def _format_function(x):
if np.abs(x) < 1:
return '.'
elif np.abs(x) < 2:
return 'o'
else:
return 'O'
def test_fabs(self):
# Test that np.abs(x +- 0j) == np.abs(x) (as mandated by C99 for cabs)
x = np.array([1+0j], dtype=complex)
assert_array_equal(np.abs(x), np.real(x))
x = np.array([complex(1, np.NZERO)], dtype=complex)
assert_array_equal(np.abs(x), np.real(x))
x = np.array([complex(np.inf, np.NZERO)], dtype=complex)
assert_array_equal(np.abs(x), np.real(x))
x = np.array([complex(np.nan, np.NZERO)], dtype=complex)
assert_array_equal(np.abs(x), np.real(x))
def trimcoef(c, tol=0):
"""
Remove "small" "trailing" coefficients from a polynomial.
"Small" means "small in absolute value" and is controlled by the
parameter `tol`; "trailing" means highest order coefficient(s), e.g., in
``[0, 1, 1, 0, 0]`` (which represents ``0 + x + x**2 + 0*x**3 + 0*x**4``)
both the 3-rd and 4-th order coefficients would be "trimmed."
Parameters
----------
c : array_like
1-d array of coefficients, ordered from lowest order to highest.
tol : number, optional
Trailing (i.e., highest order) elements with absolute value less
than or equal to `tol` (default value is zero) are removed.
Returns
-------
trimmed : ndarray
1-d array with trailing zeros removed. If the resulting series
would be empty, a series containing a single zero is returned.
Raises
------
ValueError
If `tol` < 0
See Also
--------
trimseq
Examples
--------
>>> from numpy1.polynomial import polyutils as pu
>>> pu.trimcoef((0,0,3,0,5,0,0))
array([ 0., 0., 3., 0., 5.])
>>> pu.trimcoef((0,0,1e-3,0,1e-5,0,0),1e-3) # item == tol is trimmed
array([ 0.])
>>> i = complex(0,1) # works for complex
>>> pu.trimcoef((3e-4,1e-3*(1-i),5e-4,2e-5*(1+i)), 1e-3)
array([ 0.0003+0.j , 0.0010-0.001j])
"""
if tol < 0:
raise ValueError("tol must be non-negative")
[c] = as_series([c])
[ind] = np.nonzero(np.abs(c) > tol)
if len(ind) == 0:
return c[:1] * 0
else:
return c[:ind[-1] + 1].copy()
def Calculation_Fitness(Pop, Popualation_size, K):
# 计算适应度
# Pop 种群
# Popualation_size 种群大小
FitnV = []
for iPopualation_Size in range(Popualation_Size):
# 目标函数
temp_FitnV = (5 * math.sqrt(2) * Pop[iPopualation_Size][4] /
(2 * 3.6)) * (K[0] * Pop[iPopualation_Size][0] +
K[1] * Pop[iPopualation_Size][1] +
K[2] * Pop[iPopualation_Size][2] +
K[3] * Pop[iPopualation_Size][3])
FitnV.append(np.abs(temp_FitnV))
return FitnV
def mirr(values, finance_rate, reinvest_rate):
"""
Modified internal rate of return.
Parameters
----------
values : array_like
Cash flows (must contain at least one positive and one negative
value) or nan is returned. The first value is considered a sunk
cost at time zero.
finance_rate : scalar
Interest rate paid on the cash flows
reinvest_rate : scalar
Interest rate received on the cash flows upon reinvestment
Returns
-------
out : float
Modified internal rate of return
"""
values = np.asarray(values)
n = values.size
# Without this explicit cast the 1/(n - 1) computation below
# becomes a float, which causes TypeError when using Decimal
# values.
if isinstance(finance_rate, Decimal):
n = Decimal(n)
pos = values > 0
neg = values < 0
if not (pos.any() and neg.any()):
return np.nan
numer = np.abs(npv(reinvest_rate, values * pos))
denom = np.abs(npv(finance_rate, values * neg))
return (numer / denom)**(1 / (n - 1)) * (1 + reinvest_rate) - 1
def almost(a, b, decimal=6, fill_value=True):
"""
Returns True if a and b are equal up to decimal places.
