# Source code for pylops.signalprocessing.seislet

__all__ = ["Seislet"]

from math import ceil, log
from typing import Optional, Sequence

import numpy as np

from pylops import LinearOperator
from pylops.utils.typing import DTypeLike, NDArray

def _predict_trace(
trace: NDArray,
t: NDArray,
dt: float,
dx: float,
slope: NDArray,
) -> NDArray:
r"""Slope-based trace prediction.

Resample a trace to a new time axis defined by the local slopes along the
trace. Slopes do implicitly represent a time-varying time delay
:math:\Delta t (t) = dx*s(t).

The input trace is interpolated using sinc-interpolation to a new time
axis given by the following formula: :math:t_{new} = t - dx*s(t).

Parameters
----------
trace : :obj:numpy.ndarray
Trace
t : :obj:numpy.ndarray
Time axis
dt : :obj:float
Time axis sampling
dx : :obj:float
Spatial axis sampling
slope : :obj:numpy.ndarray
Slope field
adj : :obj:bool, optional
Perform forward (False) or adjoint (True) operation

Returns
-------
tracenew : :obj:numpy.ndarray
Resampled trace

"""
newt = t - dx * slope
sinc = np.tile(newt, (len(newt), 1)) - np.tile(t[:, np.newaxis], (1, len(newt)))
tracenew = np.dot(trace, np.sinc(sinc / dt).T)
else:
tracenew = np.dot(trace, np.sinc(sinc / dt))
return tracenew

def _predict_haar(
traces: NDArray,
dt: float,
dx: float,
slopes: NDArray,
repeat: int = 0,
backward: bool = False,
) -> NDArray:
"""Predict set of traces given time-varying slopes (Haar basis function)

A set of input traces are resampled based on local slopes. If the number
of traces in slopes is twice the number of traces in traces, the
resampling is done only once per trace. If the number of traces in
slopes is a multiple of 2 of the number of traces in traces,
the prediction is done recursively or in other words the output traces
are obtained by resampling the input traces followed by repeat-1
further resampling steps of the intermediate results.

Parameters
----------
traces : :obj:numpy.ndarray
Input traces of size :math:n_x \times n_t
dt : :obj:float
Time axis sampling of the slope field
dx : :obj:float
Spatial axis sampling of the slope field
slopes: :obj:numpy.ndarray
Slope field of size :math:n_x * 2^{repeat} \times n_t
repeat : :obj:int, optional
Number of repeated predictions
backward : :obj:bool, optional
Predicted trace is on the right (False) or on the left (True)
of input trace
adj : :obj:bool, optional
Perform forward (False) or adjoint (True) operation

Returns
-------
pred : :obj:numpy.ndarray
Predicted traces

"""
if backward:
iback = 1
idir = -1
else:
iback = 0
idir = 1
slopejump = 2 ** (repeat + 1)
repeat = 2**repeat

nx, nt = traces.shape
t = np.arange(nt) * dt
pred = np.zeros_like(traces)
for ix in range(nx):
pred_tmp = traces[ix]
for irepeat in range(repeat - 1, -1, -1):
pred_tmp = _predict_trace(
pred_tmp,
t,
dt,
idir * dx,
slopes[ix * slopejump + iback * repeat + idir * irepeat],
)
else:
for irepeat in range(repeat):
pred_tmp = _predict_trace(
pred_tmp,
t,
dt,
idir * dx,
slopes[ix * slopejump + iback * repeat + idir * irepeat],
)
pred[ix] = pred_tmp
return pred

def _predict_lin(
traces: NDArray,
dt: float,
dx: float,
slopes: NDArray,
repeat: int = 0,
backward: bool = False,
) -> NDArray:
"""Predict set of traces given time-varying slopes (Linear basis function)

See _predict_haar for details.
"""
if backward:
iback = 1
idir = -1
else:
iback = 0
idir = 1
slopejump = 2 ** (repeat + 1)
repeat = 2**repeat

