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• How To: Transfer Categories in Outlook 2007 from One Installation to Another Using Google Sync

  - by nicorellius

  I use Outlook 2007 for email. I also happen to use the categories for organizing my mail. This function is cool, because you can use keyboard shortcuts that make categorizing mail quick and easy. The problem I am having is that, since I use Google Apps Sync at home and at work, I can't seem to figure out how to transfer the existing category assignments that I spent months creating at work. As my email comes and goes, I assign this and that, with thousands of messages all belonging to specific categories. When I finally installed and started using Google Apps Sync at home, the synchronization process works great, but all my messages are blank, i.e., don't have the color categories I assigned at work. What's worse, I actually changed computers at work, and now am facing the same problem on the new workstation. Using: Windows XP Pro, Outlook 2007, Google Apps. Anyone run into this and find a solution?

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• My website links now include DirectX categories

  - by Michael B. McLaughlin

  I’ve done a bit of overdue updates to my website. The links page - http://www.bobtacoindustries.com/Developers/Links  - now includes DirectX links. I’ve also updated all of the links that were broken in the recent changes to the App Hub site. If you have any good links you think I’m missing, let me know. I haven’t had a chance to do any dead link checking & pruning yet so there might be some links that go to the wrong place or go nowhere at all. That’s the problem with link repositories; maintenance of them is quite a bit of work. But hopefully I’ll find some time to do that soon.

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• How can I generate a list of appendices with LaTeX, seperate from table of contents (toc)

  - by None

  LaTeX' appendix environment formats sections differently and uses a different enumeration in the table of contents: \begin{appendix} \section{superuser} \dots{} .... Shows up as       A       superuser            11 in the \tableofcontents. Unfortunately, I have to use a different style: the appendix is not to show up in the table of contents, but in a seperate listing right before the appendix. I see the following options to solve this: suppress output in tableofcontents and somehow recreate the part on a different page generate a custom listing of specified / the following section manually create a list with the same formatting of the tableofcontents I use scrartcl as document class. PS: appedix / appendices is not a tag yet

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• Problem with implementing removeAll for List of custom object

  - by Jay

  Hello everyone, I have a scenario in my code where I need to compare two Lists and remove from the first list, objects which are present in the second list. Akin to how the "removeAll" object works for List. Since my List is created on a custom object, the removeAll method won't work for me. I have tried various methods to make this work: - implemented equals() and hashCode for the custom object comprising the list - implemented the Comparable Interface for the custom object - implemented the Comparator Interface for the custom object I've even tried using the Apache Common's CollectionUtils and ListUtils methods (subtract, intersect, removeAll). None seem to work. I understand I will perhaps need to write some custom removal code. But not sure how to go about doing that. Any pointers helping me move in the right direction will be really appreciated. Thanks, Jay

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• Convert a List of Options to an Option of List using Scalaz

  - by Rafael de F. Ferreira

  The following function transforms a list of Option[T] into a list of Some[T], in the case where all members are Some's, or None, in the case where there is at least one None member. I guess the code is clearer that this explanation: def lo2ol[T](lo: List[Option[T]]): Option[List[T]] = { lo.foldRight[Option[List[T]]](Some(Nil)){(o, ol) => (o, ol) match { case (Some(x), Some(xs)) => Some(x :: xs); case _ => None : Option[List[T]]; }}} I remember seeing somewhere a similar example, but using Scalaz to simplify the code. How would it look like?

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• converting a matrix to a list

  - by andrewj

  Suppose I have a matrix foo as follows: foo <- cbind(c(1,2,3), c(15,16,17)) > foo [,1] [,2] [1,] 1 15 [2,] 2 16 [3,] 3 17 I'd like to turn it into a list that looks like [[1]] [1] 1 15 [[2]] [1] 2 16 [[3]] [1] 3 17 You can do it as follows: lapply(apply(foo, 1, function(x) list(c(x[1], x[2]))), function(y) unlist(y)) I'm interested in an alternative method that isn't as complicated. Note, if you just do apply(foo, 1, function(x) list(c(x[1], x[2]))), it returns a list within a list, which I'm hoping to avoid.

