How do I calculate a trendline for a graph

Question

Google is not being my friend - it s been a long time since my stats class in college   I need to calculate the start and end points for a trendline on a graph - is there an easy way to do this   working in C  but whatever language works for you

User · Answer

Regarding a previous answer  if  B  y   offset   slope x  then  C  offset   y  slope x  is wrong   C  should be   offset   y- slope x   See  http   zedgraph org wiki index php title Trend

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Here is a very quick  and semi-dirty  implementation of Bedwyr Humphreys s answer   The interface should be compatible with  matt s answer as well  but uses decimal instead of int and uses more IEnumerable concepts to hopefully make it easier to use and read   Slope is b  Intercept is a  public class Trendline       public Trendline IList lt decimal gt  yAxisValues  IList lt decimal gt  xAxisValues            this yAxisValues Select  t  i    gt  new Tuple lt decimal  decimal gt  xAxisValues i   t                public Trendline IEnumerable lt Tuple lt Decimal  Decimal gt  gt  data                var cachedData   data ToList             var n   cachedData Count          var sumX   cachedData Sum x   gt  x Item1           var sumX2   cachedData Sum x   gt  x Item1   x Item1           var sumY   cachedData Sum x   gt  x Item2           var sumXY   cachedData Sum x   gt  x Item1   x Item2              b    sum x y  - sum x sum y  n                     sum x 2  - sum x  2 n          Slope    sumXY -   sumX   sumY    n                          sumX2 -  sumX   sumX   n               a   sum y  n - b sum x  n          Intercept    sumY   n  -  Slope    sumX   n             Start   GetYValue cachedData Min a   gt  a Item1            End   GetYValue cachedData Max a   gt  a Item1               public decimal Slope   get  private set        public decimal Intercept   get  private set        public decimal Start   get  private set        public decimal End   get  private set         public decimal GetYValue decimal xValue                return Intercept   Slope   xValue

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Thanks to all for your help - I was off this issue for a couple of days and just came back to it - was able to cobble this together - not the most elegant code  but it works for my purposes - thought I d share if anyone else encounters this issue   public class Statistics       public Trendline CalculateLinearRegression int   values                var yAxisValues   new List lt int gt             var xAxisValues   new List lt int gt              for  int i   0  i  lt  values Length  i                          yAxisValues Add values i                xAxisValues Add i   1                      return new Trendline yAxisValues  xAxisValues            public class Trendline       private readonly IList lt int gt  xAxisValues      private readonly IList lt int gt  yAxisValues      private int count      private int xAxisValuesSum      private int xxSum      private int xySum      private int yAxisValuesSum       public Trendline IList lt int gt  yAxisValues  IList lt int gt  xAxisValues                this yAxisValues   yAxisValues          this xAxisValues   xAxisValues           this Initialize               public int Slope   get  private set        public int Intercept   get  private set        public int Start   get  private set        public int End   get  private set         private void Initialize                 this count   this yAxisValues Count          this yAxisValuesSum   this yAxisValues Sum            this xAxisValuesSum   this xAxisValues Sum            this xxSum   0          this xySum   0           for  int i   0  i  lt  this count  i                          this xySum     this xAxisValues i  this yAxisValues i                this xxSum     this xAxisValues i  this xAxisValues i                       this Slope   this CalculateSlope            this Intercept   this CalculateIntercept            this Start   this CalculateStart            this End   this CalculateEnd               private int CalculateSlope                 try                       return   this count this xySum  -  this xAxisValuesSum this yAxisValuesSum     this count this xxSum  -  this xAxisValuesSum this xAxisValuesSum                      catch  DivideByZeroException                        return 0                       private int CalculateIntercept                 return  this yAxisValuesSum -  this Slope this xAxisValuesSum   this count             private int CalculateStart                 return  this Slope this xAxisValues First      this Intercept             private int CalculateEnd                 return  this Slope this xAxisValues Last      this Intercept

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Here s what I ended up using   public class DataPoint lt T1 T2 gt        public DataPoint T1 x  T2 y                X   x          Y   y              JsonProperty  x        public T1 X   get          JsonProperty  y        public T2 Y   get       public class Trendline       public Trendline IEnumerable lt DataPoint lt long  decimal gt  gt  dataPoints                int count   0          long sumX   0          long sumX2   0          decimal sumY   0          decimal sumXY   0           foreach  var dataPoint in dataPoints                        count                sumX    dataPoint X              sumX2    dataPoint X   dataPoint X              sumY    dataPoint Y              sumXY    dataPoint X   dataPoint Y                     Slope    sumXY -   sumX   sumY    count      sumX2 -   sumX   sumX    count            Intercept    sumY   count  -  Slope    sumX   count               public decimal Slope   get  private set        public decimal Intercept   get  private set        public decimal Start   get  private set        public decimal End   get  private set         public decimal GetYValue decimal xValue                return Slope   xValue   Intercept            My data set is using a Unix timestamp for the x-axis and a decimal for the y  Change those datatypes to fit your need  I do all the sum calculations in one iteration for the best possible performance

