Re: st: Looping over a range of observations

[Francesca Colantuoni <f.colantuoni9@gmail.com>]

Thank you!
I don't have exactly 900 stores and I have to modify your code a
little bit, but I was very far and this is extremely helpful. I also
think that 12 would be more convenient.
I'll keep you posted, thanks again Nick.

Francesca

On Fri, Aug 23, 2013 at 12:57 PM, Nick Cox <njcoxstata@gmail.com> wrote:
> Assuming your stores have identifiers 1..900 then
>
> forvalues i=1/90 {
> local first = 10 * `i' - 9
> local last = `first' + 9
> histogram price1 if inrange(store, `first',  `Iast'), by(store)
> graph save "graph`i'", replace
> }
>
> My guess is that you would find 12 or 16 graphs convenient, but the
> arithmetic is easy to modify.
>
> Nick
> njcoxstata@gmail.com
>
>
> On 23 August 2013 17:48, Francesca Colantuoni <f.colantuoni9@gmail.com> wrote:
>> Hello Statalist!
>> I am working with a panel data set. I would like to create and save
>> graphs that show
>> histograms of a variable (price1) per each store in the data set. I
>> have many stores (900), therefore if I do:
>>
>> histogram price1, by (store)
>>
>> each histogram would be microscopic and I would not be able to see anything.
>>
>> So, I tried with:
>>
>> histogram price1 in 1/2120, by (store)
>>
>> which allows me to visualize histograms for the first 10 stores only,
>> and this works way better.
>>
>> Now, though, I would like to create a loop that increments the range
>> of observations and  creates a number of graphs containing 10
>> histograms each.
>>
>> I was trying with something like:
>>
>> forvalues i=1/2120 {
>> histogram price1 in `i', by(store)
>> graph save graph "graph1.gph", replace
>> }
>>
>> Of course I need to nest the loop to update the graphs and I still
>> need to increment the range of observations, but for the moment, I
>> would like to get some help to  figure out the first part, because
>> even if the code runs no problem, the graph I get is not what I want
>> (it's only one histogram with one bar).
>>
>> Kindest Regards,
>>
>> Francesca
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