def test_float_modulus_exact(self):
# test that float results are exact for small integers. This also
# holds for the same integers scaled by powers of two.
nlst = list(range(-127, 0))
plst = list(range(1, 128))
dividend = nlst + [0] + plst
divisor = nlst + plst
arg = list(itertools.product(dividend, divisor))
tgt = list(divmod(*t) for t in arg)
a, b = np.array(arg, dtype=int).T
# convert exact integer results from Python to float so that
# signed zero can be used, it is checked.
tgtdiv, tgtrem = np.array(tgt, dtype=float).T
tgtdiv = np.where((tgtdiv == 0.0) & ((b < 0) ^ (a < 0)), -0.0, tgtdiv)
tgtrem = np.where((tgtrem == 0.0) & (b < 0), -0.0, tgtrem)
for op in [floordiv_and_mod, divmod]:
for dt in np.typecodes['Float']:
msg = 'op: %s, dtype: %s' % (op.__name__, dt)
fa = a.astype(dt)
fb = b.astype(dt)
# use list comprehension so a_ and b_ are scalars
div, rem = zip(*[op(a_, b_) for a_, b_ in zip(fa, fb)])
assert_equal(div, tgtdiv, err_msg=msg)
assert_equal(rem, tgtrem, err_msg=msg)
def test_1(self):
"""
Tests creation
"""
x = np.array([1., 1., 1., -2., pi/2.0, 4., 5., -10., 10., 1., 2., 3.])
y = np.array([5., 0., 3., 2., -1., -4., 0., -10., 10., 1., 0., 3.])
m1 = [1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0]
m2 = [0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1]
xm = self.masked_array(x, mask=m1)
ym = self.masked_array(y, mask=m2)
xf = np.where(m1, 1.e+20, x)
xm.set_fill_value(1.e+20)
assert((xm-ym).filled(0).any())
s = x.shape
assert(xm.size == reduce(lambda x, y:x*y, s))
assert(self.count(xm) == len(m1) - reduce(lambda x, y:x+y, m1))
for s in [(4, 3), (6, 2)]:
x.shape = s
y.shape = s
xm.shape = s
ym.shape = s
xf.shape = s
assert(self.count(xm) == len(m1) - reduce(lambda x, y:x+y, m1))
def setup(self):
x = np.array([1., 1., 1., -2., pi/2.0, 4., 5., -10., 10., 1., 2., 3.])
y = np.array([5., 0., 3., 2., -1., -4., 0., -10., 10., 1., 0., 3.])
a10 = 10.
m1 = [1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0]
m2 = [0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1]
xm = array(x, mask=m1)
ym = array(y, mask=m2)
z = np.array([-.5, 0., .5, .8])
zm = array(z, mask=[0, 1, 0, 0])
xf = np.where(m1, 1e+20, x)
s = x.shape
xm.set_fill_value(1e+20)
self.d = (x, y, a10, m1, m2, xm, ym, z, zm, xf, s)
def test_operate_4d_array(self):
a = np.zeros((3, 3, 3, 3), int)
fill_diagonal(a, 4)
i = np.array([0, 1, 2])
assert_equal(np.where(a != 0), (i, i, i, i))
def fv(rate, nper, pmt, pv, when='end'):
"""
Compute the future value.
Given:
* a present value, `pv`
* an interest `rate` compounded once per period, of which
there are
* `nper` total
* a (fixed) payment, `pmt`, paid either
* at the beginning (`when` = {'begin', 1}) or the end
(`when` = {'end', 0}) of each period
Return:
the value at the end of the `nper` periods
Parameters
----------
rate : scalar or array_like of shape(M, )
Rate of interest as decimal (not per cent) per period
nper : scalar or array_like of shape(M, )
Number of compounding periods
pmt : scalar or array_like of shape(M, )
Payment
pv : scalar or array_like of shape(M, )
Present value
when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
When payments are due ('begin' (1) or 'end' (0)).