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Here s a working example in golang  I searched around and found this page and converted this over to what I needed  Hope someone else can find it useful     https   classroom synonym com calculate-trendline-2709 html package main  import        quot fmt quot       quot math quot     func main          graph        float64           1  3            2  5            3  6 5              n    len graph          get the slope     var a float64     var b float64     var bx float64     var by float64     var c float64     var d float64     var slope float64      for    point    range graph            a    point 0    point 1          bx    point 0          by    point 1          c    math Pow point 0   2          d    point 0              a    float64 n               97 5     b   bx   by                  87     c    float64 n               42     d   math Pow d  2            36     slope    a - b     c - d     1 75         calculating the y-intercept  b  of the Trendline     var e float64     var f float64      e   by                               14 5     f   slope   bx                       10 5     intercept     e - f    float64 n      14 5 - 10 5    3   1 3         output     fmt Println slope      fmt Println intercept

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If you have access to Excel  look in the  Statistical Functions  section of the Function Reference within Help  For straight-line best-fit  you need SLOPE and INTERCEPT and the equations are right there   Oh  hang on  they re also defined online here  http   office microsoft com en-us excel HP052092641033 aspx for SLOPE  and there s a link to INTERCEPT  OF course  that assumes MS don t move the page  in which case try Googling for something like  SLOPE INTERCEPT EQUATION Excel site microsoft com  - the link given turned out third just now

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This is the way i calculated the slope  Source  http   classroom synonym com calculate-trendline-2709 html  class Program               public double CalculateTrendlineSlope List lt Point gt  graph                        int n   graph Count              double a   0              double b   0              double bx   0              double by   0              double c   0              double d   0              double slope   0               foreach  Point point in graph                                a    point x   point y                  bx   point x                  by   point y                  c    Math Pow point x  2                   d    point x                            a    n              b   bx   by              c    n              d   Math Pow d  2                slope    a - b     c - d               return slope                       class Point               public double x          public double y

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OK  here s my best pseudo math   The equation for your line is   Y   a   bX  Where   b    sum x y  - sum x sum y  n     sum x 2  - sum x  2 n   a   sum y  n - b sum x  n   Where sum xy  is the sum of all x y etc  Not particularly clear I concede  but it s the best I can do without a sigma symbol         and now with added Sigma  b     Sigma  xy  -   Sigma x Sigma y  n      Sigma  x 2  -   Sigma x  2 n   a     Sigma y  n - b   Sigma x  n   Where  Sigma  xy  is the sum of all x y etc  and n is the number of points

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If anyone needs the JS code for calculating the trendline of many points on a graph  here s what worked for us in the end    x000D   x000D      typedef    x000D     x  Number  x000D     y Number  x000D        Point x000D      param  Point    data x000D      returns  Function     x000D  function  getTrendlineEq data    x000D      const xySum   data reduce  acc  item    gt    x000D          const xy   item x   item y x000D          acc    xy x000D          return acc x000D         0  x000D      const xSum   data reduce  acc  item    gt    x000D          acc    item x x000D          return acc x000D         0  x000D      const ySum   data reduce  acc  item    gt    x000D          acc    item y x000D          return acc x000D         0  x000D      const aTop    data length   xySum  -  xSum   ySum  x000D      const xSquaredSum   data reduce  acc  item    gt    x000D          const xSquared   item x   item x x000D          acc    xSquared x000D          return acc x000D         0  x000D      const aBottom    data length   xSquaredSum  -  xSum   xSum  x000D      const a   aTop   aBottom x000D      const bTop   ySum -  a   xSum  x000D      const b   bTop   data length x000D      return function trendline x    x000D          return a   x   b x000D        x000D    x000D   x000D   x000D    It takes an array of  x y  points and returns the function of a y given a certain x Have fun

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Thank You so much for the solution  I was scratching my head  Here s how I applied the solution in Excel  I successfully used the two functions given by MUHD in Excel  a    sum x y  - sum x sum y  n     sum x 2  - sum x  2 n  b   sum y  n - b sum x  n   careful my a and b are the b and a in MUHD s solution    - Made 4 columns  for example    NB  my values y values are in B3 B17  so I have n 15        my x values are 1 2 3 4   15    1  Column B  Known x s   2  Column C  Known y s   3  Column D  The computed trend line   4  Column E  B values   C values  E3 B3 C3  E4 B4 C4       E17 B17 C17    5  Column F  x squared values I then sum the columns B C and E  the sums go in line 18 for me  so I have B18 as sum of Xs  C18 as sum of Ys  E18 as sum of X Y  and F18 as sum of squares  To compute a  enter the followin formula in any cell  F35 for me   F35  E18- B18 C18  15   F18- B18 B18  15  To compute b  in F36 for me   F36 C18 15-F35  B18 15  Column D values  computing the trend line according to the y   ax   b  D3  F 35 B3  F 36  D4  F 35 B4  F 36 and so on  until D17 for me    Select the column datas  C2 D17  to make the graph  HTH

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Given that the trendline is straight  find the slope by choosing any two points and calculating    A   slope    y1-y2   x1-x2   Then you need to find the offset for the line   The line is specified by the equation    B   y   offset   slope x  So you need to solve for offset   Pick any point on the line  and solve for offset    C   offset   y -  slope x   Now you can plug slope and offset into the line equation  B  and have the equation that defines your line   If your line has noise you ll have to decide on an averaging algorithm  or use curve fitting of some sort   If your line isn t straight then you ll need to look into Curve fitting  or Least Squares Fitting - non trivial  but do-able   You ll see the various types of curve fitting at the bottom of the least squares fitting webpage  exponential  polynomial  etc  if you know what kind of fit you d like   Also  if this is a one-off  use Excel

[c#] How do I calculate a trendline for a graph?

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