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  • Converting a list of an Structure into datatable

    - by strakastroukas
    I saw a lot of examples regarding conversion of a list to data-table. I would like to convert a list of structure into a data-table. How can i do that? My structure is like ... Structure MainStruct Dim Ans1 As String Dim Ans2 As String Dim Ans3 As String Dim Skipped As Boolean End Structure and... Dim St As New MainStruct Dim Build As New List(Of MainStruct) I would like to convert the Build to a datatable

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  • Efficient method to calculate the rank vector of a list in Python

    - by Tamás
    I'm looking for an efficient way to calculate the rank vector of a list in Python, similar to R's rank function. In a simple list with no ties between the elements, element i of the rank vector of a list l should be x if and only if l[i] is the x-th element in the sorted list. This is simple so far, the following code snippet does the trick: def rank_simple(vector): return [rank for rank in sorted(range(n), key=vector.__getitem__)] Things get complicated, however, if the original list has ties (i.e. multiple elements with the same value). In that case, all the elements having the same value should have the same rank, which is the average of their ranks obtained using the naive method above. So, for instance, if I have [1, 2, 3, 3, 3, 4, 5], the naive ranking gives me [0, 1, 2, 3, 4, 5, 6], but what I would like to have is [0, 1, 3, 3, 3, 5, 6]. Which one would be the most efficient way to do this in Python? Footnote: I don't know if NumPy already has a method to achieve this or not; if it does, please let me know, but I would be interested in a pure Python solution anyway as I'm developing a tool which should work without NumPy as well.

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  • A linked list with multiple heads in Java

    - by Emile
    Hi, I have a list in which I'd like to keep several head pointers. I've tried to create multiple ListIterators on the same list but this forbid me to add new elements in my list... (see Concurrent Modification exception). I could create my own class but I'd rather use a built-in implementation ;) To be more specific, here is an inefficient implementation of two basic operations and the one wich doesn't work : class MyList <E { private int[] _heads; private List<E _l; public MyList ( int nbHeads ) { _heads = new int[nbHeads]; _l = new LinkedList<E(); } public void add ( E e ) { _l.add(e); } public E next ( int head ) { return _l.get(_heads[head++]); // ugly } } class MyList <E { private Vector<ListIterator<E _iters; private List<E _l; public MyList ( int nbHeads ) { _iters = new Vector<ListIterator<E(nbHeads); _l = new LinkedList<E(); for( ListIterator<E iter : _iters ) iter = _l.listIterator(); } public void add ( E e ) { _l.add(e); } public E next ( int head ) { // ConcurrentModificationException because of the add() return _iters.get(head).next(); } } Emile

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  • Do I need to rebuild application after adding a new iPhone in ad hoc distribution?

    - by RaYell
    I'm using Ad Hoc distribution to send preview/beta version of my iPhone apps to customers for approval. I'm always sending a zipped application and mobileprovision files. Sometimes however I encountered situation when nothing in the application changed but we needed to add a new device for testing. I've added a device in Provisioning portal and assigned it to a provisioning profile my application uses for ad hoc distribution. I've downloaded new mobileprovision file and imported it to xcode now the question is do I need to rebuild the application to enable the app on a new device or would it be enough to just send old build with updated mobileprovision file?

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  • Python: Remove items from a list while iterating in Python

    - by xApple
    My problem is simple: I have a long list of elements that I want to iterate through and check every element against a condition. Depending on the outcome of the condition I would like to delete the current element of the list, and continue iterating over it as usual. I have read a few other threads on this matter. Two solutions seam to be proposed. Either make a dictionary out of the list (which implies making a copy of all the data that is already filling all the RAM in my case). Either walk the list in reverse (which breaks the concept of the alogrithm I want to implement). Is there any better or more elegant way than this to do it ? def walk_list(list_of_g): g_index = 0 while g_index < len(list_of_g): g_current = list_of_g[g_index] if subtle_condition(g_current): list_of_g.pop(g_index) else: g_index = g_index + 1

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  • Generic List created at runtime

    - by Perry
    Hi... i neeed something like this in C#.. have list in class but decide what will be in list during runtime class A { List data; Type typeOfDataInList; } public void FillData(DataTyp[] data) where DataTyp : struct { A a = new A(); A.vListuBudouDataTypu = typeof(DataTyp); A.data = new List(); A.AddRange(data); } Is this possible to do something like this ?

