问题:替换Python NumPy数组中所有大于某个值的元素

我有一个2D NumPy数组,并希望将大于或等于阈值T的所有值替换为255.0。据我所知,最基本的方法是:

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. 什么是最简洁,最pythonic的方法?

  2. 有更快的方法(可能不太简洁和/或更少的pythonic)来做到这一点吗?

这将是用于人头MRI扫描的窗口/水平调整子程序的一部分。2D numpy数组是图像像素数据。

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.


回答 0

我认为最快和最简洁的方法是使用NumPy内置的Fancy indexing。如果您具有ndarraynamed arr,则可以将所有元素替换>255为一个值x,如下所示:

arr[arr > 255] = x

我使用500 x 500随机矩阵在计算机上运行此命令,将所有> 0.5的值替换为5,平均花费了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

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

回答 1

由于您实际上想要的是arrwhere 的其他数组arr < 255255否则可以简单地完成此操作:

result = np.minimum(arr, 255)

更一般而言,对于下限和/或上限:

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

如果您只想访问超过255的值,或者更复杂的值,则@ mtitan8的回答更为笼统,但对于您的情况,np.clipand和np.minimum(或np.maximum)更好,更快:

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

如果您想就地进行操作(即修改arr而不是创建result),则可以使用out参数np.minimum

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

要么

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

out=名称是可选的,因为参数与函数定义的顺序相同。)

对于就地修改,布尔索引可以提高很多速度(无需分别制作然后修改副本),但是仍然不如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

为了进行比较,如果您想限制最小值和最大值,而无需clip两次,例如

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

要么,

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

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

回答 2

我认为您可以使用以下where功能最快地实现此目的:

例如,在numpy数组中查找大于0.2的项并将其替换为0:

import numpy as np

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

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

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)

回答 3

您可以考虑使用numpy.putmask

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

这是与Numpy内置索引的性能比较:

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

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

回答 4

另一种方法是使用np.place它进行就地替换并与多维数组一起使用:

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)

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)

回答 5

你也可以使用&|(和/或)有更多的灵活性:

介于5到10之间的值: A[(A>5)&(A<10)]

大于10或小于5的值: A[(A<5)|(A>10)]

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)]


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