Question

“缺席”可以表示nan或np.masked，以最简单的方式执行此操作。

例如：

>>> from numpy import nan
>>> do_it([1, nan, nan, 2, nan, 3, nan, nan, 4, 3, nan, 2, nan])
array([1, 1, 1, 2, 2, 3, 3, 3, 4, 3, 3, 2, 2])
# each nan is replaced with the first non-nan value before it
>>> do_it([nan, nan, 2, nan])
array([nan, nan, 2, 2])
# don't care too much about the outcome here, but this seems sensible

我可以通过for循环看到你是如何做到这一点的：

def do_it(a):
    res = []
    last_val = nan
    for item in a:
        if not np.isnan(item):
            last_val = item
        res.append(last_val)
    return np.asarray(res)

有没有更快速的矢量化方法？

Answer 1

cumsum使用一组标志提供了一种确定在NaN上写入哪些数字的好方法：

def do_it(x):
    x = np.asarray(x)

    is_valid = ~np.isnan(x)
    is_valid[0] = True

    valid_elems = x[is_valid]
    replacement_indices = is_valid.cumsum() - 1
    return valid_elems[replacement_indices]

Answer 2

假设您的数据中没有零（为了使用numpy.nan_to_num）：

b = numpy.maximum.accumulate(numpy.nan_to_num(a))
>>> array([ 1.,  1.,  1.,  2.,  2.,  3.,  3.,  3.,  4.,  4.])
mask = numpy.isnan(a)
a[mask] = b[mask]
>>> array([ 1.,  1.,  1.,  2.,  2.,  3.,  3.,  3.,  4.,  3.])

编辑：正如Eric所指出的，更好的解决方案是用-inf替换nans：

mask = numpy.isnan(a)
a[mask] = -numpy.inf
b = numpy.maximum.accumulate(a)
a[mask] = b[mask]

Answer 3

使用@ Benjamin删除的解决方案，如果使用索引

，一切都很棒

def do_it(data, valid=None, axis=0):
    # normalize the inputs to match the question examples
    data = np.asarray(data)
    if valid is None:
        valid = ~np.isnan(data)

    # flat array of the data values
    data_flat = data.ravel()

    # array of indices such that data_flat[indices] == data
    indices = np.arange(data.size).reshape(data.shape)

    # thanks to benjamin here
    stretched_indices = np.maximum.accumulate(valid*indices, axis=axis)
    return data_flat[stretched_indices]

比较解决方案运行时：

>>> import numpy as np
>>> data = np.random.rand(10000)

>>> %timeit do_it_question(data)
10000 loops, best of 3: 17.3 ms per loop
>>> %timeit do_it_mine(data)
10000 loops, best of 3: 179 µs per loop
>>> %timeit do_it_user(data)
10000 loops, best of 3: 182 µs per loop

# with lots of nans
>>> data[data > 0.25] = np.nan

>>> %timeit do_it_question(data)
10000 loops, best of 3: 18.9 ms per loop
>>> %timeit do_it_mine(data)
10000 loops, best of 3: 177 µs per loop
>>> %timeit do_it_user(data)
10000 loops, best of 3: 231 µs per loop

因此，这个和@ user2357112的解决方案都解决了问题中的解决方案，但是当nan s的数量很多时，这比@ user2357112略有优势

如何有效地“拉伸”数组中的当前值而不是不存在的值

3 个答案: