Re: Help with an algorithm...
On 8/5/19 3:20 PM, dsc--- via use-livecode wrote: Children are sub-arrays of parents. Ain't that the truth. -- Mark Wieder ahsoftw...@gmail.com ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I fixed this to use lists... Maybe. put parentArray( LISTNEW ) into aNew put parentArray( LISTOLD ) into aOld intersect aNew with aOld -- Remove from aOld that which is not in aNew union aOld with aNew recursively -- Leaves aNew unchanged except that children in aOld are brought in. I think. put parentList( aNew ) into LISTNEW Parents and children are keys in the arrays built. Children are sub-arrays of parents. > On Aug 5, 2019, at 12:31 PM, Dar Scott Consulting via use-livecode > wrote: > > I'm pretty sure I goofed somewhere, but maybe something like this? > > intersect ARRAYNEW with ARRAYOLD into temp > union temp with ARRAYNEW recursively > > >> On Aug 5, 2019, at 9:53 AM, Paul Dupuis via use-livecode >> wrote: >> >> Today is not my coding day. I have a problem I should be able to design a >> solution for an am struggling. Clearly I am missing "something" >> >> I have 2 lists (LISTNEW and LISTOLD) of the following format: >> >> ParentA >> Child 1 >> Child 2 >> etc. >> ParentB >> Child 1 >> etc. >> etc. >> >> The parents are in alphabetical sorted order, the children may not be in >> sorted order >> >> I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW >> Parents >> FOR any LISTOLD Parent present in LISTNEW, check the Children of the >> matching Parents and add any Child for the LISTOLD Parent that is not >> already under its matching LISTNEW Parent >> FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its >> Children >> >> I can not seem to write an approach to solve this today. Does any body have >> some code so solve this they may be willing to share? >> >> ___ >> use-livecode mailing list >> use-livecode@lists.runrev.com >> Please visit this url to subscribe, unsubscribe and manage your subscription >> preferences: >> http://lists.runrev.com/mailman/listinfo/use-livecode >> > > > ___ > use-livecode mailing list > use-livecode@lists.runrev.com > Please visit this url to subscribe, unsubscribe and manage your subscription > preferences: > http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I know this does not attend to the question, but my feelings are like this: if one has control, go back and use arrays from the start. Now, to your comments on robustness in parsing the files. I suppose any whitespace at the start of a line could be considered a child. Also, it is an error if the first line is a child. > On Aug 5, 2019, at 3:59 PM, dunbarxx via use-livecode > wrote: > > Hmmm. I had mentioned earlier: > > "You cannot just find matching lines between the two lists, because some > children > AND some parents may be present in both... > > The discussion has focused on array lore, and that is fine, but are we all > in agreement that the main task is to isolate the parents, regardless of > whether the names of those parents might be duplicated? I mention this > because everything depends on the reliability of those spaces. The parsing > of parent "groups", in one way or another, is essential. > > Craig > > > > -- > Sent from: > http://runtime-revolution.278305.n4.nabble.com/Revolution-User-f278306.html > > ___ > use-livecode mailing list > use-livecode@lists.runrev.com > Please visit this url to subscribe, unsubscribe and manage your subscription > preferences: > http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
Hmmm. I had mentioned earlier: "You cannot just find matching lines between the two lists, because some children AND some parents may be present in both... The discussion has focused on array lore, and that is fine, but are we all in agreement that the main task is to isolate the parents, regardless of whether the names of those parents might be duplicated? I mention this because everything depends on the reliability of those spaces. The parsing of parent "groups", in one way or another, is essential. Craig -- Sent from: http://runtime-revolution.278305.n4.nabble.com/Revolution-User-f278306.html ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
My mistake. I was thinking arrays. > On Aug 5, 2019, at 3:34 PM, Mark Wieder via use-livecode > wrote: > > On 8/5/19 2:24 PM, Dar Scott Consulting via use-livecode wrote: >> Yikes! I wasn't aware of duplicate keys being a problem. How does that >> happen? > > Marx > Groucho > Chico > etc. > Marx > Karl > etc. > > -- > Mark Wieder > ahsoftw...@gmail.com > > ___ > use-livecode mailing list > use-livecode@lists.runrev.com > Please visit this url to subscribe, unsubscribe and manage your subscription > preferences: > http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
