstack((S1, S2)).T

要么：

arr = pd.concat([S1.reset_index(drop=True), S2.reset_index(drop=True)], axis=1).values

要么：

arr = np.c_[S1, S2]

---

print (arr)
[[-0.483415 -0.961871]
 [-0.514082 -0.964762]
 [-0.515724 -0.963798]
 [-0.519375 -0.962112]
 [-0.505685 -0.962028]]

性能：

#50k values
S1 = pd.concat([S1] * 10000, ignore_index=True)
S2 = pd.concat([S2] * 10000, ignore_index=True)

In [107]: %timeit arr = np.hstack((S1.values[:, None], S2.values[:, None]))
133 µs ± 15.9 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [108]: %timeit arr = np.vstack((S1, S2)).T
176 µs ± 12 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [109]: %timeit arr = pd.concat([S1.reset_index(drop=True), S2.reset_index(drop=True)], axis=1).values
1.49 ms ± 74.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [110]: %timeit arr = np.c_[S1, S2]
320 µs ± 10.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [111]: %timeit np.array(list(zip(S1, S2)))
33 ms ± 545 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 2

把它投到ndarray：

>>> a = [1,2,3,4]
>>> b = [5,6,7,8]
>>> c = list(zip(a, b))
>>> c
[(1, 1), (2, 2), (3, 3), (4, 4)]
>>> d = np.array(c)
>>> d
array([[1, 5],
       [2, 6],
       [3, 7],
       [4, 8]])
>>> d.shape
(4, 2)

Answer 3

尝试：

numpy.hstack((S1, S2))

我认为它应该有效。