python - 如何在numpy数组中找到连续的正数，负数和零？

我使用下面的函数来查找连续的负数和正数，现在我还想添加一个条件来获得连续的零。
我该怎么做？

def consecutive_counts(arr):
    '''
    Returns number of consecutive negative and positive numbers
    arr = np.array
    negative = consecutive_counts()[0]
    positive = consecutive_counts()[1]
    '''
    pos = arr > 0
    # is used to Compute indices that are non-zero in the flattened version of arr
    idx = np.flatnonzero(pos[1:] != pos[:-1])
    count = np.concatenate(([idx[0]+1], idx[1:] - idx[:-1], [arr.size-1-idx[-1]]))
    negative = count[1::2], count[::2]
    positive = count[::2], count[1::2]
    if arr[0] < 0:
        return negative
    else:
        return positive

这是熊猫系列：

In [221]: n.temp.p['50000']
Out[221]:
name
0         0.00
1       -92.87
2       -24.01
3       -92.87
4       -92.87
5       -92.87
...       ...

我是这样用的：

arr = n.temp.p['50000'].values #Will be a numpy array as the input

预期产量：

In [225]: consecutive_counts(a)
Out[225]: (array([30, 29, 11, ...,  2,  1,  3]), array([19,  1,  1, ...,  1,  1,  2]))

谢谢：）

最佳答案

下面是一个新方法：

# create example
arr = np.random.randint(-2,3,(10))

# split into negative, zero, positive
*nzp, = map(np.flatnonzero,(arr<0,arr==0,arr>0))
# find block boundaries
*bb, = (np.flatnonzero(np.diff(x,prepend=-2,append=-2)-1) for x in nzp)
# compute block sizes
*bs, = map(np.diff,bb)

# show all
for data in (arr,nzp,bb,bs): print(data)
# [-1  1 -1  1  0  0  2 -1 -2  1]
# [array([0, 2, 7, 8]), array([4, 5]), array([1, 3, 6, 9])]
# [array([0, 1, 2, 4]), array([0, 2]), array([0, 1, 2, 3, 4])]
# [array([1, 1, 2]), array([2]), array([1, 1, 1, 1])]

关于python - 如何在numpy数组中找到连续的正数，负数和零？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/58013048/