Suppose that I have define one 2x2 matrix using numpy:
array([[1, 2],
[2, 3]])
Now the other 2x1 matrix:
array([[3],
[4]])
How can I concatenate these 2 matrix by column, so that it will become:
array([[1, 2, 3],
[2, 3, 4]])
And how can I also delete the specify column, so that it will became
array([[1],
[2]])
There is a numpy.concatenate method
import numpy as np
np.concatenate( [ np.array( [ [1,2], [2,3] ] ), np.array( [ [3],[4] ] ) ] , axis = 1)
or simply use hstack or vstack
np.hstack( [ np.array( [ [1,2], [2,3] ] ), np.array( [ [3],[4] ] ) ] )
These can be also used to remove the column (concatenate two subarrays) - this can be used to remove many columns.
To remove i'th column you can take subarrays to this column, and from the next one, and concatenate them. For example, to remove second column (index 1):
a - np.array( [ [1,2,3], [2,3,4] ] )
a1= a[:,:1]
a2= a[:,2:]
np.hstack([a1,a2])
so in general
def remove_column( a, i ):
return np.hstack( [a[:,:i], a[:,(i+1):] ] )
and then
>>> remove_column(a, 1)
array([[1, 3],
[2, 4]])
>>> remove_column(a, 0)
array([[2, 3],
[3, 4]])
Actually, as pointed out in the comment - numpy implements its own delete method
np.delete(a, 1, 1)
deleted second column
and deleting multiple ones can be performed using
np.delete(a, [column1, columne2, ..., columnK], 1)
The third argument is the axis specifier, 0 would imply rows, 1 columns, None flatterns the whole array
You can use numpy.hstack:
>>> import numpy as np
>>> a = np.array([[1,2], [2,3]])
>>> b = np.array([[3], [4]])
>>> np.hstack((a,b))
array([[1, 2, 3],
[2, 3, 4]])
Removing is even easier, just slice:
>>> c = a[:,:1]
array([[1],
[2]])
In [3]: x = np.array([[1, 2], [2, 3]]
In [4]: y = np.array([[3], [4]])
In [9]: z = np.hstack([x, y])
In [10]: z
Out[10]:
array([[1, 2, 3],
[2, 3, 4]])
In [11]: z[:,:1]
array([[1],
[2]])
Related
I am learning numpy framework.This piece of code I don't understand.
import numpy as np
a =np.array([[0,1,2],[3,4,5],[6,7,8],[9,10,11]])
print(a)
row = np.array([[0,0],[3,3]])
col = np.array([[0,2],[0,2]])
b = a[row,col]
print("This is b array:",b)
This b array returns the corner values of a array, that is, b equals [[0,2],[9,11]].
When indexing is done using an array or "array-like", to access/modify the elements of an array, then it's called advanced indexing.
In [37]: a
Out[37]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
In [38]: row
Out[38]:
array([[0, 0],
[3, 3]])
In [39]: col
Out[39]:
array([[0, 2],
[0, 2]])
In [40]: a[row, col]
Out[40]:
array([[ 0, 2],
[ 9, 11]])
That's what you got. Below is an explanation:
Indices of
`a[row, col]` row column
|| || || ||
VV VV VV VV
a[0, 0] a[0, 2]
a[3, 0] a[3, 2]
|__________| |
row-idx array |
|__________|
column-idx array
You're indexing a using two equally shaped 2d-arrays, hence you're output array will also have the same shape as col and row. To better understand how array indexing works you can check the docs, where as shown, indexing with 1d-arrays over the existing axis' of a given array works as follows:
result[i_1, ..., i_M] == x[ind_1[i_1, ..., i_M], ind_2[i_1, ..., i_M],
..., ind_N[i_1, ..., i_M]]
Where the same logic applies in the case of indexing with 2d-arrays over each axis, but instead you'd have a result array with up to i_N_M indices.
So going back to your example you are essentially selecting from the rows of a based on row, and from those rows you are selecting some columns col. You might find it more intuitive to translate the row and column indices into (x,y) coordinates:
(0,0), (0,2)
(3,0), (3,2)
Which, by accordingly selecting from a, results in the output array:
print(a[row,col])
array([[ 0, 2],
[ 9, 11]])
You can understand it by making more tries, to see more examples.
