graph for evolution variable in panel form

[Khati Zolfaghari]

hello, I have panel data for 171 countries from 1996-2019 , and want to see the "Evolution of average fertility rates over time ", and also declare some specific years as a recession in my graph. the code that I used was this:

lgraph fertilityrate year, tline(2003 2009 2012 2014 2016)

and my graph is :

if it is correct, I am confused about how can I add two other lines to my graph one "for the adavances economies" and another "for emerging market economies" with a different colors. I tried several ways.


I hope I receive your assistanse so soon.

Many thanks in advance for your valuable time.


regards,
Khatereh

[Nick Cox]

Blog . . . . . . . . . . . . Adding recession shading to time-series graphs
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D. Schenck
2/20 http://blog.stata.com/2020/02/13/add...ssion-shading-
to-time-series-graphs/


SJ-16-3 gr0067 . Speaking Stata: Shading zones on time series and other plots
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N. J. Cox
Q3/16 SJ 16(3):805--812 (no commands)
discusses background shading of time series and other plots

Otherwise you need to calculate those averages before you can plot them. EDIT: Wrong. lgraph has a syntax for specifying group mwans.

[Clyde Schechter]

-lgraph- is not part of official Stata, and I'm unfamiliar with it. Suffice it to say that what you want can be accomplished with Stata's official -graph twoway connect-. To get multiple curves on the same plot, you need to get the fertility rates for the advanced economies and emerging market economies into separate variables. As I do not know how your data is organized or what you have, I can't advise you how to extract those as separate variables from what you have, but it is likely to involve -reshape-. Once you do that:

graph twoway connect fertilityrate fertility_advanced fertility_emerging year, xline(2003 2009 2012 2014 2016)

will put all three curves on the same plot.

If you need specific advice to transform your data, please post back showing example data, and use the -dataex- command to do that. If you are running version 17, 16 or a fully updated version 15.1 or 14.2, -dataex- is already part of your official Stata installation. If not, run -ssc install dataex- to get it. Either way, run -help dataex- to read the simple instructions for using it. -dataex- will save you time; it is easier and quicker than typing out tables. It includes complete information about aspects of the data that are often critical to answering your question but cannot be seen from tabular displays or screenshots. It also makes it possible for those who want to help you to create a faithful representation of your example to try out their code, which in turn makes it more likely that their answer will actually work in your data.

Added: Crossed with #2.

[Rich Goldstein]

lgraph is a user-written command as the FAQ asks you to explain; one of its options is "addplot(plot)" and that is probably the direction you want to go; if you need more, please supply, as requested in the FAQ, a -dataex- example within CODE blocks

[Khati Zolfaghari]

hello, many thanks for your reply. @Nick Cox , @Clyde Schechter , and @Rich Goldstein.

this is part of my data:

* Example generated by -dataex-. For more info, type help dataex
clear
input str4 iso double(year fertilityrate) float(ae em)
"USA" 1996  1.976 1 0
"USA" 1997  1.971 1 0
"USA" 1998  1.999 1 0
"USA" 1999 2.0075 1 0
"USA" 2000  2.056 1 0
"USA" 2001 2.0305 1 0
"USA" 2002 2.0205 1 0
"USA" 2003 2.0475 1 0
"USA" 2004 2.0515 1 0
"USA" 2005  2.057 1 0
"USA" 2006  2.108 1 0
"USA" 2007   2.12 1 0
"USA" 2008  2.072 1 0
"USA" 2009  2.002 1 0
"USA" 2010  1.931 1 0
"USA" 2011 1.8945 1 0
"USA" 2012 1.8805 1 0
"USA" 2013 1.8575 1 0
"USA" 2014 1.8625 1 0
"USA" 2015 1.8435 1 0
"USA" 2016 1.8205 1 0
"USA" 2017 1.7655 1 0
"USA" 2018 1.7295 1 0
"USA" 2019  1.706 1 0
"GBR" 1996   1.73 1 0
"GBR" 1997   1.72 1 0
"GBR" 1998   1.71 1 0
"GBR" 1999   1.68 1 0
"GBR" 2000   1.64 1 0
"GBR" 2001   1.63 1 0
"GBR" 2002   1.63 1 0
"GBR" 2003    1.7 1 0
"GBR" 2004   1.75 1 0
"GBR" 2005   1.76 1 0
"GBR" 2006   1.82 1 0
"GBR" 2007   1.86 1 0
"GBR" 2008   1.91 1 0
"GBR" 2009   1.89 1 0
"GBR" 2010   1.92 1 0
"GBR" 2011   1.91 1 0
"GBR" 2012   1.92 1 0
"GBR" 2013   1.83 1 0
"GBR" 2014   1.81 1 0
"GBR" 2015    1.8 1 0
"GBR" 2016   1.79 1 0
"GBR" 2017   1.74 1 0
"GBR" 2018   1.68 1 0
"GBR" 2019   1.63 1 0
"AUT" 1996   1.45 1 0
"AUT" 1997   1.39 1 0
"AUT" 1998   1.37 1 0
"AUT" 1999   1.34 1 0
"AUT" 2000   1.36 1 0
"AUT" 2001   1.33 1 0
"AUT" 2002   1.39 1 0
"AUT" 2003   1.38 1 0
"AUT" 2004   1.42 1 0
"AUT" 2005   1.41 1 0
"AUT" 2006   1.41 1 0
"AUT" 2007   1.38 1 0
"AUT" 2008   1.42 1 0
"AUT" 2009   1.39 1 0
"AUT" 2010   1.44 1 0
"AUT" 2011   1.43 1 0
"AUT" 2012   1.44 1 0
"AUT" 2013   1.44 1 0
"AUT" 2014   1.46 1 0
"AUT" 2015   1.49 1 0
"AUT" 2016   1.53 1 0
"AUT" 2017   1.52 1 0
"AUT" 2018   1.47 1 0
"AUT" 2019   1.46 1 0
"BEL" 1996   1.59 1 0
"BEL" 1997    1.6 1 0
"BEL" 1998    1.6 1 0
"BEL" 1999   1.62 1 0
"BEL" 2000   1.67 1 0
"BEL" 2001   1.67 1 0
"BEL" 2002   1.65 1 0
"BEL" 2003   1.67 1 0
"BEL" 2004   1.72 1 0
"BEL" 2005   1.76 1 0
"BEL" 2006    1.8 1 0
"BEL" 2007   1.82 1 0
"BEL" 2008   1.85 1 0
"BEL" 2009   1.84 1 0
"BEL" 2010   1.86 1 0
"BEL" 2011   1.81 1 0
"BEL" 2012    1.8 1 0
"BEL" 2013   1.76 1 0
"BEL" 2014   1.74 1 0
"BEL" 2015    1.7 1 0
"BEL" 2016   1.68 1 0
"BEL" 2017   1.65 1 0
"BEL" 2018   1.62 1 0
"BEL" 2019    1.6 1 0
"DNK" 1996   1.75 1 0
"DNK" 1997   1.75 1 0
"DNK" 1998   1.72 1 0
"DNK" 1999   1.73 1 0
end

• that here I create a dummy variable for "adavances economies" and another for "emerging market economies" , name is "ae and em",respectively.
• if I understood correct based on your advice, with "lgraph" I cannot reach the specific graph that I wanted, so first of all, I generate the mean of fertility: "

  egen mean_fertility=mean( fertilityrate ), by(year)

• but now when I did "
  

  twoway conn mean_fertility year  || conn mean_fertility year if ae==1 || conn mean_fertility year if em==1, xline(2003 2009 2012 2014 2016)

  my graph is this: which is not correct. I apprriciate to recive your advice. many thanks
  

many thanks in advance for your valuable time and advice.

best regards,

[Nick Cox]

 
 egen mean_fertility=mean( fertilityrate ), by(ae em year)

is closer to what you want, I guess.

[Khati Zolfaghari]

@Nick Cox thanks for your reply. I tried based on your advice. and I did:

• egen mean_fertility=mean( fertilityrate ), by(ifscode year)

  that gives me the mean of fertility rate for the whole sample.

• egen mean_fertility_ae=mean( fertilityrate ), by(ifscode year), if ae==1

• egen mean_fertility_em=mean( fertilityrate ), by(ifscode year), if em==1

• but now when I do in the graph the results is not as "

  twoway conn mean_fertility year || conn mean_fertility_ae year|| conn mean_fertility_em year , xline(2003 2009 2012 2014 2016)

it gives me for each country. I want to get three separate lines just to have a comparison, but when I use the "lgraph" for each I can get seperatily only one line, that can show me the Evolution of average fertility rates over time.

many thanks in advance for your valuable time and advice.

regards,

[Nick Cox]

You didn't follow my advice at all. I don't know what ifscode is: it is not mentioned in your data example or in any previous post in this thread.