nx, nt = traces.shape
t = np.arange(nt) * dt
pred = np.zeros_like(traces)
for ix in range(nx):
pred_tmp = traces[ix]
if not ((ix == 0 and not backward) or (ix == nx - 1 and backward)):
pred_tmp1 = traces[ix - idir]
for irepeat in range(repeat - 1, -1, -1):
if (ix == 0 and not backward) or (ix == nx - 1 and backward):
pred_tmp = _predict_trace(
pred_tmp,
t,
dt,
idir * dx,
slopes[ix * slopejump + iback * repeat + idir * irepeat],
)
pred_tmp1 = 0
else:
pred_tmp = _predict_trace(
pred_tmp,
t,
dt,
idir * dx,
slopes[ix * slopejump + iback * repeat + idir * irepeat],
)
pred_tmp1 = _predict_trace(
pred_tmp1,
t,
dt,
(-idir) * dx,
slopes[ix * slopejump + iback * repeat - idir * irepeat],
)
else:
if not ((ix == nx - 1 and not backward) or (ix == 0 and backward)):
pred_tmp1 = traces[ix + idir]
for irepeat in range(repeat):
if (ix == nx - 1 and not backward) or (ix == 0 and backward):
pred_tmp = _predict_trace(
pred_tmp,
t,
dt,
idir * dx,
slopes[ix * slopejump + iback * repeat + idir * irepeat],
)
pred_tmp1 = 0
else:
pred_tmp = _predict_trace(
pred_tmp,
t,
dt,
idir * dx,
slopes[ix * slopejump + iback * repeat + idir * irepeat],
)
pred_tmp1 = _predict_trace(
pred_tmp1,
t,
dt,
(-idir) * dx,
slopes[
(ix + idir) * slopejump + iback * repeat - idir * irepeat
],
)

# if (adj and ((ix == 0 and not backward) or (ix == nx - 1 and backward))) or
#    (ix == nx - 1 and not backward) or (ix == 0 and backward):
#    pred[ix] = pred_tmp
# else:
if ix == nx - 1:
pred[ix] = pred_tmp + pred_tmp1 / 2.0
else:
pred[ix] = (pred_tmp + pred_tmp1) / 2.0
return pred

[docs]class Seislet(LinearOperator):
r"""Two dimensional Seislet operator.

Apply 2D-Seislet Transform to an input array given an
estimate of its local slopes. In forward mode, the input array is
reshaped into a two-dimensional array of size :math:n_x \times n_t and
the transform is performed along the first (spatial) axis (see Notes for
more details).

Parameters
----------
slopes : :obj:numpy.ndarray
Slope field of size :math:n_x \times n_t
sampling : :obj:tuple, optional
Sampling steps in x- and t-axis.
level : :obj:int, optional
Number of scaling levels (must be >=0).
kind : :obj:str, optional
Basis function used for predict and update steps: haar or
linear.
inv : :obj:int, optional
Apply inverse transform when invoking the adjoint (True)
or not (False). Note that in some scenario it may be more
appropriate to use the exact inverse as adjoint of the Seislet
operator even if this is not an orthogonal operator and the dot-test
would not be satisfied (see Notes for details). Otherwise, the user
can access the inverse directly as method of this class.
dtype : :obj:str, optional
Type of elements in input array.
name : :obj:str, optional

Name of operator (to be used by :func:pylops.utils.describe.describe)

Attributes
----------
shape : :obj:tuple
Operator shape
explicit : :obj:bool
Operator contains a matrix that can be solved explicitly
(True) or not (False)

Raises
------
NotImplementedError
If kind is different from haar or linear
ValueError
If sampling has more or less than two elements.

Notes
-----
The Seislet transform [1]_ is implemented using the lifting scheme.

In its simplest form (i.e., corresponding to the Haar basis function for
the Wavelet transform) the input dataset is separated into even
(:math:\mathbf{e}) and odd (:math:\mathbf{o}) traces. Even traces are
used to forward predict the odd traces using local slopes and the
new odd traces (also referred to as residual) is defined as:

.. math::
\mathbf{o}^{i+1} = \mathbf{r}^i = \mathbf{o}^i - P(\mathbf{e}^i)

where :math:P = P^+ is the slope-based forward prediction operator
(which is here implemented as a sinc-based resampling).
The residual is then updated and summed to the even traces to obtain the
new even traces (also referred to as coarse representation):

.. math::
\mathbf{e}^{i+1} = \mathbf{c}^i = \mathbf{e}^i + U(\mathbf{o}^{i+1})

where :math:U = P^- / 2 is the update operator which performs a
slope-based backward prediction. At this point
:math:\mathbf{e}^{i+1} becomes the new data and the procedure is repeated
level times (at maximum until :math:\mathbf{e}^{i+1} is a single trace.
The Seislet transform is effectively composed of all residuals and
the coarsest data representation.

