This question already has answers here:
check for identical rows in different numpy arrays
(7 answers)
Closed 1 year ago.
I have two arrays:
A = np.array([[3, 1], [4, 1], [1, 4]])
B = np.array([[0, 1, 5], [2, 4, 5], [2, 3, 5]])
Is it possible to use numpy.isin rowwise for 2D arrays? I want to check if A[i,j] is in B[i] and return this result into C[i,j]. At the end I would get the following C:
np.array([[False, True], [True, False], [False, False]])
It would be great, if this is also doable with the == operator, then I could use it also with PyTorch.
Edit:
I also considered check for identical rows in different numpy arrays. The question is somehow similar, but I am not able to apply its solutions to this slightly different problem.
Not sure that my code solves your problem perfectly. please run it on more test cases to confirm. but i would do smth like i've done in the following code taking advantage of numpys vector outer operations ability (similar to vector outer product). If it works as intended it should work with pytorch as well.
import numpy as np
A = np.array([[3, 1], [4, 1], [1, 4]])
B = np.array([[0, 1, 5], [2, 4, 5], [2, 3, 5]])
AA = A.reshape(3, 2, 1)
BB = B.reshape(3, 1, 3)
(AA == BB).sum(-1).astype(bool)
output:
array([[False, True],
[ True, False],
[False, False]])
Updated answer
Here is a way to do this :
(A == B[..., None]).any(axis=1).astype(bool)
# > array([[False, True],
# [ True, False],
# [False, False]])
Previous answer
You could also do it inside a list comprehension:
[np.isin(a, b) for a,b in zip(A, B)]
# > [array([False, True]), array([ True, False]), array([False, False])]
np.array([np.isin(a, b) for a,b in zip(A, B)])
# > array([[False, True],
# [ True, False],
# [False, False]])
But, as #alex said it defeats the purpose of numpy.
Related
I can perform a boolean mask on an array of arrays like this
import numpy as np
a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
b = [[True, False, False], [True, True, False], [False, False, False]]
np.array(a)[np.array(b)]
and get array([1, 4, 5])
How would I preserve the information of which numbers belonged to the same array?
something like this would work
is_in_original(1, 4)
> False
is_in_origina(5, 4)
>True
One thing I could think of is this
def is_in_original(x, y):
for arry in np.array(a):
if x in arry and y in arry:
return True
return False
I am wondering if this is the most computationally efficient method. I will be working with very large array of arrays, and need the throughput to be as fast as possible.
You can use np.where(mask, array, 0) to preserve dimensions.
import numpy as np
a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
b = [[True, False, False], [True, True, False], [False, False, False]]
ret = np.where(np.array(b), np.array(a), 0)
Output:
array([[1, 0, 0],
[4, 5, 0],
[0, 0, 0]])
In this case you can change third parameter of np.where is 0, you can change the value to any number or inf
I am given a predefined array
a = [[2 3 4]
[5 6 7]
[8 9 10]]
my issue is converting this array into a boolean array where all even numbers are true. Any help is appreciated!
I would use numpy, and it's pretty straightforward:
import numpy as np
a = [[2, 3, 4],
[5, 6, 7],
[8, 9, 10]]
a = np.asarray(a)
(a % 2) == 0
Do as follows using Numpy:
import numpy as np
a = [[2, 3, 4],
[5, 6, 7],
[8, 9, 10]]
aBool=(np.asarray(a)%2==0)
The variable aBool will contain True where there is an Even value and False for an Odd value.
array([[ True, False, True],
[False, True, False],
[ True, False, True]])
This is adapted from the answer found here:
a=np.array([1,2,3,4,5,6])
answer = (a[a%2==0])
print(answer)
which is essentially what Andrew said but without using NumPy
If you're interested in getting booleans and testing for certain conditions within a NumPy array you can find a neat post here
This can be done simply using one line (if you don't want to use numpy):
[[i % 2 == 0 for i in sublist] for sublist in a]
>>> [[True, False, True], [False, True, False], [True, False, True]]
Here, i % 2 denotes the modulus operator, which gives the remainder when the first number is divided by the second. In this case, for even numbers i % 2 = 0, and for odd numbers i % 2 = 1. The two = signs means that the expression is evaluated as a boolean.
