How to select a random sample from all your data [trick]

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The other day, I got a text message (SMS) from one of our readers. It read,

How to select random samples from data in Excel?

So today, let us learn a very easy trick to select random sample from your data.

Lets take a look at the data

Since the text message has no actual data, I made up this.

Random samples needed from this data set

Now, if you just want to select any 10 (or x number of) random items from this list, then your job would very simple.

  1. Shuffle (or randomly arrange) this list
  2. Just pick first 10 items

But our problem is to get 2 random samples per user.

Selecting random samples from data

Follow below steps.

  1. Add an extra column and fill it with =RAND() formula. This generates random fraction between 0 and 1.

    Add RAND() function to the adjacent column

  2. Create a pivot table from this data (tutorial: How to create a pivot table?)
  3. Add User ID & Case ID as Row labels and Random as value field.
    Pivot table layout for selecting random samples
  4. Click on the filter icon on Case ID column, choose Value filter > Top 10
  5. Filter for top 2 random values. (related: Filter top 10 values in pivot tables – how to?)

    Pivot Table value filters - filtering top 2 values

  6. Adjust report layout (Table layout, no sub-totals, no grand totals)

    Report layout to show just the samples and nothing else

  7. Done!

Final random samples - easy and awesome.

To see new samples

Just select any cell in the pivot table, press ALT+F5. Your pivot table will be refreshed and you get new samples.

That is just easy and awesome!

Download Example Workbook

Click here to download the example file. Refresh the pivot table (ALT+F5) to see fresh samples.

Do you sample your data?

Drawing samples, running experiments, analyzing results are life breath for many businesses. As business data is growing, these analytical skills  are becoming important.

How do you draw samples? What techniques you use when analyzing the data? Please share your stories, experiences & tips using comments.

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14 Responses to “How many ‘Friday the 13th’s are in this year? [Formula fun + challenge]”

  1. in C3=2016
    in C4=3
    in C5=1 (the first next year with three Friday the 13ths)

    =SMALL(IF(MMULT(--(MOD(DATE(C3+ROW(1:1000),COLUMN(A:L),13),7)=6),ROW(1:12)^0)=C4,C3+ROW(1:1000)),C5)

    formula check in the next 1000 years

  2. Brian says:

    This will generate a table of counts of Friday the 13th's by year. If I didn't screw it up the next year with three is 2026.

    I created a simple parameter table with a start date and end date that I wanted to evaluate. That calculates the number of days and generates a list of those days. Then filter and group. The generation of the list in power query (i.e. without populating a date table in excel) is pretty cool, otherwise this isn't really doing anything than creating a big date and filtering/counting.

    let
    Source = List.Dates(StartDateAsDate, Days2, #duration(1,0,0,0)),
    ConvertDateListToTable = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
    AddDayOfMonthColumn = Table.AddColumn(ConvertDateListToTable, "DayOfMonth", each Date.Day([Column1])),
    AddYearColumn = Table.AddColumn(AddDayOfMonthColumn, "Year", each Date.Year([Column1])),
    AddDayOfWeekColumn = Table.AddColumn(AddYearColumn, "Day of Week", each Date.DayOfWeek([Column1])),
    FilterFriday13 = Table.SelectRows(AddDayOfWeekColumn, each ([DayOfMonth] = 13) and ([Day of Week] = 5)),
    Friday13thsByYear = Table.Group(FilterFriday13, {"Year"}, {{"Number of Friday the 13ths!", each Table.RowCount(_), type number}})
    in
    Friday13thsByYear

    • Brian says:

      With the parameters replaced by values should you want to play along at home. This runs for 20 years starting on 1/1/2016.

      let
      Source = List.Dates(#date(2016,1,1), 7300, #duration(1,0,0,0)),
      ConvertDateListToTable = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
      AddDayOfMonthColumn = Table.AddColumn(ConvertDateListToTable, "DayOfMonth", each Date.Day([Column1])),
      AddYearColumn = Table.AddColumn(AddDayOfMonthColumn, "Year", each Date.Year([Column1])),
      AddDayOfWeekColumn = Table.AddColumn(AddYearColumn, "Day of Week", each Date.DayOfWeek([Column1])),
      FilterFriday13 = Table.SelectRows(AddDayOfWeekColumn, each ([DayOfMonth] = 13) and ([Day of Week] = 5)),
      Friday13thsByYear = Table.Group(FilterFriday13, {"Year"}, {{"Number of Friday the 13ths!", each Table.RowCount(_), type number}})
      in
      Friday13thsByYear

  3. Alex Groberman says:

    =MATCH(3,MMULT(N(WEEKDAY(DATE(C3+ROW(1:100)-1,COLUMN(A:L),13))=6),1^ROW(1:12)),)+C3-1

    • David N says:

      It should be pointed out that Alex's solution, unlike some others, has the additional advantage of being non-array. My solution was nearly identical but with -- and SIGN instead of N and 1^.

