Replace all elements of Python NumPy Array that are greater than some value

I have a 2D NumPy array and would like to replace all values in it greater than or equal to a threshold T with 255.0. To my knowledge, the most fundamental way would be:

shape = arr.shape
result = np.zeros(shape)
for x in range(0, shape[0]):
    for y in range(0, shape[1]):
        if arr[x, y] >= T:
            result[x, y] = 255
  1. What is the most concise and pythonic way to do this?
  2. Is there a faster (possibly less concise and/or less pythonic) way to do this?

This will be part of a window/level adjustment subroutine for MRI scans of the human head. The 2D numpy array is the image pixel data.

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

I think both the fastest and most concise way to do this is to use NumPy’s built-in Fancy indexing. If you have an ndarray named arr, you can replace all elements >255 with a value x as follows:

arr[arr > 255] = x

I ran this on my machine with a 500 x 500 random matrix, replacing all values >0.5 with 5, and it took an average of 7.59ms.

In [1]: import numpy as np
In [2]: A = np.random.rand(500, 500)
In [3]: timeit A[A > 0.5] = 5
100 loops, best of 3: 7.59 ms per loop

Method 2

Since you actually want a different array which is arr where arr < 255, and 255 otherwise, this can be done simply:

result = np.minimum(arr, 255)

More generally, for a lower and/or upper bound:

result = np.clip(arr, 0, 255)

If you just want to access the values over 255, or something more complicated, @mtitan8’s answer is more general, but np.clip and np.minimum (or np.maximum) are nicer and much faster for your case:

In [292]: timeit np.minimum(a, 255)
100000 loops, best of 3: 19.6 µs per loop

In [293]: %%timeit
   .....: c = np.copy(a)
   .....: c[a>255] = 255
   .....: 
10000 loops, best of 3: 86.6 µs per loop

If you want to do it in-place (i.e., modify arr instead of creating result) you can use the out parameter of np.minimum:

np.minimum(arr, 255, out=arr)

or

np.clip(arr, 0, 255, arr)

(the out= name is optional since the arguments in the same order as the function’s definition.)

For in-place modification, the boolean indexing speeds up a lot (without having to make and then modify the copy separately), but is still not as fast as minimum:

In [328]: %%timeit
   .....: a = np.random.randint(0, 300, (100,100))
   .....: np.minimum(a, 255, a)
   .....: 
100000 loops, best of 3: 303 µs per loop

In [329]: %%timeit
   .....: a = np.random.randint(0, 300, (100,100))
   .....: a[a>255] = 255
   .....: 
100000 loops, best of 3: 356 µs per loop

For comparison, if you wanted to restrict your values with a minimum as well as a maximum, without clip you would have to do this twice, with something like

np.minimum(a, 255, a)
np.maximum(a, 0, a)

or,

a[a>255] = 255
a[a<0] = 0

Method 3

I think you can achieve this the quickest by using the where function:

For example looking for items greater than 0.2 in a numpy array and replacing those with 0:

import numpy as np

nums = np.random.rand(4,3)

print np.where(nums > 0.2, 0, nums)

Method 4

Another way is to use np.place which does in-place replacement and works with multidimentional arrays:

import numpy as np

# create 2x3 array with numbers 0..5
arr = np.arange(6).reshape(2, 3)

# replace 0 with -10
np.place(arr, arr == 0, -10)

Method 5

You can consider using numpy.putmask:

np.putmask(arr, arr>=T, 255.0)

Here is a performance comparison with the Numpy’s builtin indexing:

In [1]: import numpy as np
In [2]: A = np.random.rand(500, 500)

In [3]: timeit np.putmask(A, A>0.5, 5)
1000 loops, best of 3: 1.34 ms per loop

In [4]: timeit A[A > 0.5] = 5
1000 loops, best of 3: 1.82 ms per loop

Method 6

You can also use &, | (and/or) for more flexibility:

values between 5 and 10: A[(A>5)&(A<10)]

values greater than 10 or smaller than 5: A[(A<5)|(A>10)]

Method 7

Lets us assume you have a numpy array that has contains the value from 0 all the way up to 20 and you want to replace numbers greater than 10 with 0

import numpy as np

my_arr = np.arange(0,21) # creates an array
my_arr[my_arr > 10] = 0 # modifies the value

Note this will however modify the original array to avoid overwriting the original array try using arr.copy() to create a new detached copy of the original array and modify that instead.

import numpy as np

my_arr = np.arange(0,21)
my_arr_copy = my_arr.copy() # creates copy of the orignal array

my_arr_copy[my_arr_copy > 10] = 0

Method 8

np.where() works great!

np.where(arr > 255, 255, arr)

example:

FF = np.array([[0, 0],
              [1, 0],
              [0, 1],
              [1, 1]])
np.where(FF == 1, '+', '-')
Out[]: 
array([['-', '-'],
       ['+', '-'],
       ['-', '+'],
       ['+', '+']], dtype='<U1')


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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