遍历一个numpy数组-Python 实用宝典

遍历一个numpy数组

有没有那么冗长的替代方案: for x in xrange(array.shape[0]): for y in xrange(array.shape[1]): do_stuff(x, y) 我想出了这个: for x, y in itertools.product(map(xrange, array.shape)): do_stuff(x, y) 这节省了一个缩进,但仍然很丑陋。 我希望看起来像这样的伪代码: for x, y in array.indices: do_stuff(x, y) 有没有类似的东西存在?

问题:遍历一个numpy数组

有没有那么冗长的替代方案:

for x in xrange(array.shape[0]):
    for y in xrange(array.shape[1]):
        do_stuff(x, y)

我想出了这个:

for x, y in itertools.product(map(xrange, array.shape)):
    do_stuff(x, y)

这节省了一个缩进,但仍然很丑陋。

我希望看起来像这样的伪代码:

for x, y in array.indices:
    do_stuff(x, y)

有没有类似的东西存在?

Is there a less verbose alternative to this:

for x in xrange(array.shape[0]):
    for y in xrange(array.shape[1]):
        do_stuff(x, y)

I came up with this:

for x, y in itertools.product(map(xrange, array.shape)):
    do_stuff(x, y)

Which saves one indentation, but is still pretty ugly.

I'm hoping for something that looks like this pseudocode:

for x, y in array.indices:
    do_stuff(x, y)

Does anything like that exist?


回答 0

我认为您正在寻找ndenumerate

>>> a =numpy.array([[1,2],[3,4],[5,6]])
>>> for (x,y), value in numpy.ndenumerate(a):
...  print x,y
... 
0 0
0 1
1 0
1 1
2 0
2 1

关于性能。它比列表理解要慢一些。

X = np.zeros((100, 100, 100))

%timeit list([((i,j,k), X[i,j,k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k in range(X.shape[2])])
1 loop, best of 3: 376 ms per loop

%timeit list(np.ndenumerate(X))
1 loop, best of 3: 570 ms per loop

如果您担心性能,可以通过查看实现来进一步优化ndenumerate,它实现了两件事,转换为数组并循环。如果知道有数组,则可以调用.coords平面迭代器的属性。

a = X.flat
%timeit list([(a.coords, x) for x in a.flat])
1 loop, best of 3: 305 ms per loop

I think you're looking for the ndenumerate.

>>> a =numpy.array([[1,2],[3,4],[5,6]])
>>> for (x,y), value in numpy.ndenumerate(a):
...  print x,y
... 
0 0
0 1
1 0
1 1
2 0
2 1

Regarding the performance. It is a bit slower than a list comprehension.

X = np.zeros((100, 100, 100))

%timeit list([((i,j,k), X[i,j,k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k in range(X.shape[2])])
1 loop, best of 3: 376 ms per loop

%timeit list(np.ndenumerate(X))
1 loop, best of 3: 570 ms per loop

If you are worried about the performance you could optimise a bit further by looking at the implementation of ndenumerate, which does 2 things, converting to an array and looping. If you know you have an array, you can call the .coords attribute of the flat iterator.

a = X.flat
%timeit list([(a.coords, x) for x in a.flat])
1 loop, best of 3: 305 ms per loop

回答 1

如果只需要索引,可以尝试numpy.ndindex

>>> a = numpy.arange(9).reshape(3, 3)
>>> [(x, y) for x, y in numpy.ndindex(a.shape)]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

If you only need the indices, you could try numpy.ndindex:

>>> a = numpy.arange(9).reshape(3, 3)
>>> [(x, y) for x, y in numpy.ndindex(a.shape)]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

回答 2

nditer

import numpy as np
Y = np.array([3,4,5,6])
for y in np.nditer(Y, op_flags=['readwrite']):
    y += 3

Y == np.array([6, 7, 8, 9])

y = 3将无法使用y *= 0y += 3而是使用。

see nditer

import numpy as np
Y = np.array([3,4,5,6])
for y in np.nditer(Y, op_flags=['readwrite']):
    y += 3

Y == np.array([6, 7, 8, 9])

y = 3 would not work, use y *= 0 and y += 3 instead.


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