I'd like to turn an open mesh returned by the numpy ix_ routine to a list of coordinates
eg, for:
In[1]: m = np.ix_([0, 2, 4], [1, 3])
In[2]: m
Out[2]:
(array([[0],
[2],
[4]]), array([[1, 3]]))
What I would like is:
([0, 1], [0, 3], [2, 1], [2, 3], [4, 1], [4, 3])
I'm pretty sure I could hack it together with some iterating, unpacking and zipping, but I'm sure there must be a smart numpy way of achieving this...
Approach #1 Use np.meshgrid and then stack -
r,c = np.meshgrid(*m)
out = np.column_stack((r.ravel('F'), c.ravel('F') ))
Approach #2 Alternatively, with np.array() and then transposing, reshaping -
np.array(np.meshgrid(*m)).T.reshape(-1,len(m))
For a generic case with for generic number of arrays used within np.ix_, here are the modifications needed -
p = np.r_[2:0:-1,3:len(m)+1,0]
out = np.array(np.meshgrid(*m)).transpose(p).reshape(-1,len(m))
Sample runs -
Two arrays case :
In [376]: m = np.ix_([0, 2, 4], [1, 3])
In [377]: p = np.r_[2:0:-1,3:len(m)+1,0]
In [378]: np.array(np.meshgrid(*m)).transpose(p).reshape(-1,len(m))
Out[378]:
array([[0, 1],
[0, 3],
[2, 1],
[2, 3],
[4, 1],
[4, 3]])
Three arrays case :
In [379]: m = np.ix_([0, 2, 4], [1, 3],[6,5,9])
In [380]: p = np.r_[2:0:-1,3:len(m)+1,0]
In [381]: np.array(np.meshgrid(*m)).transpose(p).reshape(-1,len(m))
Out[381]:
array([[0, 1, 6],
[0, 1, 5],
[0, 1, 9],
[0, 3, 6],
[0, 3, 5],
[0, 3, 9],
[2, 1, 6],
[2, 1, 5],
[2, 1, 9],
[2, 3, 6],
[2, 3, 5],
[2, 3, 9],
[4, 1, 6],
[4, 1, 5],
[4, 1, 9],
[4, 3, 6],
[4, 3, 5],
[4, 3, 9]])
Related
I have two numpy matrices (6 rows and 3 columns) :
a = np.array([[1,2,4],[3,6,2],[3,4,7],[9,7,7],[6,3,1],[3,5,9]])
b = np.array([[4,5,2],[9,2,5],[1,5,6],[4,5,6],[1,2,6],[6,4,3]])
a = array([[1, 2, 4],
[3, 6, 2],
[3, 4, 7],
[9, 7, 7],
[6, 3, 1],
[3, 5, 9]])
b = array([[4, 5, 2],
[9, 2, 5],
[1, 5, 6],
[4, 5, 6],
[1, 2, 6],
[6, 4, 3]])
I would like to calculate the pearson correlation coefficient between the first column of a and b, the second column of a and b and the third column of a and b.
The result would be a vector of 3 (3 correlation coeff).
One way using numpy.corrcoef and diagonal:
corr = np.corrcoef(a.T, b.T).diagonal(a.shape[1])
corr
Output:
array([-0.2324843 , -0.03631365, -0.18057878])
I'm trying to concatenate 2 arrays element wise. I have the concatenation working to produce the correct shape but it has not been applied element wise.
So i have this array
[0, 1]
[2, 3]
[4, 5]
I want to append each element in the array with each element. the target result would be
[0, 1, 0, 1]
[0, 1, 2, 3]
[0, 1, 4, 5]
[2, 3, 0, 1]
[2, 3, 2, 3]
[2, 3, 4, 5]
[4, 5, 0, 1]
[4, 5, 2, 3]
[4, 5, 4, 5]
i think i may need to change an axis but then i can't get the broadcasting to work.
any help would be greatly appreciated. lots to learn in numpy !
a = np.arange(6).reshape(3, 2))
b = np.concatenate((a, a), axis=1)
One way would be stacking replicated versions created with np.repeat and np.tile -
In [52]: n = len(a)
In [53]: np.hstack((np.repeat(a,n,axis=0),np.tile(a,(n,1))))
Out[53]:
array([[0, 1, 0, 1],
[0, 1, 2, 3],
[0, 1, 4, 5],
[2, 3, 0, 1],
[2, 3, 2, 3],
[2, 3, 4, 5],
[4, 5, 0, 1],
[4, 5, 2, 3],
[4, 5, 4, 5]])
Another would be with broadcasted-assignment, since you mentioned broadcasting -
def create_mesh(a):
m,n = a.shape
out = np.empty((m,m,2*n),dtype=a.dtype)
out[...,:n] = a[:,None]
out[...,n:] = a
return out.reshape(-1,2*n)
One solution is to build on senderle's cartesian_product to extend this to 2D arrays. Here's how I usually do this:
# Your input array.
arr
# array([[0, 1],
# [2, 3],
# [4, 5]])
idxs = cartesian_product(*[np.arange(len(arr))] * 2)
arr[idxs].reshape(idxs.shape[0], -1)
# array([[0, 1, 0, 1],
# [0, 1, 2, 3],
# [0, 1, 4, 5],
# [2, 3, 0, 1],
# [2, 3, 2, 3],
# [2, 3, 4, 5],
# [4, 5, 0, 1],
# [4, 5, 2, 3],
# [4, 5, 4, 5]])
Suppose I have a numpy array as follows:
data = np.array([[1, 3, 8, np.nan], [np.nan, 6, 7, 9], [np.nan, 0, 1, 2], [5, np.nan, np.nan, 2]])
I would like to randomly select n-valid items from the array, including their indices.