If fill_value is True, masked values considered equal. Otherwise,
masked values are considered unequal.
"""
m = mask_or(getmask(a), getmask(b))
d1 = filled(a)
d2 = filled(b)
if d1.dtype.char == "O" or d2.dtype.char == "O":
return np.equal(d1, d2).ravel()
x = filled(masked_array(d1, copy=False, mask=m), fill_value).astype(float_)
y = filled(masked_array(d2, copy=False, mask=m), 1).astype(float_)
d = np.around(np.abs(x - y), decimal) <= 10.0 ** (-decimal)
return d.ravel()
def irr(values):
"""
Return the Internal Rate of Return (IRR).
This is the "average" periodically compounded rate of return
that gives a net present value of 0.0; for a more complete explanation,
see Notes below.
:class:`decimal.Decimal` type is not supported.
Parameters
----------
values : array_like, shape(N,)
Input cash flows per time period. By convention, net "deposits"
are negative and net "withdrawals" are positive. Thus, for
example, at least the first element of `values`, which represents
the initial investment, will typically be negative.
Returns
-------
out : float
Internal Rate of Return for periodic input values.
Notes
-----
The IRR is perhaps best understood through an example (illustrated
using np.irr in the Examples section below). Suppose one invests 100
units and then makes the following withdrawals at regular (fixed)
intervals: 39, 59, 55, 20. Assuming the ending value is 0, one's 100
unit investment yields 173 units; however, due to the combination of
compounding and the periodic withdrawals, the "average" rate of return
is neither simply 0.73/4 nor (1.73)^0.25-1. Rather, it is the solution
(for :math:`r`) of the equation:
.. math:: -100 + \\frac{39}{1+r} + \\frac{59}{(1+r)^2}
+ \\frac{55}{(1+r)^3} + \\frac{20}{(1+r)^4} = 0
In general, for `values` :math:`= [v_0, v_1, ... v_M]`,
irr is the solution of the equation: [G]_
.. math:: \\sum_{t=0}^M{\\frac{v_t}{(1+irr)^{t}}} = 0
References
----------
.. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
Addison-Wesley, 2003, pg. 348.
Examples
--------
>>> round(irr([-100, 39, 59, 55, 20]), 5)
0.28095
>>> round(irr([-100, 0, 0, 74]), 5)
-0.0955
>>> round(irr([-100, 100, 0, -7]), 5)
-0.0833
>>> round(irr([-100, 100, 0, 7]), 5)
0.06206
>>> round(irr([-5, 10.5, 1, -8, 1]), 5)
0.0886
(Compare with the Example given for numpy.lib.financial.npv)
"""
# `np.roots` call is why this function does not support Decimal type.
#
# Ultimately Decimal support needs to be added to np.roots, which has
# greater implications on the entire linear algebra module and how it does
# eigenvalue computations.
res = np.roots(values[::-1])
mask = (res.imag == 0) & (res.real > 0)
if not mask.any():
return np.nan
res = res[mask].real
# NPV(rate) = 0 can have more than one solution so we return
# only the solution closest to zero.
rate = 1 / res - 1
rate = rate.item(np.argmin(np.abs(rate)))
return rate
def g(a, b):
return np.abs(complex(a, b))
def f(a):
return np.abs(np.conj(a))
def test_simple(self):
x = np.array([1+1j, 0+2j, 1+2j, np.inf, np.nan])
y_r = np.array([np.sqrt(2.), 2, np.sqrt(5), np.inf, np.nan])
y = np.abs(x)
for i in range(len(x)):
assert_almost_equal(y[i], y_r[i])
def test_simple(self):
x = np.array([1+0j, 1+2j])
y_r = np.log(np.abs(x)) + 1j * np.angle(x)
y = np.log(x)
for i in range(len(x)):
assert_almost_equal(y[i], y_r[i])
