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我正在寻找一个简单的 javascript 财务 RATE 函数，我找到了这个。但这似乎太难理解了。我想简化这个功能，我需要你的帮助。如果有人有最简单的功能，请回答。（这是一个 excel RATE 函数等价物。）

var rate = function(nper, pmt, pv, fv, type, guess) {
  if (guess == null) guess = 0.01;
  if (fv == null) fv = 0;
  if (type == null) type = 0;

  var FINANCIAL_MAX_ITERATIONS = 128;//Bet accuracy with 128
  var FINANCIAL_PRECISION = 0.0000001;//1.0e-8

  var y, y0, y1, x0, x1 = 0, f = 0, i = 0;
  var rate = guess;
  if (Math.abs(rate) < FINANCIAL_PRECISION) {
     y = pv * (1 + nper * rate) + pmt * (1 + rate * type) * nper + fv;
  } else {
     f = Math.exp(nper * Math.log(1 + rate));
     y = pv * f + pmt * (1 / rate + type) * (f - 1) + fv;
  }
  y0 = pv + pmt * nper + fv;
  y1 = pv * f + pmt * (1 / rate + type) * (f - 1) + fv;

  // find root by Newton secant method
  i = x0 = 0.0;
  x1 = rate;
  while ((Math.abs(y0 - y1) > FINANCIAL_PRECISION) && (i < FINANCIAL_MAX_ITERATIONS)) {
     rate = (y1 * x0 - y0 * x1) / (y1 - y0);
     x0 = x1;
     x1 = rate;

     if (Math.abs(rate) < FINANCIAL_PRECISION) {
        y = pv * (1 + nper * rate) + pmt * (1 + rate * type) * nper + fv;
     } else {
        f = Math.exp(nper * Math.log(1 + rate));
        y = pv * f + pmt * (1 / rate + type) * (f - 1) + fv;
     }

     y0 = y1;
     y1 = y;
     ++i;
  }
  return rate;}

谢谢！

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1 回答 1

3

数学太复杂了，我无法理解，但这对您来说可能更容易阅读。一些变量已被重命名以更有意义，并且它的格式更容易让您看到

function rate(paymentsPerYear, paymentAmount, presentValue, futureValue, dueEndOrBeginning, interest)
{
    //If interest, futureValue, dueEndorBeginning was not set, set now
    if (interest == null)
        interest = 0.01;

    if (futureValue == null)
        futureValue = 0;

    if (dueEndOrBeginning == null)
        dueEndOrBeginning = 0;

    var FINANCIAL_MAX_ITERATIONS = 128;//Bet accuracy with 128
    var FINANCIAL_PRECISION = 0.0000001;//1.0e-8

    var y, y0, y1, x0, x1 = 0, f = 0, i = 0;
    var rate = interest;
    if (Math.abs(rate) < FINANCIAL_PRECISION)
    {
        y = presentValue * (1 + paymentsPerYear * rate) + paymentAmount * (1 + rate * dueEndOrBeginning) * paymentsPerYear + futureValue;
    }
    else
    {
        f = Math.exp(paymentsPerYear * Math.log(1 + rate));
        y = presentValue * f + paymentAmount * (1 / rate + dueEndOrBeginning) * (f - 1) + futureValue;
    }
    y0 = presentValue + paymentAmount * paymentsPerYear + futureValue;
    y1 = presentValue * f + paymentAmount * (1 / rate + dueEndOrBeginning) * (f - 1) + futureValue;

    // find root by Newton secant method
    i = x0 = 0.0;
    x1 = rate;
    while ((Math.abs(y0 - y1) > FINANCIAL_PRECISION)
        && (i < FINANCIAL_MAX_ITERATIONS))
    {
        rate = (y1 * x0 - y0 * x1) / (y1 - y0);
        x0 = x1;
        x1 = rate;

        if (Math.abs(rate) < FINANCIAL_PRECISION)
        {
            y = presentValue * (1 + paymentsPerYear * rate) + paymentAmount * (1 + rate * dueEndOrBeginning) * paymentsPerYear + futureValue;
        }
        else
        {
            f = Math.exp(paymentsPerYear * Math.log(1 + rate));
            y = presentValue * f + paymentAmount * (1 / rate + dueEndOrBeginning) * (f - 1) + futureValue;
        }

        y0 = y1;
        y1 = y;
        ++i;
    }
    return rate;
}
于 2012-08-22T00:37:41.840 回答