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• sharepoint list relationship question

  - by Korey

  I have two custom lists in sharepoint with a one to many relationship on a column in both lists. So list A has text column 1 and list B has a lookup column 1' on list A's column 1. I want to display the rows in list A, where column 1 has a match on column 1' and a 2nd column in list B with a specific text string. I have Sharepoint Designer and am fumbling along with linked sources described here http://office.microsoft.com/en-us/sharepointdesigner/HA100991441033.aspx, but can't get exactly what I'm looking for.

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• Upgrading a SharePoint list instance that was deployed via feature

  - by Goldmember

  I'm curious how others address this issue. Using VSEWSS 1.3, I have created a site content type, a list definition (w event receivers), and a list instance. All of them are in the same WSP solution and each is activated individually via features. Now let's assume that all the features have been activated for some time, and the list instance contains a number of items (that can't be deleted). Now suppose I need to make a change to the schema.xml (inject some javascript, modify views, whatever) of the list. Is it even possible to "upgrade" the schema of the existing list instance? Otherwise I would think I'm stuck creating a new instance and copying items over.

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• Fill a array with List data with one more element

  - by marionmaiden

  Hello, By a question that I made, I figured out that tho copy elements from one list to an array I just need to use the method toArray() for this. But let's suppose I have a List with n objects. I want to copy then into a array sized n+1 and add into the first position another object and in the other n positions the n data of the list. This is the way I'm doing it for now, but I'm just wondering if there is a better way for do that: Object array[] = new Object[list.size() + 1]; Object chk = new Object(); array[0] = chk; for(int i = 1; i < array.length; i++){ array[i] = list.get(i); }

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• just-in-time list

  - by intuited

  I'd like to know if there is a class available, either in the standard library or in pypi, that fits this description. The constructor would take an iterator. It would implement the container protocol (ie _getitem_, _len_, etc), so that slices, length, etc., would work. In doing so, it would iterate and retain just enough values from its constructor argument to provide whatever information was requested. So if jitlist[6] was requested, it would call self.source.next() 7 times, save those elements in its list, and return the last one. This would allow downstream code to use it as a list, but avoid unnecessarily instantiating a list for cases where list functionality was not needed, and avoid allocating memory for the entire list if only a few members ended up being requested. It seems like a pretty easy one to write, but it also seems useful enough that it's likely that someone would have already made it available in a module.

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• Python: Repeat elements in a list comprehension?

  - by User

  I have the following list comprehension which returns a list of coordinate objects for each location. coordinate_list = [Coordinates(location.latitude, location.longitude) for location in locations] This works. Now suppose the location object has a number_of_times member. I want a list comprehension to generate n Coordinate objects where n is the number_of_times for the particular location. So if a location has number_of_times = 5 then the coordinates for that location will be repeated 5 times in the list. (Maybe this is a case for a for-loop but I'm curious if it can be done via list comprehensions)

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• python conditional list creation from 2D lists

  - by dls

  Say I've got a list of lists. Say the inner list of three elements in size and looks like this: ['apple', 'fruit', 1.23] The outer list looks like this data = [['apple', 'fruit', 1.23], ['pear', 'fruit', 2.34], ['lettuce', 'vegetable', 3.45]] I want to iterate through the outer list and cull data for a temporary list only in the case that element 1 matches some keyword (aka: 'fruit'). So, if I'm matching fruit, I would end up with this: tempList = [('apple', 1.23), ('pear', 2.34)] This is one way to accomplish this: tempList = [] for i in data: if i[1] == 'fruit': tempList.append(i[0], i[2]) is there some 'Pythonic' way to do this in fewer lines?

