stepchart graph with time series values by year

[Fernando Greve]

Hi stata users,
I have some time series values and I want a graph with steps like that in the link:

http://peltiertech.com/line-chart-without-risers/

Do you know if there is a graph function?
Regards

[Nick Cox]

This can be done with the graph command (not a function). The data structure needed in Stata is similar to that in your source. Try

* Example generated by -dataex-. To install: ssc install dataex
clear
input str9 date float rate
"14-Sep-75"  .1
"31-Dec-75"  .1
"31-Dec-75"   .
"31-Dec-75" .13
"29-May-78" .13
"29-May-78"   .
"29-May-78" .15
"22-Mar-81" .15
"22-Mar-81"   .
"22-Mar-81" .18
"1-Nov-81"  .18
"1-Nov-81"    .
"1-Nov-81"   .2
"17-Feb-85"  .2
"17-Feb-85"   .
"17-Feb-85" .22
"3-Apr-88"  .22
"3-Apr-88"    .
"3-Apr-88"  .25
"3-Feb-91"  .25
"3-Feb-91"    .
"3-Feb-91"  .29
"1-Jan-95"  .29
"1-Jan-95"    .
"1-Jan-95"  .32
"10-Jan-99" .32
"10-Jan-99"   .
"10-Jan-99" .33
"7-Jan-01"  .33
"7-Jan-01"    .
"7-Jan-01"  .34
"30-Jun-02" .34
"30-Jun-02"   .
"30-Jun-02" .37
"8-Jan-06"  .37
"8-Jan-06"    .
"8-Jan-06"  .39
"14-May-07" .39
"14-May-07"   .
"14-May-07" .41
"12-May-08" .41
"12-May-08"   .
"12-May-08" .42
end
gen ndate = daily(date, "DMY", 2050)
format ndate %td
twoway line rate ndate, cmissing(no) xla(, format(%tdCY)) xtitle("") yla(, ang(h))

[Fernando Greve]

Thank you for your answer, Nick.
there exist a way to: duplicate my yearly data; for every duplicated obs, gen a var with 1-jan-"year" and 31-dec-"year"; and finally, get a null obs between those pairs?
If I get that I could use your code.
Regards

[Nick Cox]

help expand

[Fernando Greve]

This is the code:

*ssc install expgen
clear
input str9 date float rate
1960    
1961    1.3
1962    1.3
1963    1.3
1964    1.3
1965    1.3
1966    3.1
1967    3.1
1968    3.1
1969    3.1
1970    3.1
1971    -3.2
1972    -3.2
1973    -3.2
1974    -3.2
1975    -3.2
1976    5.1
1977    5.1
1978    5.1
1979    5.1
1980    5.1
1981    -2.3
1982    -2.3
1983    -2.3
1984    -2.3
1985    -2.3
1986    2.1
1987    2.1
1988    2.1
1989    2.1
1990    2.1
1991    4.9
1992    4.9
1993    4.9
1994    4.9
1995    4.9
1996    0.9
1997    0.9
1998    0.9
1999    0.9
2000    0.9
2001    1
2002    1
2003    1
2004    1
2005    
2006    
2007    
2008    
2009    
2010    
2011    
2012    
2013    
2014    
2015    
end


expgen nreps=2
gen ndate = daily(date, "Y", 2050)
format ndate %td

forvalues i = 2(2)88 {
    replace ndate = ndate[`i'+1] - 1 in `i'/`i'
    }

twoway line rate ndate, cmissing(no) xla(, format(%tdCY)) xtitle("") yla(, ang(h))

[Nick Cox]

It was late my time zone when I posted, so I confined myself to a hint. For your data, this code suffices. I see no obvious point to importing a string year and converting it to a daily date.

clear
input date rate
1960    
1961    1.3
1962    1.3
1963    1.3
1964    1.3
1965    1.3
1966    3.1
1967    3.1
1968    3.1
1969    3.1
1970    3.1
1971    -3.2
1972    -3.2
1973    -3.2
1974    -3.2
1975    -3.2
1976    5.1
1977    5.1
1978    5.1
1979    5.1
1980    5.1
1981    -2.3
1982    -2.3
1983    -2.3
1984    -2.3
1985    -2.3
1986    2.1
1987    2.1
1988    2.1
1989    2.1
1990    2.1
1991    4.9
1992    4.9
1993    4.9
1994    4.9
1995    4.9
1996    0.9
1997    0.9
1998    0.9
1999    0.9
2000    0.9
2001    1
2002    1
2003    1
2004    1
end 
expand 2 if rate != rate[_n+1] 
sort date 
replace rate = . if date == date[_n-1] 
line rate date, cmissing(no)