In the inverse transform the two operations are reverted. Starting from the
coarsest scale data representation :math:\mathbf{c} and residual
:math:\mathbf{r}, the even and odd parts of the previous scale are
reconstructed as:

.. math::
\mathbf{e}^i = \mathbf{c}^i - U(\mathbf{r}^i)
= \mathbf{e}^{i+1} - U(\mathbf{o}^{i+1})

and:

.. math::
\mathbf{o}^i  = \mathbf{r}^i + P(\mathbf{e}^i)
= \mathbf{o}^{i+1} + P(\mathbf{e}^i)

A new data is formed by interleaving :math:\mathbf{e}^i and
:math:\mathbf{o}^i and the procedure repeated until the new data as the
same number of traces as the original one.

Finally the adjoint operator can be easily derived by writing the lifting
scheme in a matricial form:

.. math::
\begin{bmatrix}
\mathbf{r}_1  \\ \mathbf{r}_2  \\ \vdots \\ \mathbf{r}_N \\
\mathbf{c}_1  \\ \mathbf{c}_2  \\ \vdots \\ \mathbf{c}_N
\end{bmatrix} =
\begin{bmatrix}
\mathbf{I} & \mathbf{0} & \ldots & \mathbf{0} & -\mathbf{P} & \mathbf{0}  & \ldots & \mathbf{0}  \\
\mathbf{0} & \mathbf{I} & \ldots & \mathbf{0} & \mathbf{0}  & -\mathbf{P} & \ldots & \mathbf{0}  \\
\vdots     & \vdots     & \ddots & \vdots     & \vdots      & \vdots      & \ddots & \vdots      \\
\mathbf{0} & \mathbf{0} & \ldots & \mathbf{I} & \mathbf{0}  & \mathbf{0}  & \ldots & -\mathbf{P} \\
\mathbf{U} & \mathbf{0} & \ldots & \mathbf{0} & \mathbf{I}-\mathbf{UP} & \mathbf{0}  & \ldots & \mathbf{0} \\
\mathbf{0} & \mathbf{U} & \ldots & \mathbf{0} & \mathbf{0}  & \mathbf{I}-\mathbf{UP} & \ldots & \mathbf{0} \\
\vdots     & \vdots     & \ddots & \vdots     & \vdots      & \vdots      & \ddots   & \vdots    \\
\mathbf{0} & \mathbf{0} & \ldots & \mathbf{U} & \mathbf{0}  & \mathbf{0}  & \ldots & \mathbf{I}-\mathbf{UP}
\end{bmatrix}
\begin{bmatrix}
\mathbf{o}_1  \\ \mathbf{o}_2  \\ \vdots \\ \mathbf{o}_N \\
\mathbf{e}_1  \\ \mathbf{e}_2  \\ \vdots \\ \mathbf{e}_N
\end{bmatrix}

.. math::
\begin{bmatrix}
\mathbf{o}_1  \\ \mathbf{o}_2  \\ \vdots \\ \mathbf{o}_N \\
\mathbf{e}_1  \\ \mathbf{e}_2  \\ \vdots \\ \mathbf{e}_N
\end{bmatrix} =
\begin{bmatrix}
\mathbf{I} & \mathbf{0} & \ldots & \mathbf{0} & -\mathbf{U^T} & \mathbf{0}    & \ldots & \mathbf{0}  \\
\mathbf{0} & \mathbf{I} & \ldots & \mathbf{0} & \mathbf{0}    & -\mathbf{U^T} & \ldots & \mathbf{0}  \\
\vdots     & \vdots     & \ddots & \vdots     & \vdots        & \vdots        & \ddots & \vdots      \\
\mathbf{0} & \mathbf{0} & \ldots & \mathbf{I} & \mathbf{0} & \mathbf{0} & \ldots & -\mathbf{U^T} \\
\mathbf{P^T} & \mathbf{0} & \ldots & \mathbf{0} & \mathbf{I}-\mathbf{P^T U^T} & \mathbf{0} & \ldots & \mathbf{0}  \\
\mathbf{0} & \mathbf{P^T} & \ldots & \mathbf{0} & \mathbf{0} & \mathbf{I}-\mathbf{P^T U^T} & \ldots & \mathbf{0}  \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
\mathbf{0} & \mathbf{0} & \ldots & \mathbf{P^T} & \mathbf{0} & \mathbf{0} & \ldots & \mathbf{I}-\mathbf{P^T U^T}
\end{bmatrix}
\begin{bmatrix}
\mathbf{r}_1  \\ \mathbf{r}_2  \\ \vdots \\ \mathbf{r}_N \\
\mathbf{c}_1  \\ \mathbf{c}_2  \\ \vdots \\ \mathbf{c}_N
\end{bmatrix}

which can be written more easily in the following two steps:

.. math::
\mathbf{o} = \mathbf{r} + \mathbf{U}^H\mathbf{c}

and:

.. math::
\mathbf{e} = \mathbf{c} - \mathbf{P}^H(\mathbf{r} + \mathbf{U}^H(\mathbf{c})) =
\mathbf{c} - \mathbf{P}^H\mathbf{o}

Similar derivations follow for more complex wavelet bases.

.. [1] Fomel, S.,  Liu, Y., "Seislet transform and seislet frame",
Geophysics, 75, no. 3, V25-V38. 2010.

"""

def __init__(
self,
slopes: NDArray,
sampling: Sequence[float] = (1.0, 1.0),
level: Optional[int] = None,
kind: str = "haar",
inv: bool = False,
dtype: DTypeLike = "float64",
name: str = "S",
) -> None:
if len(sampling) != 2:
raise ValueError("provide two sampling steps")

# define predict and update steps
if kind == "haar":
self.predict = _predict_haar
elif kind == "linear":
self.predict = _predict_lin
else:
raise NotImplementedError("kind should be haar or linear")

# define padding for length to be power of 2
dims = slopes.shape
ndimpow2 = 2 ** ceil(log(dims[0], 2))
dimsd = [ndimpow2] + list(dims[1:])
super().__init__(dtype=np.dtype(dtype), dims=dims, dimsd=dimsd, name=name)

pad = [(0, ndimpow2 - self.dims[0])] + [(0, 0)] * (len(self.dims) - 1)
self.nx, self.nt = self.dimsd

# define levels
nlevels_max = int(np.log2(self.dimsd[0]))
levels_size = np.flip(np.array([2**i for i in range(nlevels_max)]))
if level is None:
levels_size = levels_size[:-1]
level = nlevels_max - 1
else:
levels_size = levels_size[:level]
self.level = level
self.levels_size = levels_size
self.levels_cum = np.insert(np.cumsum(self.levels_size), 0, 0)

self.dx, self.dt = sampling
self.inv = inv

def _matvec(self, x: NDArray) -> NDArray:
x = np.reshape(x, self.dimsd)
y = np.zeros((np.sum(self.levels_size) + self.levels_size[-1], self.nt))
for ilevel in range(self.level):
odd = x[1::2]
even = x[::2]
res = odd - self.predict(
even, self.dt, self.dx, self.slopes, repeat=ilevel, backward=False
)
x = (
even
+ self.predict(
res, self.dt, self.dx, self.slopes, repeat=ilevel, backward=True
)
/ 2.0
)
y[self.levels_cum[ilevel] : self.levels_cum[ilevel + 1]] = res
y[self.levels_cum[-1] :] = x
return y.ravel()

def _rmatvec(self, x: NDArray) -> NDArray:
if not self.inv:
x = np.reshape(x, self.dimsd)
y = x[self.levels_cum[-1] :]
for ilevel in range(self.level, 0, -1):
res = x[self.levels_cum[ilevel - 1] : self.levels_cum[ilevel]]
odd = (
res
+ self.predict(
y,
self.dt,
self.dx,
self.slopes,
repeat=ilevel - 1,
backward=True,
)
/ 2.0
)
even = y - self.predict(
odd,
self.dt,
self.dx,
self.slopes,
repeat=ilevel - 1,
backward=False,
)
y = np.zeros((2 * even.shape[0], self.nt))
y[1::2] = odd
y[::2] = even
else:
y = self.inverse(x)
return y

def inverse(self, x: NDArray) -> NDArray:
x = np.reshape(x, self.dimsd)
y = x[self.levels_cum[-1] :]
for ilevel in range(self.level, 0, -1):
res = x[self.levels_cum[ilevel - 1] : self.levels_cum[ilevel]]
even = (
y
- self.predict(
res, self.dt, self.dx, self.slopes, repeat=ilevel - 1, backward=True
)
/ 2.0
)
odd = res + self.predict(
even, self.dt, self.dx, self.slopes, repeat=ilevel - 1, backward=False
)
y = np.zeros((2 * even.shape[0], self.nt))
y[1::2] = odd
y[::2] = even