How can the elements from (take only range 3 to 8)
a = np.array([1,2,3,4,5,6,7,8,9])
go to
A = np.array([[0,0,0],
[0,0,0]])
Ideal output would be:
A = ([[3,4,5],
[6,7,8]])
np.arange(3, 9).reshape((2, 3))
outputs
array([[3, 4, 5],
[6, 7, 8]])
A possible technique, presuming that you have an existing numpy array a is to use slicing and reshaping:
Starting array
>>> a = np.array([1,2,3,4,5,6,7,8,9])
>>> a
array([1, 2, 3, 4, 5, 6, 7, 8, 9])
Slicing
>>> A = a[2:-1]
>>> A
array([3, 4, 5, 6, 7, 8])
Reshaping
>>> A.reshape((2, 3))
>>> A
array([[3, 4, 5],
[6, 7, 8]])
The above solution presumes that you know which index to choose when doing the slicing. In this case, I presumed that we knew that the element 3 occurs at the second index position and I presumed that we knew that the last desired element 8 occurred in the second to last position in the array (at index -1). For clarity's sake: slicing starts at the given index, but goes up to and not including the second index position AND it is often easier to find the index position close to the end of the list by counting backwards using negative index numbers as I have done here. An alternate would be to use the index position of the last element which is an 8:
A = a[2:8].
A one-liner solution would be to daisy-chain the method calls together:
Starting array
>>> a = np.array([1,2,3,4,5,6,7,8,9])
>>> a
array([1, 2, 3, 4, 5, 6, 7, 8, 9])
Slicing and reshaping
>>> A = a[2:-1].reshape((2, 3))
>>> A
array([[3, 4, 5],
[6, 7, 8]])
Related
if i have an array
a = np.array([[1,2,3],[4,5,6],[7,8,9]])
output would be a =[1,2,3],[4,5,6],[7,8,9]
using slice [start:endindex:stepindex],
how could i retrieve 3 and 7?
is it possible?
I have tried
a[:3:2]
this gave me 1rst row and third row
In [928]: a = np.array([[1,2,3],[4,5,6],[7,8,9]])
In [929]: a
Out[929]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
[3,7] isn't regular pattern in this 2d array. But its flattened view:
In [931]: a.ravel()
Out[931]: array([1, 2, 3, 4, 5, 6, 7, 8, 9])
In [932]: a.ravel()[2::4]
Out[932]: array([3, 7])
In [933]: a.flat[2::4]
Out[933]: array([3, 7])
Now guarantee that it can be extended for larger arrays and selections.
I'm new in python, I was looking into a code which is similar to as follows,
import numpy as np
a = np.ones([1,1,5,5], dtype='int64')
b = np.ones([11], dtype='float64')
x = b[a]
print (x.shape)
# (1, 1, 5, 5)
I looked into the python numpy documentation I didn't find anything related to such case. I'm not sure what's going on here and I don't know where to look.
Edit
The actual code
def gausslabel(length=180, stride=2):
gaussian_pdf = signal.gaussian(length+1, 3)
label = np.reshape(np.arange(stride/2, length, stride), [1,1,-1,1])
y = np.reshape(np.arange(stride/2, length, stride), [1,1,1,-1])
delta = np.array(np.abs(label - y), dtype=int)
delta = np.minimum(delta, length-delta)+length/2
return gaussian_pdf[delta]
I guess that this code is trying to demonstrate that if you index an array with an array, the result is an array with the same shape as the indexing array (in this case a) and not the indexed array (i.e. b)
But it's confusing because b is full of 1s. Rather try this with a b full of different numbers:
>> a = np.ones([1,1,5,5], dtype='int64')
>> b = np.arange(11) + 3
array([ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13])
>>> b[a]
array([[[[4, 4, 4, 4, 4],
[4, 4, 4, 4, 4],
[4, 4, 4, 4, 4],
[4, 4, 4, 4, 4],
[4, 4, 4, 4, 4]]]])
because a is an array of 1s, the only element of b that is indexed is b[1] which equals 4. The shape of the result though is the shape of a, the array used as the index.
