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# Some simple financial calculations
#  patterned after spreadsheet computations.

# There is some complexity in each function
#  so that the functions behave like ufuncs with
#  broadcasting and being able to be called with scalars
#  or arrays (or other sequences).
import numpy as np

__all__ = ['fv', 'pmt', 'nper', 'ipmt', 'ppmt', 'pv', 'rate',
           'irr', 'npv', 'mirr']

_when_to_num = {'end':0, 'begin':1,
                'e':0, 'b':1,
                0:0, 1:1,
                'beginning':1,
                'start':1,
                'finish':0}

def _convert_when(when):
    try:
        return _when_to_num[when]
    except KeyError:
        return [_when_to_num[x] for x in when]


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
    miter = np.broadcast(rate, nper, pmt, pv, when)
    zer = np.zeros(miter.shape)
    fact = np.where(rate==zer, nper+zer, (1+rate*when)*(temp-1)/rate+zer)
    return -(pv*temp + pmt*fact)

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.asarray, [rate, nper, pv, fv, when])
    temp = (1+rate)**nper
    miter = np.broadcast(rate, nper, pv, fv, when)
    zer = np.zeros(miter.shape)
    fact = np.where(rate==zer, nper+zer, (1+rate*when)*(temp-1)/rate+zer)
    return -(fv + pv*temp) / fact

def nper(rate, pmt, pv, fv=0, when='end'):
    """
    Compute the number of periods.

    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

    or, when ``rate == 0``::

      fv + pv + pmt * nper == 0

    Examples
    --------
    If you only had $150 to spend as payment, how long would it take to pay-off
    a loan of $8,000 at 7% annual interest?

    >>> np.nper(0.07/12, -150, 8000)
    64.073348770661852

    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[0.06/12:0.071/12:0.01/12, -200:-99:100, 6000:7001:1000]))
    array([[[ 32.58497782,  38.57048452],
            [ 71.51317802,  86.37179563]],
           [[ 33.07413144,  39.26244268],
            [ 74.06368256,  90.22989997]]])

    """
    when = _convert_when(when)
    rate, pmt, pv, fv, when = map(np.asarray, [rate, pmt, pv, fv, when])
    try:
        z = pmt*(1.0+rate*when)/rate
    except ZeroDivisionError:
        z = 0.0
    A = -(fv + pv)/(pmt+0.0)
    B = np.log((-fv+z) / (pv+z))/np.log(1.0+rate)
    miter = np.broadcast(rate, pmt, pv, fv, when)
    zer = np.zeros(miter.shape)
    return np.where(rate==zer, A+zer, B+zer) + 0.0

def ipmt(rate, per, nper, pv, fv=0.0, when='end'):
    """
    Not implemented. Compute the payment portion for loan interest.

    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``

    """
    total = pmt(rate, nper, pv, fv, when)
    # Now, compute the nth step in the amortization
    raise NotImplementedError

def ppmt(rate, per, nper, pv, fv=0.0, when='end'):
    """
    Not implemented. Compute the payment against loan principal.

    Parameters
    ----------
    rate : array_like
        Rate of interest (per period)
    per : array_like, int
        Amount paid against the loan changes.  The `per` is the period of
        interest.
    nper : array_like
        Number of compounding periods
    pv : array_like
        Present value
    fv : array_like, optional
        Future value
    when : {{'begin', 1}, {'end', 0}}, {string, int}
        When payments are due ('begin' (1) or 'end' (0))

    See Also
    --------
    pmt, pv, ipmt

    """
    total = pmt(rate, nper, pv, fv, when)
    return total - ipmt(rate, per, nper, pv, fv, when)

def pv(rate, nper, pmt, fv=0.0, 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
    miter = np.broadcast(rate, nper, pmt, fv, when)
    zer = np.zeros(miter.shape)
    fact = np.where(rate == zer, nper+zer, (1+rate*when)*(temp-1)/rate+zer)
    return -(fv + pmt*fact)/temp