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  • Reverse a singly linked list

    - by Madhan
    I would be wondered if there exists some logic to reverse the linked list using only two pointers. The following is used to reverse the single linked list using three pointers namely p, q, r: struct node { int data; struct node *link; }; void reverse() { struct node *p = first, *q = NULL, *r; while (p != NULL) { r = q; q = p; p = p->link; q->link = r; } q = first; } Is there any other alternate to reverse the linked list? what would be the best logic to reverse a singly linked list, in terms of time complexity?

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  • Get the first and second objects from a list using LINQ

    - by Vahid
    I have a list of Person objects. How can I get the first and second Person objects that meet a certain criteria from List<Person> People using LINQ? Let's say here is the list I've got. How can I get the first and second persons that are over 18 that is James and Jodie. public class Person { public string Name; public int age; } var People = new List<Person> { new Person {Name = "Jack", Age = 15}, new Person {Name = "James" , Age = 19}, new Person {Name = "John" , Age = 14}, new Person {Name = "Jodie" , Age = 21}, new Person {Name = "Jessie" , Age = 19} }

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  • Multiple objects in list, C#

    - by anon
    I'm looking for something similar to List<T>, that would allow me to have multiple T. For example: List<TabItem, DataGrid, int, string, ...> = new List<TabItem, DataGrid, int, string, ...>().

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  • Algorithm - combine multiple lists, resulting in unique list and retaining order

    - by hitch
    I want to combine multiple lists of items into a single list, retaining the overall order requirements. i.e.: 1: A C E 2: D E 3: B A D result: B A C D E above, starting with list 1, we have ACE, we then know that D must come before E, and from list 3, we know that B must come before A, and D must come after B and A. If there are conflicting orderings, the first ordering should be used. i.e. 1: A C E 2: B D E 3: F D result: A C B D E F 3 conflicts with 2, therefore requirements for 2 will be used. If ordering requirements mean an item must come before or after another, it doesn't matter if it comes immediately before or after, or at the start or end of the list, as long as overall ordering is maintained. This is being developed using VB.Net, so a LINQy solution (or any .Net solution) would be nice - otherwise pointers for an approach would be good.

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  • Deleting a node from linked list in C

    - by LuckySlevin
    My problem is deleting a node from linked list. I have two structs : typedef struct inner_list { int count; char word[100]; inner_list*next; } inner_list; typedef struct outer_list { char word [100]; inner_list * head; int count; outer_list * next; } outer_list; My problem is in deleting a node from outer_list linked list. For example when user entered aaa to delete, delete function should find the node with outer_list->word = aaa and delete this node and reconnect the list again. I tried the below code to do this. but After finding and deleting I'm losing the list. I don't know what's wrong. Please notice that outer_list have also a linked list of inner_list inside. void delnode(outer_list *head,char num[100]) { outer_list *temp, *m; temp=head; while(temp!=NULL) { if(strcmp(temp->word==num)==0) { if(temp==head) { head=temp->next; free(temp); return; } else { m->next=temp->next; free(temp); return; } }else { m=temp; temp= temp->next; } } printf(" ELEMENT %s NOT FOUND ", num); } What are your ideas about this?

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  • Sum up values in an array list in java

    - by user1449997
    Hello to all tired and frustrated programmers, There is a problem I can't solve: In my array list there are Object arrays with the length of three and always on the last cell of the array there is a double. I need a loop which goes through the array list and sums up all doubles. (I need also some other thing from the arrays so a general idea how to loop through an array list would be perfect :-)) Any ideas? Thanks for your help, Greetings Demian

