英文:
Locating indices with element 1 and converting to a list in Python
问题
我有一个数组```A```。我想要找出所有包含元素1的索引并以列表形式打印出来。但是我遇到了一个错误。以下是预期的输出。
```python
import numpy as np
A = np.array([[1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0]])
A1 = np.where(A[0] == 1)
A1.tolist()
print(A1)
错误是
在<module>中
A1.tolist()
AttributeError: 'tuple' object has no attribute 'tolist'
预期的输出是
[[0, 2, 3, 5]]
英文:
I have an array A. I want to identify all indices with element 1 and print as a list. But I am getting an error. I present the expected output.
import numpy as np
A=np.array([[1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0]])
A1=np.where(A[0]==1)
A1.tolist()
print(A1)
The error is
in <module>
A1.tolist()
AttributeError: 'tuple' object has no attribute 'tolist'
The expected output is
[[0, 2, 3, 5]]
答案1
得分: 3
使用np.argwhere函数找到条件为True的索引（数组中非零的元素）：
res = np.argwhere(A[0] == 1).T
[[0 2 3 5]]
英文:
Use np.argwhere to find indices where condition is True (array elements that are non-zero):
res = np.argwhere(A[0] == 1).T
[[0 2 3 5]]
答案2
得分: 1
A1 = [i for i, x in enumerate(A[0]) if x == 1]
or
A1 = list(np.where(A[0] == 1)[0])
You have an array inside an array hence A[0]
英文:
A1 = [i for i,x in enumerate(A[0]) if x == 1]
or
A1=list(np.where(A[0]==1)[0])
You have an array inside an array hence A[0]
答案3
得分: 0
数组位于元组的零索引位置，因此执行 [A1[0].tolist()]，您将获得您期望的输出。
英文:
The array is at the zeroth index of the tuple, so do
[A1[0].tolist()] and you will have your expected output.
答案4
得分: 0
正如其他人指出的，你有一个二维数组，所以该函数返回每个维度的数组元组。然而，这并不能得到你期望的输出。有两种可能的方法可以实现这一点，一种是经典版本，另一种是基于你的输入数据（0和1）可以使用nonzero方法来完成的numpy版本：
# 经典版本
A1 = [idx for idx, v in enumerate(A[0]) if v == 1]
# numpy版本
A1 = np.nonzero(A[0])
nonzero也可以用于应用不同的条件来获取索引，比如等于2的情况：
np.nonzero(A[0] == 2)
你可以在这里了解更多关于nonzero的用法：numpy.nonzero文档。
英文:
As others have pointed out, you have a 2 dimensions array, so the function returns a tuple of array for each dimension. Still, this does not bring your desired output.
There 2 possible ways to achieve this, a classic version and the numpy version - which based on your input data (0 and 1) can be done with nonzero method:
# the classic version
A1 = [idx for idx, v in enumerate(A[0]) if v == 1]
# the numpy version
A1 = np.nonzero(A[0])
The nonzero can also be used to apply a different condition to get the indexes, let's say equal to 2.
np.nonzero(A[0] == 2)
You can read more about the nonzero use cases here: numpy.nonzero doc.
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