我有一个数组数组，如下所示：
segments[:30]
array([array([ 131.2]), array([ 124.1]), 0.23679025210440158,
array([ 133.65]), array([ 123.3]), 0.3221912760287523,
array([ 116.7]), array([ 147.7]), 0.24318619072437286,
array([ 102.3]), array([ 120.55]), 0.07436020392924547,
array([ 130.25]), array([ 100.5625]), 0.029634355247253552,
array([ 143.6]), array([ 132.4]), 0.5843092009425164,
array([ 151.65]), array([ 131.6]), 0.4865431547164917,
array([ 143.3]), array([ 152.05]), 0.2774583905003965,
array([ 111.65]), array([ 125.]), 0.23880321211181582,
array([ 123.1875]), array([ 79.5625]), 0.1562070251966361], dtype=object)
我想摆脱array([ 131.2])并仅提取值131.2。
我的预期输出是：
array([131.2, 124.1, 0.23679025210440158,
133.65, 123.3, 0.3221912760287523,
116.7,147.7, 0.24318619072437286,
102.3, 120.55, 0.07436020392924547,....])
我尝试了以下方法：
np.array(segments)
但是它并没有改变我的数据。
答案 0 :(得分：2)
concatenate使所有元素都变成数组，但是尺寸存在问题。有些是1d，有些是0d：
In [109]: np.concatenate(arr)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-109-062a20dcc2f7> in <module>()
----> 1 np.concatenate(arr)
ValueError: all the input arrays must have same number of dimensions
hstack之所以有效，是因为它首先使用[atleast_1d(_m) for _m in tup]将所有内容转换为一维数组：
In [110]: np.hstack(arr)
Out[110]:
array([1.31200000e+02, 1.24100000e+02, 2.36790252e-01, 1.33650000e+02,
1.23300000e+02, 3.22191276e-01, 1.16700000e+02, 1.47700000e+02,
2.43186191e-01, 1.02300000e+02, 1.20550000e+02, 7.43602039e-02,
1.30250000e+02, 1.00562500e+02, 2.96343552e-02, 1.43600000e+02,
1.32400000e+02, 5.84309201e-01, 1.51650000e+02, 1.31600000e+02,
4.86543155e-01, 1.43300000e+02, 1.52050000e+02, 2.77458391e-01,
1.11650000e+02, 1.25000000e+02, 2.38803212e-01, 1.23187500e+02,
7.95625000e+01, 1.56207025e-01])
结果是数字dtype（不是对象）。
处理对象数组需要某种python级别的迭代-除了像reshape这样的有限操作实际上并不操纵元素之外。而且，对象的迭代比列表的迭代慢（但比数字数组的Python级别的迭代快）。
In [113]: [np.atleast_1d(i) for i in arr] # consistent dimensions
Out[113]:
[array([131.2]),
array([124.1]),
array([0.23679025]),
array([133.65]),
array([123.3]),
...]
In [116]: [np.asarray(i) for i in arr] # mixed dimensions
Out[116]:
[array([131.2]),
array([124.1]),
array(0.23679025),
array([133.65]),
array([123.3]),...]
内部atleast_1d对尺寸进行一些测试。它也可以与*args一起使用，因此我们可以编写
In [123]: np.atleast_1d(*arr)
Out[123]:
[array([131.2]),
array([124.1]),
array([0.23679025]),
array([133.65]),
array([123.3]),
...]
因此
np.concatenate(np.atleast_1d(*arr))
定时测试表明@sacul's的“天真”列表理解最快：np.array([i[0] if isinstance(i, np.ndarray) else i for i in segments])
答案 1 :(得分：1)
一种方法是在列表理解为np.arrays时迭代并提取浮点数：
np.array([i[0] if isinstance(i, np.ndarray) else i for i in segments])
哪个返回：
array([1.31200000e+02, 1.24100000e+02, 2.36790252e-01, 1.33650000e+02,
1.23300000e+02, 3.22191276e-01, 1.16700000e+02, 1.47700000e+02,
2.43186191e-01, 1.02300000e+02, 1.20550000e+02, 7.43602039e-02,
1.30250000e+02, 1.00562500e+02, 2.96343552e-02, 1.43600000e+02,
1.32400000e+02, 5.84309201e-01, 1.51650000e+02, 1.31600000e+02,
4.86543155e-01, 1.43300000e+02, 1.52050000e+02, 2.77458391e-01,
1.11650000e+02, 1.25000000e+02, 2.38803212e-01, 1.23187500e+02,
7.95625000e+01, 1.56207025e-01])
这是一种幼稚但直接的做事方式。但这在非常大的阵列上可能会非常慢。
如果您的结构始终与示例相同，即 ie 2个数组，后跟一个浮点数，那么您可以调整数组的形状，从每3个值中提取2个浮点数，然后进行连接数据以相同顺序一起返回：
x = segments.reshape(-1,3)
f = np.concatenate(x[:,[0,1]].flatten()).reshape(-1,2)
l = x[:,2].reshape(-1,1)
np.concatenate((f,l),1).flatten()
哪个返回：
array([131.2, 124.1, 0.23679025210440158, 133.65, 123.3,
0.3221912760287523, 116.7, 147.7, 0.24318619072437286, 102.3,
120.55, 0.07436020392924547, 130.25, 100.5625,
0.029634355247253552, 143.6, 132.4, 0.5843092009425164, 151.65,
131.6, 0.4865431547164917, 143.3, 152.05, 0.2774583905003965,
111.65, 125.0, 0.23880321211181582, 123.1875, 79.5625,
0.1562070251966361], dtype=object)
说明
仅是为了帮助直观了解此处发生的情况，您可以查看一下在重新连接在一起之前提取的重塑数据。
>>> x
array([[array([131.2]), array([124.1]), 0.23679025210440158],
[array([133.65]), array([123.3]), 0.3221912760287523],
[array([116.7]), array([147.7]), 0.24318619072437286],
[array([102.3]), array([120.55]), 0.07436020392924547],
[array([130.25]), array([100.5625]), 0.029634355247253552],
[array([143.6]), array([132.4]), 0.5843092009425164],
[array([151.65]), array([131.6]), 0.4865431547164917],
[array([143.3]), array([152.05]), 0.2774583905003965],
[array([111.65]), array([125.]), 0.23880321211181582],
[array([123.1875]), array([79.5625]), 0.1562070251966361]],
dtype=object)
>>> f
array([[131.2 , 124.1 ],
[133.65 , 123.3 ],
[116.7 , 147.7 ],
[102.3 , 120.55 ],
[130.25 , 100.5625],
[143.6 , 132.4 ],
[151.65 , 131.6 ],
[143.3 , 152.05 ],
[111.65 , 125. ],
[123.1875, 79.5625]])
>>> l
array([[0.23679025210440158],
[0.3221912760287523],
[0.24318619072437286],
[0.07436020392924547],
[0.029634355247253552],
[0.5843092009425164],
[0.4865431547164917],
[0.2774583905003965],
[0.23880321211181582],
[0.1562070251966361]], dtype=object)