      =C3-1+MATCH(3,MMULT(--(WEEKDAY(DATE(C3-1+ROW(1:25),COLUMN(A:L),13))=6),SIGN(ROW(1:12))),0)

  4. SunnyKow says:

    Sub Friday13()

    Dim StartDate As Date
    Dim EndDate As Date
    Dim x As Long
    Dim r As Long

    Range("C7:C12").ClearContents
    StartDate = CDate("01/01/" & Range("C3"))
    EndDate = CDate("31/12/" & Range("C3"))
    r = 7
    For x = StartDate To EndDate
    If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
    Cells(r, 3) = Month(x)
    r = r + 1
    End If
    Next
    End Sub

    • SunnyKow says:

      Calculate next year with 3 Friday 13th. Good for 100 years different from year entered in cell C3

      Sub ThreeFriday13()

      Dim StartDate As Date
      Dim EndDate As Date
      Dim x As Long
      Dim WhatYear As Integer
      Dim Counter As Integer

      Range("E7").ClearContents
      StartDate = CDate("01/01/" & Range("C3") + 1)
      EndDate = CDate("31/12/" & Range("C3") + 100)
      Counter = 0

      For x = StartDate To EndDate
      If WhatYear Year(x) Then
      WhatYear = Year(x)
      'Different year so reset counter
      Counter = 0
      End If
      If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
      Counter = Counter + 1
      If Counter = 3 Then
      WhatYear = Year(x)
      Exit For
      End If
      End If
      Next
      Range("E7") = WhatYear

      End Sub

      • SunnyKow says:

        *RE-POST as not equal did not show earliuer
        Calculate next year with 3 Friday 13th. Good for 100 years different from year entered in cell C3

        Sub ThreeFriday13()

        Dim StartDate As Date
        Dim EndDate As Date
        Dim x As Long
        Dim WhatYear As Integer
        Dim Counter As Integer

        Range("E7").ClearContents
        StartDate = CDate("01/01/" & Range("C3") + 1)
        EndDate = CDate("31/12/" & Range("C3") + 100)
        Counter = 0

        For x = StartDate To EndDate
        If WhatYear NE Year(x) Then
        WhatYear = Year(x)
        'Different year so reset counter
        Counter = 0
        End If
        If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
        Counter = Counter + 1
        If Counter = 3 Then
        WhatYear = Year(x)
        Exit For
        End If
        End If
        Next
        Range("E7") = WhatYear

        End Sub

  5. Devesh says:

    I've a doubt with using array formula here.
    In sample workbook, I tried to replicate the formula again.
    =IFERROR(SMALL(IF(WEEKDAY(DATE($C$3,ROW($A$1:$A$12),13))=6,ROW($A$1:$A$12)),$B7),"")
    For this I selected C7 to C12, and typed the same formula and pressed ctrl+alt+Enter. But in all cells it is taking $B7 (and not $B7, $B8, $B9.... etc)
    and since it is array formula I can't edit individual cell.
    Please guide.
    Thanks

  6. Pablo says:

    Hi Chandoo,
    Cool stuff. You need to clarify that the answer of 5 represents the 1st month in the year that has a Friday the 13th, and not the number of Fridays the 13th in the year. Subtle, but important difference.
    Thanks,
    Pablo

  7. Micah Dail says:

    I like the MMULT() function far more, but here's how I would have tackled it. It uses an EDATE() base and MODE() over 100 years. I'm assuming that 100 years is enough time to catch the next year with 3 friday 13th's. Array entered, of course.

    {=MODE(IFERROR(YEAR(IF((WEEKDAY(EDATE(DATE(C3, 1, 13), ROW(INDIRECT("1:1200"))))=6), EDATE(DATE(C3, 1, 13), ROW(INDIRECT("1:1200"))), "")), ""))}

  8. Jason Morin says:

    Finding all the Friday the 13ths in a Year:

    =SUMPRODUCT((DAY(ROW(INDIRECT(DATE(C3,1,1)&":"&DATE(C3,12,31))))=13)*(TEXT(ROW(INDIRECT(DATE(C3,1,1)&":"&DATE(C3,12,31))),"ddd")="Fri"))

  9. jmdias says:

    {=sum(if(day.of.week(DATe($YEAR;{1;2;3;4;5;6;7;8;9;10;11;12};13);1)=6;1;0))}
    just list the years

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