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  • Pairs from single list

    - by Apalala
    Often enough, I've found the need to process a list by pairs. I was wondering which would be the pythonic and efficient way to do it, and found this on Google: pairs = zip(t[::2], t[1::2]) I thought that was pythonic enough, but after a recent discussion involving idioms versus efficiency, I decided to do some tests: import time from itertools import islice, izip def pairs_1(t): return zip(t[::2], t[1::2]) def pairs_2(t): return izip(t[::2], t[1::2]) def pairs_3(t): return izip(islice(t,None,None,2), islice(t,1,None,2)) A = range(10000) B = xrange(len(A)) def pairs_4(t): # ignore value of t! t = B return izip(islice(t,None,None,2), islice(t,1,None,2)) for f in pairs_1, pairs_2, pairs_3, pairs_4: # time the pairing s = time.time() for i in range(1000): p = f(A) t1 = time.time() - s # time using the pairs s = time.time() for i in range(1000): p = f(A) for a, b in p: pass t2 = time.time() - s print t1, t2, t2-t1 These were the results on my computer: 1.48668909073 2.63187503815 1.14518594742 0.105381965637 1.35109519958 1.24571323395 0.00257992744446 1.46182489395 1.45924496651 0.00251388549805 1.70076990128 1.69825601578 If I'm interpreting them correctly, that should mean that the implementation of lists, list indexing, and list slicing in Python is very efficient. It's a result both comforting and unexpected. Is there another, "better" way of traversing a list in pairs? Note that if the list has an odd number of elements then the last one will not be in any of the pairs. Which would be the right way to ensure that all elements are included? I added these two suggestions from the answers to the tests: def pairwise(t): it = iter(t) return izip(it, it) def chunkwise(t, size=2): it = iter(t) return izip(*[it]*size) These are the results: 0.00159502029419 1.25745987892 1.25586485863 0.00222492218018 1.23795199394 1.23572707176 Results so far Most pythonic and very efficient: pairs = izip(t[::2], t[1::2]) Most efficient and very pythonic: pairs = izip(*[iter(t)]*2) It took me a moment to grok that the first answer uses two iterators while the second uses a single one. To deal with sequences with an odd number of elements, the suggestion has been to augment the original sequence adding one element (None) that gets paired with the previous last element, something that can be achieved with itertools.izip_longest().

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  • Is the m2eclipse mailing list being managed?

    - by sdoca
    Hi, I tried to subscribe to the m2eclipse mailing list by sending an email to [email protected] but it never sent me back any confirmation email. I sent an email to [email protected] asking if the mailing list was still running a couple of weeks ago and haven't received any response back to that either. Then I found another way to subscribe via http://xircles.codehaus.org/lists/info/search. So I did and am now receiving emails sent to the list. However, I can't send the list emails. They get bounced back with a failure notice: Hi. This is the qmail-send program at mail.codehaus.org. I'm afraid I wasn't able to deliver your message to the following addresses. This is a permanent error; I've given up. Sorry it didn't work out. <[email protected]>: Sorry, only subscribers may post. If you are a subscriber, please forward this message to [email protected] to get your new address included. (#5.7.2) I've forwarded this message as instructed twice now (once yesterday and once this morning) without result so far. Does anyone know if this mailing list is being managed? It sucks to be only able to read messages and not to send them. Thanks in advance!

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  • Wordpress/CSS: List Style Problem.

    - by Ben
    Hi all! I am reasonably new to Wordpress custom themes but I do have some knowlege of CSS, although I am terrible at customising other peoples code which is why I have the problem I am about to explain. I have started a website that my girlfriend and I are doing together. It uses the Simplo Wordpress theme which is pretty cool but when you add a post that has a list in it. It doesn't show the bullets on each of the items. I have tried adding 'list-style:circle;' to a the.postItem class and a couple of other divs and classes that I thought were relevant but it doesn't work. You can take a look for yourself, it may be easier to understand my problem that way. If you go to: http://brokeandstarving.com/234/bennos-shepherds-pie/ there is a list of ingredients and also the steps under the method is a list as well. My question is, where do I edit the CSS to make the bullets appear next to the list items? I hope that makes sense. Thanks in advance, - Ben

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  • Counting entries in a list of dictionaries: for loop vs. list comprehension with map(itemgetter)

    - by Dennis Williamson
    In a Python program I'm writing I've compared using a for loop and increment variables versus list comprehension with map(itemgetter) and len() when counting entries in dictionaries which are in a list. It takes the same time using a each method. Am I doing something wrong or is there a better approach? Here is a greatly simplified and shortened data structure: list = [ {'key1': True, 'dontcare': False, 'ignoreme': False, 'key2': True, 'filenotfound': 'biscuits and gravy'}, {'key1': False, 'dontcare': False, 'ignoreme': False, 'key2': True, 'filenotfound': 'peaches and cream'}, {'key1': True, 'dontcare': False, 'ignoreme': False, 'key2': False, 'filenotfound': 'Abbott and Costello'}, {'key1': False, 'dontcare': False, 'ignoreme': True, 'key2': False, 'filenotfound': 'over and under'}, {'key1': True, 'dontcare': True, 'ignoreme': False, 'key2': True, 'filenotfound': 'Scotch and... well... neat, thanks'} ] Here is the for loop version: #!/usr/bin/env python # Python 2.6 # count the entries where key1 is True # keep a separate count for the subset that also have key2 True key1 = key2 = 0 for dictionary in list: if dictionary["key1"]: key1 += 1 if dictionary["key2"]: key2 += 1 print "Counts: key1: " + str(key1) + ", subset key2: " + str(key2) Output for the data above: Counts: key1: 3, subset key2: 2 Here is the other, perhaps more Pythonic, version: #!/usr/bin/env python # Python 2.6 # count the entries where key1 is True # keep a separate count for the subset that also have key2 True from operator import itemgetter KEY1 = 0 KEY2 = 1 getentries = itemgetter("key1", "key2") entries = map(getentries, list) key1 = len([x for x in entries if x[KEY1]]) key2 = len([x for x in entries if x[KEY1] and x[KEY2]]) print "Counts: key1: " + str(key1) + ", subset key2: " + str(key2) Output for the data above (same as before): Counts: key1: 3, subset key2: 2 I'm a tiny bit surprised these take the same amount of time. I wonder if there's something faster. I'm sure I'm overlooking something simple. One alternative I've considered is loading the data into a database and doing SQL queries, but the data doesn't need to persist and I'd have to profile the overhead of the data transfer, etc., and a database may not always be available. I have no control over the original form of the data. The code above is not going for style points.

