Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 3 is different from 100)
时间: 2024-03-30 16:33:47 浏览: 24
这个错误通常出现在使用numpy的ufunc(通用函数)时,它表示你正在尝试使用不兼容的尺寸计算。其中一个操作数的形状(shape)不符合ufunc的预期输入形状。在你的情况下,这个错误可能是由于你正在尝试使用大小为3的数组与大小为100的数组进行计算。
要解决这个问题,你需要检查你的代码并确保你的输入数组的形状(shape)相同。你可以使用numpy的reshape函数来改变数组的形状。如果你无法找到错误所在,你可以将你的代码和输入数据提供给我,我可以帮你找出问题并提供更具体的解决方案。
相关问题
matmul: Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 10 is different from 1)
This error message suggests that there is a problem with the dimensions of the input matrices in a matrix multiplication operation. Specifically, the first input matrix has a mismatch in its first dimension, which should match the second dimension of the second input matrix.
The gufunc signature (n?,k),(k,m?)->(n?,m?) indicates that the first input matrix has shape (n,k), the second input matrix has shape (k,m), and the output matrix should have shape (n,m).
The "size 10 is different from 1" part of the error message suggests that the first input matrix has a size of 10 in its first dimension, while the second input matrix has a size of 1 in its second dimension. This is not compatible with the gufunc signature, which requires the first dimension of the first matrix to match the second dimension of the second matrix.
To fix this error, you need to ensure that the shapes of the input matrices are compatible with the gufunc signature. This may involve reshaping or transposing the matrices, or modifying the code that generates them.
python报错ValueError: matmul: Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 2 is different from 1)
这个错误通常是由于矩阵乘法中输入矩阵的维数不匹配引起的。更具体地说,错误信息中的“核心维度0”指的是第一个输入矩阵的列数,而“大小2与1不同”则表示第二个输入矩阵的行数与第一个矩阵的列数不匹配。
例如,如果你尝试执行以下代码:
```python
import numpy as np
a = np.array([[1, 2], [3, 4], [5, 6]])
b = np.array([7, 8])
c = np.matmul(a, b)
```
就会出现上述错误,因为矩阵a的列数为2,而向量b的长度为1,无法进行矩阵乘法。
要解决这个问题,你需要确保输入矩阵的维度是正确的。根据矩阵乘法的规则,第一个矩阵的列数必须等于第二个矩阵的行数,否则无法进行乘法运算。
如果你不确定输入矩阵的维度,可以使用numpy的shape属性来查看其形状。例如,对于上述代码,你可以使用以下语句来查看a和b的形状:
```python
print(a.shape) # 输出(3, 2)
print(b.shape) # 输出(2,)
```
在这个例子中,你可以通过将向量b转换为一个2x1的矩阵来解决问题,如下所示:
```python
import numpy as np
a = np.array([[1, 2], [3, 4], [5, 6]])
b = np.array([[7], [8]])
c = np.matmul(a, b)
print(c) # 输出[[23], [53], [83]]
```
这里,向量b被转换为一个2x1的矩阵,使得它的行数与矩阵a的列数匹配。这样,你就可以成功地执行矩阵乘法并获得结果。