Question

从以下数组开始

array([ nan,  nan,  nan,   1.,  nan,  nan,   0.,  nan,  nan])

生成如下：

import numpy as np
row = np.array([ np.nan,  np.nan,  np.nan,   1.,  np.nan,  np.nan,   0.,  np.nan,  np.nan])

我想获取已排序数组的索引，然后排除nans。在这种情况下，我想获得[6,3]。

我想出了以下方法来做到这一点：

vals = np.sort(row)
inds = np.argsort(row)

def select_index_by_value(indices, values):
    selected_indices = []
    for i in range(len(indices)):
        if not np.isnan(values[i]):
            selected_indices.append(indices[i])
    return selected_indices

selected_inds = select_index_by_value(inds, vals)

现在selected_inds是[6,3]。但是，这似乎有很多行代码可以实现简单的操作。这可能是一种更短的方式吗？

Answer 1

你可以这样做 -

# Store non-NaN indices
idx = np.where(~np.isnan(row))[0]

# Select non-NaN elements, perform argsort and use those argsort       
# indices to re-order non-NaN indices as final output
out = idx[row[idx].argsort()]

Answer 2

另一种选择：

row.argsort()[~np.isnan(np.sort(row))]
# array([6, 3])

Answer 3

还有另一种更快的解决方案（对于OP数据）。

Psidom的解决方案

%timeit row.argsort()[~np.isnan(np.sort(row))]

The slowest run took 31.23 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.16 µs per loop

Divakar的解决方案

%timeit idx = np.where(~np.isnan(row))[0]; idx[row[idx].argsort()]

The slowest run took 35.11 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 4.73 µs per loop

基于Divakar的解决方案

%timeit np.where(~np.isnan(row))[0][::-1]

The slowest run took 9.42 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.86 µs per loop

我认为这是有效的，因为np.where(~np.isnan(row))保留了订单。

根据应用于另一个数组的布尔表达式选择一个数组的值

3 个答案: