[英]Column_match([[1],[1,1]]) <— how to make dimensions match with NA values?
Any flag for this? 为此有任何标志吗? Please, see the intended.
请看预期的。
>>> numpy.column_stack([[1], [1,2]])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/pymodules/python2.7/numpy/lib/shape_base.py", line 296, in column_stack
return _nx.concatenate(arrays,1)
ValueError: array dimensions must agree except for d_0
Input 输入项
[[1],[1,2]]
Intended Output 预期输出
[[NA,1], [1,2]]
In general 一般来说
[[1],[2,2],[3,3,3],...,[n,n,n,n,n...,n]]
to 至
[[NA, NA, NA,..., NA,1], [NA, NA, ..., 2, 2], ...[n,n,n,n,n]]
where the columns may be a triangluar zero matrix initially. 其中列最初可能是三角三角零矩阵。 Yes you can understand the term NA as None.
是的,您可以将术语“ NA”理解为“无”。 I got the triangular matrix almost below.
我几乎在下面得到了三角形矩阵。
>>> a=[[1],[2,2],[3,3,3]]
>>> a
[[1], [2, 2], [3, 3, 3]]
>>> len(a)
3
>>> [aa+['']*(N-len(aa)) for
...
KeyboardInterrupt
>>> N=len(a)
>>> [aa+['']*(N-len(aa)) for aa in a]
[[1, '', ''], [2, 2, ''], [3, 3, 3]]
>>> transpose([aa+['']*(N-len(aa)) for aa in a])
array([['1', '2', '3'],
['', '2', '3'],
['', '', '3']],
dtype='|S4')
a pure numpy solution: 一个纯粹的numpy解决方案:
>>> lili = [[1],[2,2],[3,3,3],[4,4,4,4]]
>>> y = np.nan*np.ones((4,4))
>>> y[np.tril_indices(4)] = np.concatenate(lili)
>>> y
array([[ 1., nan, nan, nan],
[ 2., 2., nan, nan],
[ 3., 3., 3., nan],
[ 4., 4., 4., 4.]])
>>> y[:,::-1]
array([[ nan, nan, nan, 1.],
[ nan, nan, 2., 2.],
[ nan, 3., 3., 3.],
[ 4., 4., 4., 4.]])
I'm not sure which triangular array you want, there is also np.triu_indices 我不确定您想要哪个三角形数组,也有np.triu_indices
(maybe not always faster, but easy to read) (也许并不总是更快,但易于阅读)
column_stack adds a column to an array. column_stack将一列添加到数组。 That column is supposed to be a smalled (1D) array.
该列应该是缩小(1D)数组。
When I try : 当我尝试:
from numpy import *
x = array([0])
z = array([1, 2])
if you do this : 如果您这样做:
r = column_stack ((x,z))
You'll get this : 您将得到:
>>> array([0,1,2])
So, in order to add a column to your first array, maybe this : 因此,为了将列添加到您的第一个数组,也许可以这样:
n = array([9])
arr = ([column_stack((n, x))], z)
It shows up this : 它显示如下:
>>> arr
([array([[9, 0]])], array([[1, 2]]))
It has the same look as your "intended output" 它的外观与您的“预期输出”相同
Hope this was helpful ! 希望这对您有所帮助!
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