Name	Size	Mode	Actions
arraypad.cpython-36.pyc	21827	0644	editdlrm
arraysetops.cpython-36.pyc	21808	0644	editdlrm
arrayterator.cpython-36.pyc	7021	0644	editdlrm
financial.cpython-36.pyc	30060	0644	editdlrm
format.cpython-36.pyc	25179	0644	editdlrm
function_base.cpython-36.pyc	137226	0644	editdlrm
histograms.cpython-36.pyc	32023	0644	editdlrm
index_tricks.cpython-36.pyc	27910	0644	editdlrm
mixins.cpython-36.pyc	7112	0644	editdlrm
nanfunctions.cpython-36.pyc	54460	0644	editdlrm
npyio.cpython-36.pyc	67062	0644	editdlrm
polynomial.cpython-36.pyc	38177	0644	editdlrm
recfunctions.cpython-36.pyc	47325	0644	editdlrm
scimath.cpython-36.pyc	15349	0644	editdlrm
setup.cpython-36.pyc	550	0644	editdlrm
shape_base.cpython-36.pyc	35848	0644	editdlrm
stride_tricks.cpython-36.pyc	8315	0644	editdlrm
twodim_base.cpython-36.pyc	28023	0644	editdlrm
type_check.cpython-36.pyc	19237	0644	editdlrm
ufunclike.cpython-36.pyc	7841	0644	editdlrm
user_array.cpython-36.pyc	11338	0644	editdlrm
utils.cpython-36.pyc	24641	0644	editdlrm
_datasource.cpython-36.pyc	20400	0644	editdlrm
_iotools.cpython-36.pyc	25903	0644	editdlrm
_version.cpython-36.pyc	4819	0644	editdlrm
__init__.cpython-36.pyc	1570	0644	editdlrm

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}tj|dk||�SdS)a  Compute the number of periodic payments. .. deprecated:: 1.18 `nper` is deprecated; for details, see NEP 32 [1]_. Use the corresponding function in the numpy-financial library, https://pypi.org/project/numpy-financial. :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 References ---------- .. [1] NumPy Enhancement Proposal (NEP) 32, https://numpy.org/neps/nep-0032-remove-financial-functions.html 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(np.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]]]) F�raise)
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|||�|}y>> 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 rr
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|�}||||S)a� Compute the present value. .. deprecated:: 1.18 `pv` is deprecated; for details, see NEP 32 [1]_. Use the corresponding function in the numpy-financial library, https://pypi.org/project/numpy-financial. 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 ---------- .. [1] NumPy Enhancement Proposal (NEP) 32, https://numpy.org/neps/nep-0032-remove-financial-functions.html .. [2] 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]) # may vary 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. rr
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}|S)a Return the Internal Rate of Return (IRR). .. deprecated:: 1.18 `irr` is deprecated; for details, see NEP 32 [1]_. Use the corresponding function in the numpy-financial library, https://pypi.org/project/numpy-financial. 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: [2]_ .. math:: \sum_{t=0}^M{\frac{v_t}{(1+irr)^{t}}} = 0 References ---------- .. [1] NumPy Enhancement Proposal (NEP) 32, https://numpy.org/neps/nep-0032-remove-financial-functions.html .. [2] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed., Addison-Wesley, 2003, pg. 348. Examples -------- >>> round(np.irr([-100, 39, 59, 55, 20]), 5) 0.28095 >>> round(np.irr([-100, 0, 0, 74]), 5) -0.0955 >>> round(np.irr([-100, 100, 0, -7]), 5) -0.0833 >>> round(np.irr([-100, 100, 0, 7]), 5) 0.06206 >>> round(np.irr([-5, 10.5, 1, -8, 1]), 5) 0.0886 Nrr
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