Defaults to {'end', 0}.
Returns
-------
out : ndarray
Future values. If all input is scalar, returns a scalar float. If
any input is array_like, returns future values for each input element.
If multiple inputs are array_like, they all must have the same shape.
Notes
-----
The future value is computed by solving the equation::
fv +
pv*(1+rate)**nper +
pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) == 0
or, when ``rate == 0``::
fv + pv + pmt * nper == 0
References
----------
.. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
Open Document Format for Office Applications (OpenDocument)v1.2,
Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
Pre-Draft 12. Organization for the Advancement of Structured Information
Standards (OASIS). Billerica, MA, USA. [ODT Document].
Available:
http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
OpenDocument-formula-20090508.odt
Examples
--------
What is the future value after 10 years of saving $100 now, with
an additional monthly savings of $100. Assume the interest rate is
5% (annually) compounded monthly?
>>> np.fv(0.05/12, 10*12, -100, -100)
15692.928894335748
By convention, the negative sign represents cash flow out (i.e. money not
available today). Thus, saving $100 a month at 5% annual interest leads
to $15,692.93 available to spend in 10 years.
If any input is array_like, returns an array of equal shape. Let's
compare different interest rates from the example above.
>>> a = np.array((0.05, 0.06, 0.07))/12
>>> np.fv(a, 10*12, -100, -100)
array([ 15692.92889434, 16569.87435405, 17509.44688102])
"""
when = _convert_when(when)
(rate, nper, pmt, pv, when) = map(np.asarray, [rate, nper, pmt, pv, when])
temp = (1 + rate)**nper
fact = np.where(rate == 0, nper, (1 + rate * when) * (temp - 1) / rate)
return -(pv * temp + pmt * fact)
def pv(rate, nper, pmt, fv=0, when='end'):
"""
Compute the present value.
Given:
* a future value, `fv`
* an interest `rate` compounded once per period, of which
there are
* `nper` total
* a (fixed) payment, `pmt`, paid either
* at the beginning (`when` = {'begin', 1}) or the end
(`when` = {'end', 0}) of each period
Return:
the value now
Parameters
----------
rate : array_like
Rate of interest (per period)
nper : array_like
Number of compounding periods
pmt : array_like
Payment
fv : array_like, optional
Future value
when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
When payments are due ('begin' (1) or 'end' (0))
Returns
-------
out : ndarray, float
Present value of a series of payments or investments.
Notes
-----
The present value is computed by solving the equation::
fv +
pv*(1 + rate)**nper +
pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) = 0
or, when ``rate = 0``::
fv + pv + pmt * nper = 0
for `pv`, which is then returned.
References
----------
.. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
Open Document Format for Office Applications (OpenDocument)v1.2,
Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
Pre-Draft 12. Organization for the Advancement of Structured Information
Standards (OASIS). Billerica, MA, USA. [ODT Document].
Available:
http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
OpenDocument-formula-20090508.odt
Examples
--------
What is the present value (e.g., the initial investment)
of an investment that needs to total $15692.93
after 10 years of saving $100 every month? Assume the
interest rate is 5% (annually) compounded monthly.
>>> np.pv(0.05/12, 10*12, -100, 15692.93)
-100.00067131625819
By convention, the negative sign represents cash flow out
(i.e., money not available today). Thus, to end up with
$15,692.93 in 10 years saving $100 a month at 5% annual
interest, one's initial deposit should also be $100.
If any input is array_like, ``pv`` returns an array of equal shape.
Let's compare different interest rates in the example above:
>>> a = np.array((0.05, 0.04, 0.03))/12
>>> np.pv(a, 10*12, -100, 15692.93)
array([ -100.00067132, -649.26771385, -1273.78633713])
So, to end up with the same $15692.93 under the same $100 per month
"savings plan," for annual interest rates of 4% and 3%, one would
need initial investments of $649.27 and $1273.79, respectively.
