I have an array of arrays of size (4,1) like:
ar1 = np.array([[1,1],
[2,2],
[3,3],
[4,4]])
and a second array of size (1,2) like:
ar2 = np.array([2,3])
I'm trying to multiply every first item of the first array times the first item of the second array, and every second item of the first array times the second item of the second array, such as the result is:
ar_result = np.array([[2,3],
[4,6],
[6,9],
[8,12]])
Is there a way to do this in an easy and vectorized way?
When I try to ar1*ar2 it I'm getting this error:
ValueError: operands could not be broadcast together with shapes (4,) (2,)
Thanks
EDIT: To clarify, in my case ar1 is a DataFrame df1 column or row, and ar2 is the content of a cell in another DataFrame df2 (df2.loc[x,y] = [2,3])
Have you tried ar1 * ar2?
ar_result = ar1 * ar2
#array([[ 2, 3],
# [ 4, 6],
# [ 6, 9],
# [ 8, 12]])
Related
I have a list of 2-d numpy arrays, and I wish to create one array consisting of the non-zero values (or-wise) of each array set to 1. For example
arr1 = np.array([[1,0],[0,0]])
arr2 = np.array([[0,10],[0,0]])
arr3 = np.array([[0,0],[0,8]])
arrs = [arr1, arr2, arr3]
And so my op would yield
op(arrs) = [[1, 1], [0, 1]]]
What is an efficient way to do this in numpy (for about 8 image arrays of 600 by 600)?
Took me a while to understand. Try just summing all the arrays keeping their dimensions and then replace non-zero values with 1 as follows-
def op(arrs):
return np.where(np.add.reduce(arrs) != 0, 1, 0)
I'm trying to build an array of some given shape in which all elements are given by another array. Is there a function in numpy which does that efficiently, similar to np.full(), or any other elegant way, without simply employing for loops?
Example: Let's say I want an array with shape
(dim1,dim2) filled with a given, constant scalar value. Numpy has np.full() for this:
my_array = np.full((dim1,dim2),value)
I'm looking for an analog way of doing this, but I want the array to be filled with another array of shape (filldim1,filldim2) A brute-force way would be this:
my_array = np.array([])
for i in range(dim1):
for j in range(dim2):
my_array = np.append(my_array,fill_array)
my_array = my_array.reshape((dim1,dim2,filldim1,filldim2))
EDIT
I was being stupid, np.full() does take arrays as fill value if the shape is modified accordingly:
my_array = np.full((dim1,dim2,filldim1,filldim2),fill_array)
Thanks for pointing that out, #Arne!
You can use np.tile:
>>> shape = (2, 3)
>>> fill_shape = (4, 5)
>>> fill_arr = np.random.randn(*fill_shape)
>>> arr = np.tile(fill_arr, [*shape, 1, 1])
>>> arr.shape
(2, 3, 4, 5)
>>> np.all(arr[0, 0] == fill_arr)
True
Edit: better answer, as suggested by #Arne, directly using np.full:
>>> arr = np.full([*shape, *fill_shape], fill_arr)
>>> arr.shape
(2, 3, 4, 5)
>>> np.all(arr[0, 0] == fill_arr)
True
I am (re)building up my knowledge of numpy, having used it a little while ago.
I have a question about fancy indexing with multidimenional (in this case 2D) arrays.
Given the following snippet:
>>> a = np.arange(12).reshape(3,4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> i = np.array( [ [0,1], # indices for the first dim of a
... [1,2] ] )
>>> j = np.array( [ [2,1], # indices for the second dim
... [3,3] ] )
>>>
>>> a[i,j] # i and j must have equal shape
array([[ 2, 5],
[ 7, 11]])
Could someone explain in simple English, the logic being applied to give the results produced. Ideally, the explanation would be applicable for 3D and higher rank arrays being used to index an array.
Conceptually (in terms of restrictions placed on "rows" and "columns"), what does it mean to index using a 2D array?
Conceptually (in terms of restrictions placed on "rows" and "columns"), what does it mean to index using a 2D array?
It means you are constructing a 2d array R, such that R=A[B, C]. This means that the value for rij=abijcij.
So it means that the item located at R[0,0] is the item in A with as row index B[0,0] and as column index C[0,0]. The item R[0,1] is the item in A with row index B[0,1] and as column index C[0,1], etc.
So in this specific case:
>>> b = a[i,j]
>>> b
array([[ 2, 5],
[ 7, 11]])
b[0,0] = 2 since i[0,0] = 0, and j[0,0] = 2, and thus a[0,2] = 2. b[0,1] = 5 since i[0,0] = 1, and j[0,0] = 1, and thus a[1,1] = 5. b[1,0] = 7 since i[0,0] = 1, and j[0,0] = 3, and thus a[1,3] = 7. b[1,1] = 11 since i[0,0] = 2, and j[0,0] = 3, and thus a[2,3] = 11.
So you can say that i will determine the "row indices", and j will determine the "column indices". Of course this concept holds in more dimensions as well: the first "indexer" thus determines the indices in the first index, the second "indexer" the indices in the second index, and so on.
How to understand the difference between a[:,:-1] and a[:,-1]?
a = np.array([[1,2],[3,4],[5,6]])
b = a[:,:-1]
print b
The output for this is:
[[1]
[3]
[5]]
And for the following code-
b = a[:,-1]
print b
The output is:
[2 4 6]
Let's create another numpy array for understanding.
my_array = np.array([[1,2,3],[4,5,6],[7,8,9]])
This array contains three different arrays. That is, my_array is an array of arrays.
type(my_array) and type(my_array[0]) will both return numpy.ndarray
When you execute my_array[:,-1], this means goto every element in my_array and print the last item in that element. The : before , means all and -1 means the last element.
So the output of my_array[:,-1] will be
array([3, 6, 9])
meaning- The last element of each array inside my_array.
Now, when you execute my_array[:,:-1], the output is:
array([[1, 2],
[4, 5],
[7, 8]])
Meaning- print all the items in all arrays of my_array except the last item.
Here : means goto all elements and :-1 means exclude the last item.
I have a numpy array, a:
a = np.array([[-21.78878256, 97.37484004, -11.54228119],
[ -5.72592375, 99.04189958, 3.22814204],
[-19.80795922, 95.99377136, -10.64537733]])
I have another array, b:
b = np.array([[ 54.64642121, 64.5172014, 44.39991983],
[ 9.62420892, 95.14361441, 0.67014312],
[ 49.55036427, 66.25136632, 40.38778238]])
I want to extract minimum value indices from the array, b.
ixs = [[2],
[2],
[2]]
Then, want to extract elements from the array, a using the indices, ixs:
The expected answer is:
result = [[-11.54228119]
[3.22814204]
[-10.64537733]]
I tried as:
ixs = np.argmin(b, axis=1)
print ixs
[2,2,2]
result = np.take(a, ixs)
print result
Nope!
Any ideas are welcomed
You can use
result = a[np.arange(a.shape[0]), ixs]
np.arange will generate indices for each row and ixs will have indices for each column. So effectively result will have required result.
You can try using below code
np.take(a, ixs, axis = 1)[:,0]
The initial section will create a 3 by 3 array and slice the first column
>>> np.take(a, ixs, axis = 1)
array([[-11.54228119, -11.54228119, -11.54228119],
[ 3.22814204, 3.22814204, 3.22814204],
[-10.64537733, -10.64537733, -10.64537733]])