# Licensed under a 3-clause BSD style license - see LICENSE.rst
"""
Convenience functions for `astropy.cosmology`.
"""
import warnings
import numpy as np
from .core import CosmologyError
from astropy.units import Quantity
__all__ = ['z_at_value']
__doctest_requires__ = {'*': ['scipy.integrate']}
[docs]def z_at_value(func, fval, zmin=1e-8, zmax=1000, ztol=1e-8, maxfun=500):
""" Find the redshift ``z`` at which ``func(z) = fval``.
This finds the redshift at which one of the cosmology functions or
methods (for example Planck13.distmod) is equal to a known value.
.. warning::
Make sure you understand the behavior of the function that you
are trying to invert! Depending on the cosmology, there may not
be a unique solution. For example, in the standard Lambda CDM
cosmology, there are two redshifts which give an angular
diameter distance of 1500 Mpc, z ~ 0.7 and z ~ 3.8. To force
``z_at_value`` to find the solution you are interested in, use the
``zmin`` and ``zmax`` keywords to limit the search range (see the
example below).
Parameters
----------
func : function or method
A function that takes a redshift as input.
fval : astropy.Quantity instance
The value of ``func(z)``.
zmin : float, optional
The lower search limit for ``z``. Beware of divergences
in some cosmological functions, such as distance moduli,
at z=0 (default 1e-8).
zmax : float, optional
The upper search limit for ``z`` (default 1000).
ztol : float, optional
The relative error in ``z`` acceptable for convergence.
maxfun : int, optional
The maximum number of function evaluations allowed in the
optimization routine (default 500).
Returns
-------
z : float
The redshift ``z`` satisfying ``zmin < z < zmax`` and ``func(z) =
fval`` within ``ztol``.
Notes
-----
This works for any arbitrary input cosmology, but is inefficient
if you want to invert a large number of values for the same
cosmology. In this case, it is faster to instead generate an array
of values at many closely-spaced redshifts that cover the relevant
redshift range, and then use interpolation to find the redshift at
each value you're interested in. For example, to efficiently find
the redshifts corresponding to 10^6 values of the distance modulus
in a Planck13 cosmology, you could do the following:
>>> import astropy.units as u
>>> from astropy.cosmology import Planck13, z_at_value
Generate 10^6 distance moduli between 24 and 43 for which we
want to find the corresponding redshifts:
>>> Dvals = (24 + np.random.rand(1e6) * 20) * u.mag
Make a grid of distance moduli covering the redshift range we
need using 50 equally log-spaced values between zmin and
zmax. We use log spacing to adequately sample the steep part of
the curve at low distance moduli:
>>> zmin = z_at_value(Planck13.distmod, Dvals.min())
>>> zmax = z_at_value(Planck13.distmod, Dvals.max())
>>> zgrid = np.logspace(np.log10(zmin), np.log10(zmax), 50)
>>> Dgrid = Planck13.distmod(zgrid)
Finally interpolate to find the redshift at each distance modulus:
>>> zvals = np.interp(Dvals.value, zgrid, Dgrid.value)
Examples
--------
>>> import astropy.units as u
>>> from astropy.cosmology import Planck13, z_at_value
The age and lookback time are monotonic with redshift, and so a
unique solution can be found:
>>> z_at_value(Planck13.age, 2 * u.Gyr)
3.19812268...
The angular diameter is not monotonic however, and there are two
redshifts that give a value of 1500 Mpc. Use the zmin and zmax keywords
to find the one you're interested in:
>>> z_at_value(Planck13.angular_diameter_distance, 1500 * u.Mpc, zmax=1.5)
0.6812769577...
>>> z_at_value(Planck13.angular_diameter_distance, 1500 * u.Mpc, zmin=2.5)
3.7914913242...
Also note that the luminosity distance and distance modulus (two
other commonly inverted quantities) are monotonic in flat and open
universes, but not in closed universes.
"""
from scipy.optimize import fminbound
fval_zmin = func(zmin)
fval_zmax = func(zmax)
if np.sign(fval - fval_zmin) != np.sign(fval_zmax - fval):
warnings.warn("""\
fval is not bracketed by func(zmin) and func(zmax). This means either
there is no solution, or that there is more than one solution between
zmin and zmax satisfying fval = func(z).""")
if isinstance(fval_zmin, Quantity):
val = fval.to_value(fval_zmin.unit)
f = lambda z: abs(func(z).value - val)
else:
f = lambda z: abs(func(z) - fval)
zbest, resval, ierr, ncall = fminbound(f, zmin, zmax, maxfun=maxfun,
full_output=1, xtol=ztol)
if ierr != 0:
warnings.warn('Maximum number of function calls ({}) reached'.format(
ncall))
if np.allclose(zbest, zmax):
raise CosmologyError("Best guess z is very close the upper z limit.\n"
"Try re-running with a different zmax.")
elif np.allclose(zbest, zmin):
raise CosmologyError("Best guess z is very close the lower z limit.\n"
"Try re-running with a different zmin.")
return zbest