Sort file by length of lines
Sort file by length of lines
cat $@ | awk '{ print length, $0 }' | sort -n | awk '{$1=""; print $0}'
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cmp-length - compare the length of two series and return -1, 0, or 1, to support stable sorts by length
; Support func for stable sort comparator
cmp-length: func [a [series!] b [series!] /local len-a len-b] [
len-a: length? a
len-b: length? b
case [
len-a < len-b [-1]
len-a > len-b [1]
len-a = len-b [0]
]
]
one-item? - simple wrapper to aid readability
one-item?: func [series] [1 = length? series]
Adjust Page Number when Page Length Changes
In a web page you are display a certain page worth of results. If you allow the user to change the page length, you'll need to find the new page number that contains the first item on the current page. This is VB.Net code.
The index of the first item on the current page happens to be a count of the items up to that point. So assume we are paging that amount and use the last page as our new current page. We apply standard paging logic of items integer divided by page length to determine the number of pages and add one if there is a remainder.
There are a couple formulas at work here worth noting.
Index of Last Item on page:
page number * page length
Index of First Item on page:
(index of last item of previous page) + 1
((page number - 1) * page length) + 1
Page Count based on page length:
number of items [integer divide] page length
if there is a remainder to the above then add 1
intOldPageLength = GetPageLengthValue() ' if there is a new page length value If intOldPageLength <> m_intPageLength Then ' adjust the page number to compensate intItemCount = ((m_intPage - 1) * intOldPageLength) + 1 m_intPage = intItemCount \ m_intPageLength If intItemCount - (m_intPage * m_intPageLength) > 0 Then m_intPage = m_intPage + 1 End If End If
The index of the first item on the current page happens to be a count of the items up to that point. So assume we are paging that amount and use the last page as our new current page. We apply standard paging logic of items integer divided by page length to determine the number of pages and add one if there is a remainder.
There are a couple formulas at work here worth noting.
Index of Last Item on page:
page number * page length
Index of First Item on page:
(index of last item of previous page) + 1
((page number - 1) * page length) + 1
Page Count based on page length:
number of items [integer divide] page length
if there is a remainder to the above then add 1
Lenght between a dot and a line //JavaScript Function
Given a dot and a line, it returns the distance between them, the last parameter tells if the line should be considered infinite or if the function should respect its limits.
[UPDATED CODE AND HELP CAN BE FOUND HERE]
//+ Jonas Raoni Soares Silva //@ http://jsfromhell.com/math/dot-line-length [v1.0] dotLineLength = function( x, y, x0, y0, x1, y1, o ){ function lineLength( x, y, x0, y0 ){ return Math.sqrt( ( x -= x0 ) * x + ( y -= y0 ) * y ); } if( o && !( o = function( x, y, x0, y0, x1, y1 ){ if( !( x1 - x0 ) ) return { x: x0, y: y }; else if( !( y1 - y0 ) ) return { x: x, y: y0 }; var left, tg = -1 / ( ( y1 - y0 ) / ( x1 - x0 ) ); return { x: left = ( x1 * ( x * tg - y + y0 ) + x0 * ( x * - tg + y - y1 ) ) / ( tg * ( x1 - x0 ) + y0 - y1 ), y: tg * left - tg * x + y }; }( x, y, x0, y0, x1, y1 ), o.x >= Math.min( x0, x1 ) && o.x <= Math.max( x0, x1 ) && o.y >= Math.min( y0, y1 ) && o.y <= Math.max( y0, y1 ) ) ){ var l1 = lineLength( x, y, x0, y0 ), l2 = lineLength( x, y, x1, y1 ); return l1 > l2 ? l2 : l1; } else { var a = y0 - y1, b = x1 - x0, c = x0 * y1 - y0 * x1; return Math.abs( a * x + b * y + c ) / Math.sqrt( a * a + b * b ); } };
Line Length //JavaScript Function
Returns the length of a line.
[UPDATED CODE AND HELP CAN BE FOUND HERE]
//+ Jonas Raoni Soares Silva //@ http://jsfromhell.com/math/line-length [v1.0] lineLength = function( x, y, x0, y0 ){ return Math.sqrt( ( x -= x0 ) * x + ( y -= y0 ) * y ); };
