Change every single element in Numpy object - numpy

Question

I have a Numpy object with random N*M elements, and I also have two numbers A and B.
Now I want to access every element in this N*M array and make a change, i.e., if the element > 0, replace this element to A (i.e., element <- A), and if this element < 0, replace this element to B (i.e., element <- B).
I know there is a naive way to implement this method, that is accessing every single element using for loop, but it is very slow.
Can we use more fancy code to implement this ?

Answer 1

Boolean masked assignment will change values in place:
In [493]: arr = np.random.randint(-10,10,(5,7))
In [494]: arr
Out[494]:
array([[ -5, -6, -7, -1, -8, -8, -10],
[ -9, 1, -3, -9, 3, 8, -1],
[ 6, -7, 4, 0, -4, 4, -2],
[ -3, -10, -2, 7, -4, 2, 2],
[ -5, 5, -1, -7, 7, 5, -7]])
In [495]: arr[arr>0] = 100
In [496]: arr[arr<0] = -50
In [497]: arr
Out[497]:
array([[-50, -50, -50, -50, -50, -50, -50],
[-50, 100, -50, -50, 100, 100, -50],
[100, -50, 100, 0, -50, 100, -50],
[-50, -50, -50, 100, -50, 100, 100],
[-50, 100, -50, -50, 100, 100, -50]])
I just gave a similar answer in
python numpy: iterate for different conditions without using a loop

Answer 2

IIUC:
narr = np.random.randint(-100,100,(10,5))
array([[ 70, -20, 96, 73, -94],
[ 42, 35, -55, 56, 54],
[ 97, -16, 24, 32, 78],
[ 49, 49, -11, -82, 82],
[-10, 59, -42, -68, -70],
[ 95, 23, 22, 58, -38],
[ -2, -64, 27, -33, -95],
[ 98, 42, 8, -83, 85],
[ 23, 51, -99, -82, -7],
[-28, -11, -44, 95, 93]])
A = 1000
B = -999
Use np.where:
np.where(narr > 0, A, np.where(narr < 0, B , narr))
Output:
array([[1000, -999, 1000, 1000, -999],
[1000, 1000, -999, 1000, 1000],
[1000, -999, 1000, 1000, 1000],
[1000, 1000, -999, -999, 1000],
[-999, 1000, -999, -999, -999],
[1000, 1000, 1000, 1000, -999],
[-999, -999, 1000, -999, -999],
[1000, 1000, 1000, -999, 1000],
[1000, 1000, -999, -999, -999],
[-999, -999, -999, 1000, 1000]])

Answer 3

Because you mentioned that you're interested in the speed of the computation, I made a speed comparision of several different approaches for your problem.
test.py:
import numpy as np
A = 100
B = 50
def createArray():
array = np.random.randint(-100,100,(500,500))
return array
def replace(x):
return A if x > 0 else B
def replace_ForLoop():
"""Simple for-loop."""
array = createArray()
for i in range(array.shape[0]):
for j in range(array.shape[1]):
array[i][j] = replace(array[i][j])
def replace_nditer():
"""Use numpy.nditer to iterate over values."""
array = createArray()
for elem in np.nditer(array, op_flags=['readwrite']):
elem[...] = replace(elem)
def replace_masks():
"""Use boolean masks."""
array = createArray()
array[array>0] = A
array[array<0] = B
def replace_vectorize():
"""Use numpy.vectorize"""
array = createArray()
vectorfunc = np.vectorize(replace)
array = vectorfunc(array)
def replace_where():
"""Use numpy.where"""
array = createArray()
array = np.where(array > 0, A, np.where(array < 0, B , array))
Note: The variants using nested for-loops, np.nditer and boolean masks work inplace, the last two do not.
Timing comparision:
> python -mtimeit -s'import test' 'test.replace_ForLoop()'
10 loops, best of 3: 185 msec per loop
> python -mtimeit -s'import test' 'test.replace_nditer()'
10 loops, best of 3: 294 msec per loop
> python -mtimeit -s'import test' 'test.replace_masks()'
100 loops, best of 3: 5.8 msec per loop
> python -mtimeit -s'import test' 'test.replace_vectorize()'
10 loops, best of 3: 55.3 msec per loop
> python -mtimeit -s'import test' 'test.replace_where()'
100 loops, best of 3: 5.42 msec per loop
Using loops is indeed quite slow. numpy.nditer is even slower, which comes as a surprise to me, because the doc calls it an efficient multi-dimensional iterator object to iterate over arrays. numpy.vectorize is essentially a for-loop, but still manages to be thrice as fast as the naive implementation.
The np.where variant proposed by Scott Boston is slightly faster than using boolean masks as per hpaulj's answer. However, it does need more memory because it does not modify inplace.