def __rsub__(s, t):
cls, new, (prec, rounding) = s._ctxdata
if type(t) in int_types:
v = new(cls)
v._mpf_ = mpf_sub(from_int(t), s._mpf_, prec, rounding)
return v
t = s.mpf_convert_lhs(t)
if t is NotImplemented:
return t
return t - s
def __rsub__(s, t):
cls, new, (prec, rounding) = s._ctxdata
if type(t) in int_types:
v = new(cls)
v._mpf_ = mpf_sub(from_int(t), s._mpf_, prec, rounding)
return v
t = s.mpf_convert_lhs(t)
if t is NotImplemented:
return t
return t - s
def fsub(ctx, x, y, **kwargs):
"""
Subtracts the numbers *x* and *y*, giving a floating-point result,
optionally using a custom precision and rounding mode.
See the documentation of :func:`fadd` for a detailed description
of how to specify precision and rounding.
**Examples**
Using :func:`fsub` with precision and rounding control::
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = False
>>> fsub(2, 1e-20)
mpf('2.0')
>>> fsub(2, 1e-20, rounding='d')
mpf('1.9999999999999998')
>>> nprint(fsub(2, 1e-20, prec=100), 25)
1.99999999999999999999
>>> nprint(fsub(2, 1e-20, dps=15), 25)
2.0
>>> nprint(fsub(2, 1e-20, dps=25), 25)
1.99999999999999999999
>>> nprint(fsub(2, 1e-20, exact=True), 25)
1.99999999999999999999
Exact subtraction avoids cancellation errors, enforcing familiar laws
of numbers such as `x-y+y = x`, which don't hold in floating-point
arithmetic with finite precision::
>>> x, y = mpf(2), mpf('1e1000')
>>> print x - y + y
0.0
>>> print fsub(x, y, prec=inf) + y
2.0
>>> print fsub(x, y, exact=True) + y
2.0
Exact addition can be inefficient and may be impossible to perform
with large magnitude differences::
>>> fsub(1, '1e-100000000000000000000', prec=inf)
Traceback (most recent call last):
...
OverflowError: the exact result does not fit in memory
"""
prec, rounding = ctx._parse_prec(kwargs)
x = ctx.convert(x)
y = ctx.convert(y)
try:
if hasattr(x, '_mpf_'):
if hasattr(y, '_mpf_'):
return ctx.make_mpf(mpf_sub(x._mpf_, y._mpf_, prec, rounding))
if hasattr(y, '_mpc_'):
return ctx.make_mpc(mpc_sub((x._mpf_, fzero), y._mpc_, prec, rounding))
if hasattr(x, '_mpc_'):
if hasattr(y, '_mpf_'):
return ctx.make_mpc(mpc_sub_mpf(x._mpc_, y._mpf_, prec, rounding))
if hasattr(y, '_mpc_'):
return ctx.make_mpc(mpc_sub(x._mpc_, y._mpc_, prec, rounding))
except (ValueError, OverflowError):
raise OverflowError(ctx._exact_overflow_msg)
raise ValueError("Arguments need to be mpf or mpc compatible numbers")
def fsub(ctx, x, y, **kwargs):
"""
Subtracts the numbers *x* and *y*, giving a floating-point result,
optionally using a custom precision and rounding mode.
See the documentation of :func:`~mpmath.fadd` for a detailed description
of how to specify precision and rounding.
**Examples**
Using :func:`~mpmath.fsub` with precision and rounding control::
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = False
>>> fsub(2, 1e-20)
mpf('2.0')
>>> fsub(2, 1e-20, rounding='d')
mpf('1.9999999999999998')
>>> nprint(fsub(2, 1e-20, prec=100), 25)
1.99999999999999999999
>>> nprint(fsub(2, 1e-20, dps=15), 25)
2.0
>>> nprint(fsub(2, 1e-20, dps=25), 25)
1.99999999999999999999
>>> nprint(fsub(2, 1e-20, exact=True), 25)
1.99999999999999999999
Exact subtraction avoids cancellation errors, enforcing familiar laws
of numbers such as `x-y+y = x`, which don't hold in floating-point
arithmetic with finite precision::
>>> x, y = mpf(2), mpf('1e1000')
>>> print x - y + y
0.0
>>> print fsub(x, y, prec=inf) + y
2.0
>>> print fsub(x, y, exact=True) + y
2.0
Exact addition can be inefficient and may be impossible to perform
with large magnitude differences::
>>> fsub(1, '1e-100000000000000000000', prec=inf)
Traceback (most recent call last):
...
OverflowError: the exact result does not fit in memory
"""
prec, rounding = ctx._parse_prec(kwargs)
x = ctx.convert(x)
y = ctx.convert(y)
try:
if hasattr(x, '_mpf_'):
if hasattr(y, '_mpf_'):
return ctx.make_mpf(mpf_sub(x._mpf_, y._mpf_, prec, rounding))
if hasattr(y, '_mpc_'):
return ctx.make_mpc(mpc_sub((x._mpf_, fzero), y._mpc_, prec, rounding))
if hasattr(x, '_mpc_'):
if hasattr(y, '_mpf_'):
return ctx.make_mpc(mpc_sub_mpf(x._mpc_, y._mpf_, prec, rounding))
if hasattr(y, '_mpc_'):
return ctx.make_mpc(mpc_sub(x._mpc_, y._mpc_, prec, rounding))
except (ValueError, OverflowError):
raise OverflowError(ctx._exact_overflow_msg)
raise ValueError("Arguments need to be mpf or mpc compatible numbers")