"""
when = _convert_when(when)
(rate, nper, pmt, fv, when) = map(np.asarray, [rate, nper, pmt, fv, when])
temp = (1 + rate)**nper
fact = np.where(rate == 0, nper, (1 + rate * when) * (temp - 1) / rate)
return -(fv + pmt * fact) / temp
def ipmt(rate, per, nper, pv, fv=0, when='end'):
"""
Compute the interest portion of a payment.
Parameters
----------
rate : scalar or array_like of shape(M, )
Rate of interest as decimal (not per cent) per period
per : scalar or array_like of shape(M, )
Interest paid against the loan changes during the life or the loan.
The `per` is the payment period to calculate the interest amount.
nper : scalar or array_like of shape(M, )
Number of compounding periods
pv : scalar or array_like of shape(M, )
Present value
fv : scalar or array_like of shape(M, ), optional
Future value
when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
When payments are due ('begin' (1) or 'end' (0)).
Defaults to {'end', 0}.
Returns
-------
out : ndarray
Interest portion of payment. If all input is scalar, returns a scalar
float. If any input is array_like, returns interest payment for each
input element. If multiple inputs are array_like, they all must have
the same shape.
See Also
--------
ppmt, pmt, pv
Notes
-----
The total payment is made up of payment against principal plus interest.
``pmt = ppmt + ipmt``
Examples
--------
What is the amortization schedule for a 1 year loan of $2500 at
8.24% interest per year compounded monthly?
>>> principal = 2500.00
The 'per' variable represents the periods of the loan. Remember that
financial equations start the period count at 1!
>>> per = np.arange(1*12) + 1
>>> ipmt = np.ipmt(0.0824/12, per, 1*12, principal)
>>> ppmt = np.ppmt(0.0824/12, per, 1*12, principal)
Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal
'pmt'.
>>> pmt = np.pmt(0.0824/12, 1*12, principal)
>>> np.allclose(ipmt + ppmt, pmt)
True
>>> fmt = '{0:2d} {1:8.2f} {2:8.2f} {3:8.2f}'
>>> for payment in per:
... index = payment - 1
... principal = principal + ppmt[index]
... print(fmt.format(payment, ppmt[index], ipmt[index], principal))
1 -200.58 -17.17 2299.42
2 -201.96 -15.79 2097.46
3 -203.35 -14.40 1894.11
4 -204.74 -13.01 1689.37
5 -206.15 -11.60 1483.22
6 -207.56 -10.18 1275.66
7 -208.99 -8.76 1066.67
8 -210.42 -7.32 856.25
9 -211.87 -5.88 644.38
10 -213.32 -4.42 431.05
11 -214.79 -2.96 216.26
12 -216.26 -1.49 -0.00
>>> interestpd = np.sum(ipmt)
>>> np.round(interestpd, 2)
-112.98
"""
when = _convert_when(when)
rate, per, nper, pv, fv, when = np.broadcast_arrays(
rate, per, nper, pv, fv, when)
total_pmt = pmt(rate, nper, pv, fv, when)
ipmt = _rbl(rate, per, total_pmt, pv, when) * rate
try:
ipmt = np.where(when == 1, ipmt / (1 + rate), ipmt)
ipmt = np.where(np.logical_and(when == 1, per == 1), 0, ipmt)
except IndexError:
pass
return ipmt
def nper(rate, pmt, pv, fv=0, when='end'):
"""
Compute the number of periodic payments.
:class:`decimal.Decimal` type is not supported.
Parameters
----------
rate : array_like
Rate of interest (per period)
pmt : array_like
Payment
pv : array_like
Present value
fv : array_like, optional
Future value
when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
When payments are due ('begin' (1) or 'end' (0))
Notes
-----
The number of periods ``nper`` is computed by solving the equation::
fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate*((1+rate)**nper-1) = 0
but if ``rate = 0`` then::
fv + pv + pmt*nper = 0
Examples
--------
If you only had $150/month to pay towards the loan, how long would it take
to pay-off a loan of $8,000 at 7% annual interest?