If you have one dimensional index:
In [58]: np.arange(10)[np.array([1,3,4,6])]
Out[58]: array([1, 3, 4, 6])
In case of two dimensional index:
In [57]: np.arange(10)[np.array([[1,3],[4,6]])]
Out[57]:
array([[1, 3],
[4, 6]])
If you use 3 dimensional index:
In [59]: np.arange(10)[np.array([[[1],[3]],[[4],[6]]])]
Out[59]:
array([[[1],
[3]],
[[4],
[6]]])
As you can see, if you make hierarchy in indexing, you will get it in the output as well.
Proceeding by steps:
import numpy as np
a = np.array([[0,1,2],[3,4,5],[6,7,8],[9,10,11]])
print(a)
gives 2d array a:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
Then:
row = np.array([[0,0],[3,3]])
assigns to 2d array row values [0,0] and [3,3]:
array([[0, 0],
[3, 3]])
Then:
col = np.array([[0,2],[0,2]])
assigns to 2d array col values [0,2] and [0,2]:
array([[0, 2],
[0, 2]])
Finally:
b = a[row,col]
assigns to b values given by a[0,0], a[0,2] for the first row, a[3,0], a[3,2] for the second row, that is:
array([[ 0, 2],
[ 9, 11]])
Where does b[0,0] <-- a[0,0] come from? It comes from the combination of row[0,0] which is 0 and col[0,0] which is 0.
What about b[0,1] <-- a[0,2]? It comes from the combination of row[0,1] which is 0 and col[0,1] which is 2.
And so forth.
I have a 2D array:
>>> in_arr = np.array([[1,2],[4,3]])
array([[1, 2],
[4, 3]])
and I find the sorted indices by columns to yield another 2D array:
>>> col_sort = np.argsort(in_arr, axis=1)
array([[0, 1],
[1, 0]])
I would like to know the efficient numpy slice to index the first by the second:
>>> redordered_in_arr = np.*SOME_SLICE_METHOD*(in_arr, col_sort, axis=1)
array([[1, 2],
[3, 4]])
The intention is to then perform a (more complicated) function on the array by column, e.g.:
>>> arr_with_function = reordered_in_arr ** np.array([1,2])
array([[1, 4],
[3, 16]])
and return the elements to their original position in the array
>>> return_order = np.argsort(col_sort, axis=1)
>>> redordered_in_arr = np.*SOME_SLICE_METHOD*(arr_with_function, return_order, axis=1)
array([[1, 4],
[16, 3]])
Ok so thinking about it as I type I might just use apply_over_axis, but I would still like know how to the above efficiently in case it is of value later..
If you want to do all those operations in-place then you don't need argsort(). Numpy supports in-place operations in such situations:
In [12]: in_arr = np.array([[1,2],[4,3]])
In [13]: in_arr.sort(axis=1)
In [14]: in_arr **= [1, 2]
In [15]: in_arr
Out[15]:
array([[ 1, 4],
[ 3, 16]])
But if you need the indices of the sorted items you can get the expected result with a simple indexing.
In [18]: in_arr[np.arange(2)[:,None], col_sort]
Out[18]:
array([[1, 2],
[3, 4]])
I want to create a numpy array within a numpy array. If i do it with normal python its something like
a = [[1,2], [3,4]]
a[0][1] = [1,1,1]
print a
The result is [[1, [1, 1, 1]], [3, 4]]
How can I achieve the same using numpy arrays? The code I have is:
a = np.array([(1, 2, 3),(4, 5, 6)])
b = np.array([1,1,1])
a[0][1] = b
a as created is dtype int. Each element can only be another integer:
In [758]: a = np.array([(1, 2, 3),(4, 5, 6)])
...: b = np.array([1,1,1])
...:
In [759]: a
Out[759]:
array([[1, 2, 3],
[4, 5, 6]])
In [760]: b
Out[760]: array([1, 1, 1])
In [761]: a[0,1]=b
...
ValueError: setting an array element with a sequence.
You can make another dtype of array, one that holds pointers to objects, much as list does:
In [762]: aO = a.astype(object)
In [763]: aO
Out[763]:
array([[1, 2, 3],
[4, 5, 6]], dtype=object)
Now it is possible to replace one of those element pointers with a pointer to b array:
In [765]: aO[0,1]=b
In [766]: aO
Out[766]:
array([[1, array([1, 1, 1]), 3],
[4, 5, 6]], dtype=object)
But as asked in the comments - why do you want/need to do this? What are you going to do with such an array? It is possible to do some numpy math on such an array, but as shown in some recent SO questions, it is hit-or-miss. It is also slower.
