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<DIV><FONT color=#0000ff size=2 face=Verdana>Thanks, David.</FONT></DIV>
<DIV><FONT color=#0000ff size=2 face=Verdana></FONT> </DIV>
<DIV><FONT color=#0000ff size=2 face=Verdana>I think I understand now. It seems
the the stride should always be positive. But when execute NCL can automatically
determine to increase or to decrease the loop indentifier by the stride
depending on whether the start index is smaller than the end index or the start
index is greater than the end index.</FONT></DIV>
<DIV><FONT color=#0000ff size=2 face=Verdana></FONT> </DIV>
<DIV><FONT color=#0000ff size=2 face=Verdana>Thanks again and all my
best!</FONT></DIV>
<DIV> </DIV>
<DIV><FONT color=#0000ff size=2 face=Verdana></FONT> </DIV>
<DIV><FONT color=#0000ff size=2 face=Verdana>Shawn</FONT></DIV>
<DIV><FONT size=2 face=Verdana></FONT> </DIV>
<DIV align=left>
<DIV align=left><FONT size=2 face=Verdana>
<HR style="WIDTH: 122px; HEIGHT: 2px" SIZE=2>
</FONT></DIV>
<DIV><FONT color=#c0c0c0><FONT size=2 face=Verdana>Wen.J.Qu</FONT></DIV>
<DIV><FONT size=2 face=Verdana>2012-10-22</FONT></FONT></DIV></DIV>
<DIV><FONT size=2 face=Verdana>
<HR>
</FONT></DIV>
<DIV><FONT face=Verdana><FONT size=2><STRONG>发件人：</STRONG> David
Brown</FONT></FONT></DIV>
<DIV><FONT face=Verdana><FONT size=2><STRONG>发送时间：</STRONG>
2012-10-22 17:36:28</FONT></FONT></DIV>
<DIV><FONT face=Verdana><FONT size=2><STRONG>收件人：</STRONG>
Wen.J.Qu@gmail.com</FONT></FONT></DIV>
<DIV><FONT face=Verdana><FONT size=2><STRONG>抄送：</STRONG>
ncl-talk</FONT></FONT></DIV>
<DIV><FONT face=Verdana><FONT size=2><STRONG>主题：</STRONG> Re: [ncl-talk] How to
accomplish a loop with negative stride?</FONT></FONT></DIV>
<DIV><FONT size=2 face=Verdana></FONT> </DIV>
<DIV><FONT size=2 face=Verdana>See the reference manual at <A
href="http://www.ncl.ucar.edu/Document/Manuals/Ref_Manual/NclStatements.shtml#Loops">http://www.ncl.ucar.edu/Document/Manuals/Ref_Manual/NclStatements.shtml#Loops</A>
<DIV><BR></DIV>
<DIV>Here's what it says about looping backwards:</DIV>
<DIV><BR></DIV>
<DIV><SPAN
style="LINE-HEIGHT: 16px; FONT-FAMILY: verdana, sans-serif; FONT-SIZE: 13px"
class=Apple-style-span>
<UL
style="LIST-STYLE-POSITION: inside; PADDING-BOTTOM: 0px; MARGIN: 0px 0px 5px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; PADDING-TOP: 0px">
<P
style="PADDING-BOTTOM: 0px; LINE-HEIGHT: 12pt; FONT-VARIANT: normal; MARGIN: 10px 0px 20px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; FONT-FAMILY: verdana, sans-serif; COLOR: rgb(51,51,51); FONT-SIZE: 10pt; FONT-WEIGHT: normal; PADDING-TOP: 0px">The
following shows how to specify a loop with an explicit stride. The following
form is also necessary to loop backwards. With this form, if start is greater
than end, then the identifier is decremented by the stride value, which must
always be positive. If the start is less than the end, then the identifier is
incremented by the stride value.</P>
<DIV
style="PADDING-BOTTOM: 0px; LINE-HEIGHT: 12pt; FONT-VARIANT: normal; MARGIN: 10px 0px 20px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; FONT-FAMILY: verdana, sans-serif; COLOR: rgb(51,51,51); FONT-SIZE: 10pt; FONT-WEIGHT: normal; PADDING-TOP: 0px"><BR
class=webkit-block-placeholder></DIV><PRE style="BORDER-BOTTOM: transparent 15px solid; BORDER-LEFT: transparent 0px solid; PADDING-BOTTOM: 0px; LINE-HEIGHT: 12pt; TEXT-INDENT: 0em; MARGIN: 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; FONT-FAMILY: courier; FONT-SIZE: 10pt; BORDER-TOP: transparent 0px solid; BORDER-RIGHT: transparent 0px solid; PADDING-TOP: 0px"> do <I style="PADDING-BOTTOM: 0px; MARGIN: 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; PADDING-TOP: 0px">loop_identifier</I> = <I style="PADDING-BOTTOM: 0px; MARGIN: 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; PADDING-TOP: 0px">scalar_start_expr</I> , <I style="PADDING-BOTTOM: 0px; MARGIN: 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; PADDING-TOP: 0px">scalar_end_expr</I> , <I style="PADDING-BOTTOM: 0px; MARGIN: 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; PADDING-TOP: 0px"> scalar_stride_expr</I>
<I style="PADDING-BOTTOM: 0px; MARGIN: 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; PADDING-TOP: 0px"> statement list </I>
end do</PRE></UL></SPAN>
<DIV><BR></DIV></DIV>
<DIV><BR>
<DIV>
<DIV>On Oct 22, 2012, at 4:15 PM, Wen.J.Qu wrote:</DIV><BR
class=Apple-interchange-newline>
<BLOCKQUOTE type="cite">
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<DIV><FONT face=Verdana><FONT size=2>Hi, all</FONT></FONT></DIV>
<DIV><FONT size=2 face=Verdana></FONT> </DIV>
<DIV><FONT size=2 face=Verdana>Would you please give some suggestions on how
to accomplish a loop with negative stride? Just as below?</FONT></DIV>
<DIV><FONT size=2 face=Verdana></FONT> </DIV>
<DIV><FONT size=2 face=Verdana>Thanks a lot.</FONT></DIV>
<DIV><FONT size=2 face=Verdana></FONT> </DIV>
<DIV>Shawn</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=Verdana><FONT size=2></FONT></FONT> </DIV>
<DIV><FONT face=Verdana><FONT
size=2>**************************************************</FONT></FONT></DIV>
<DIV> do
i=lon90,lon180<BR>
n588(year_index,day_index,i)=0.0<BR>
do
j=lat80,lat10,-1<BR>
if((h(year_index,day_index,j,i)/10.0 .le. 588.0) .and.
(h(year_index,day_index,j-1,i)/10.0 .le.
588.0))then<BR>
n588(year_index,day_index,i)=j<BR>
break<BR>
end if<BR>
end do<BR> end do</DIV>
<DIV><FONT face=Verdana><FONT
size=2>**************************************************</FONT></FONT></DIV><FONT
face=Verdana>
<DIV><FONT size=2></FONT> </DIV>
<DIV align=left><FONT size=2>
<HR style="WIDTH: 122px; HEIGHT: 2px" SIZE=2>
</FONT></DIV>
<DIV><FONT color=#c0c0c0><FONT size=2>Wen.J.Qu</FONT></FONT></DIV><FONT
color=#c0c0c0></FONT>
<DIV><FONT color=#c0c0c0><FONT
size=2>2012-10-22</FONT></FONT></DIV></FONT></DIV>_______________________________________________<BR>ncl-talk
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