View Single Post
  #2   Report Post  
Ed Ferrero
 
Posts: n/a
Default

Hi Chris,

You have not had an answer yet, so obviously I am not the only person in the
gorup that has no idea what you're talking about. Let's see if I can
guess...

Your data is laid out like this...

Specie Dog Cat Pup Kitten
Colour 1 2 1 1
Measure2

You could add a row above your data and some columns at the end like this...

ColourNo 1 2
Specie Dog Cat Pup Kitten ="Colour" & F1 ="Colour" & G1
Colour 1 2 1 1 =COUNTIF($B$3:$E$3,F1) =COUNTIF($B$3:$E$3,F1)
Measure2

Does that make sense? Or are you trying to do something else?

Ed Ferrero
http://edferrero.m6.net/


Hello,

Thought I was done with my charting questions, but have one more... I have
a
spreadsheet with ~100 columns (individual animals, divided equally into 5
species) and ~100 rows (different morphological measurements). When I
create
my graphs, I have one graph for each measurement. The species are the
x-axis
(1, 2, 3, 4, 5) and the measurement range is the y-axis.

Some of my measurements are categorical, such as color (black = 1, white =
2, yellow = 3, etc.). When I graph my 100 individuals, I see a point for
each
color for each species type. This is not surprising as there is a lot of
variation in the color of my species. However, I want to get a better idea
of
the number of black individuals, brown individuals, etc. for each species.
Thus, is there a way to have Excel count the number of individuals of each
color & species and then use those values as data labels?

As we are constantly adding characters and individuals, and we have a
bunch
of continuous variables within the spreadsheet, I'd rather not deal with
PivotTables or adding the counts to a separate column to label my data. If
labeling in this way is not possible, is it possible to add extra data
points
to a graph. For example, one point for every 10 occurences of a variable.
Thus if there were 50 black individuals of species 1, rather than seeing
one
data point, the chart would show 5 points next to each other.

Hopefully my question(s) make sense. Thanks for any ideas you might have,

Chris