The two for loops iterate over each list inside of the 2D list.
This can be extended if you find this format easier to understand, but it essentially does the same thing:
>>> newlist = []
>>> for sublist in a:
partial_list = []
for i in sublist:
partial_list.append(i % 2 == 0)
newlist.append(partial_list)
>>> newlist
[[True, False, True], [False, True, False], [True, False, True]]
Why not to mention negations:
a = [[2, 3, 4],
[5, 6, 7],
[8, 9, 10]]
>>> ~np.array(a)%2
array([[1, 0, 1],
[0, 1, 0],
[1, 0, 1]], dtype=int32)
This is a boolean form:
>>> ~(np.array(a)%2).astype(bool)
array([[ True, False, True],
[False, True, False],
[ True, False, True]])
With 2 numpy 2-d arrays, I'd like a result array filled with row-by-row equality results. For example:
>>> a = np.array([[1, 2], [3, 4], [5, 6]])
>>> b = np.array([[5, 6], [3, 4], [1, 2]])
>>> a == b # not quite what I want
array([[False, False],
[ True, True],
[False, False]])
>>> np.equal(a, b) # also not quite
array([[False, False],
[ True, True],
[False, False]])
The result I want, each row's equality as one element, would be:
array([False, True, False])
What's the compact/idiomatic way to get this result?
You can use all() over axis=1 to turn each row of [bool, ...] into True if all-true, and False if any false.
For example:
>>> a = np.array([[1, 2], [3, 4], [5, 6]])
>>> b = np.array([[5, 6], [3, 4], [1, 2]])
>>> np.all(a == b, axis=1)
array([False, True, False])
I'm trying to find the row in which a 2d array appears in a 3d numpy ndarray. Here's an example of what I mean. Give:
arr = [[[0, 3], [3, 0]],
[[0, 0], [0, 0]],
[[3, 3], [3, 3]],
[[0, 3], [3, 0]]]
I'd like to find all occurrences of:
[[0, 3], [3, 0]]
The result I'd like is:
[0, 3]
I tried to use argwhere but that unfortunately got me nowhere. Any ideas?
Try
np.argwhere(np.all(arr==[[0,3], [3,0]], axis=(1,2)))
How it works:
arr == [[0,3], [3,0]] returns
array([[[ True, True],
[ True, True]],
[[ True, False],
[False, True]],
[[False, True],
[ True, False]],
[[ True, True],
[ True, True]]], dtype=bool)
This is a three dimensional array where the innermost axis is 2. The values at this axis are:
[True, True]
[True, True]
[True, False]
[False, True]
[False, True]
[True, False]
[True, True]
[True, True]
Now with np.all(arr==[[0,3], [3,0]], axis=2) you are checking if both elements on a row are True and its shape will be reduced to (4, 2) from (4, 2, 2). Like this:
array([[ True, True],
[False, False],
[False, False],
[ True, True]], dtype=bool)
You need one more step of reducing as you want both of them to be the same (both [0, 3] and [3, 0]. You can do it either by reducing on the result (now the innermost axis is 1):
np.all(np.all(test, axis = 2), axis=1)
Or you can also do it by giving a tuple for the axis parameter to do the same thing step by step (first innermost, then one step higher). The result will be:
array([ True, False, False, True], dtype=bool)
The 'contains' function in the numpy_indexed package (disclaimer: I am its author) can be used to make queries of this kind. It implements a solution similar to the one offered by Saullo.
