python有内置函数可以将矩阵转换为行阶梯形（也称为上三角）吗？
sympyMatrix.rref()In [8]: sympy.Matrix(np.random.random((4,4))).rref()
Out[8]:
([1, 1.42711055402454e-17, 0, -1.38777878078145e-17]
[0, 1.0, 0, 2.22044604925031e-16]
[0, -2.3388341405089e-16, 1, -2.22044604925031e-16]
[0, 3.65674099486992e-17, 0, 1.0],
[0, 1, 2, 3])
我同意@Mile的评论对@WinstonEwert的回答计算机没有理由不能以给定的精度执行RREF。
RREF的实现应该不会很复杂，而且matlab不知何故设法拥有这个功能，所以numpy也应该有。
我做了一个非常简单直接的实现，效率很低。但对于简单的矩阵来说它效果很好：
更新：
似乎，@JustMe 在这个
rreffrom numpy import *
def rref(mat,precision=0,GJ=False):
m,n = mat.shape
p,t = precision, 1e-1**precision
A = around(mat.astype(float).copy(),decimals=p )
if GJ:
A = hstack((A,identity(n)))
pcol = -1 #pivot colum
for i in range(m):
pcol += 1
if pcol >= n : break
#pivot index
pid = argmax( abs(A[i:,pcol]) )
#Row exchange
A[i,:],A[pid+i,:] = A[pid+i,:].copy(),A[i,:].copy()
#pivot with given precision
while pcol < n and abs(A[i,pcol]) < t:
pcol += 1
if pcol >= n : break
#pivot index
pid = argmax( abs(A[i:,pcol]) )
#Row exchange
A[i,:],A[pid+i,:] = A[pid+i,:].copy(),A[i,:].copy()
if pcol >= n : break
pivot = float(A[i,pcol])
for j in range(m):
if j == i: continue
mul = float(A[j,pcol])/pivot
A[j,:] = around(A[j,:] - A[i,:]*mul,decimals=p)
A[i,:] /= pivot
A[i,:] = around(A[i,:],decimals=p)
if GJ:
return A[:,:n].copy(),A[:,n:].copy()
else:
return A
这里有一些简单的测试
print("/*--------------------------------------/")
print("/ Simple TEST /")
print("/--------------------------------------*/")
A = array([[1,2,3],[4,5,6],[-7,8,9]])
R = rref(A,precision=6)
print("A:\n",A)
print("R:\n",R)
print()
print("With GJ ")
R,E = rref(A,precision=6,GJ=True)
print("R:\n",R)
print("E:\n",E)
print("AdotE:\n",around( dot(A,E), decimals=0))
print()
A = array([[0, 0, 1], [0, 1, 0]])
R = rref(A, precision=1)
print("A:\n",A)
print("R:\n",R)
print()
A = array([[1,2,2,2],[2,4,6,8],[3,6,8,10]])
R = rref(A,precision=6)
print("A:\n",A)
print("R:\n",around(R, decimals=0))
print()
print("/*--------------------------------------/")
print( "/ Not Invertable TEST /")
print( "/--------------------------------------*/")
A = array([
[2,2,4, 4],
[3,1,6, 2],
[5,3,10,6]])
R = rref(A,precision=2)
print("A:\n",A)
print("R:\n",R)
print()
print("A^{T}:\n",A.T)
R = rref(A.T,precision=10)
print("R:\n",R)
/*--------------------------------------/
/ Simple TEST /
/--------------------------------------*/
A:
[[ 1 2 3]
[ 4 5 6]
[-7 8 9]]
R:
[[1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]
With GJ
R:
[[1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]
E:
[[-0.071428 0.142857 -0.071429]
[-1.857142 0.714285 0.142857]
[ 1.595237 -0.523809 -0.071428]]
AdotE:
[[ 1. 0. 0.]
[ 0. 1. 0.]
[-0. 0. 1.]]
A:
[[0 0 1]
[0 1 0]]
R:
[[0. 1. 0.]
[0. 0. 1.]]