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• Fastest way to check a List<T> for a date

  - by fishhead

  I have a list of dates that a machine has worked on, but it doesn't include a date that machine was down. I need to create a list of days worked and not worked. I am not sure of the best way to do this. I have started by incrementing through all the days of a range and checking to see if the date is in the list by iterating through the entire list each time. I am looking for a more efficient means of finding the dates. class machineday { datetime WorkingDay; } class machinedaycollection : List<machineday> { }

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• Display/Hide part of a list

  - by superexsl

  Hi I have an (unordered)list (generated by a repeater) of items. However, I'd like to show the first three items, and have the rest hidden by the main content div. When a button is pressed, I would like the list's div to expand, pushing the main content div down and showing the rest of the list. I was thinking of using slideDown(), but that closes the entire div ( and I would like to show the first 3 items of the list). What would be the best way to achieve an effect like this? Is there a plugin that can easily show X items of a list and the display the rest upon request? Thanks

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• Best way to implement a List(Of) with a maximum number of items

  - by Ben

  I'm trying to figure out a good way of implementing a List(Of) that holds a maximum number of records. e.g. I have a List(Of Int32) - it's being populated every 2 seconds with a new Int32 item. I want to store only the most current 2000 items. How can I make the list hold a maximum of 2000 items, then when the 2001'th item is attempted to be added, the List drops the 2000'th item (resulting in the current total being 1999). Thing is, I need to make sure I'm dropping only the oldest item and adding a new item into the List. Ben

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• Trim email list into domain list

  - by hanjaya

  The function below is part of a script to trim email list from a file into domain list and removes duplicates. /* define a function that can accept a list of email addresses */ function getUniqueDomains($list) { // iterate over list, split addresses and add domain part to another array $domains = array(); foreach ($list as $l) { $arr = explode("@", $l); $domains[] = trim($arr[1]); } // remove duplicates and return return array_unique($domains); } What does $domains[] = trim($arr[1]); mean? Specifically the $arr[1]. What does [1] mean in this context? How come variable $arr becomes an array variable?

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• Using a Loop to add objects to a list(python)

  - by Will

  Hey guys so im trying to use a while loop to add objects to a list. Heres bascially what i want to do: (ill paste actually go after) class x: blah blah choice = raw_input(pick what you want to do) while(choice!=0): if(choice==1): Enter in info for the class: append object to list (A) if(choice==2): print out length of list(A) if(choice==0): break ((((other options)))) as im doing this i can get the object to get added to the list, but i am stuck as to how to add multiple objects to the list in the loop. Here is my actual code i have so far... print "Welcome to the Student Management Program" class Student: def init (self, name, age, gender, favclass): self.name = name self.age = age self.gender = gender self.fac = favclass choice = int(raw_input("Make a Choice: " )) while (choice !=0): if (guess==1): print("STUDENT") namer = raw_input("Enter Name: ") ager = raw_input("Enter Age: ") sexer = raw_input("Enter Sex: ") faver = raw_input("Enter Fav: ") elif(guess==2): print "TESTING LINE" elif(guess==3): print(len(a)) guess=int(raw_input("Make a Choice: ")) s = Student(namer, ager, sexer, faver) a =[]; a.append(s) raw_input("Press enter to exit") any help would be greatly appreciated!

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• Java, searching within a list of objects?

  - by Rick

  I'm a bit lost on the way to make this happen the fastest. I have a large list of objects that have basic variable attributes (with getters / setters) and I need to do a search in this list to find the objects within the list that match a given parameter I have found how to do a regular list search but I need to, for example search for the value of the result of doing a call getName() for each object in the list and get objects that have a result that matches my input. Something like below where the third argument is the result of the method call and the second is what I am trying to find. int index = Collections.binarySearch(myList, "value", getName()); Any advice is appreciated

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• C++ Generic List Assignment

  - by S73417H

  I've clearly been stuck in Java land for too long... Is it possible to do the C++ equivalent of the following Java code: // Method List<Bar> getBars() { return new LinkedList<Bar>(); } // Assignment statement. List<Foo> stuff = getBars(); Where Foo is a sub-class of Bar. So in C++.... std::list<Bar> & getBars() { std::list<Bar> bars; return bars; } std::list<Foo> stuff = getBars(); Hope that makes sense....

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• Flex list-controls - maintain remote data

  - by artemb

  Hello. I have a TileList which represents some remote data. I also have a form which allows me to change the data. And the data may be changed by someone else too. What is the best way to maintain data in the list in an up-to-date state? The simplest option I see is the following: Select an item in the list Edit it in the form Save it. The form submits the data to the server When the server reports success the list re-fetches it's data The very bad thing about this workflow is that the list loses selection (a tree would also loose the nodes' expanded/collapsed state). I would really love to find out another option which would enable the list to maintain it's selection state. Any guesses on how it may be done?

