用相同的值在非零元素之间的numpy数组中填充零 [英] Filling zeros in numpy array that are between non-zero elements with the same value

问题描述

我有一个带有整数的一维numpy numpy数组，在这里，当且仅当下一个非零值相同时，我才想用先前的非零值替换零.

I have a 1D numpy numpy array with integers, where I want to replace zeros with the previous non-zero value if and only if the next non-zero value is the same.

例如，一个数组:

in: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,3,1,0,1])
out: [1,0,1,1,0,0,2,0,3,0,0,0,3,1,0,1]

应该成为

out: [1,1,1,1,0,0,2,0,3,3,3,3,3,1,1,1]

有矢量化的方法可以做到这一点吗?我找到了一种填充零值的方法

Is there a vectorized way to do this? I found some way to fill values of zeros here, but not how to do it with exceptions, i.e. to not fill the zeros that are within integers with different value.

推荐答案

以下是一种矢量化方法，灵感来自 NumPy based forward-filling 用于此解决方案中的前填充部分以及masking和slicing-

Here's a vectorized approach taking inspiration from NumPy based forward-filling for the forward-filling part in this solution alongwith masking and slicing -

def forward_fill_ifsame(x):
    # Get mask of non-zeros and then use it to forward-filled indices
    mask = x!=0
    idx = np.where(mask,np.arange(len(x)),0)
    np.maximum.accumulate(idx,axis=0, out=idx)

    # Now we need to work on the additional requirement of filling only
    # if the previous and next ones being same
    # Store a copy as we need to work and change input data
    x1 = x.copy()

    # Get non-zero elements
    xm = x1[mask]

    # Off the selected elements, we need to assign zeros to the previous places
    # that don't have their correspnding next ones different
    xm[:-1][xm[1:] != xm[:-1]] = 0

    # Assign the valid ones to x1. Invalid ones become zero.
    x1[mask] = xm

    # Use idx for indexing to do the forward filling
    out = x1[idx]

    # For the invalid ones, keep the previous masked elements
    out[mask] = x[mask]
    return out

样品运行-

In [289]: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,3,1,0,1])

In [290]: np.vstack((x, forward_fill_ifsame(x)))
Out[290]: 
array([[1, 0, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 3, 1, 0, 1],
       [1, 1, 1, 1, 0, 0, 2, 0, 3, 3, 3, 3, 3, 1, 1, 1]])

In [291]: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,1,1,0,1])

In [292]: np.vstack((x, forward_fill_ifsame(x)))
Out[292]: 
array([[1, 0, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 1],
       [1, 1, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 1, 1]])

In [293]: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,1,1,0,2])

In [294]: np.vstack((x, forward_fill_ifsame(x)))
Out[294]: 
array([[1, 0, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2],
       [1, 1, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2]])

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