基于布尔条件将np数组转换为锯齿状np数组

Say I have an array x equal to np.array(0 0 0 0 0 0 0 0 1000 0 0 0 0 1000 1000 1000)

假设我的数组x等于np.array(0 0 0 0 0 0 0 0 1000 0 0 0 0 1000 1000 1000)

and I want to turn it into a matrix array([array([0 0 0 0 0 0 0 0]), array([1000]), array([0 0 0 0]), array([1000 1000 1000])]). How would I do that?

我想把它变成一个矩阵数组([array([0 0 0 0 0 0 0 0]),array([1000]),array([0 0 0 0]),array([1000 1000 1000] )])。我该怎么办?

The boolean conditions would be, if it's a string of0's, segment it so that it's one array inside the matrix. If it's a string of 1000's segment it the same way.

布尔条件是,如果它是0的字符串,则将其分段,使其成为矩阵内的一个数组。如果它是一个1000字节的字符串,它的方式相同。

2 个解决方案

#1

Here's one approach with np.split -

这是np.split的一种方法 -

m = x!=0
out = np.split(x,np.flatnonzero(m[1:] != m[:-1])+1)

Sample run -

样品运行 -

In [53]: x = np.array([0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0, 1000, 1000, 1000])

In [54]: m = x!=0

In [55]: np.split(x,np.flatnonzero(m[1:] != m[:-1])+1)
Out[55]: 
[array([0, 0, 0, 0, 0, 0, 0, 0]),
 array([1000]),
 array([0, 0, 0, 0]),
 array([1000, 1000, 1000])]

---

To have an array of arrays as output, wrap it with np.array() -

要将数组数组作为输出,请使用np.array()包装它 -

In [56]: np.array(np.split(x,np.flatnonzero(m[1:] != m[:-1])+1))
Out[56]: 
array([array([0, 0, 0, 0, 0, 0, 0, 0]), array([1000]), array([0, 0, 0, 0]),
       array([1000, 1000, 1000])], dtype=object)

---

For a list of lists as output, here's one way -

对于列表作为输出,这是一种方式 -

m = x!=0
idx = np.r_[0, np.flatnonzero(m[1:] != m[:-1])+1, x.size]
out = [x.tolist()[idx[i]:idx[i+1]] for i in range(len(idx)-1)]

Sample run -

样品运行 -

In [94]: m = x!=0

In [95]: idx = np.r_[0, np.flatnonzero(m[1:] != m[:-1])+1, x.size]

In [96]: [x.tolist()[idx[i]:idx[i+1]] for i in range(len(idx)-1)]
Out[96]: [[0, 0, 0, 0, 0, 0, 0, 0], [1000], [0, 0, 0, 0], [1000, 1000, 1000]]

#2

Try this in python:

在python中尝试这个:

a = [0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0, 1000, 1000, 1000]
out = []
flag = 0
tmp = []
for i in a:
    if not flag and i==0:
        tmp.append(i)
    elif not flag and i==1000:
        out.append(tmp)
        tmp=[i]
        flag=1
    elif flag and i==1000:
        tmp.append(i)
    else:
        out.append(tmp)
        tmp=[i]
        flag=0
out.append(tmp)
print out