creating a graph with volume of procedures and average line

[Martin Imelda Borg]

I would like to create a graph depicting volume of use of the procedure: hysteroscopy vs ID

I have an extremely large database, tried to plot a histogram and it crashed stata. The problem being as I have a million records.

With this data I would like to plot a graph (see picture) showing the volume of procedures (hysteroscopy) done per individual (represented by ID)
with an average line representing the average number of hysteroscopies (procedures) between them all individuals are done

* Example generated by -dataex-. For more info, type help dataex
clear
input float(ID hyseteroscopy)
1 1
2 0
3 1
2 0
2 0
2 1
3 1
2 1
1 1
end

[Hemanshu Kumar]

I do think you need a histogram rather than the graph you have drawn. Here is some sample code. You can ignore much of the initial part, which is only designed to create a dataset with 100 million observations.

clear
set obs 1000000
set seed 12345

gen `c(obs)' ID = _n
expand 100
gen byte hysteroscopy = rbinomial(1,0.1)

* start your code here
collapse (sum) hysteroscopy,  by(ID)

sum hysteroscopy, meanonly
local mean = r(mean)

#delimit ;
twoway histogram hysteroscopy,
        discrete frac horizontal ||
        pci `mean' 0 `mean' 0.15,
        legend(off)
        ytitle("Volume of hysteroscopies")
        xtitle("Fraction")
        text(`mean' 0.14 "Average", place(6))
        scheme(s1color)
        ;
#delimit cr

[Martin Imelda Borg]

I will look into this

having a read through your code

expand 100 what does this mean? gen byte hysteroscopy = rbinomial(1,0.1) What does rbinomial (1,0.1) — represent ?

[Hemanshu Kumar]

Again, those are bits of code that you can ignore, because I just wanted to create a large dataset of the same scale you have.

But to explain:

I first create a database with 1 million observations, each with a unique ID. Then I expand it so that there are 100 observations for each ID. Then I create a random variable hysteroscopy by drawing from a Binomial distribution with 1 trial and 10% probability of success (i.e. a binary variable with a 10% chance of being 1, else 0).

[Martin Imelda Borg]

Thanks for this, three other questions, with reference to the code below. What do they refer to ....

pci `mean' 0 `mean' 0.15, // what does pci mean ? why mean 0 and mean 0.15?
text(`mean' 0.14 "Average", place(6)) // Here I understand you're inserting average line, but the average line is at 10 so why is it being reference to place 6? mean 0.14?
LAST QUESTION:
Step 1:
collapse(sum) hyseteroscopy, by (ID) ///creates a sum of all the ID numbers --> all ok

But I can't get the following code below to work , the values remain the same as (step 1)

STEP 2: CREATING THE AVERAGE

. sum hyseteroscopy, meanonly


. dataex in 1/3

----------------------- copy starting from the next line -----------------------

* Example generated by -dataex-. For more info, type help dataex
clear
input float ID double hyseteroscopy
1 2
2 2
3 2
end

------------------ copy up to and including the previous line ------------------

Listed 3 out of 3 observations

. local mean = r(mean)

. dataex in 1/3

----------------------- copy starting from the next line -----------------------

* Example generated by -dataex-. For more info, type help dataex
clear
input float ID double hyseteroscopy
1 2
2 2
3 2
end

------------------ copy up to and including the previous line ------------------

Listed 3 out of 3 observations

The sum code is working , but stata is not interpreting the mean code . As you know The values should appear as below, but I remain with the sum only

1 - Sum 2 --> Average 1
2- Sum of 2 --> Average 0.4
3 - Sum of 2 --> Average 1

[Martin Imelda Borg]

Ah this is a local macro using r scalars needs to run en block and not line by line

sum hyseteroscopy, meanonly
local mean = r(mean)

// i think i found the reason why, the above is a local macro whereby meanonly is a quick command to generate the mean. This is then stored as a local macro ‘mean’ generated by code r(mean)

the histogram code should be highlighed incorporating with the local macro
however the pci i supose refers to the numbers you would like the y axis to start with ie 0 (min) and 0.15 is the max no?

however, i don’t know what place(6) refers to….

pci `mean' 0 `mean' 0.15, // what does pci mean ? why mean 0 and mean 0.15?
text(`mean' 0.14 "Average", place(6)) // Here I understand you're inserting average line, but the average line is at 10 so why is it being reference to place 6? mean 0.14?

[Hemanshu Kumar]

Some of this is about the way locations are specified in different contexts in Stata graphs. See

help twoway pci
help added_text_options

within the text location, you can specify the placement around the coordinates provided using compass- or clock-style directions:

help compassdirstyle

"6" is effectively centre-bottom / south.