python中按功能分组的紧凑1-0数组

与 a 作为输入数组/列表,我们可以-

# Compare against 1 to get a mask. Append on either sides with False
# so that when do consecutive comparison next, we will catch the
# transitions including leading and trailing islands that might be 
# starting at the first element of the array or ending as the last one. 
# These transitions are signal the start and end of each island of 1s.
m = np.r_[False,np.asarray(a)==1,False]
idx = np.flatnonzero(m[:-1]!=m[1:])

# After catching those start,end indices, simply subtract between start
# and end indices to get island lengths. That's our o/p.
out = idx[1::2]-idx[::2]

如果 一 已经是一个数组,我们也可以使用 a.astype(bool) 代替 np.asarray(a)==1 .

样本运行-

In [81]: a
Out[81]: [0, 0, 1, 1, 0, 1, 0, 1, 0, 0]

In [82]: m = np.r_[False,np.asarray(a)==1,False]
    ...: idx = np.flatnonzero(m[:-1]!=m[1:])
    ...: out = idx[1::2]-idx[::2]

In [83]: out
Out[83]: array([2, 1, 1])

这里有一个解决方案,我还没有和你的时间对上,但可能更接近你想要的

a = [0, 0, 1, 1, 0, 1, 0, 1, 0, 0]
# Convert the array to a string of 1s and 0s
a = ''.join([str(x) for x in a])
# Split on the 0s to 'remove them'
a = a.split('0')
# Count the length of the arrays that are greater than 0
b = [len(x) for x in a if len(x) > 0]
print(b)

输出: [2, 1, 1]