On 8/5/19 2:24 PM, Dar Scott Consulting via use-livecode wrote: Yikes! I wasn't aware of duplicate keys being a problem. How does that happen? Marx Groucho Chico etc. Marx Karl etc. -- Mark Wieder ahsoftw...@gmail.com ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
Yikes! I wasn't aware of duplicate keys being a problem. How does that happen? > On Aug 5, 2019, at 2:28 PM, Alex Tweedly via use-livecode > wrote: > > I'm a great fan of, and user of, arrays - but we always need to be careful of > losing data with duplicated keys. > > You haven't said that the parent names in LISTNEW are guaranteed to be > unique. This simple code assumes they are - if they're not, it's easy to add > a check ... (And it also assumes the inputs are properly formatted!!) > > (inputs are in fields "FNew" and "FOld", output in field "FOut") > > on mouseUp >local tOld, tNew, tOut >local tA >local tLastParent > >put the text of fld "FNew" into tNew >repeat for each line L in tNew -- convert to array > if L begins with space then -- a child > put L after tA[tLastParent] > else > put L into tLastParent > end if >end repeat > >put the text of fld "FOld" into tOld >put empty into tLastParent >repeat for each line L in tOld > if L begins with space then -- a child > if (tLastParent is not empty) \ > AND (L is not among the lines of tA[tLastParent]) then > put L after tA[tLastParent] > end if > else > if L is among the keys of tA then -- a Parent we need to deal with > put L into tLastParent > else > put empty into tLastParent > end if > end if >end repeat > >-- and then collect the expanded outptu >local tKeys >put the keys of tA into tKeys >sort lines of tKeys >repeat for each line K in tKeys > put K & tA[K] after tOut >end repeat >put tOut into fld "fOut" > end mouseUp > > Alex. > > > On 05/08/2019 16:53, Paul Dupuis via use-livecode wrote: >> Today is not my coding day. I have a problem I should be able to design a >> solution for an am struggling. Clearly I am missing "something" >> >> I have 2 lists (LISTNEW and LISTOLD) of the following format: >> >> ParentA >> Child 1 >> Child 2 >> etc. >> ParentB >> Child 1 >> etc. >> etc. >> >> The parents are in alphabetical sorted order, the children may not be in >> sorted order >> >> I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW >> Parents >> FOR any LISTOLD Parent present in LISTNEW, check the Children of the >> matching Parents and add any Child for the LISTOLD Parent that is not >> already under its matching LISTNEW Parent >> FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its >> Children >> >> I can not seem to write an approach to solve this today. Does any body have >> some code so solve this they may be willing to share? >> >> ___ >> use-livecode mailing list >> use-livecode@lists.runrev.com >> Please visit this url to subscribe, unsubscribe and manage your subscription >> preferences: >> http://lists.runrev.com/mailman/listinfo/use-livecode > > ___ > use-livecode mailing list > use-livecode@lists.runrev.com > Please visit this url to subscribe, unsubscribe and manage your subscription > preferences: > http://lists.runrev.com/mailman/listinfo/use-livecode > ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I'm a great fan of, and user of, arrays - but we always need to be careful of losing data with duplicated keys. You haven't said that the parent names in LISTNEW are guaranteed to be unique. This simple code assumes they are - if they're not, it's easy to add a check ... (And it also assumes the inputs are properly formatted!!) (inputs are in fields "FNew" and "FOld", output in field "FOut") on mouseUp local tOld, tNew, tOut local tA local tLastParent put the text of fld "FNew" into tNew repeat for each line L in tNew -- convert to array if L begins with space then -- a child put L after tA[tLastParent] else put L into tLastParent end if end repeat put the text of fld "FOld" into tOld put empty into tLastParent repeat for each line L in tOld if L begins with space then -- a child if (tLastParent is not empty) \ AND (L is not among the lines of tA[tLastParent]) then put L after tA[tLastParent] end if else if L is among the keys of tA then -- a Parent we need to deal with put L into tLastParent else put empty into tLastParent end if end if end repeat -- and then collect the expanded outptu local tKeys put the keys of tA into tKeys sort lines of tKeys repeat for each line K in tKeys put K & tA[K] after tOut end repeat put tOut into fld "fOut" end mouseUp Alex. On 05/08/2019 16:53, Paul Dupuis via use-livecode wrote: Today is not my coding day. I have a problem I should be able to design a solution for an am struggling. Clearly I am missing "something" I have 2 lists (LISTNEW and LISTOLD) of the following format: ParentA Child 1 Child 2 etc. ParentB Child 1 etc. etc. The parents are in alphabetical sorted order, the children may not be in sorted order I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW Parents FOR any LISTOLD Parent present in LISTNEW, check the Children of the matching Parents and add any Child for the LISTOLD Parent that is not already under its matching LISTNEW Parent FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its Children I can not seem to write an approach to solve this today. Does any body have some code so solve this they may be willing to share? ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I'm pretty sure I goofed somewhere, but maybe something like this? intersect ARRAYNEW with ARRAYOLD into temp union temp with ARRAYNEW recursively > On Aug 5, 2019, at 9:53 AM, Paul Dupuis via use-livecode > wrote: > > Today is not my coding day. I have a problem I should be able to design a > solution for an am struggling. Clearly I am missing "something" > > I have 2 lists (LISTNEW and LISTOLD) of the following format: > > ParentA > Child 1 > Child 2 > etc. > ParentB > Child 1 > etc. > etc. > > The parents are in alphabetical sorted order, the children may not be in > sorted order > > I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW > Parents > FOR any LISTOLD Parent present in LISTNEW, check the Children of the > matching Parents and add any Child for the LISTOLD Parent that is not already > under its matching LISTNEW Parent > FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its > Children > > I can not seem to write an approach to solve this today. Does any body have > some code so solve this they may be willing to share? > > ___ > use-livecode mailing list > use-livecode@lists.runrev.com > Please visit this url to subscribe, unsubscribe and manage your subscription > preferences: > http://lists.runrev.com/mailman/listinfo/use-livecode > ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
use-list formatting (was Re: Help with an algorithm...)
I agree that the lack of formatting makes it hard to communicate. I would favor changing the list settings to allow for it. I don't think we have a spam problem that would discourage that. And to help support this, LC copy should include types easily pasted into mail clients. > On Aug 5, 2019, at 11:49 AM, dunbarxx via use-livecode > wrote: > > I really do not like the use-list. It is difficult to format one's answers. ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I saw your post preceded my second one. Besides the fact that this works as well and is faster: repeat for each line tLine in tAll put tLine into myArray[ tLine] end repeat The array thing is the easy part. The real working of this gadget depends on being able to isolate the children, and the only way I can see doing that easily to rely on those spaces. This may be fragile, depending on how consistent that format sticks. You cannot find matching lines between the two lists, because some children AND some parents may be present in both, and you would not know which "class" of data you were dealing with. I would format such lists in the future more robustly, perhaps with tabs. Craig -- Sent from: http://runtime-revolution.278305.n4.nabble.com/Revolution-User-f278306.html ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I'd tend to look for ways to do this with functions that work on whole collections and avoid loops. If that is not found, or is hard to work with, I'd change the lists to be arrays. Each array is is keyed by parents. Each parent is an array of children. Children can be represented as an arrayed indexed by children. The elements of those might be ignored or have data. Loop on keys in LISTNEW... > On Aug 5, 2019, at 9:53 AM, Paul Dupuis via use-livecode > wrote: > > Today is not my coding day. I have a problem I should be able to design a > solution for an am struggling. Clearly I am missing "something" > > I have 2 lists (LISTNEW and LISTOLD) of the following format: > > ParentA > Child 1 > Child 2 > etc. > ParentB > Child 1 > etc. > etc. > > The parents are in alphabetical sorted order, the children may not be in > sorted order > > I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW > Parents > FOR any LISTOLD Parent present in LISTNEW, check the Children of the > matching Parents and add any Child for the LISTOLD Parent that is not already > under its matching LISTNEW Parent > FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its > Children > > I can not seem to write an approach to solve this today. Does any body have > some code so solve this they may be willing to share? > > ___ > use-livecode mailing list > use-livecode@lists.runrev.com > Please visit this url to subscribe, unsubscribe and manage your subscription > preferences: > http://lists.runrev.com/mailman/listinfo/use-livecode > ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I really do not like the use-list. It is difficult to format one's answers. Here is a handler that places a few pieces of data into two variables. These would be the two isolated lists of children derived from the parents as described earlier. There is one difference between the two lists, the child "xxx'. on mouseUp put "aaa" & return & "bbb" & return & "ccc" into tNew put "aaa" & return & "xxx" & return & "ccc" into tOld put tOld & return & tnew into tAll repeat with y = 1 to the number of lines of tAll put line y of tAll into myArray[ line y of tAll] end repeat breakpoint end mouseUp Straightforward array stuff. The final list has four children, "xxx" being added to the mix, as can be seen in the debugger at the breakpoint. Craig -- Sent from: http://runtime-revolution.278305.n4.nabble.com/Revolution-User-f278306.html ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
