Question

我想以正确的方式做到这一点：

data = np.array(data)
data =[
[1, 1, 2, 1],
[0, 1, 3, 2],
[0, 2, 3, 2],
[2, 4, 3, 1],
[0, 2, 1, 4],
[3, 1, 4, 1]]

这应该成为（删除以0开头的行）：

[1, 1, 2, 1]
[2, 4, 3, 1]
[3, 1, 4, 1]

到目前为止，我这样做了：

lines = []
for i in range(0, len(data[0])):
    if data[0,i] != 0:
        lines.append(data[:,i])
lines = np.array(lines)

然后我找到了这个好方法：

mask = 1 <= data[0,:]

现在我想将该掩码应用于该数组。该掩码显示：[True, False, False, True, False, True]。我该怎么做？

Answer 1

为什么不呢：

[ar for ar in data if ar[0] != 0]

这假设数组不为空。

Answer 2

I presume you have a numpy array based on the data[0,:] and data[0,i] you have in your question and you mean data[:, 0] :

import numpy as np

data = np.array([
    [1, 1, 2, 1],
    [0, 1, 3, 2],
    [0, 2, 3, 2],
    [2, 4, 3, 1],
    [0, 2, 1, 4],
    [3, 1, 4, 1]])

data = data[data[:,0] != 0]
print(data)

Output:

[[1 1 2 1]
 [2 4 3 1]
 [3 1 4 1]]

data[0,:] is the first row [1 1 2 1] not the first column

Answer 3

使用列表理解

In [56]: [elem for elem in data if elem[0] !=0]
Out[56]: [[1, 1, 2, 1], [2, 4, 3, 1], [3, 1, 4, 1]]

将蒙版应用于多维数组

3 个答案: