Find unique columns and column membership

问题

I went through these threads:

Find unique rows in numpy.array
Removing duplicates in each row of a numpy array
Pandas: unique dataframe

and they all discuss several methods for computing the matrix with unique rows and columns.

However, the solutions look a bit convoluted, at least to the untrained eye. Here is for example top solution from the first thread, which (correct me if I am wrong) I believe it is the safest and fastest:

np.unique(a.view(np.dtype((np.void, a.dtype.itemsize*a.shape[1])))).view(a.dtype).reshape(-1, 
a.shape[1])

Either way, the above solution only returns the matrix of unique rows. What I am looking for is something along the original functionality of np.unique

u, indices = np.unique(a, return_inverse=True)

which returns, not only the list of unique entries, but also the membership of each item to each unique entry found, but how can I do this for columns?

Here is an example of what I am looking for:

array([[0, 2, 0, 2, 2, 0, 2, 1, 1, 2],
       [0, 1, 0, 1, 1, 1, 2, 2, 2, 2]])

We would have:

u       = array([0,1,2,3,4])
indices = array([0,1,0,1,1,3,4,4,3])

Where the different values in u represent the set of unique columns in the original array:

0 -> [0,0]
1 -> [2,1]
2 -> [0,1]
3 -> [2,2]
4 -> [1,2]

回答1:

Essentially, you want np.unique to return the indexes of the unique columns, and the indices of where they're used? This is easy enough to do by transposing the matrix and then using the code from the other question, with the addition of return_inverse=True.

at = a.T
b = np.ascontiguousarray(at).view(np.dtype((np.void, at.dtype.itemsize * at.shape[1])))
_, u, indices = np.unique(b, return_index=True, return_inverse=True)

With your a, this gives:

In [35]: u
Out[35]: array([0, 5, 7, 1, 6])

In [36]: indices
Out[36]: array([0, 3, 0, 3, 3, 1, 4, 2, 2, 4])

It's not entirely clear to me what you want u to be, however. If you want it to be the unique columns, then you could use the following instead:

at = a.T
b = np.ascontiguousarray(at).view(np.dtype((np.void, at.dtype.itemsize * at.shape[1])))
_, idx, indices = np.unique(b, return_index=True, return_inverse=True)
u = a[:,idx]

This would give

In [41]: u
Out[41]:
array([[0, 0, 1, 2, 2],
       [0, 1, 2, 1, 2]])

In [42]: indices
Out[42]: array([0, 3, 0, 3, 3, 1, 4, 2, 2, 4])

回答2:

First lets get the unique indices, to do so we need to start by transposing your array:

>>> a=a.T

Using a modified version of the above to get unique indices.

>>> ua, uind = np.unique(np.ascontiguousarray(a).view(np.dtype((np.void,a.dtype.itemsize * a.shape[1]))),return_inverse=True)

>>> uind
array([0, 3, 0, 3, 3, 1, 4, 2, 2, 4])

#Thanks to @Jamie
>>> ua = ua.view(a.dtype).reshape(ua.shape + (-1,))
>>> ua
array([[0, 0],
       [0, 1],
       [1, 2],
       [2, 1],
       [2, 2]])

For sanity:

>>> np.all(a==ua[uind])
True

To reproduce your chart:

>>> for x in range(ua.shape[0]):
...     print x,'->',ua[x]
...
0 -> [0 0]
1 -> [0 1]
2 -> [1 2]
3 -> [2 1]
4 -> [2 2]

To do exactly what you ask, but will be a bit slower if it has to convert the array:

>>> b=np.asfortranarray(a).view(np.dtype((np.void,a.dtype.itemsize * a.shape[0])))
>>> ua,uind=np.unique(b,return_inverse=True)
>>> uind
array([0, 3, 0, 3, 3, 1, 4, 2, 2, 4])
>>> ua.view(a.dtype).reshape(ua.shape+(-1,),order='F')
array([[0, 0, 1, 2, 2],
       [0, 1, 2, 1, 2]])

#To return this in the previous order.
>>> ua.view(a.dtype).reshape(ua.shape + (-1,))

回答3:

Not entirely sure what you are after, but have a look at the numpy_indexed package (disclaimer: I am its author); it is sure to make problems of this kind easier:

import numpy_indexed as npi
unique_columns = npi.unique(A, axis=1)
# or perhaps this is what you want?
unique_columns, indices = npi.group_by(A.T, np.arange(A.shape[1])))

来源：https://stackoverflow.com/questions/18197071/find-unique-columns-and-column-membership

标签

python

numpy

unique