Search Results

Search found 9501 results on 381 pages for 'word 2007'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 110/381 | < Previous Page | 106 107 108 109 110 111 112 113 114 115 116 117  | Next Page >

  • find whether the string starts and ends with the same word

    - by Ajax
    I am trying to check whether the string starts and ends with the same word. egearth. s=raw_input(); m=re.search(r"^(earth).*(earth)$",s) if m is not None: print "found" my problem is when the string consists only of one word eg: earth At present I have hard coded this case by if m is not None or s=='earth': print "found" Is there any other way to do this? EDIT: words in a string are separated by spaces. looking for a regex solution some examples: "earth is earth" ,"earth", -- valid "earthearth", "eartheeearth", "earth earth mars" -- invalid

    Read the article

  • How can Excel 2007 / 2010 consume a REST web service?

    - by jallen
    What options exist to consume a REST web service from within Excel 2007 / 2010? I can use XML Maps to consume a basic XML list, but that doesn't let me build a dynamic URL (so I could include parameters). For example, I can add an XML Map to Excel for http://machine/service/level/5 and display the values in the workbook just fine - no problem there. The real question is, how can I dynamically change the /5 part of the URL to come from another cell in excel? That way I can have a couple of cells that have the options (what ID, what name, etc.) and whenever those values change (ideally) a new dynamic URL would be constructed and the XML map would be refreshed. Is such a thing possible? Does anyone else have a better way to take some parameters, call a web service (REST or SOAP, I'm not picky) and shove the results back into excel for further manipulation?

    Read the article

  • MS Word to Stylesheet

    - by Chris Johnson
    Is there an easy way to automatically convert a bunch of MS Word documents to xslt stylesheets that can be displayed in the browser? What I have is a large collection of forms in Word format that have to be displayed in the browser, or sent to the user, with known fields populated from a data source, edited by a user and, finally, printed (including the original headers and footers). The data entered by the user will not need to be saved. I'm not sure if converting the documents to stylesheets is even feasible. Maybe someone has a better idea of how to achieve this? Installing Office on the server is not an option in my case.

    Read the article

  • Next/Last word with Netbeans in Mac OSX

    - by Phuong Nguy?n
    I've just switched to Mac OSX from Ubuntu. The development on Mac OSX using Netbeans is really a joy (of hell, though) I don't know how to go to next word, last word using keyboard. In windows or ubuntu, this should be Ctrl+Left/Right, but definely not in Mac. In Mac, such key combinations result in go to begin/end of line. Futhermore, the use of Home and End is for something else, not go to begin/end of line (What a cool hell). Is there another key combination? I prefer default key settings, please. I'm sick of going around with custom script.

    Read the article

  • Repurpose builtin Word commands - access original command within repurposed function

    - by Aurril
    It is possible to repurpose a Word builtin command in Word 2007 using the customUI.xml file. Example: <customUI xmlns="http://schemas.microsoft.com/office/2009/07/customui"> <commands> <command idMso="Save" onAction="MySave"/> </commands> </customUI> I then have to define a callback function in VBA which is used instead of the builtin function: 'Callback for Save onAction Sub MySave(control As IRibbonControl, ByRef cancelDefault) someFancyPreparationFunction oldSaveFunction ' Where, how? someOtherFancyAfterWorkFunction End Sub And here is my problem, if I want to reuse internal functionality in my repurposed Function, I don't know how to access it. My example is trivial and the save could be easily rebuilt by a call to ThisDocument.save but as I said, it is just an example to show the problem. A call to CommandBars.ExecuteMso('Save') would call my repurposed function and not the original one. Any ideas on how to access the internal functionality after repurposing would be very appreciated!

    Read the article

  • Word on HTML document point.

    - by leolobato
    Hey guys, How can I figure out which word is at the point where the user tapped on a UIWebView? I am able to detect the CGPoint for the tap (subclassing UIWindow like this), and I can actually get the DOM element on that point using javascript. But I know very little of javascript and DOM to figure out how can I actually get which word the user tapped on. Is that possible? Here's what I have right now: int scrollPosition = [[webView stringByEvaluatingJavaScriptFromString:@"window.pageYOffset"] intValue]; NSString *js = [NSString stringWithFormat:@"document.elementFromPoint(%f, %f).tagName", point.x, point.y+scrollPosition]; NSString *value = [webView stringByEvaluatingJavaScriptFromString:js]; NSLog(@"element: %@", value);

    Read the article

  • Sort List by occurrence of a word by LINQ C#

    - by Thomas
    i have stored data in list like List<SearchResult> list = new List<SearchResult>(); SearchResult sr = new SearchResult(); sr.Description = "sample description"; list.Add(sr); suppose my data is stored in description field like "JCB Excavator - ECU P/N: 728/35700" "Geo Prism 1995 - ABS #16213899" "Geo Prism 1995 - ABS #16213899" "Geo Prism 1995 - ABS #16213899" "Wie man BBA reman erreicht" "this test JCB" "Ersatz Airbags, Gurtstrammer und Auto Körper Teile" now i want to query the list with my search term like geo jcb if you look then the word geo has stored many times in the description field. so i want to sort my list in such way that the word in search term found maximum that data will come first. please help me to do so. thanks

