Using np.where to find element from sub-arrays
I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:
X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
Y = np.array([1, 2, 3, 4, 4, 5])
a = [2, 3, 4]
matchedX = X[np.where(Y == a)]
I am expecting to get the result like this:
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
but I got different results:
array(, shape=(0, 2), dtype=int64)
So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.
matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]
python numpy
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
add a comment |
I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:
X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
Y = np.array([1, 2, 3, 4, 4, 5])
a = [2, 3, 4]
matchedX = X[np.where(Y == a)]
I am expecting to get the result like this:
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
but I got different results:
array(, shape=(0, 2), dtype=int64)
So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.
matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]
python numpy
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
add a comment |
I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:
X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
Y = np.array([1, 2, 3, 4, 4, 5])
a = [2, 3, 4]
matchedX = X[np.where(Y == a)]
I am expecting to get the result like this:
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
but I got different results:
array(, shape=(0, 2), dtype=int64)
So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.
matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]
python numpy
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:
X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
Y = np.array([1, 2, 3, 4, 4, 5])
a = [2, 3, 4]
matchedX = X[np.where(Y == a)]
I am expecting to get the result like this:
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
but I got different results:
array(, shape=(0, 2), dtype=int64)
So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.
matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]
python numpy
python numpy
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
asked Nov 17 '18 at 0:04
Esraa MohamedEsraa Mohamed
234
234
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can use the set functions of numpy:
X[np.where(np.isin(Y, a))]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
You can skip the np.where, which is redundant here, and just index using np.isin:
X[np.isin(Y,a)]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
This is because np.isin gives you a boolean array of where Y is in a:
array([False, True, True, True, True, False])
So by indexing with this array, it only selects those rows where True
answered Nov 17 '18 at 0:16
saculsacul
30k41740
1
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use the set functions of numpy:
X[np.where(np.isin(Y, a))]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
You can use the set functions of numpy:
X[np.where(np.isin(Y, a))]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
You can use the set functions of numpy:
X[np.where(np.isin(Y, a))]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
You can use the set functions of numpy:
X[np.where(np.isin(Y, a))]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
answered Nov 17 '18 at 0:12
Matthias OssadnikMatthias Ossadnik
57427
57427
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
You can skip the np.where, which is redundant here, and just index using np.isin:
X[np.isin(Y,a)]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
This is because np.isin gives you a boolean array of where Y is in a:
array([False, True, True, True, True, False])
So by indexing with this array, it only selects those rows where True
answered Nov 17 '18 at 0:16
saculsacul
30k41740
1
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
You can skip the np.where, which is redundant here, and just index using np.isin:
X[np.isin(Y,a)]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
This is because np.isin gives you a boolean array of where Y is in a:
array([False, True, True, True, True, False])
So by indexing with this array, it only selects those rows where True
answered Nov 17 '18 at 0:16
saculsacul
30k41740
1
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
You can skip the np.where, which is redundant here, and just index using np.isin:
X[np.isin(Y,a)]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
This is because np.isin gives you a boolean array of where Y is in a:
array([False, True, True, True, True, False])
So by indexing with this array, it only selects those rows where True
answered Nov 17 '18 at 0:16
saculsacul
30k41740
You can skip the np.where, which is redundant here, and just index using np.isin:
X[np.isin(Y,a)]
array([[2, 1],
[1, 3],
[5, 9],
[6, 7]])
This is because np.isin gives you a boolean array of where Y is in a:
array([False, True, True, True, True, False])
So by indexing with this array, it only selects those rows where True
answered Nov 17 '18 at 0:16
saculsacul
30k41740
edited Nov 17 '18 at 0:20
answered Nov 17 '18 at 0:16
saculsacul
30k41740
answered Nov 17 '18 at 0:16
saculsacul
30k41740
answered Nov 17 '18 at 0:16
saculsacul
30k41740
30k41740
1
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
1
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
1
1
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
It works fine, thank you :)
– Esraa Mohamed
Nov 17 '18 at 0:20
add a comment |
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