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• Linked List Inserting in sorted format

  - by user2738718

  package practise; public class Node { public int data; public Node next; public Node (int data, Node next) { this.data = data; this.next = next; } public int size (Node list) { int count = 0; while(list != null){ list = list.next; count++; } return count; } public static Node insert(Node head, int value) { Node T; if (head == null || head.data <= value) { T = new Node(value,head); return T; } else { head.next = insert(head.next, value); return head; } } } This work fine for all data values less than the first or the head. anything greater than than doesn't get added to the list.please explain in simple terms thanks.

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• python: creating a list inside a dictionary

  - by user1871081

  I just started using python and I'm trying to create a program that will read a file that looks like this: AAA x 111 AAB x 111 AAA x 112 AAC x 123 ... the file is 50 lines long and I'm trying to make the letters into keys in a dictionary and the numbers lists that correspond with the keys. I want the output to look like this: {AAA: ['111', '112'], AAB: ['111'], AAC: [123], ...} This is what I've tried file = open("filename.txt", "r") readline = file.readline().rstrip() while readline!= "": list = [] list = readline.split(" ") j = list.index("x") k = list[0:j] v = list[p + 1:] d = {} if k in d == False d[k] = [] d[k].append(v) else d[k].append(v) readline = file.readline().rstrip() I keep getting syntax errors on my if statement and I can't figure out what I've done wrong.

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• improve my code for collapsing a list of data.frames

  - by romunov

  Dear StackOverFlowers (flowers in short), I have a list of data.frames (walk.sample) that I would like to collapse into a single (giant) data.frame. While collapsing, I would like to mark (adding another column) which rows have came from which element of the list. This is what I've got so far. This is the data.frame that needs to be collapsed/stacked. > walk.sample [[1]] walker x y 1073 3 228.8756 -726.9198 1086 3 226.7393 -722.5561 1081 3 219.8005 -728.3990 1089 3 225.2239 -727.7422 1032 3 233.1753 -731.5526 [[2]] walker x y 1008 3 205.9104 -775.7488 1022 3 208.3638 -723.8616 1072 3 233.8807 -718.0974 1064 3 217.0028 -689.7917 1026 3 234.1824 -723.7423 [[3]] [1] 3 [[4]] walker x y 546 2 629.9041 831.0852 524 2 627.8698 873.3774 578 2 572.3312 838.7587 513 2 633.0598 871.7559 538 2 636.3088 836.6325 1079 3 206.3683 -729.6257 1095 3 239.9884 -748.2637 1005 3 197.2960 -780.4704 1045 3 245.1900 -694.3566 1026 3 234.1824 -723.7423 I have written a function to add a column that denote from which element the rows came followed by appending it to an existing data.frame. collapseToDataFrame <- function(x) { # collapse list to a dataframe with a twist walk.df <- data.frame() for (i in 1:length(x)) { n.rows <- nrow(x[[i]]) if (length(x[[i]])>1) { temp.df <- cbind(x[[i]], rep(i, n.rows)) names(temp.df) <- c("walker", "x", "y", "session") walk.df <- rbind(walk.df, temp.df) } else { cat("Empty list", "\n") } } return(walk.df) } > collapseToDataFrame(walk.sample) Empty list Empty list walker x y session 3 1 -604.5055 -123.18759 1 60 1 -562.0078 -61.24912 1 84 1 -594.4661 -57.20730 1 9 1 -604.2893 -110.09168 1 43 1 -632.2491 -54.52548 1 1028 3 240.3905 -724.67284 1 1040 3 232.5545 -681.61225 1 1073 3 228.8756 -726.91980 1 1091 3 209.0373 -740.96173 1 1036 3 248.7123 -694.47380 1 I'm curious whether this can be done more elegantly, with perhaps do.call() or some other more generic function?

