[英]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.我有一个 2D NumPy 数组,想用 255.0 替换其中大于或等于阈值 T 的所有值。 To my knowledge, the most fundamental way would be:
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
What is the most concise and pythonic way to do this?最简洁和最pythonic的方法是什么?
Is there a faster (possibly less concise and/or less pythonic) way to do this?是否有更快(可能不那么简洁和/或不那么 Pythonic)的方式来做到这一点?
This will be part of a window/level adjustment subroutine for MRI scans of the human head.这将是人体头部 MRI 扫描的窗口/水平调整子程序的一部分。 The 2D numpy array is the image pixel data.
2D numpy 数组是图像像素数据。
I think both the fastest and most concise way to do this is to use NumPy's built-in Fancy indexing.我认为最快和最简洁的方法是使用 NumPy 的内置 Fancy 索引。 If you have an
ndarray
named arr
, you can replace all elements >255
with a value x
as follows:如果你有一个名为
arr
的ndarray
,你可以用值x
替换所有>255
元素,如下所示:
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.我用 500 x 500 随机矩阵在我的机器上运行它,用 5 替换所有 >0.5 的值,平均需要 7.59 毫秒。
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
Since you actually want a different array which is arr
where arr < 255
, and 255
otherwise, this can be done simply:由于您实际上想要一个不同的数组,即
arr
where arr < 255
,否则为255
,这可以简单地完成:
result = np.minimum(arr, 255)
More generally, for a lower and/or upper bound:更一般地,对于下限和/或上限:
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:如果您只想访问超过 255 的值或更复杂的值,@mtitan8 的答案更笼统,但
np.clip
和np.minimum
(或np.maximum
)对您的情况更好更快:
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
If you want to do it in-place (ie, modify arr
instead of creating result
) you can use the out
parameter of np.minimum
:如果您想就地进行(即修改
arr
而不是创建result
),您可以使用np.minimum
的out
参数:
np.minimum(arr, 255, out=arr)
or或者
np.clip(arr, 0, 255, arr)
(the out=
name is optional since the arguments in the same order as the function's definition.) (
out=
名称是可选的,因为参数与函数定义的顺序相同。)
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
:对于就地修改,布尔索引加快了很多(无需单独制作然后修改副本),但仍然不如
minimum
快:
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为了比较,如果你想用最小值和最大值来限制你的值,如果没有
clip
你必须这样做两次,比如
np.minimum(a, 255, a)
np.maximum(a, 0, a)
or,或者,
a[a>255] = 255
a[a<0] = 0
I think you can achieve this the quickest by using the where
function:我认为您可以通过使用
where
函数最快地实现这一点:
For example looking for items greater than 0.2 in a numpy array and replacing those with 0:例如,在 numpy 数组中查找大于 0.2 的项目并将其替换为 0:
import numpy as np
nums = np.random.rand(4,3)
print np.where(nums > 0.2, 0, nums)
Another way is to use np.place
which does in-place replacement and works with multidimentional arrays:另一种方法是使用
np.place
进行就地替换并使用多维数组:
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)
You can consider using numpy.putmask :您可以考虑使用numpy.putmask :
np.putmask(arr, arr>=T, 255.0)
Here is a performance comparison with the Numpy's builtin indexing:这是与 Numpy 的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 also use &
, |
您也可以使用
&
, |
(and/or) for more flexibility: (和/或)以获得更大的灵活性:
values between 5 and 10: A[(A>5)&(A<10)]
5 到 10 之间的值:
A[(A>5)&(A<10)]
values greater than 10 or smaller than 5: A[(A<5)|(A>10)]
大于 10 或小于 5 的值:
A[(A<5)|(A>10)]
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让我们假设您有一个
numpy
数组,其中包含从 0 一直到 20 的值,并且您想用 0 替换大于 10 的数字
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.请注意,这将修改原始数组以避免覆盖原始数组尝试使用
arr.copy()
创建原始数组的新分离副本并修改它。
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
np.where() works great! np.where() 效果很好!
np.where(arr > 255, 255, arr)
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