这是Matlab向量：a = [inf(m,1);ones(m,1)]
我试图以python方式创建类似的对象。
我尝试过：
import numpy as np
a = np.stack((np.tile(np.array([np.inf]), (m, 0))),np.ones((m,0)))
通过在控制台中对此进行测试，例如将m更改为5，然后尝试使用PyCharm进行查看，我得到了以下消息：
Traceback (most recent call last):
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\pydev\_pydev_comm\server.py", line 34, in handle
self.processor.process(iprot, oprot)
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\third_party\thriftpy\_shaded_thriftpy\thrift.py", line 266, in process
self.handle_exception(e, result)
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\third_party\thriftpy\_shaded_thriftpy\thrift.py", line 254, in handle_exception
raise e
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\third_party\thriftpy\_shaded_thriftpy\thrift.py", line 263, in process
result.success = call()
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\third_party\thriftpy\_shaded_thriftpy\thrift.py", line 228, in call
return f(*(args.__dict__[k] for k in api_args))
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\pydev\_pydev_bundle\pydev_console_utils.py", line 359, in getArray
return pydevd_thrift.table_like_struct_to_thrift_struct(array, name, roffset, coffset, rows, cols, format)
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\pydev\_pydevd_bundle\pydevd_thrift.py", line 602, in table_like_struct_to_thrift_struct
return TYPE_TO_THRIFT_STRUCT_CONVERTERS[type_name](array, name, roffset, coffset, rows, cols, format)
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\pydev\_pydevd_bundle\pydevd_thrift.py", line 377, in array_to_thrift_struct
array, array_chunk, r, c, f = array_to_meta_thrift_struct(array, name, format)
File "C:\Program Files\JetBrains\PyCharm 2019.1.3\helpers\pydev\_pydevd_bundle\pydevd_thrift.py", line 476, in array_to_meta_thrift_struct
bounds = (array.min(), array.max())
File "C:\Users\Azerty\PycharmProjects\OptionsHedgeFund\venv37\lib\site-packages\numpy\core\_methods.py", line 32, in _amin
return umr_minimum(a, axis, None, out, keepdims, initial)
ValueError: zero-size array to reduction operation minimum which has no identity
怎么了？
答案 0 :(得分：2)
因此，如果我正确理解了您的预期结果，您可能会这样做以生成矩阵。
m = 5 #or whatever
a = np.ones((2, m)) #create a matrix 2 x m of ones
a[0] = np.inf #replace first row with infinites
a是：
array([[inf, inf, inf, inf, inf],
[ 1., 1., 1., 1., 1.]])
答案 1 :(得分：2)
在八度会话中：
>> m = 5;
>> a = [inf(m,1); ones(m,1)];
>> size(a)
ans =
10 1
>> a
a =
Inf
Inf
Inf
Inf
Inf
1
1
1
1
1
在ipython会话中：
In [21]: m=5
In [22]: np.vstack([np.ones((m,1))*np.inf, np.ones((m,1))])
Out[22]:
array([[inf],
[inf],
[inf],
[inf],
[inf],
[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.]])
In [23]: _.shape
Out[23]: (10, 1)
具有相同功能的变体：`
np.concatenate([np.full((m,1), np.inf), np.ones((m,1))], axis=0)
===
对于两行，以（1，m）形状开始：
>> a = [inf(1,m); ones(1,m)];
>> size(a)
ans =
2 5
>> a
a =
Inf Inf Inf Inf Inf
1 1 1 1 1
In [26]: np.concatenate([np.full((1,m), np.inf), np.ones((1,m))], axis=0)
Out[26]:
array([[inf, inf, inf, inf, inf],
[ 1., 1., 1., 1., 1.]])
===
对于您的错误，np.stack在连接两个（5,0）形状的数组时遇到问题（我不确定为什么）。
In [27]: a = np.stack((np.tile(np.array([np.inf]), (m, 0))),np.ones((m,0)))
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-27-c6f8ec8462c9> in <module>
----> 1 a = np.stack((np.tile(np.array([np.inf]), (m, 0))),np.ones((m,0)))
/usr/local/lib/python3.6/dist-packages/numpy/core/shape_base.py in stack(arrays, axis, out)
417
418 result_ndim = arrays[0].ndim + 1
--> 419 axis = normalize_axis_index(axis, result_ndim)
420
421 sl = (slice(None),) * axis + (_nx.newaxis,)
TypeError: only size-1 arrays can be converted to Python scalars
您的错误有所不同；看起来它是由pydev而不是Python本身产生的。
但是检查碎片：
In [28]: np.tile(np.array([np.inf]), (m, 0))
Out[28]: array([], shape=(5, 0), dtype=float64)
In [29]: np.ones((m,0))
Out[29]: array([], shape=(5, 0), dtype=float64)
In [31]: a = np.vstack([(np.tile(np.array([np.inf]), (m, 0))),np.ones((m,0))])
In [32]: a
Out[32]: array([], shape=(10, 0), dtype=float64)
将（m，0）替换为（m，1），我们得到所需的（10,1）数组：
In [33]: a = np.vstack([(np.tile(np.array([np.inf]), (m, 1))),np.ones((m,1))])
stack添加一个尺寸-这不是您想要的尺寸：
In [35]: a = np.stack([(np.tile(np.array([np.inf]), (m, 1))),np.ones((m,1))])
In [36]: a.shape
Out[36]: (2, 5, 1)
尽管起始形状很简单，但我们得到了2行：
In [37]: a = np.stack([(np.tile(np.array([np.inf]), (m,))),np.ones((m))])
In [38]: a.shape
Out[38]: (2, 5)
实际上，您的stack的问题是放错位置的)。这是正确的：
In [50]: np.stack((np.tile(np.array([np.inf]), (m, 0)),np.ones((m,0))))
Out[50]: array([], shape=(2, 5, 0), dtype=float64)
您的结束）位于tile之后，因此np.ones()表达式位于axis参数位置。
答案 2 :(得分：0)
Numpy具有专用于填充数组的功能。参见“ full（）”，“ fill（）”。
a = np.full((2, 2), np.inf)
array([[inf, inf],
[inf, inf]])