import numpy_indexed as npi
test = [[[0, 3], [3, 0]]]
# check which elements of arr are present in test (checked along axis=0 by default)
flags = npi.contains(test, arr)
# if you want the indexes:
idx = np.flatnonzero(flags)
In you can use np.in1d after defining a new data type which will have the memory size of each row in your arr. To define such data type:
mydtype = np.dtype((np.void, arr.dtype.itemsize*arr.shape[1]*arr.shape[2]))
then you have to convert your arr to a 1-D array where each row will have arr.shape[1]*arr.shape[2] elements:
aView = np.ascontiguousarray(arr).flatten().view(mydtype)
You are now ready to look for your 2-D array pattern [[0, 3], [3, 0]] which also has to be converted to dtype:
bView = np.array([[0, 3], [3, 0]]).flatten().view(mydtype)
You can now check the occurrencies of bView in aView:
np.in1d(aView, bView)
#array([ True, False, False, True], dtype=bool)
This mask is easily converted to indices using np.where, for example.
Timings (updated)
THe following function is used to implement this approach:
def check2din3d(b, a):
"""
Return where `b` (2D array) appears in `a` (3D array) along `axis=0`
"""
mydtype = np.dtype((np.void, a.dtype.itemsize*a.shape[1]*a.shape[2]))
aView = np.ascontiguousarray(a).flatten().view(mydtype)
bView = np.ascontiguousarray(b).flatten().view(mydtype)
return np.in1d(aView, bView)
The updated timings considering #ayhan comments showed that this method can be faster the np.argwhere, but the different is not significant and for large arrays like below, #ayhan's approach is considerably faster:
arrLarge = np.concatenate([arr]*10000000)
arrLarge = np.concatenate([arrLarge]*10, axis=2)
pattern = np.ascontiguousarray([[0,3]*10, [3,0]*10])
%timeit np.argwhere(np.all(arrLarger==pattern, axis=(1,2)))
#1 loops, best of 3: 2.99 s per loop
%timeit check2din3d(pattern, arrLarger)
#1 loops, best of 3: 4.65 s per loop
This question already has answers here:
find and delete from more-dimensional numpy array
(3 answers)
Closed 9 years ago.
I have two numpy arrays that represent 2D coordinates. Each row represents (x, y) pairs:
a = np.array([[1, 1], [2, 1], [3, 1], [3, 2], [3, 3], [5, 5]])
b = np.array([[1, 1], [5, 5], [3, 2]])
I would like to remove elements from a which are in b efficiently. So the result would be:
array([[2, 1], [3, 1], [3, 3]])
I can do it by looping and comparing, I was hoping I could do it easier.
Python sets does a nice job of giving differences. It doesn't, though, maintain order
np.array(list(set(tuple(x) for x in a.tolist()).difference(set(tuple(x) for x in b.tolist()))))
Or to use boolean indexing, use broadcasting to create an outer equals, and sum with any and all
A = np.all((a[None,:,:]==b[:,None,:]),axis=-1)
A = np.any(A,axis=0)
a[~A,:]
Or make a and b complex:
ac = np.dot(a,[1,1j])
bc = np.dot(b,[1,1j])
A = np.any(ac==bc[:,None],axis=0)
a[~A,:]
or to use setxor1d
xx = np.setxor1d(ac,bc)
# array([ 2.+1.j, 3.+1.j, 3.+3.j])
np.array([xx.real,xx.imag],dtype=int).T
=================
In [222]: ac = np.dot(a,[1,1j])
...: bc = np.dot(b,[1,1j])
In [223]: ac
Out[223]: array([ 1.+1.j, 2.+1.j, 3.+1.j, 3.+2.j, 3.+3.j, 5.+5.j])
In [225]: bc
Out[225]: array([ 1.+1.j, 5.+5.j, 3.+2.j])
In [226]: ac == bc[:,None]
Out[226]:
array([[ True, False, False, False, False, False],
[False, False, False, False, False, True],
[False, False, False, True, False, False]], dtype=bool)