merge/zip 两个系列变成 ndarray 的 ndarray

merge/zip two series into ndarray of ndarray

我有两个 pandas series 长度相同,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 numpy ndarray of ndarray,这样它看起来像这样:

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

如果我想要 tuplelist,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),
  ...
]

我如何做同样的事 "zip" 但取回一个 ndarray of ndarray?我不想要循环。

Zip这里不是必须的,为了更好的性能使用numpy或者pandas:

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

或:

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

或:

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

或:

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)

只需将其转换为 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)

尝试:

numpy.hstack((S1, S2))

我觉得应该可以。