execution of math on a list of numbers using applescript
I have three columns of numbers in a text file (using BBEdit, impossible to use numbers as there are more than 1 million lines):
299400 3275700 - 57.9788
299500-3275700 - 57.9811
299300 3275600 - 57.9346
299400 3275600 - 58.0249
299500-3275600 - 58.0026
...
I simply multiply the last column (what happens to start by "-") by the value of 0.75, but I am confused.
Beginning AppleScript, so I looked everywhere for example scripts manipulation of numbers but not lucky to have.
So far, I have this:
"But I get this error message: error "cannot make characters 16 through 23 of \"tmp_ITFR2000.txt\ text document" application \"BBEdit\" number. "- 1700 characters 16 to 23 of 'class TxtD '"tmp_ITFR2000.txt"to the number of number
I have to somehow dimension
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app.findGrepPreferences=null;
app.changeGrepPreferences=null;
app.findGrepPreferences.findWhat="^\\d+?\\.\\t+(.)";
app.changeGrepPreferences.changeTo="$1";
app.changeGrepPreferences.appliedParagraphStyle="NL";
app.activeDocument.changeGrep();
I need the list of numbers between two columns.
My source table contains the intervals with different ranges and I'm currently thinking dividing them by size.
I was able to get a solution following the information contained in the Web page:
http://StackOverflow.com/questions/3971981/get-list-of-numbers-in-between-two-columns
The solution I've used is based on the following example:
But this doesn't work in my table if I have few and modest intervals (maximum 5 digits range).with t as (
select 10 startnum, 15 endnum from dual
union all
select 18, 22 from dual
union all
select 10000, 19999999999 from dual
)
select lvl
from (select level lvl
from dual
connect by level <= (select max(endnum) from t)) a
where exists (
select 1
from t
where lvl between t.startnum and t.endnum )
In the example, I was trying to understand why this might be failing, but I do not have a conclusion yet.
There is something I noticed and that I try to understand, that is:
If I change this query to:
It will increase from 2 to 3 seconds to return results to a few minutes at least (canceled after a few minutes!)with t as (
select 10 startnum, 15 endnum from dual
union all
select 18, 22 from dual
union all
select 19999999990, 19999999999 from dual
)
select lvl
from (select level lvl
from dual
connect by level <= (select max(endnum) from t)) a
where exists (
select 1
from t
where lvl between t.startnum and t.endnum )
Thus, using this example to understand if my problem is related to having many between start and end number or if it is linked with having small intervals between large numbers.
Can someone explain to me why I have this speed of different treatment depending on the value of the interval? And why it takes longer with the small interval?
I need to apply it to different beaches (from 2 ranges of numbers to 16 range) and in all ranges, I will have small intervals to check.
For example:
endnum startnum
7341 7342
7422-7423
7439 7456
2522200050 2522200054
2522200146 2522200150
4216969839880 4216969839884
4216969839893 4216969839897
Having this problem I can only ask the first values I have in the picture!
Best regards
Ricardo Tomas
WITH cntr AS
(
SELECT LEVEL - 1 AS n
FROM (
SELECT MAX ( endnum - startnum) AS max_diff
FROM t
)
CONNECT BY LEVEL <= max_diff + 1
)
SELECT startnum + c.n
FROM t
JOIN cntr c ON t.endnum >= t.startnum + c.n
;
Joins are usually faster than EXISTS subqueries.Acrobat - Create TOC bookmarks
var myDocFonts=app.documents[0].stories.everyItem().textStyleRanges.everyItem().appliedFont,
l = myDocFonts.length, myFonts="";
uniqueFonts = {}, n;
while (l--) uniqueFonts[myDocFonts[l].name] = myDocFonts[l].name;
for (n in uniqueFonts) myFonts+=(uniqueFonts[n])+"\r";
alert(myFonts)
myList.itemRenderer = new ClassFactory(MyItemRenderer);
Thank you
public static final long MILLIS_IN_YEAR = 1000L * 60L * 60L * 24L * 365L
Why all numbers use symbol 'L' in the following code?
Thank you
public static final long MILLIS_IN_YEAR = 1000L * 60L * 60L * 24L * 365L
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