A:
[[ 1 2 2 2]
[ 2 4 6 8]
[ 3 6 8 10]]
R:
[[ 1. 2. 0. -2.]
[ 0. 0. 1. 2.]
[ 0. 0. 0. 0.]]
/*--------------------------------------/
/ Not Invertable TEST /
/--------------------------------------*/
A:
[[ 2 2 4 4]
[ 3 1 6 2]
[ 5 3 10 6]]
R:
[[ 1. 0. 2. 0.]
[-0. 1. -0. 2.]
[ 0. 0. 0. 0.]]
A^{T}:
[[ 2 3 5]
[ 2 1 3]
[ 4 6 10]
[ 4 2 6]]
R:
[[ 1. 0. 1.]
[-0. 1. 1.]
[ 0. 0. 0.]
[ 0. 0. 0.]]
参见 http://mail.scipy.org/pipermail/numpy-discussion/2008-November/038705.html
基本上：不要这样做。
rref 算法在计算机上实现时会产生太多不准确性。所以你要么想用另一种方式解决问题，要么使用 @aix 建议的符号。
您可以自己定义：
def rref(matrix):
A = np.array(matrix, dtype=np.float64)
i = 0 # row
j = 0 # column
while True:
# find next nonzero column
while all(A.T[j] == 0.0):
j += 1
# if reached the end, break
if j == len(A[0]) - 1 : break
# if a_ij == 0 find first row i_>=i with a
# nonzero entry in column j and swap rows i and i_
if A[i][j] == 0:
i_ = i
while A[i_][j] == 0:
i_ += 1
# if reached the end, break
if i_ == len(A) - 1 : break
A[[i, i_]] = A[[i_, i]]
# divide ith row a_ij to make it a_ij == 1
A[i] = A[i] / A[i][j]
# eliminate all other entries in the jth column by subtracting
# multiples of of the ith row from the others
for i_ in range(len(A)):
if i_ != i:
A[i_] = A[i_] - A[i] * A[i_][j] / A[i][j]
# if reached the end, break
if (i == len(A) - 1) or (j == len(A[0]) - 1): break
# otherwise, we continue
i += 1
j += 1
return A
这是一个工作版本，它几乎只是 MATLAB rref 函数的 numpy 版本：
def rref(A, tol=1.0e-12):
m, n = A.shape
i, j = 0, 0
jb = []
while i < m and j < n:
# Find value and index of largest element in the remainder of column j
k = np.argmax(np.abs(A[i:m, j])) + i
p = np.abs(A[k, j])
if p <= tol:
# The column is negligible, zero it out
A[i:m, j] = 0.0
j += 1
else:
# Remember the column index
jb.append(j)
if i != k:
# Swap the i-th and k-th rows
A[[i, k], j:n] = A[[k, i], j:n]
# Divide the pivot row i by the pivot element A[i, j]
A[i, j:n] = A[i, j:n] / A[i, j]
# Subtract multiples of the pivot row from all the other rows
for k in range(m):
if k != i:
A[k, j:n] -= A[k, j] * A[i, j:n]
i += 1
j += 1
# Finished
return A, jb
示例：
A = np.array([[16.0, 2, 3, 13], [5, 11, 10, 8],
[9, 7, 6, 12], [4, 14, 15, 1]])
Areduced, jb = rref(A)
print(f"The matrix as rank {len(jb)}")
print(Areduced)
对于任何矩阵（具有满秩或不足），您可以循环遍历行（或列，这是等效的）并查看哪一行对排名有贡献。 选择前 n 行，您将得到一个最多具有 n 阶的矩阵。
from numpy.linalg import matrix_rank
def get_linind_rows(M):
Mrows=[]
Mranks=[]
for i,Mrow in enumerate(M):
Mrows.append(Mrow)
Mrank = matrix_rank(Mrows) # this rank can be max i+1
#print(i,Mrow,Mrank)
Mranks.append(Mrank)
Mranks_diff = np.diff(Mranks,prepend=0) # see where the rank increases by 1
return M[Mranks_diff > 0] # select these rows as they are linear independent