Say I have the following 5x5 numpy array called A
array([[6, 7, 7, 7, 8],
[4, 2, 5, 5, 9],
[1, 2, 4, 7, 4],
[0, 7, 3, 6, 8],
[4, 9, 6, 1, 6]])
and this 5x5 array called F
array([[1,0,0,0,0],
[1,0,0,0,0],
[1,0,0,0,0],
[1,0,0,0,0],
[0,0,0,0,0]])
I've been trying to use np.copyto, but I can't wrap my head around why it is not working/how it works.ValueError: could not broadcast input array from shape (5,5) into shape (2)
Is there a easy way to get the values of only the matching integers that have a corresponding 1 in F when laid over A? e.i it would return, 6,4,1,0
you can just do this little trick: A[F==1]
In [8]:
A[F==1]
Out[8]:
array([6, 4, 1, 0])
Check out Boolean indexing
To use np.copyto make sure that the destination array is np.empty.
This basically solved my problem.
I have two arrays of same size. In general dtype of these arrays is object (dtype = 'O'). What is the best way to access elements with same indicies from both arrays.
Possibility 1:
remove_indices = [i for i in range(len(array1)) if value in array1]
array1 = np.delete(array1, remove_indices, 0)
array2 = np.delete(array2, remove_indices, 0)
Possibility 2:
array3 = np.array([[array1[i], array2[i]] for i in range(len(array1))
if value not in array1[i]])
array1 = array3[:,0]
array2 = array3[:,1]
Note that Possibility 2 is faster. Is there any other solution with similar execution time (or faster)? How could I make Possiblity 2 more readable?
Not sure to understand well your examples, but sticking to What is the best way to access elements with same indicies from both arrays. make me think about zip. But using numpy why not using transpose ?
Like:
>>> array1 = numpy.array([0, 1, 2, 3, 4])
>>> array2 = numpy.array([5, 6, 7, 8, 9])
>>> numpy.array([array1, array2])
array([[0, 1, 2, 3, 4],
[5, 6, 7, 8, 9]])
>>> numpy.array([array1, array2]).T
array([[0, 5],
[1, 6],
[2, 7],
[3, 8],
[4, 9]])
For example, I have a ndarray that is:
a = np.array([1, 3, 5, 7, 2, 4, 6, 8])
Now I want to split a into two parts, one is all numbers <5 and the other is all >=5:
[array([1,3,2,4]), array([5,7,6,8])]
Certainly I can traverse a and create two new array. But I want to know does numpy provide some better ways?
Similarly, for multidimensional array, e.g.
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
[2, 4, 7]])
I want to split it according to the first column <3 and >=3, which result is:
[array([[1, 2, 3],
[2, 4, 7]]),
array([[4, 5, 6],
[7, 8, 9]])]
Are there any better ways instead of traverse it? Thanks.
import numpy as np
def split(arr, cond):
return [arr[cond], arr[~cond]]
a = np.array([1,3,5,7,2,4,6,8])
print split(a, a<5)
a = np.array([[1,2,3],[4,5,6],[7,8,9],[2,4,7]])
print split(a, a[:,0]<3)
This produces the following output:
[array([1, 3, 2, 4]), array([5, 7, 6, 8])]
[array([[1, 2, 3],
[2, 4, 7]]), array([[4, 5, 6],
[7, 8, 9]])]
It might be a quick solution
a = np.array([1,3,5,7])
b = a >= 3 # variable with condition
a[b] # to slice the array
len(a[b]) # count the elements in sliced array
1d array
a = numpy.array([2,3,4,...])
a_new = a[(a < 4)] # to get elements less than 5
2d array based on column(consider value of column i should be less than 5,
a = numpy.array([[1,2],[5,6],...]
a = a[(a[:,i] < 5)]
if your condition is multicolumn based, then you can make a new array applying the conditions on the columns. Then you can just compare the new array with value 5(according to my assumption) to get indexes and follow above codes.
Note that, whatever i have written in (), returns the index array.