def fit(self, data: pd.DataFrame, seed: Optional[int] = None):
# Given a d-dimensional random vector x and its (d,n) observed data matrix X,
# apply an ICA algorithm to obtain an estimate of A.
d = len(data.columns)
B = FastICA(random_state=seed).fit(data).components_
# Find the unique permutation of the rows of W = A^-1 that yields a matrix W'
# without any zeros on the main diagonal. The permutation is sought by minimizing
# sum_i (1/|W'_ii|). This minimization problem is the classical linear assignment
# problem, and here the Hungarian algorithm (Kuhn, 1955) is used.
_, K = hungarian(1 / np.abs(B))
B = B.take(K, 0)
# Divide each row of W' by its corresponding diagonal element in order to
# yield a new matrix W'' with a diagonal consisting entirely of 1s.
B /= B.diagonal()[..., None]
# Compute an estimate B' of B by using B' = I - W''.
B = np.identity(d) - B
# Finally, to estimate a causal order k(i), determine the permutation matrix
# K of B', obtaining the matrix B' = PB'K^T that is as close as possible
# to having a strictly lower triangular structure.
K = None
if d < 8:
# For a small number of variables, i.e., fewer than 8, the lower triangularity
# of B' can be measured by using the sum of squared bij in its upper triangular
# section sum_i<=j (b'_ij^2). In addition, an exhaustive search over all possible
# permutations is feasible and is hence performed.
vmin = np.inf
for p in permutations(range(d)):
score = np.sum(np.square(np.triu(B.take(p, 0))))
if score < vmin:
vmin = score
K = p
K = np.array(K)
else:
# For higher-dimensional data, the following approximate algorithm is used,
# which sets small absolute valued elements in B' to zero, and whereby it can be
# determined whether it is possible to permute the resulting matrix to become
# strictly lower triangular:
# (a) Set the d(d+1)/2 smallest (in absolute value) elements of B' to zero.
i = round(d * (d + 1) / 2)
pmin = np.argsort(np.abs(np.ravel(B)))
B.flat[pmin[:i]] = 0
# (b) Repeat
while K is None:
# i. Determine whether B' can be permuted to become strictly lower triangular.
# If this is possible, stop and return the permuted B'.
K, A, L = np.zeros(d, int), np.arange(d), B
while len(A) > 0:
# Find a row where all elements are zero, if any.
j = np.where(np.sum(np.abs(L), axis=1) == 0)
# If there is no row with zero elements, exit.
if len(j[0]) == 0:
K = None
break
# Select the first row with zero elements.
j = j[0][0]
# Add original index to permutation matrix.
K[d - len(A)] = A[j]
A = np.delete(A, j)
# Remove selected row and columns.
mask = np.delete(np.arange(len(L)), j)
L = L[mask][:, mask]
# ii. In addition, set the next smallest (in absolute value) element of Bb to zero.
if K is None:
B.flat[pmin[i]] = 0
i += 1
return K
