次の構文は、例えば(例多くを解決するため、間違いなくすべてではない彼らのだろう「1.7」または "。 2014年に3年」)。私はで動作するようにあなたのサンプルデータを再作成まず
...あなたはそれに多くの作業を行う必要がありますが、これはあなたがうまく始める必要があります。
今
data list list/age (a30).
begin data
"2"
"5 months"
"11 days"
"1.7"
"13 yr"
"22 yrs"
"13 Months"
"10 mo"
"6/19/2016"
"3y10m"
"10m"
"12y"
"3.5 years"
"3 YEARS"
"11 mos"
"1 year 10 months"
"1 year, two months"
"20 Y"
"13 y/o"
"3 years in 2014"
end data.
仕事に:
* some necessary definitions.
string ageCleaned (a30) chr (a1) nm d m y (a5).
compute ageCleaned="".
* my first step is to create a "cleaned" age variable (it's possible to
manage without this variable but using this is better for debugging and
improving the method).
* in the `ageCleaned` variable I only keep digits, periods (for decimal
point) and the characters "d", "m", "y".
do if CHAR.INDEX(lower(age),'ymd',1)>0.
loop #chrN=1 to char.length(age).
compute chr=lower(char.substr(age,#chrN,1)).
if CHAR.INDEX(chr,'ymd.',1)>0 ageCleaned=concat(rtrim(ageCleaned),chr).
end loop.
end if.
* the following line accounts for the word "days" which in the `ageCleaned`
variable has turned into the characters "dy".
compute ageCleaned=replace(ageCleaned,"dy","d").
exe.
* now I can work through the `ageCleaned` variable, accumulating digits
until I meet a character, then assigning the accumulated number to the
right variable according to that character ("d", "m" or "y").
compute nm="".
loop #chrN=1 to char.length(ageCleaned).
compute chr=char.substr(ageCleaned,#chrN,1).
do if CHAR.INDEX(chr,'.',1)>0.
compute nm=concat(rtrim(nm),chr).
else.
if chr="y" y=nm.
if chr="m" m=nm.
if chr="d" d=nm.
compute nm="".
end if.
end loop.
exe.
* we now have the numbers in string format, so after turning them into
numbers they are ready for use in calculations.
alter type d m y (f8.2).
compute DaysOld=sum(365*y, 30.5*m, d).
はい、本当にありがとうございました!残されたのは、数字だけのインスタンス、すなわち非常に簡単な「27」のインスタンスを転送することでした。あなたは私に1トンの時間を救った! – Rozo