# Computed with Sage
#  (y + (r + 1)^n*x + p*((r + 1)^n - 1)*(r*w + 1)/r)/(n*(r + 1)^(n - 1)*x - p*((r + 1)^n - 1)*(r*w + 1)/r^2 + n*p*(r + 1)^(n - 1)*(r*w + 1)/r + p*((r + 1)^n - 1)*w/r)

def _g_div_gp(r, n, p, x, y, w):
    t1 = (r+1)**n
    t2 = (r+1)**(n-1)
    return (y + t1*x + p*(t1 - 1)*(r*w + 1)/r)/(n*t2*x - p*(t1 - 1)*(r*w + 1)/(r**2) + n*p*t2*(r*w + 1)/r + p*(t1 - 1)*w/r)

# Use Newton's iteration until the change is less than 1e-6
#  for all values or a maximum of 100 iterations is reached.
#  Newton's rule is
#  r_{n+1} = r_{n} - g(r_n)/g'(r_n)
#     where
#  g(r) is the formula
#  g'(r) is the derivative with respect to r.
def rate(nper, pmt, pv, fv, when='end', guess=0.10, tol=1e-6, maxiter=100):
    """
    Compute the rate of interest per period.

    Parameters
    ----------
    nper : array_like
        Number of compounding periods
    pmt : array_like
        Payment
    pv : array_like
        Present value
    fv : array_like
        Future value
    when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
        When payments are due ('begin' (1) or 'end' (0))
    guess : float, optional
        Starting guess for solving the rate of interest
    tol : float, optional
        Required tolerance for the solution
    maxiter : int, optional
        Maximum iterations in finding the solution

    Notes
    -----
    The rate of interest is computed by iteratively solving the
    (non-linear) equation::

     fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1) = 0

    for ``rate``.

    References
    ----------
    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

    """
    when = _convert_when(when)
    nper, pmt, pv, fv, when = map(np.asarray, [nper, pmt, pv, fv, when])
    rn = guess
    iter = 0
    close = False
    while (iter < maxiter) and not close:
        rnp1 = rn - _g_div_gp(rn, nper, pmt, pv, fv, when)
        diff = abs(rnp1-rn)
        close = np.all(diff<tol)
        iter += 1
        rn = rnp1
    if not close:
        # Return nan's in array of the same shape as rn
        return np.nan + rn
    else:
        return rn

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.

    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
    --------
    >>> np.irr([-100, 39, 59, 55, 20])
    0.2809484211599611

    (Compare with the Example given for numpy.lib.financial.npv)

    """
    res = np.roots(values[::-1])
    # Find the root(s) between 0 and 1
    mask = (res.imag == 0) & (res.real > 0) & (res.real <= 1)
    res = res[mask].real
    if res.size == 0:
        return np.nan
    rate = 1.0/res - 1
    if rate.size == 1:
        rate = rate.item()
    return rate

def npv(rate, values):
    """
    Returns the NPV (Net Present Value) of a cash flow series.

    Parameters
    ----------
    rate : scalar
        The discount rate.
    values : array_like, shape(M, )
        The values of the time series of cash flows.  The (fixed) time
        interval between cash flow "events" must be the same as that
        for which `rate` is given (i.e., if `rate` is per year, then
        precisely a year is understood to elapse between each cash flow
        event).  By convention, investments or "deposits" are negative,
        income or "withdrawals" are positive; `values` must begin with
        the initial investment, thus `values[0]` will typically be
        negative.

    Returns
    -------
    out : float
        The NPV of the input cash flow series `values` at the discount `rate`.

    Notes
    -----
    Returns the result of: [G]_

    .. math :: \\sum_{t=0}^M{\\frac{values_t}{(1+rate)^{t}}}

    References
    ----------
    .. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
       Addison-Wesley, 2003, pg. 346.

    Examples
    --------
    >>> np.npv(0.281,[-100, 39, 59, 55, 20])
    -0.0066187288356340801

    (Compare with the Example given for numpy.lib.financial.irr)

    """
    values = np.asarray(values)
    return (values / (1+rate)**np.arange(1,len(values)+1)).sum(axis=0)

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, dtype=np.double)
    n = values.size
    pos = values > 0
    neg = values < 0
    if not (pos.any() and neg.any()):
        return np.nan
    numer = np.abs(npv(reinvest_rate, values*pos))*(1 + reinvest_rate)
    denom = np.abs(npv(finance_rate, values*neg))*(1 + finance_rate)
    return (numer/denom)**(1.0/(n - 1))*(1 + reinvest_rate) - 1