Blog merge

I'm glad to announce that this blog will be merged to Apps, games, and more.. All upcoming blog posts will be posted in that blog. seeyou.. :)

Topic #3: Trendline Analysis

Our next topic will be Trendline Analysis. In this topic, we will be making a program that would require us to input the size and the data. Then the outpout would be the equation of the trendline. Assuming that there is  no seasons involved. To start, we need to declare all the variables we will be using. Here are the variables:

    Dim x As Single

    Dim y() As Single

    Dim xsum As Single

    Dim ysum As Single

    Dim xsqsum As Single

    Dim xysum As Single

    Dim n As Integer

    Dim b1 As Single

    Dim b0 As Single

    Dim response As Single

Then, we need to put a code in order for us to input the data and the data size.

n = Application.InputBox("Enter number of data.")   

    ReDim y(1 To n) As Single

    For x = 1 To n

        y(x) = Application.InputBox("enter data " & x)

    Next

After the data input, we need to do the necessary solutions to solve for the trendline equation. In my work, i used the normal equation method instead of the conventional way to minimize rounding errors. Here is how normal equation works:

Now, if we solve it manually, you will be using systems of equations involving two variables to determine the value of b₀ and b₁.

Now for solving using visual basic, we need to put the code below:

        xsum = 0

        ysum = 0

        xsq = 0

        xysum = 0

       

    For x = 1 To n

        xsum = xsum + x

        ysum = ysum + y(x)

        xsqsum = xsqsum + x ^ 2

        xysum = xysum + y(x) * x

    Next

    b1 = (ysum * xsum - n * xysum) / (xsum ^ 2 - n * xsqsum)

    b0 = (ysum - b1 * xsum) / n

Now for the output, we need to put this code:

response = MsgBox("the equation is y=" & b1 & "x + " & b0 & ".", vbOKOnly)

Topic #2: Mode

The next thing we would be doing is to create a program that will generate the mode of a given data. By definition, mode is the number which appears most often in a set of numbers.. We would be using the code for arranging data from the previous topic. I’ll put the code below:

    countmin = 0

    For indexdata = 1 To sizedata

        If min = data(indexdata) Then

            countmin = countmin + 1

        End If

    Next

   

    For indexsort = 1 To countmin

        For indexdata = 1 To sizedata

            If min = data(indexdata) Then

                sorteddata(indexsort) = data(indexdata)

            End If

        Next

    Next

   

    datanum = countmin

    For indexrange = 1 To range

        For indexdata = 1 To (sizedata)

            If data(indexdata) = min + indexrange Then

                sorteddata(datanum + 1) = data(indexdata)

                datanum = datanum + 1

            End If

        Next

    Next

The code above will be be sorting the data. Don’t forget to declare every variables. Next is to count every repeatitions in the data. All we have to do is to input the code below:

    For indexcount = 1 To sizedata

        If (indexcount + 1) <= sizedata Then

            If sorteddata(indexcount) = sorteddata(indexcount + 1) Then

            Else

                For indexdata = 1 To sizedata

                    If sorteddata(indexcount) = sorteddata(indexdata) Then

                       count(indexcount) = count(indexcount) + 1

                    End If

                Next

            End If

        Else

            For indexdata = 1 To sizedata

                    If sorteddata(indexcount) = sorteddata(indexdata) Then

                       count(indexcount) = count(indexcount) + 1

                    End If

            Next

        End If

    Next

After getting the count of every value in the data, we would be getting the most count. We would be using the code below:

    For indexcount = 1 To sizedata

        If (indexcount + 1) <= sizedata Then

            If sorteddata(indexcount) = sorteddata(indexcount + 1) Then

            Else

                For indexdata = 1 To sizedata

                    If sorteddata(indexcount) = sorteddata(indexdata) Then

                       count(indexcount) = count(indexcount) + 1

                    End If

                Next

            End If

        Else

            For indexdata = 1 To sizedata

                    If sorteddata(indexcount) = sorteddata(indexdata) Then

                       count(indexcount) = count(indexcount) + 1

                    End If

            Next

        End If

    Next

There would be a problem if we have multiple modes. Because the codes given will only show only one mode. To fix the problem, we will be inserting the code below:

    If maxcount > 1 Then

        countmode = 0

        For indexdata = 1 To sizedata

            If maxcount = count(indexdata) Then

                countmode = countmode + 1

            End If

        Next

       

        ReDim mode(1 To countmode) As Single

       

        modenum = 0

        For indexdata = 1 To sizedata

            If maxcount = count(indexdata) Then

                modenum = modenum + 1

                mode(modenum) = sorteddata(indexdata)

            End If

        Next

    End If

Next, we would be applying the appropriate words to our output. We just have to put the code below:

    If countmode = 1 Then

        isare = "is"

        modesmode = "mode"

    Else

        isare = "are"

        modesmode = "modes"

    End If

The last step is to show our the results. The code below also is flexible for bimodal and trimodal problems.

    response = MsgBox("there " & isare & " " & countmode & " " & modesmode)

    For indexmode = 1 To countmode

        response = MsgBox("the " & modesmode & " " & isare & ": " & mode(indexmode))

    Next

Topic #1: Median

Welcome to my blog. Here in my blog, we will be converting mathematical equations to computer programs, specifically into visual basic for applications codes, even more specifically, using only few statements like loop, if, and operational statements. Note that the codes in the blog is just one of the possible codes we can use.