>>> print(round(np.nper(0.07/12, -150, 8000), 5))
64.07335
So, over 64 months would be required to pay off the loan.
The same analysis could be done with several different interest rates
and/or payments and/or total amounts to produce an entire table.
>>> np.nper(*(np.ogrid[0.07/12: 0.08/12: 0.01/12,
... -150 : -99 : 50 ,
... 8000 : 9001 : 1000]))
array([[[ 64.07334877, 74.06368256],
[ 108.07548412, 127.99022654]],
[[ 66.12443902, 76.87897353],
[ 114.70165583, 137.90124779]]])
"""
when = _convert_when(when)
(rate, pmt, pv, fv, when) = map(np.asarray, [rate, pmt, pv, fv, when])
use_zero_rate = False
with np.errstate(divide="raise"):
try:
z = pmt * (1 + rate * when) / rate
except FloatingPointError:
use_zero_rate = True
if use_zero_rate:
return (-fv + pv) / pmt
else:
A = -(fv + pv) / (pmt + 0)
B = np.log((-fv + z) / (pv + z)) / np.log(1 + rate)
return np.where(rate == 0, A, B)
def pmt(rate, nper, pv, fv=0, when='end'):
"""
Compute the payment against loan principal plus interest.
Given:
* a present value, `pv` (e.g., an amount borrowed)
* a future value, `fv` (e.g., 0)
* an interest `rate` compounded once per period, of which
there are
* `nper` total
* and (optional) specification of whether payment is made
at the beginning (`when` = {'begin', 1}) or the end
(`when` = {'end', 0}) of each period
Return:
the (fixed) periodic payment.
Parameters
----------
rate : array_like
Rate of interest (per period)
nper : array_like
Number of compounding periods
pv : array_like
Present value
fv : array_like, optional
Future value (default = 0)
when : {{'begin', 1}, {'end', 0}}, {string, int}
When payments are due ('begin' (1) or 'end' (0))
Returns
-------
out : ndarray
Payment against loan plus interest. If all input is scalar, returns a
scalar float. If any input is array_like, returns payment for each
input element. If multiple inputs are array_like, they all must have
the same shape.
Notes
-----
The payment is computed by solving the equation::
fv +
pv*(1 + rate)**nper +
pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) == 0
or, when ``rate == 0``::
fv + pv + pmt * nper == 0
for ``pmt``.
Note that computing a monthly mortgage payment is only
one use for this function. For example, pmt returns the
periodic deposit one must make to achieve a specified
future balance given an initial deposit, a fixed,
periodically compounded interest rate, and the total
number of periods.
References
----------
.. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
Open Document Format for Office Applications (OpenDocument)v1.2,
Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
Pre-Draft 12. Organization for the Advancement of Structured Information
Standards (OASIS). Billerica, MA, USA. [ODT Document].
Available:
http://www.oasis-open.org/committees/documents.php
?wg_abbrev=office-formulaOpenDocument-formula-20090508.odt
Examples
--------
What is the monthly payment needed to pay off a $200,000 loan in 15
years at an annual interest rate of 7.5%?
>>> np.pmt(0.075/12, 12*15, 200000)
-1854.0247200054619
In order to pay-off (i.e., have a future-value of 0) the $200,000 obtained
today, a monthly payment of $1,854.02 would be required. Note that this
example illustrates usage of `fv` having a default value of 0.
"""
when = _convert_when(when)
(rate, nper, pv, fv, when) = map(np.array, [rate, nper, pv, fv, when])
temp = (1 + rate)**nper
mask = (rate == 0)
masked_rate = np.where(mask, 1, rate)
fact = np.where(mask != 0, nper,
(1 + masked_rate * when) * (temp - 1) / masked_rate)
return -(fv + pv * temp) / fact
