替换 Python NumPy 数组中大于某个值的所有元素
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Replace all elements of Python NumPy Array that are greater than some value
提问by NLi10Me
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:
我有一个 2D NumPy 数组，并希望将其中大于或等于阈值 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
What is the most concise and pythonic way to do this?
Is there a faster (possibly less concise and/or less pythonic) way to do this?
执行此操作的最简洁和 Pythonic 的方法是什么？
有没有更快（可能不那么简洁和/或不那么 Pythonic）的方法来做到这一点？
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.
这将是用于人体头部 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。
采纳答案by mdml
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 ndarraynamed arr, you can replace all elements >255with a value xas follows:
我认为最快和最简洁的方法是使用 NumPy 的内置 Fancy 索引。如果你有一个ndarraynamed arr，你可以>255用一个值替换所有元素，x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
回答by askewchan
Since you actually want a different array which is arrwhere arr < 255, and 255otherwise, this can be done simply:
由于您实际上想要一个不同的数组arrwhere arr < 255，255否则，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.clipand 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，但是np.clipand np.minimum(or 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 (i.e., modify arrinstead of creating result) you can use the outparameter of np.minimum:
如果您想就地进行（即修改arr而不是创建result），您可以使用以下out参数np.minimum：
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），但仍然没有那么快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 clipyou would have to do this twice, with something like
为了进行比较，如果您想用最小值和最大值来限制您的值，clip那么您将不得不这样做两次，例如
np.minimum(a, 255, a)
np.maximum(a, 0, a)
or,
或者，
a[a>255] = 255
a[a<0] = 0
回答by lev
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
回答by Amir F
I think you can achieve this the quickest by using the wherefunction:
我认为您可以通过使用该where功能最快地实现这一点：
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)
回答by Shital Shah
Another way is to use np.placewhich does in-place replacement and works with multidimentional arrays:
另一种方法是使用np.placewhich 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)
回答by Mahdi Shahbaba
You can also use &, |(and/or) for more flexibility:
您还可以使用&, |（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)]