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  • Check if something is a list

    - by 8EM
    What is the easiest way to check if something is a list? A method doSomething has the parameters a and b. In the method, it will loop through the list a and do something. I'd like a way to make sure a is a list, before looping through - thus avoiding an error or the unfortunate circumstance of passing in a string then getting back each letter. This question must have been asked before - however my googles failed me. Cheers.

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  • Iterate over a dict or list in Python

    - by Chris Dutrow
    Just wrote some nasty code that iterates over a dict or a list in Python. I have a feeling this was not the best way to go about it. The problem is that in order to iterate over a dict, this is the convention: for key in dict_object: dict_object[key] = 1 But modifying the object properties by key does not work if the same thing is done on a list: # Throws an error because the value of key is the property value, not # the list index: for key in list_object: list_object[key] = 1 The way I solved this problem was to write this nasty code: if isinstance(obj, dict): for key in obj: do_loop_contents(obj, key) elif isinstance(obj, list): for i in xrange(0, len(obj)): do_loop_contents(obj, i) def do_loop_contents(obj, key): obj[key] = 1 Is there a better way to do this? Thanks!

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  • Adding an existing site column to a custom list

    - by Graeme
    I think I'm going mad - this seemed like an easy thing to do but I can't find any info on it at all. I have created a Custom List and added 4 columns to it. Created By and Modified By are already in the list but hidden from the view. I want to add a Date Modified column (which is a built in field) to this Custom List. How do I do this programmatically?

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  • Creating a circually linked list in C#?

    - by Dexter
    What would be the best way to create a circually linked list in C#. Should I derive it from the LinkedList< T collection? I'm planning on creating a simple address book using this Linked List to store my contacts (it's gonna be a suck-y address book, but I don't care cause I'll be the only one to use it). I mainly just want to create the crucially linked list so that I can use it again in other projects. If you don't think the Linked List is the right way to go let me know which way would be better.

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  • c++ : list(vector) definition with array

    - by Meloun
    I have Class Email, there is parameter "bcc" in her construktor. Its actually list of emails for copies. There is no fixed number of these emails and later i have to have possibility to extend this list. //construktor prototype Email::Email(vector<string> bcc) So i want to use type vector or list for that and function push_back(). How can i make a new instance with bcc emails? I need actually declaration with definition for my list. I've found this definition with iterator for integer type: int myints[] = {16,2,77,29}; Email myEmail(vector<int> (myints, myints + sizeof(myints) / sizeof(int) )); , but its not very user friend and i need it with strings. Is there something like this? Email myEmail(vector<string> ("first","second","third"));

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  • Random List of millions of elements in Python Efficiently

    - by eWizardII
    Hello, I have read this answer potentially as the best way to randomize a list of strings in Python. I'm just wondering then if that's the most efficient way to do it because I have a list of about 30 million elements via the following code: import json from sets import Set from random import shuffle a = [] for i in range(0,193): json_data = open("C:/Twitter/user/user_" + str(i) + ".json") data = json.load(json_data) for j in range(0,len(data)): a.append(data[j]['su']) new = list(Set(a)) print "Cleaned length is: " + str(len(new)) ## Take Cleaned List and Randomize it for Analysis shuffle(new) If there is a more efficient way to do it, I'd greatly appreciate any advice on how to do it. Thanks,

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