Question

什么是获得nr行直到numpy中的下一个信号值的有效方法？

我有一个信号值列表（-1，nan，1），它看起来类似于下表，并且希望得到另一个列表，其中nr为行值直到下一个信号。考虑到消极和积极的价值观。

鉴于此表的第二列signal，我想生成第三列backward：

+-------+--------+----------+
| index | signal | backward |
+-------+--------+----------+
|     0 |        |          |
|     1 |        |          |
|     2 |        |          |
|     3 |      1 |        4 |
|     4 |        |        3 |
|     5 |        |        2 |
|     6 |        |        1 |
|     7 |     -1 |       -3 |
|     8 |        |       -2 |
|     9 |        |       -1 |
|    10 |      1 |        3 |
|    11 |        |        2 |
|    12 |        |        1 |
|    13 |      1 |        5 |
|    14 |        |        4 |
|    15 |        |        3 |
|    16 |        |        2 |
|    17 |        |        1 |
|    18 |     -1 |       -3 |
|    19 |        |       -2 |
|    20 |        |       -1 |
|    21 |     -1 |       -5 |
|    22 |        |       -4 |
|    23 |        |       -3 |
|    24 |        |       -2 |
|    25 |        |       -1 |
|    26 |      1 |        4 |
|    27 |        |        3 |
|    28 |        |        2 |
|    29 |        |        1 |
+-------+--------+----------+

原始numpy的形状看起来像这样。请原谅我创建这个随机列表的方式，我不知道更好的方法:)这只是为了演示目的

import numpy as np
data = np.random.randint(-4, 4, (1000,)).astype(float)
data[data == -2] = 'nan'
data[data == -3] = 'nan'
data[data == -4] = 'nan'
data[data == 0] = 'nan'
data[data == 2] = 'nan'
data[data == 3] = 'nan'
print(data)

它的大小是几百万，所以它必须尽可能高效

Answer 1

这是一种基于累积求和的方法 -

def seq_descending(a):
    mask = ~np.isnan(a)
    idx = np.flatnonzero(mask)
    shift_idx = np.hstack((idx[1:] - idx[:-1], a.size - idx[-1] ))

    out = -np.ones(a.size, dtype=int)
    out[idx] = shift_idx-1
    idx0 = idx[0]

    out[:idx0] = 0
    out[idx0] += 1

    cumsums = out.cumsum()
    signs = np.repeat(a[idx].astype(int), shift_idx)
    cumsums[idx0:] *= signs

    return cumsums

示例运行 -

1）设置输入数组：

In [82]: a = np.full((30,), np.nan)
    ...: a[[3,7,10,13,18,21,26]] = [1,-1,1,1,-1,-1,1]
    ...:

2）根据输入获取输出数组和堆栈以进行比较：

In [83]: np.column_stack((a, seq_descending(a) ))
Out[83]: 
array([[ nan,   0.],
       [ nan,   0.],
       [ nan,   0.],
       [  1.,   4.],
       [ nan,   3.],
       [ nan,   2.],
       [ nan,   1.],
       [ -1.,  -3.],
       [ nan,  -2.],
       [ nan,  -1.],
       [  1.,   3.],
       [ nan,   2.],
       [ nan,   1.],
       [  1.,   5.],
       [ nan,   4.],
       [ nan,   3.],
       [ nan,   2.],
       [ nan,   1.],
       [ -1.,  -3.],
       [ nan,  -2.],
       [ nan,  -1.],
       [ -1.,  -5.],
       [ nan,  -4.],
       [ nan,  -3.],
       [ nan,  -2.],
       [ nan,  -1.],
       [  1.,   4.],
       [ nan,   3.],
       [ nan,   2.],
       [ nan,   1.]])

Answer 2

数据：

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

您可以使用pandas：

df = pd.DataFrame({'id':np.square(np.nan_to_num(data)).cumsum(),'signal':data}) df['backward'] = df.groupby('id')['id'].transform(lambda x: np.arange(1, len(x)+1)[::-1]) df['backward'] = df['backward']*df.signal.fillna(method='ffill') >>> df id signal backward 0 0 NaN NaN 1 1 -1 -11 2 1 NaN -10 3 1 NaN -9 4 1 NaN -8 5 1 NaN -7 6 1 NaN -6 7 1 NaN -5 8 1 NaN -4 9 1 NaN -3 10 1 NaN -2 11 1 NaN -1 12 2 1 1 13 3 -1 -2 14 3 NaN -1 15 4 -1 -1 16 5 -1 -3 17 5 NaN -2 18 5 NaN -1 19 6 -1 -1

如何获得行的nr直到下一个值

2 个答案: