如何在Numpy中将索引数组转换为掩码数组？

Is it possible to convert an array of indices to an array of ones and zeros, given the range? i.e. [2,3] -> [0, 0, 1, 1, 0], in range of 5

在给定范围的情况下，是否可以将索引数组转换为1和0的数组？即[2,3] - > [0,0,1,1,0]，范围为5

I'm trying to automate something like this:

我正在尝试自动化这样的事情：

>>> index_array = np.arange(200,300)
array([200, 201, ... , 299])

>>> mask_array = ???           # some function of index_array and 500
array([0, 0, 0, ..., 1, 1, 1, ... , 0, 0, 0])

>>> train(data[mask_array])    # trains with 200~299
>>> predict(data[~mask_array]) # predicts with 0~199, 300~499

4 个解决方案

#1

Here's one way:

这是一种方式：

In [1]: index_array = np.array([3, 4, 7, 9])

In [2]: n = 15

In [3]: mask_array = np.zeros(n, dtype=int)

In [4]: mask_array[index_array] = 1

In [5]: mask_array
Out[5]: array([0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0])

If the mask is always a range, you can eliminate index_array, and assign 1 to a slice:

如果掩码始终是范围，则可以消除index_array，并将1分配给切片：

In [6]: mask_array = np.zeros(n, dtype=int)

In [7]: mask_array[5:10] = 1

In [8]: mask_array
Out[8]: array([0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0])

If you want an array of boolean values instead of integers, change the dtype of mask_array when it is created:

如果需要布尔值数组而不是整数，请在创建时更改mask_array的dtype：

In [11]: mask_array = np.zeros(n, dtype=bool)

In [12]: mask_array
Out[12]: 
array([False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False], dtype=bool)

In [13]: mask_array[5:10] = True

In [14]: mask_array
Out[14]: 
array([False, False, False, False, False,  True,  True,  True,  True,
        True, False, False, False, False, False], dtype=bool)

#2

For a single dimension, try:

对于单个维度，请尝试：

n = (15,)
index_array = [2, 5, 7]
mask_array = numpy.zeros(n)
mask_array[index_array] = 1

For more than one dimension, convert your n-dimensional indices into one-dimensional ones, then use ravel:

对于多个维度，将n维索引转换为一维索引，然后使用ravel：

n = (15, 15)
index_array = [[1, 4, 6], [10, 11, 2]] # you may need to transpose your indices!
mask_array = numpy.zeros(n)
flat_index_array = np.ravel_multi_index(
    index_array,
    mask_array.shape)
numpy.ravel(mask_array)[flat_index_array] = 1

#3

As requested, here it is in an answer. The code:

根据要求，这里是一个答案。代码：

[x in index_array for x in range(500)]

will give you a mask like you asked for, but it will use Bools instead of 0's and 1's.

会给你一个像你要求的面具，但它会使用Bools而不是0和1。

#4

There's a nice trick to do this as a one-liner, too - use the numpy.in1d and numpy.arange functions like this (the final line is the key part):

有一个很好的技巧可以做一个单线程 - 使用像这样的numpy.in1d和numpy.arange函数（最后一行是关键部分）：

>>> x = np.linspace(-2, 2, 10)
>>> y = x**2 - 1
>>> idxs = np.where(y<0)

>>> np.in1d(np.arange(len(x)), idxs)
array([False, False, False,  True,  True,  True,  True, False, False, False], dtype=bool)

The downside of this approach is that it's ~10-100x slower than the appropch Warren Weckesser gave... but it's a one-liner, which may or may not be what you're looking for.

这种方法的缺点是它比适当的Warren Weckesser给出的速度慢了10-100倍......但它是一个单线，可能是也可能不是你想要的。

#1