[Solved] (Was: Re: Need to advice: i want to reduce code lines)

Brandon Weaver <[email protected]> writes:

> The main thing you can do is not use regex to get each consecutive two digits of a string. Just use indexes on strings.
>
> An extract_date function might look something like this once cleaned up:
>
> def extract_date(date_string)
>   year  = "20#{date_string[0..1]}".to_i
>   month = date_string[2..3].to_i
>   day   = date_string[4..5].to_i
>
>   DateTime.new(day, month, year, 0, 0, 0, 0.375)
> end

Above code are very useful, thanks!!!


> I would leave the offsetting (21 and 35) outside of this function, as it's a separate concern. I would also give names to those values, which I'd guess are 3 and 5 week intervals respectively.

You are right!!! 21 -> 3 week, 35 -> 5 week.

> Now on to the unix code you have above, there are ways to get the same data in Ruby:
>
> def load_data(source_url: FULL_URL)
>   Net::HTTP
>     .get(URI(source_url))
>     # Gets the lines of a String
>     .lines
>     # Lines that start with 4 digits and have NA in there, avoids blanks
>     .grep(/^\d{4}.*NA/)
>     # Use the | delimiter to get the 1st and 3rd value
>     .map { |line| line.split(/ | /).values_at(0, 2) }
>     # Since you want in groups of the 1st and 3rd this would line them
>     # back up
>     .transpose
> end

Are yes, but i need to study more for understanding your codes.
Anyway thanks a lot!

> There are some which would (*fairly) say that transposing and extracting the data isn't the concern of that function. That'd be a refactoring left as an exercise to the reader, as I'm not aiming for perfect code
> in this, just demonstrating ideas.

Thanks again for detail comments ;;;

Sincerely,

Byung-Hee from South Korea

-- 
^고맙습니다 _地平天成_ 감사합니다_^))//


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--- z001.rb.soyeomul_version	2019-01-22 22:18:50.048000179 +0900
+++ z001.rb	2019-01-22 22:28:21.740000213 +0900
@@ -19,34 +19,16 @@
 
 
 def _make_oday(n)
-  exp = Regexp.new("^([0-9]{2})([0-9]{2})([0-9]{2})$")
-  
-  rs = n.to_s
-  
-  ix = exp.match(rs)[1]
-  iy = exp.match(rs)[2]
-  iz = exp.match(rs)[3]
-
-  sx = "20"+ix.to_s
-  x = sx.to_i
-
-  if iy[0].chr == '0'
-    y = iy[1].chr.to_i
-  else
-    y = iy.to_i
-  end
-
-  if iz[0].chr == '0'
-    z = iz[1].chr.to_i
-  else
-    z = iz.to_i
-  end
-
+  x = "20#{n[0..1]}".to_i
+  y = n[2..3].to_i
+  z = n[4..5].to_i
+    
   oday = DateTime.new(x, y, z, 0, 0, 0, 0.375)
 
   return oday
 end
 # 분만 예정일 6자리 숫자를 분석하여 날짜양식으로 형변환
+# Thanks to: Brandon Weaver on gmane.comp.lang.ruby.general
 
 def the_day_1(n)
   _make_oday(n) - 35
@@ -85,4 +67,4 @@
 
 # 편집: Emacs 26.1 (Ubuntu 18.04)
 # 최초 작성일: 2019년 1월 8일
-# 마지막 갱신: 2019년 1월 21일
+# 마지막 갱신: 2019년 1월 22일