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• latex list environment inside the tabular environment: extra line at top preventing alignment

  - by Usagi

  Hello good people of stackoverflow. I have a LaTeX question that is bugging me. I have been trying to get a list environment to appear correctly inside the tabular environment. So far I have gotten everything to my liking except one thing: the top of the list does not align with other entries in the table, in fact it looks like it adds one line above the list... I would like to have these lists at the top. This is what I have, a custom list environment: \newenvironment{flushemize}{ \begin{list}{$\bullet$} {\setlength{\itemsep}{1pt} \setlength{\parskip}{0pt} \setlength{\parsep}{0pt} \setlength{\partopsep}{0pt} \setlength{\topsep}{0pt} \setlength{\leftmargin}{12pt}}}{\end{list}} Renamed ragged right: \newcommand{\rr}{\raggedright} and here is my table: \begin{table}[H]\caption{Tank comparisons}\label{tab:tanks} \centering \rowcolors{2}{white}{tableShade} \begin{tabular}{p{1in}p{1.5in}p{1.5in}rr} \toprule {\bf Material} & {\bf Pros} & {\bf Cons} & {\bf Size} & {\bf Cost} \\ \midrule \rr Reinforced concrete &\rr \begin{flushemize}\item Strong \item Secure \end{flushemize}&\rr \begin{flushemize}\item Prone to leaks \item Relatively expensive to install \item Heavy \end{flushemize} & 100,000 gal & \$299,400 \\ \rr Steel & \begin{flushemize}\item Strong \item Secure \end{flushemize} & \begin{flushemize}\item Relatively expensive to install \item Heavy \item Require painting to prevent rusting \end{flushemize} & 100,000 gal & \$130,100 \\ \rr Polypropylene & \begin{flushemize}\item Easy to install \item Mobile \item Inexpensive \item Prefabricated \end{flushemize} & \begin{flushemize}\item Relatively insecure \item Max size available 10,000 gal \end{flushemize} & 10,000 gal & \$5,000 \\ \rr Wood & \begin{flushemize}\item Easy to install \item Mobile \item Cheap to install \end{flushemize} & \begin{flushemize}\item Prone to rot \item Must remain full once constructed \end{flushemize} & 100,000 gal & \$86,300\\ \bottomrule \end{tabular} \end{table} Thank you for any advice :)

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• Subset a data.frame by list and apply function on each part, by rows

  - by aL3xa

  This may seem as a typical plyr problem, but I have something different in mind. Here's the function that I want to optimize (skip the for loop). # dummy data set.seed(1985) lst <- list(a=1:10, b=11:15, c=16:20) m <- matrix(round(runif(200, 1, 7)), 10) m <- as.data.frame(m) dfsub <- function(dt, lst, fun) { # check whether dt is `data.frame` stopifnot (is.data.frame(dt)) # check if vectors in lst are "whole" / integer # vector elements should be column indexes is.wholenumber <- function(x, tol = .Machine$double.eps^0.5) abs(x - round(x)) < tol # fall if any non-integers in list idx <- rapply(lst, is.wholenumber) stopifnot(idx) # check for list length stopifnot(ncol(dt) == length(idx)) # subset the data subs <- list() for (i in 1:length(lst)) { # apply function on each part, by row subs[[i]] <- apply(dt[ , lst[[i]]], 1, fun) } # preserve names names(subs) <- names(lst) # convert to data.frame subs <- as.data.frame(subs) # guess what =) return(subs) } And now a short demonstration... actually, I'm about to explain what I primarily intended to do. I wanted to subset a data.frame by vectors gathered in list object. Since this is a part of code from a function that accompanies data manipulation in psychological research, you can consider m as a results from personality questionnaire (10 subjects, 20 vars). Vectors in list hold column indexes that define questionnaire subscales (e.g. personality traits). Each subscale is defined by several items (columns in data.frame). If we presuppose that the score on each subscale is nothing more than sum (or some other function) of row values (results on that part of questionnaire for each subject), you could run: > dfsub(m, lst, sum) a b c 1 46 20 24 2 41 24 21 3 41 13 12 4 37 14 18 5 57 18 25 6 27 18 18 7 28 17 20 8 31 18 23 9 38 14 15 10 41 14 22 I took a glance at this function and I must admit that this little loop isn't spoiling the code at all... BUT, if there's an easier/efficient way of doing this, please, let me know!

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