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从元组生成二维布尔数组

[英]Generate a 2D boolean array from tuples

How can I generate a 2D boolean array using a list of tuples that shows the indices of the True values? 如何使用显示True值索引的元组列表生成2D布尔数组?

For example I have the following list of tuples: 例如,我有以下元组列表:

lst = [(0,1), (0, 2), (1, 0), (1, 3), (2,1)]

What I do is, I first generate an array of False's: 我要做的是,我首先生成一个False数组:

arr = np.repeat(False, 12).reshape(3, 4)

Then, iterate over the list to assign True values: 然后,遍历列表以分配True值:

for tup in lst:
    arr[tup] = True
print(arr)
array([[False,  True,  True, False],
       [ True, False, False,  True],
       [False,  True, False, False]], dtype=bool)

It seems like a common use case to me so I was wondering if there is a built-in method for this, without the loops. 对我来说,这似乎是一个常见的用例,所以我想知道是否有一个内置方法,没有循环。

zip(*...) is a handy way of 'transposing' a list of lists (or tuples). zip(*...)是“转置”列表列表(或元组)的便捷方法。 And A[x,y] is the same as A[(x,y)] . 并且A[x,y]A[(x,y)]

In [397]: lst = [(0,1), (0, 2), (1, 0), (1, 3), (2,1)]

In [398]: tuple(zip(*lst))    # make a tuple of tuples (or lists)
Out[398]: ((0, 0, 1, 1, 2), (1, 2, 0, 3, 1))

In [399]: A=np.zeros((3,4),dtype=bool)  # make an array of False

In [400]: A[tuple(zip(*lst))] = True  # assign True to the 5 values

In [401]: A
Out[401]: 
array([[False,  True,  True, False],
       [ True, False, False,  True],
       [False,  True, False, False]], dtype=bool)

You can use multidimensional indexing for this: 您可以为此使用多维索引

>>> lst = np.array(lst)
>>> arr = np.repeat(False, 12).reshape(3, 4)
>>> arr[lst[:,0], lst[:,1]] = True
>>> arr
array([[False,  True,  True, False],
       [ True, False, False,  True],
       [False,  True, False, False]], dtype=bool)

Here's an approach using NumPy's linear indexing that works for tuples of any lengths intended to generate multi-dimensional arrays - 这是一种使用NumPy's linear indexing的方法,该方法适用于旨在生成多维数组的任何长度的元组-

# Convert list of indices to a 2D array version
idx = np.array(lst)

# Decide on the shape of output array based on the extents, then initialize
shp = idx.max(0)+1
out = np.zeros(shp,dtype=bool)

# Using np.put insert 1s in out at places specified by linear indices version
np.put(out,np.ravel_multi_index(idx.T,shp),1)

Sample input, output - 样本输入,输出-

In [54]: lst
Out[54]: [(0, 1, 3), (0, 2, 2), (1, 0, 0), (1, 3, 1), (2, 1, 3)]

In [55]: out
Out[55]: 
array([[[False, False, False, False],
        [False, False, False,  True],
        [False, False,  True, False],
        [False, False, False, False]],

       [[ True, False, False, False],
        [False, False, False, False],
        [False, False, False, False],
        [False,  True, False, False]],

       [[False, False, False, False],
        [False, False, False,  True],
        [False, False, False, False],
        [False, False, False, False]]], dtype=bool)

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