As far as I know, you cannot do this. Numpy arrays cannot have entries of varying shape. Your request to make an array like [[1, [1, 1, 1]], [3, 4]] is impossible. However, you could make a numpy matrix of dimensions (3x2x3) to get
[
[
[1,0,0],
[1,1,1],
[0,0,0],
]
[
[3,0,0],
[4,0,0],
[0,0,0]
]
]
Your only option is to pad empty elements with some number (I used 0s above) or use another data structure.
problem is very simple: I have two 2d np.array and I want to get a third array that only contains the rows that are not in common with the latter twos.
for example:
X = np.array([[0,1],[1,2],[4,5],[5,6],[8,9],[9,10]])
Y = np.array([[5,6],[9,10]])
Z = function(X,Y)
Z = array([[0, 1],
[1, 2],
[4, 5],
[8, 9]])
I tried np.delete(X,Y,axis=0) but it doesn't work...
Z = np.vstack(row for row in X if row not in Y)
The numpy_indexed package (disclaimer: I am its author) extends the standard numpy array set operations to multi-dimensional use cases such as these, with good efficiency:
import numpy_indexed as npi
Z = npi.difference(X, Y)
Here's a views based approach -
# Based on http://stackoverflow.com/a/41417343/3293881 by #Eric
def setdiff2d(a, b):
# check that casting to void will create equal size elements
assert a.shape[1:] == b.shape[1:]
assert a.dtype == b.dtype
# compute dtypes
void_dt = np.dtype((np.void, a.dtype.itemsize * np.prod(a.shape[1:])))
orig_dt = np.dtype((a.dtype, a.shape[1:]))
# convert to 1d void arrays
a = np.ascontiguousarray(a)
b = np.ascontiguousarray(b)
a_void = a.reshape(a.shape[0], -1).view(void_dt)
b_void = b.reshape(b.shape[0], -1).view(void_dt)
# Get indices in a that are also in b
return np.setdiff1d(a_void, b_void).view(orig_dt)
Sample run -
In [81]: X
Out[81]:
array([[ 0, 1],
[ 1, 2],
[ 4, 5],
[ 5, 6],
[ 8, 9],
[ 9, 10]])
In [82]: Y
Out[82]:
array([[ 5, 6],
[ 9, 10]])
In [83]: setdiff2d(X,Y)
Out[83]:
array([[0, 1],
[1, 2],
[4, 5],
[8, 9]])
Z = np.unique([tuple(row) for row in X + Y])
How do I select a row from a NxM numpy array as an array of size 1xM:
> import numpy
> a = numpy.array([[1,2], [3,4], [5,6]])
> a
array([[1, 2],
[3, 4],
[5, 6]])
> a.shape
(e, 2)
> a[0]
array([1, 2])
> a[0].shape
(2,)
I'd like
a[0].shape == (1,2)
I'm doing this because a library I want to use seems to require this.
If you have something of shape (2,) and you want to add a new axis so that the shape is (1,2), the easiest way is to use np.newaxis:
a = np.array([1,2])
a.shape
#(2,)
b = a[np.newaxis, :]
print b
#array([[1,2]])
b.shape
#(1,2)
If you have something of shape (N,2) and want to slice it with the same dimensionality to get a slice with shape (1,2), then you can use a range of length 1 as your slice instead of one index:
a = np.array([[1,2], [3,4], [5,6]])
a[0:1]
#array([[1, 2]])
a[0:1].shape
#(1,2)
Another trick is that some functions have a keepdims option, for example:
a
#array([[1, 2],
# [3, 4],
# [5, 6]])
a.sum(1)
#array([ 3, 7, 11])
a.sum(1, keepdims=True)
#array([[ 3],
# [ 7],
# [11]])
If you already have it, call .reshape():
>>> a = numpy.array([[1, 2], [3, 4]])
>>> b = a[0]
>>> c = b.reshape((1, -1))
>>> c
array([[1, 2]])
>>> c.shape
(1, 2)
You can also use a range to keep the array two-dimensional in the first place:
>>> b = a[0:1]
>>> b
array([[1, 2]])
>>> b.shape
(1, 2)
Note that all of these will have the same backing store.