Question

我有一个带有整数的1 n numpy numpy数组，我想用以前的非零值替换零，当且仅当下一个非零值相同时。

例如，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]

有没有矢量化的方法来做到这一点？我找到了一些方法来填充零here的值，但不是如何使用异常来填充它，即不填充具有不同值的整数内的零。

Answer 1

这是一个矢量化方法，从NumPy based forward-filling获取灵感，用于此解决方案中的前向填充部分以及masking和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]])

在numpy数组中填充零，这些零在具有相同值的非零元素之间

1 个答案: