Python 基于应用于另一个数组的布尔表达式选择一个数组的值

从以下数组开始

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

它是这样生成的：

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

我想获得排序数组的索引，然后排除

nan

。在这种情况下，我想得到

[6,3]

我想出了以下方法：

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)

---

现在所选的

inds

是

[6,3]

。然而，这似乎需要相当多的代码来实现一些简单的东西。有没有一种更简单的方法

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()]

---

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()]

另一种选择：

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

另一种选择：

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

---

还有另一种更快的解决方案（针对OP数据）

Psidom解

%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

%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

%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

迪瓦卡溶液

%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

%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

%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

基于Divakar的解

%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

%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

%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

---

我认为这是可行的，因为

np.where（~np.isnan（row））

保留了顺序

还有另一个更快的解决方案（针对OP数据）

Psidom解

%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

%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

%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

迪瓦卡溶液

%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

%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

%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

基于Divakar的解

%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

%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

%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

---

我认为这是可行的，因为

np.where（~np.isnan（row））

保留了顺序

这两种解决方案都有效，但我发现布尔索引的使用比Numpy的

where

更优雅。谢谢这两种解决方案都有效，但我发现布尔索引的使用比Numpy的

where

更优雅。谢谢