I had not considered using arrays. I have no idea why, as it provides a mechanism. Just not thinking well today. Thanks for the tip! On 8/5/2019 1:40 PM, dunbarx--- via use-livecode wrote: Hi. This seems like a good case for arrays (pseudo): You have to create an array for each parent. The only way i see to find those is to go through the list looking for lines that DO NOT start with a space. If you work from the bottom up, since children are always at the bottom and they always have spaces in front, you can find the lastMost parent. Set the itemDel to that parent, and isolate the last item. This will be a list of children of that parent. See if there is a corresponding parent in the new list, and isolate that "item" list of children. Run through the children of the old list and see if they live in the new list. If they do, add them:The array part is something like: put oldChildren & return & newChildren into allChildrenrepeat for each line tLine in allChildrenput tLine into myArray[tLine] Delete the last item of the old list and do it again. This way the last item can be isolated directly. Craig -Original Message- From: Paul Dupuis via use-livecode To: How to use LiveCode Cc: Paul Dupuis Sent: Mon, Aug 5, 2019 11:53 am Subject: Help with an algorithm... Today is not my coding day. I have a problem I should be able to design a solution for an am struggling. Clearly I am missing "something" I have 2 lists (LISTNEW and LISTOLD) of the following format: ParentA Child 1 Child 2 etc. ParentB Child 1 etc. etc. The parents are in alphabetical sorted order, the children may not be in sorted order I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW Parents FOR any LISTOLD Parent present in LISTNEW, check the Children of the matching Parents and add any Child for the LISTOLD Parent that is not already under its matching LISTNEW Parent FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its Children I can not seem to write an approach to solve this today. Does any body have some code so solve this they may be willing to share? ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
Re: Help with an algorithm...
Hi. This seems like a good case for arrays (pseudo): You have to create an array for each parent. The only way i see to find those is to go through the list looking for lines that DO NOT start with a space. If you work from the bottom up, since children are always at the bottom and they always have spaces in front, you can find the lastMost parent. Set the itemDel to that parent, and isolate the last item. This will be a list of children of that parent. See if there is a corresponding parent in the new list, and isolate that "item" list of children. Run through the children of the old list and see if they live in the new list. If they do, add them:The array part is something like: put oldChildren & return & newChildren into allChildrenrepeat for each line tLine in allChildrenput tLine into myArray[tLine] Delete the last item of the old list and do it again. This way the last item can be isolated directly. Craig -Original Message- From: Paul Dupuis via use-livecode To: How to use LiveCode Cc: Paul Dupuis Sent: Mon, Aug 5, 2019 11:53 am Subject: Help with an algorithm... Today is not my coding day. I have a problem I should be able to design a solution for an am struggling. Clearly I am missing "something" I have 2 lists (LISTNEW and LISTOLD) of the following format: ParentA Child 1 Child 2 etc. ParentB Child 1 etc. etc. The parents are in alphabetical sorted order, the children may not be in sorted order I need to hunt through LISTOLD comparing the LISTOLD Parents to the LISTNEW Parents FOR any LISTOLD Parent present in LISTNEW, check the Children of the matching Parents and add any Child for the LISTOLD Parent that is not already under its matching LISTNEW Parent FOR any LISTOLD Parent NOT in LISTNEW, I can ignore the Parent and its Children I can not seem to write an approach to solve this today. Does any body have some code so solve this they may be willing to share? ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode ___ use-livecode mailing list use-livecode@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-livecode
