cf.Field.laplacian_xy¶

Field.
laplacian_xy
(x_wrap=None, one_sided_at_boundary=False, radius=None)[source]¶ Calculate the Laplacian in XY coordinates.
The horizontal Laplacian of a scalar function is calculated from a field that has dimension coordinates of X and Y, in either Cartesian (e.g. plane projection) or spherical polar coordinate systems.
The horizontal Laplacian in Cartesian coordinates is given by:
\[\nabla^2 f(x, y) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\]The horizontal Laplacian in spherical polar coordinates is given by:
\[\nabla^2 f(\theta, \phi) = \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2 f}{\partial \phi^2}\]where r is radial distance to the origin, \(\theta\) is the polar angle with respect to polar axis, and \(\phi\) is the azimuthal angle.
The Laplacian is calculated using centred finite differences apart from at the boundaries (see the x_wrap and one_sided_at_boundary parameters). If missing values are present then missing values will be returned at all points where a centred finite difference could not be calculated.
New in version 3.12.0.
See also
 Parameters
 x_wrap:
bool
, optional Whether the X axis is cyclic or not. By default x_wrap is set to the result of this call to the field construct’s
iscyclic
method:f.iscyclic('X')
. If the X axis is cyclic then centred differences at one boundary will always use values from the other boundary, regardless of the setting of one_sided_at_boundary.The cyclicity of the Y axis is always set to the result of
f.iscyclic('Y')
. one_sided_at_boundary:
bool
, optional If True then onesided finite differences are calculated at the noncyclic boundaries. By default missing values are set at noncyclic boundaries.
 radius: optional
Specify the radius of the latitudelongitude plane defined in spherical polar coordinates. The radius is that which would be returned by this call of the field construct’s
radius
method:f.radius(default=radius)
. The radius is defined by the datum of a coordinate reference construct, and if and only if no such radius is found then the default value given by the radius parameter is used instead. A value of'earth'
is equivalent to a default value of 6371229 metres.
 x_wrap:
 Returns
>>> f = cf.example_field(0) >>> print(f) Field: specific_humidity (ncvar%q)  Data : specific_humidity(latitude(5), longitude(8)) 1 Cell methods : area: mean Dimension coords: latitude(5) = [75.0, ..., 75.0] degrees_north : longitude(8) = [22.5, ..., 337.5] degrees_east : time(1) = [20190101 00:00:00] >>> f[...] = 0.1 >>> print(f.array) [[0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1] [0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1] [0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1] [0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1] [0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1]] >>> lp = f.laplacian_xy(radius='earth') >>> lp <CF Field: long_name=XY Laplacian of specific_humidity(latitude(5), longitude(8)) m2.rad2> >>> print(lp.array) [[       ] [       ] [0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0] [       ] [       ]] >>> lp = f.laplacian_xy(radius='earth', one_sided_at_boundary=True) >>> print(lp.array) [[0. 0. 0. 0. 0. 0. 0. 0.] [0. 0. 0. 0. 0. 0. 0. 0.] [0. 0. 0. 0. 0. 0. 0. 0.] [0. 0. 0. 0. 0. 0. 0. 0.] [0. 0. 0. 0. 0. 0. 0. 0.]]