我试图获得在找到矩阵的`k`个最小特征值之前需要的迭代次数。尽管Scipy可以很容易地找到特征值，但我没有找到一种可以轻松获取所需迭代次数的方法。
在如下的代码块中，我认为将变量*i*中的计数器添加进去会解决我的问题，但这不可能是真的，因为*i*可能大于maxiter，这是一个明显的矛盾。
```python
import scipy.sparse.linalg._eigen.arpack as bark
#%%
def ct(v):    
    return np.conj(v.T)
def eigsb(A, k=6, M=None, sigma=None, which='LM', v0=None,
         ncv=None, maxiter=None, tol=0, return_eigenvectors=True,
         Minv=None, OPinv=None, OPpart=None):
        n = A.shape[0]
        mode = 1
        M_matvec = None
        Minv_matvec = None
        matvec = bark._aslinearoperator_with_dtype(A).matvec
        params = bark._UnsymmetricArpackParams(n, k, A.dtype.char, matvec, mode,
                                      M_matvec, Minv_matvec, sigma,
                                      ncv, v0, maxiter, which, tol)
        i=0
        while not params.converged:
            params.iterate()
            i=i+1
        return i,params.extract(return_eigenvectors)
###制作Hermitian矩阵
d = 50
mat = np.random.rand(d,d)+ 1j * (np.random.rand(d,d))
mat = mat - (np.random.rand(d,d)+ 1j * (np.random.rand(d,d)))
mat = mat + ct(mat)
###
i,(vals, vecs) = eigsb(mat, k=1, which='SR', ncv=6, maxiter=50)

<details>
<summary>英文:</summary>
I&#39;m trying to get a number of iterations needed until convergence for finding the `k` smallest eigenvalues of a matrix. While scipy can find the eigenvalues without much difficulty, there&#39;s no way that I see to easily get out the number of iterations needed.
In the code block as shown, my thought was that the addition of the counter in variable *i* would solve my problem, but this can&#39;t be the case as *i* can be greater than maxiter, a clear contradiction. 

    i=0
    while not params.converged:
        params.iterate()
        i=i+1
    return i,params.extract(return_eigenvectors)

Is there a different way I should be going about this? Or is there some conversion between the count *i* and the true iteration value I'm not seeing?
</details>
# 答案1
**得分**: 1
tol在这里起到了作用。我尝试将tol设置为非零正值，无论d或maxiter的值如何，它都会迭代相同的次数。尝试将tol设置为非零正值并检查。
参考链接：
https://sisl.readthedocs.io/en/v0.9.2/_modules/scipy/sparse/linalg/eigen/arpack/arpack.html
```python
class IterInv(LinearOperator):
    """
    IterInv:
       辅助类，使用迭代方法重复解决M*x=b。
    """
    def __init__(self, M, ifunc=gmres, tol=0):
        if tol <= 0:
            # 当tol=0时，ARPACK使用由LAPACK的_LAMCH函数计算的机器精度。
            # 我们应该匹配这个值。
            tol = 2 * np.finfo(M.dtype).eps
tol: float，可选
    特征值的相对精度（停止准则）
    默认值0表示机器精度。