用列表的值替换numpy索引数组的值

Question

Suppose you have a numpy array and a list:

>>> a = np.array([1,2,2,1]).reshape(2,2)
>>> a
array([[1, 2],
       [2, 1]])
>>> b = [0, 10]

I'd like to replace values in an array, so that 1 is replaced by 0, and 2 by 10.

I found a similar problem here - http://mail.python.org/pipermail//tutor/2011-September/085392.html

But using this solution:

for x in np.nditer(a):
    if x==1:
        x[...]=x=0
    elif x==2:
        x[...]=x=10

Throws me an error:

ValueError: assignment destination is read-only

I guess that's because I can't really write into a numpy array.

PS The actual size of the numpy array is 514 by 504 and of the list is 8.

Answer 1

好吧，我想你需要的是什么

a[a==2] = 10 #replace all 2's with 10's

Answer 2

Read-only array in numpy can be made writable:

nArray.flags.writeable = True

This will then allow assignment operations like this one:

nArray[nArray == 10] = 9999 # replace all 10's with 9999's

The real problem was not assignment itself but the writable flag.

Answer 3

Instead of replacing the values one by one, it is possible to remap the entire array like this:

import numpy as np
a = np.array([1,2,2,1]).reshape(2,2)
# palette must be given in sorted order
palette = [1, 2]
# key gives the new values you wish palette to be mapped to.
key = np.array([0, 10])
index = np.digitize(a.ravel(), palette, right=True)
print(key[index].reshape(a.shape))

yields

[[ 0 10]
 [10  0]]

---

Credit for the above idea goes to @JoshAdel . It is significantly faster than my original answer:

import numpy as np
import random
palette = np.arange(8)
key = palette**2
a = np.array([random.choice(palette) for i in range(514*504)]).reshape(514,504)

def using_unique():
    palette, index = np.unique(a, return_inverse=True)
    return key[index].reshape(a.shape)

def using_digitize():
    index = np.digitize(a.ravel(), palette, right=True)
    return key[index].reshape(a.shape)

if __name__ == '__main__':
    assert np.allclose(using_unique(), using_digitize())

I benchmarked the two versions this way:

In [107]: %timeit using_unique()
10 loops, best of 3: 35.6 ms per loop
In [112]: %timeit using_digitize()
100 loops, best of 3: 5.14 ms per loop

Answer 4

I found another solution with the numpy function place . (Documentation here )

Using it on your example:

>>> a = np.array([1,2,2,1]).reshape(2,2)
>>> a
array([[1, 2],
   [2, 1]])
>>> np.place(a, a==1, 0)
>>> np.place(a, a==2, 10)
>>> a
array([[ 0, 10],
       [10,  0]])

Answer 5

You can also use np.choose(idx, vals) , where idx is an array of indices that indicate which value of vals should be put in their place. The indices must be 0-based, though. Also make sure that idx has an integer datatype. So you would only need to do:

np.choose(a.astype(np.int32) - 1, b)

Answer 6

I was unable to set the flags, or use a mask to modify the value. In the end I just made a copy of the array.

a2 = np.copy(a)