In Numpy I have a boolean array of equal length of a matrix. I want to run a calculation on the matrix elements that correspond to the boolean array. How do I do this?
a: [true, false, true]
b: [[1,1,1],[2,2,2],[3,3,3]]
Say the function was to sum the elements of the sub arrays
index 0 is True: thus I add 3 to the summation (Starts at zero)
index 1 is False: thus summation remains at 3
index 2 is True: thus I add 9 to the summation for a total of 12
How do I do this (the boolean and summation part; I don't need how to add up each individual sub array)?
You can simply use your boolean array a to index into the rows of b, then take the sum of the resulting (2, 3) array:
import numpy as np
a = np.array([True, False, True])
b = np.array([[1,1,1],[2,2,2],[3,3,3]])
# index rows of b where a is True (i.e. the first and third row of b)
print(b[a])
# [[1 1 1]
# [3 3 3]]
# take the sum over all elements in these rows
print(b[a].sum())
# 12
It sounds like you would benefit from reading the numpy documentation on array indexing.
Related
I have a 6×6 matrix A, i'm trying to indexing the matrix A using two 2×2 matrices B and C. Each row of B and C specify a pair of indices for row and column in A. In detail, each row of B will specify the row needed to be indexed and each row of C specify the column.
For example,
A = np.arange(0,36).reshape(6,6)
B = np.array([[0,1],
[2,4]])
C = np.array([[1,2],
[3,4]])
I need to get a 2×2×2 matrix like this:
results =
[[[ 1 2]
[7 8]]
[[15 16]
[27 28]]]
example of indexing
If just get one matrix using index like B=[0,1] and C=[1,2], it can be done with:
d = A[B,:]
results = d[:,C]
But things different when I need to get two 2×2 matrices (2×2×2)， and each matrix is index using each row of B and C.
p.s. Please change the title of this question if you can think of a more precise one.
I have two 2D arrays like:
A=array[[4,5,6],
[0,7,8],
[0,9,0]]
B = array[[11,12,13],
[14,15,16],
[17,18,19]]
In array A where element value is 0 i want to replace same value in array B by 0 and store the changed matrix in a new variable and retain the old B matrix.
Thanks in advance.
import numpy as np
A=np.array([[4,5,6],
[0,7,8],
[0,9,0]])
B =np.array([[11,12,13],
[14,15,16],
[17,18,19]])
C = B.copy()
B[A == 0] = 0
C, B = B, C
The line B[A == 0] basically first gets all the the values where the array A is 0 by the line A == 0 . It return a boolean array with true at the position where value is zero in array A. This boolean array is then used to mask the array B and assigns 0 to indices the boolean values is True.
I have a numpy array M of shape (n, 1000, 6). This can be thought of as n matrices with 1000 rows and 6 columns. For each matrix I would like to reverse the order of the rows (i.e. the top row is now at the bottom and vice versa) and then reverse the order of just the first 4 columns (so column 0 is now column 3, column 1 is column 2, column 2 is column 1 and column 3 is column 0 but column 4 is still column 4 and column 5 is still column 5). I would like to do this in a single operation, without doing indexing on the left side of the expression, so this would not be acceptable:
M[:,0:4,:] = M[:,0:4,:][:,::-1,:]
M[:,:,:] = M[:,:,::-1]
The operation needs to be achieveable using Keras backend which disallowes this. It must be of the form
M = M[indexing here that solves the task]
If I wanted to reverse the order of all the columns instead of just the first 4 this could easily be achieved with M = M[:,::-1,::-1] so I've being trying to modify this to achieve my goal but unfortunately can't work out how. Is this even possible?
M[:, ::-1, [3, 2, 1, 0, 4, 5]]
What is the most efficient numpy way to replace masked values in an array with the average of the closest unmasked values next to them?
eg:
a = np.array([2,6,4,8])
b = np.ma.masked_where(a>5,a)
print b
masked_array(data = [2 -- 4 --],
mask = [False True False True],
fill_value = 999999)
I want the masked values in b to be replaced with the average of values just next to them. Boundaries can repeat the closest unmasked value. So in this example, b will be the following:
b = [2,3,4,4]
The main reason for this question is to see whether this can be done efficiently without the use of an iterator.
you can use np.interp and np.where
import numpy as np
a = np.array([2,6,4,8])
mymask = a>5
b = np.ma.masked_where(mymask,a)
print b
# [2 -- 4 --]
c = np.interp(np.where(mymask)[0],np.where(~mymask)[0],b[np.where(~mymask)[0]])
b[np.where(mymask)[0]] = c
print b
# [2 3 4 4]
Let's say:
import numpy as np
f=np.matrix("1 2; 3 4 ; 5 6")
Is retrieving number of column which have maximum sum of column from matrix possible? How?
You could write:
>>> f.sum(axis=0).argmax()
1
So column 1 sums to the greatest value.
To clarify what this does: f.sum(axis=0) sums the columns of the matrix f, returning the matrix matrix([[ 9, 12]]). Then argmax() is used to find the index of the maximum value in this matrix of sums.