Does numpy provide an efficient way of doing this?
Example
data = np.array([[1, 3, 8, np.nan], [np.nan, 6, 7, 9], [np.nan, 0, 1, 2], [5, np.nan, np.nan, 2]])
n = 5
Get valid indices
y_val, x_val = np.where(~np.isnan(data))
n_val = y_val.size
Pick random subset of size n by index
pick = np.random.choice(n_val, n)
Apply index to valid coordinates
y_pick, x_pick = y_val[pick], x_val[pick]
Get corresponding data
data_pick = data[y_pick, x_pick]
Admire
data_pick
# array([2., 8., 1., 1., 2.])
y_pick
# array([3, 0, 0, 2, 3])
x_pick
# array([3, 2, 0, 2, 3])
Find nonzeros by :
In [37]: a = np.array(np.nonzero(data)).reshape(-1,2)
In [38]: a
Out[38]:
array([[0, 0],
[0, 0],
[1, 1],
[1, 1],
[2, 2],
[2, 3],
[3, 3],
[3, 0],
[1, 2],
[3, 0],
[1, 2],
[3, 0],
[2, 3],
[0, 1],
[2, 3]])
Now pick a random choice :
In [44]: idx = np.random.choice(np.arange(len(a)))
In [45]: data[a[idx][0],a[idx][1]]
Out[45]: 2.0
Can someone explain to me what the second line of this code does?
objp = np.zeros((48,3), np.float32)
objp[:,:2] = np.mgrid[0:8,0:6].T.reshape(-1,2)
Can someone explain to me what exactly the np.mgrid[0:8,0:6] part of the code is doing and what exactly the T.reshape(-1,2) part of the code is doing?
Thanks and good job!
The easiest way to see these is to use smaller values for mgrid:
In [11]: np.mgrid[0:2,0:3]
Out[11]:
array([[[0, 0, 0],
[1, 1, 1]],
[[0, 1, 2],
[0, 1, 2]]])
In [12]: np.mgrid[0:2,0:3].T # (matrix) transpose
Out[12]:
array([[[0, 0],
[1, 0]],
[[0, 1],
[1, 1]],
[[0, 2],
[1, 2]]])
In [13]: np.mgrid[0:2,0:3].T.reshape(-1, 2) # reshape to an Nx2 matrix
Out[13]:
array([[0, 0],
[1, 0],
[0, 1],
[1, 1],
[0, 2],
[1, 2]])
Then objp[:,:2] = sets the 0th and 1th columns of objp to this result.
The second line creates a multi-dimensional mesh grid, transposes it, reshapes it so that it represents two columns and inserts it into the first two columns of the objp array.
Breakdown:
np.mgrid[0:8,0:6] creates the following mgrid:
>> np.mgrid[0:8,0:6]
array([[[0, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1],
[2, 2, 2, 2, 2, 2],
[3, 3, 3, 3, 3, 3],
[4, 4, 4, 4, 4, 4],
[5, 5, 5, 5, 5, 5],
[6, 6, 6, 6, 6, 6],
[7, 7, 7, 7, 7, 7]],
[[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5],
[0, 1, 2, 3, 4, 5]]])
The .T transposes the matrix, and the .reshape(-1,2) then reshapes it into two a two-column array shape. These two columns are then the correct shape to replace two columns in the original array.
How the result can be obtained from given NumPy array (xx and yy)?
>>> xx, yy = np.mgrid[0:2, 5:7]
>>> xx
array([[0, 0],
[1, 1]])
>>> yy
array([[5, 6],
[5, 6]])
>>> result = [(0,5), (1,5), (1,6), (0,6)]
>>> result
[(0, 5), (1, 5), (1, 6), (0, 6)]
>>>
The order in your example requires some fancy indexing of xx. I had to reverse the order of the 2nd column.
In [243]: np.array([np.array([xx[:,0], xx[::-1,1]]).flatten(), yy.T.flatten()]).T.tolist()
Out[243]: [[0, 5], [1, 5], [1, 6], [0, 6]]
If the order isn't so important, then we can treat xx just like yy:
In [256]: xx, yy = np.mgrid[0:3, 5:8]
In [257]: np.array([xx.T.flatten(),yy.T.flatten()]).T.tolist()
Out[257]: [[0, 5], [1, 5], [2, 5], [0, 6], [1, 6], [2, 6], [0, 7], [1, 7], [2, 7]]
In [258]: np.array([xx.flatten(),yy.flatten()]).T.tolist()
Out[258]: [[0, 5], [0, 6], [0, 7], [1, 5], [1, 6], [1, 7], [2, 5], [2, 6], [2, 7]]
In [264]: np.array([xx,yy]).reshape(2,-1).T.tolist()
Out[264]: [[0, 5], [0, 6], [0, 7], [1, 5], [1, 6], [1, 7], [2, 5], [2, 6], [2, 7]]
In [272]: np.dstack([xx,yy]).reshape(-1,2).tolist()
Out[272]: [[0, 5], [0, 6], [0, 7], [1, 5], [1, 6], [1, 7], [2, 5], [2, 6], [2, 7]]
In [302]: list(np.broadcast(*np.ogrid[0:3,5:8]))
Out[302]: [(0, 5), (0, 6), (0, 7), (1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)]