如果数据很大，Apache POI 率公式不起作用答案

【问题标题】：Apache POI rate formula not working if data is big如果数据很大，Apache POI 率公式不起作用
【发布时间】：2017-11-09 04:53:47
【问题描述】：

对于大值，费率公式无法按预期工作......

RATE(85.77534246575343, -1589.0, -18664.0, 5855586.0) 在物理文件中返回 0.05819488005
如果我们尝试通过 POI 设置的相同公式返回 0.009056339275922086..
1. 即使我们尝试保存 excel 并打开相同的 0.009056339275922086 也会返回..

用于在 POI 中设置的代码：

XSSFWorkbook workbook = new XSSFWorkbook(); 
XSSFRow  row = sheet.createRow(1);
XSSFCell  cell = row.createCell(1);
cell.setCellType(CellType.NUMERIC);
cell.setCellFormula("RATE(85.77534246575343, -1589.0, -18664.0, 5855586.0)");
FormulaEvaluator evaluator = workbook.getCreationHelper().createFormulaEvaluator();
evaluator.evaluateInCell(cell);
cell.getNumericCellValue();

【问题讨论】：

标签： function apache-poi rate

【解决方案1】：

apache poi 的 Rate 函数声明它“// 通过牛顿正割法求根”。这是无稽之谈，因为Secant method 只是一种准牛顿方法。还有“If the initial values are not close enough to the root, then there is no guarantee that the secant method converges.”。

所以0.1 的默认guess 似乎不够“接近”，所以如果我们使用cell.setCellFormula("RATE(85.77534246575343, -1589.0, -18664.0, 5855586.0, 0, 0.06)"); - 请注意type 和guess 属性的显式设置并具有guess 属性“足够接近”到0.05819488005- 的结果，然后公式计算正确。

如果apache poi 真的会使用Newton's method，那么该函数也会使用0.1 的默认guess 正确评估。牛顿方法的缺点是它需要在每一步都评估f 及其导数f′。所以在某些情况下它可能比割线法慢。

例子：

import java.io.FileOutputStream;

import org.apache.poi.ss.usermodel.*;
import org.apache.poi.xssf.usermodel.XSSFWorkbook;
import org.apache.poi.xssf.usermodel.XSSFSheet;

public class ExcelRATEFunction {

 private static double calculateRateNewton(double nper, double pmt, double pv, double fv, double type, double guess) {

  int FINANCIAL_MAX_ITERATIONS = 20;
  double FINANCIAL_PRECISION = 0.0000001;

  double y, y1, xN = 0, f = 0, i = 0;

  double rate = guess;

  //find root by Newtons method (https://en.wikipedia.org/wiki/Newton%27s_method), not secant method!
  //Formula see: https://wiki.openoffice.org/wiki/Documentation/How_Tos/Calc:_Derivation_of_Financial_Formulas#PV.2C_FV.2C_PMT.2C_NPER.2C_RATE

  f = Math.pow(1 + rate, nper);
  y = pv * f + pmt * ((f - 1) / rate) * (1 + rate * type) + fv;

  //first derivative:
  //y1 = (pmt * nper * type * Math.pow(rate,2) * f - pmt * f - pmt * rate * f + pmt * nper * rate * f + pmt * rate + pmt + nper * pv * Math.pow(rate,2) * f) / (Math.pow(rate,2) * (rate+1));
  y1 = (f * ((pmt * nper * type + nper * pv) * Math.pow(rate,2) + (pmt * nper - pmt) * rate - pmt) + pmt * rate + pmt) / (Math.pow(rate,3) + Math.pow(rate,2));

  xN = rate - y/y1;

  while ((Math.abs(rate - xN) > FINANCIAL_PRECISION) && (i < FINANCIAL_MAX_ITERATIONS)) {

   rate = xN;

   f = Math.pow(1 + rate, nper);
   y = pv * f + pmt * ((f - 1) / rate) * (1 + rate * type) + fv;

   //first derivative:
   //y1 = (pmt * nper * type * Math.pow(rate,2) * f - pmt * f - pmt * rate * f + pmt * nper * rate * f + pmt * rate + pmt + nper * pv * Math.pow(rate,2) * f) / (Math.pow(rate,2) * (rate+1));
   y1 = (f * ((pmt * nper * type + nper * pv) * Math.pow(rate,2) + (pmt * nper - pmt) * rate - pmt) + pmt * rate + pmt) / (Math.pow(rate,3) + Math.pow(rate,2));

   xN = rate - y/y1;
   ++i;

   System.out.println(rate+", "+xN+", "+y+", "+y1);
  }

  rate = xN;    
  return rate;

 }

 public static void main(String[] args) throws Exception {

  Workbook workbook = new XSSFWorkbook();
  Sheet sheet = workbook.createSheet();
  Row  row = sheet.createRow(1);
  Cell  cell = row.createCell(1);

  cell.setCellFormula("RATE(85.77534246575343, -1589.0, -18664.0, 5855586.0, 0, 0.06)");
  FormulaEvaluator evaluator = workbook.getCreationHelper().createFormulaEvaluator();
  CellType celltype = evaluator.evaluateFormulaCellEnum(cell);

  double value = 0.0;
  if (celltype == CellType.NUMERIC) {
   value = cell.getNumericCellValue();
   System.out.println(value);
  }

  workbook.setForceFormulaRecalculation(true);

  value = calculateRateNewton(85.77534246575343, -1589.0, -18664.0, 5855586.0, 0, 0.1);
  System.out.println(value);

  workbook.write(new FileOutputStream("ExcelRATEFunction.xlsx"));
  workbook.close();

 }

}

【讨论】：