Questions, suggestions and comments are highly appreciated. You may want to post any questions regarding anything related to the topic.

Well, let’s start from statistics. In statistics, we have a term called median. And by definition, median is the middle number (in a sorted list of numbers). To find the Median, place the numbers you are given in value order and find the middle number. We will be doing a program where it will generate the median of a given group of data.

First, let’s start from the usual steps we do when doing a program using visual basic for applications. Let’s declare every variable we will be using. I will be discussing the variables as we go along. For now lets start from declaring the following variables:

Dim data() As Single

Dim sizedata As Integer

The variable ‘data’ represents the array of data and ‘sizedata’ represents the size. As you can see, ‘data’ still don’t have an exact array.

Next, we will input the size of the data. We won’t have too much problem if the size is known. But not all data groups have the same size. So we will be making a computer program where the user will input the size. In order for us to do that, we will be using the code below:

    Do

        sizedata = InputBox(prompt:="enter the size", Default:=0)

    Loop While sizedata = 0

I added a loop statement in the code to minimize bugs. Also, I have set the default value to zero (0).

Here’s what the code will show you:

After entering the data size. We will need to redimension the variable ‘data’. So we will be using the following statement:

ReDim data(1 To sizedata) As Single

Notice that we have set the dimension of the variable ‘data’. Example if ‘sizedata=3’, then we will have an array of data from 1 to 3.

data(1)

data(2)

data(3)

Next, we need to input the data in order for us to solve the median. So I’ll be using the following codes:

For indexdata = 1 To sizedata

        data(indexdata) = InputBox("enter data " & indexdata)

Next

Here’s what will be shown on the screen:

Next, we need to determine the data with the lowest value. Now we will be using conditional statements.

min = data(1)

For indexdata = 1 To sizedata

    If min > data(indexdata) Then

        min = data(indexdata)

    End If

Next

Then we will be determine the element with the highest value. Here’s the code:

    max = data(1)

    For indexdata = 1 To sizedata

        If max < data(indexdata) Then

            max = data(indexdata)

        End If

    Next

Next we determin the range:

range = max - min

Remember to declare the variables ‘min’, ‘max’, and ‘range’.

Dim min as single

Dim max as single

Dim range as single

Now, we will be introducing more variables. The next step that we need to do is to count the number of elements with the lowest value. And in order for us to do that, we need to enter this code:

    countmin = 0

    For indexdata = 1 To sizedata

        If min = data(indexdata) Then

            countmin = countmin + 1

        End If

    Next

Now we need to sort the data. Let’s start from the lowest. Then sort all the other elements using this code:

    For indexsort = 1 To countmin

        For indexdata = 1 To sizedata

            If min = data(indexdata) Then

                sorteddata(indexsort) = data(indexdata)

            End If

        Next

    Next

    datanum = countmin

    For indexrange = 1 To range

        For indexdata = 1 To (sizedata)

            If data(indexdata) = min + indexrange Then

                sorteddata(datanum + 1) = data(indexdata)

                datanum = datanum + 1

            End If

        Next

    Next

Now we are done sorting the data. Just don’t forget to declare all the variables we have used. Now we can get the median already. But first, we need to settle a problem. We can have the median immediately if the data size is an odd number. But what if it’s an even? We just have to determine if the size is even or odd using this code:

    halfint = sizedata / 2

    halfsin = sizedata / 2

The variables represent the half of the value of the size. Basically, there would be no difference between the 2 except their data type. The first will be an integer and the other will be single. By doing so, there would be a difference in the 2 if the size is an odd number. Example:

sizedata= 5

halfint=3 (5/2)

halfsin=2.5 (5/2)

If halfsin is not equal to halfint, then the size is an odd number. Now just declare the previous variables.

    Dim halfint As Integer

    Dim halfsin As Single

We can now determine the median. I’ve decided to use a function for that. Just follow the code below:

    median = getmedian(sorteddata(), sizedata, halfint, halfsin)

And for the function:

Function getmedian(sorteddata() As Single, sizedata As Integer, halfint As Integer, halfsin As Single) As Single

   

    'logical test: if true then sizedata is an even number else sizedata is odd

    If halfint = halfsin Then

        getmedian = (sorteddata(sizedata / 2) + sorteddata(sizedata / 2 + 1)) / 2

    Else

        getmedian = sorteddata(sizedata / 2 + 0.5)

    End If

       

End Function

There you go. We now have a program that will automatically solve the median of a group of data. The last thing to do is to show the resuls, use the code below:

    response = MsgBox("the median is " & median)