This shows that my idea works. You have three groups: ae == 1, em == 1, ae == 0 & em == 0. Naturally, your graph will be different. I can't use your data example as it will lead to one curve only.

webuse grunfeld, clear 
gen fertility_rate = ln(invest)
gen ae = company <= 3 
gen em = inrange(company, 4, 7)

* you start here 

egen mean_fertility = mean(fertility_rate), by(ae em year)

line mean_fertility year if ae == 1, sort || line mean_fertility year if em == 1, sort || line mean_fertility year if em == 0 & ae == 0 , sort legend(order(1 "ae"  2 "em" 3 "others") pos(3) col(1))

[Khati Zolfaghari]

@Nick Cox thanks for your reply, and I really appologize if I bother you and I did not explain good in the past.

• " ifscode" in my data the id code that I get to all of my data, so it is for whole sample. then I create base on the dummy variable=1 if my countries were advanced economy and zero otherwise.
• I did the same for the emerging market.
• I try to follow your advice , and the graph in the panel data set that I have it seems is not correct. now I did again:
• [/QUOTE]egen mean_fertility = mean(fertilityrate), by(ae em year)
  [/QUOTE]

• line mean_fertility year if ae == 1, sort || line mean_fertility year if em == 1, sort || line mean_fertility year if em == 0 & ae == 0 , sort legend(order(1 "ae" 2 "em" 3 "others") pos(3) col(1)) , tline(2003 2009 2012 2014 2016)

and I attached the graph.

• but when I did based on the lgraph and seperately:

  lgraph fertility year if em==1, tline(2003 2009 2012 2014 2016)

• lgraph fertility year if em==1, tline(2003 2009 2012 2014 2016)

• lgraph fertility year , tline(2003 2009 2012 2014 2016)

my graph it seems make sense.

again I apologize for my simple problem , many thanks in advance or your valuable time and advice.

best regards,

[Nick Cox]

Clearly #8 worked for me. There is some other story behind why it didn't work for you, possibly missing values, You don't give a data example that allows your latest to be checked. However,

tab ae em, missing

may help illumine the problem, as might

1. explaining exactly how you calculated ae and em.

2. trying

 
 egen mean_fertility = mean(fertility_rate) if !missing(ae, em), by(ae em year)

I really won't try to advise on use of lgraph, which I don't use myself.

[Khati Zolfaghari]

@Nick Cox may thanks for your reply.
1)

gen ae=0

replace ae=1 if ifscode == 111
replace ae=1 if ifscode == 112
replace ae=1 if ifscode == 122
replace ae=1 if ifscode == 124
replace ae=1 if ifscode == 128
replace ae=1 if ifscode == 132
replace ae=1 if ifscode == 134
replace ae=1 if ifscode == 136
replace ae=1 if ifscode == 138
replace ae=1 if ifscode == 144
replace ae=1 if ifscode == 146
replace ae=1 if ifscode == 156
replace ae=1 if ifscode == 158
replace ae=1 if ifscode == 172
replace ae=1 if ifscode == 174
replace ae=1 if ifscode == 178
replace ae=1 if ifscode == 182
replace ae=1 if ifscode == 184
replace ae=1 if ifscode == 193
replace ae=1 if ifscode == 196
replace ae=1 if ifscode == 436
replace ae=1 if ifscode == 528
replace ae=1 if ifscode == 532
replace ae=1 if ifscode == 542
replace ae=1 if ifscode == 576
replace ae=1 if ifscode == 142

1-1)

gen em=0

replace em=1 if ifscode == 233
replace em=1 if ifscode == 228
replace em=1 if ifscode ==964
replace em=1 if ifscode == 469
replace em=1 if ifscode ==944
replace em=1 if ifscode == 935
replace em=1 if ifscode == 199
replace em=1 if ifscode == 223
replace em=1 if ifscode == 566
replace em=1 if ifscode ==273
replace em=1 if ifscode ==542
replace em=1 if ifscode ==922
replace em=1 if ifscode ==186
replace em=1 if ifscode ==293
replace em=1 if ifscode ==578
replace em=1 if ifscode ==548
replace em=1 if ifscode ==536
replace em=1 if ifscode ==686
replace em=1 if ifscode ==924

2)

| em
ae | 0 1 | Total
-----------+----------------------+----------
0 | 3,576 432 | 4,008
1 | 600 24 | 624
-----------+----------------------+----------
Total | 4,176 456 | 4,632

but the graph did not change.




I apologized to take your time.
Many thanks in advance for your valuable time and advice.


best regards,

[Nick Cox]

Your code and results do show the problem. Look again: you have some observations with ae == 1 andem == 1, namely ifscode 542, Hence fix ae and em accordingly and re-calculate the means.

[Khati Zolfaghari]

@Nick Cox Many thanks. rally thanks and I apologize to take your time.
yes, it works,

best regards,

[Nick Cox]

The legend would be better reversed, but you're on a better road now.

[Khati Zolfaghari]

Nick Cox many thanks.


Best regards,