    Read the article

  • Finding position of each word in a sub-array of a multidimensional array

    - by Shreyas Satish
    I have an array: tokens = [["hello","world"],["hello","ruby"]] all_tokens = tokens.flatten.uniq # all_tokens=["hello","world","ruby"] Now I need to create two arrays corresponding to all_tokens, where the first array will contain the position of each word in sub-array of tokens. I.E Output: [[0,0],[1],[1]] # (w.r.t all_tokens) To make it clear it reads, The index of "hello" is 0 and 0 in the 2 sub-arrays of tokens. And second array contains index of each word w.r.t tokens.I.E Output: [[0,1],[0],[1]] To make it clear it reads,the index of hello 0,1. I.E "hello" is in index 0 and 1 of tokens array. Cheers!

    Read the article

  • Reading Microsoft word document in iphone

    - by Arnieterm
    Hi All Using CGPDFDocument class we can read pdf document, get number of pages, get a page by number. Is there any other library that helps us reading word documents like microsoft word doc where we can get number of pages and get a page by its number etc? Do not want to use safari to view the document. I have seen some apps on app store that claims to read MS Office docs like docx, xlsx etc. Does there exists any library [even static] that we can make use of? Any idea? Thanks Arnieterm

    Read the article

  • Regular Expression for accurate word-count using JavaScript

    - by Haidon
    I'm trying to put together a regular expression for a JavaScript command that accurately counts the number of words in a textarea. One solution I had found is as follows: document.querySelector("#wordcount").innerHTML = document.querySelector("#editor").value.split(/\b\w+\b/).length -1; But this doesn't count any non-Latin characters (eg: Cyrillic, Hangul, etc); it skips over them completely. Another one I put together: document.querySelector("#wordcount").innerHTML = document.querySelector("#editor").value.split(/\s+/g).length -1; But this doesn't count accurately unless the document ends in a space character. If a space character is appended to the value being counted it counts 1 word even with an empty document. Furthermore, if the document begins with a space character an extraneous word is counted. Is there a regular expression I can put into this command that counts the words accurately, regardless of input method?

    Read the article

  • Extract Bullets and Tables information in Word doc from c#

    - by Siva
    Hi All, I need to create an word document based on the template in c#. I have tags for only the paragraphs. Is there any way to replace the bullets and tables that are already available in the template based on the user input. I was able to replace the paragraph with input text using the Replace command in the Word InterOp. Need help to do the rest of the items. Replace the bullets based on the user input Fill the tables with the input values Code for replacing the Paragraph based on the tag: FindAndReplace(wordApplication, "/date/", DateTime.Now.Date.ToString("MMM dd, yyyy")); FindAndReplace(){ wordApplication.Selection.Find.Execute(ref findText, ref matchCase, ref matchWholeWord, ref matchWildCards, ref matchSoundsLike, ref matchAllWordsForms, ref forward, ref wrap, ref format, ref replaceWithText, ref replace, ref matchKashida, ref matchDiacritics, ref matchAlefHamsa, ref matchControl); } Thanks in Advance. ASAP

    Read the article

  • VBA - Prevent Excel 2007 from showing a defined names message box?

    - by John M
    I am working on a Excel 2007 workbook that will contain a macro to save the current sheet (a template) as a PDF file (no problem) a Excel 97-2003 file (problem) When saving the Excel file a messagebox appears asking about "Defined names of formulas in this workbook may display different values when they are recalculated...Do you want Excel to recalculate all formulas when this workbook is opened?". The user can then select Yes/No and then the file will save. How do I disable the messagebox from appearing? The default answer would be 'No'. My code for saving: Sub saveAs_97_2003_Workbook(tempFilePath As String, tempFileName As String) Dim Destwb As Workbook Dim SaveFormat As Long 'Remember the users setting SaveFormat = Application.DefaultSaveFormat 'Set it to the 97-2003 file format Application.DefaultSaveFormat = 56 ActiveSheet.Copy Set Destwb = ActiveWorkbook Destwb.CheckCompatibility = False With Destwb .SaveAs tempFilePath & tempFileName & ".xls", FileFormat:=56 .Close SaveChanges:=False End With 'Set DefaultSaveFormat back to the users setting Application.DefaultSaveFormat = SaveFormat End Sub

    Read the article

  • sIFR works, except for one word in my navigation bar

    - by Jasper
    Hi there, My website uses sIFR and it all seems to work great, except for one word in my navigation bar. The navigation links all work, but the word "Zoeken" (= Search) in front of the search form doesn't get changed to the desired font type. I have checked the header.php of my site, as well as the CSS, but any of the changes that I made, don't seem to work. Could anyone help me out? I am hoping it's just a "piece of cake", but my knowledge of sIFR is close to non-existent. Many thanks! Jasper

    Read the article

  • display alert when mouse hovers over word in text

    - by user1672790
    I have been struggling with this for a few days. I need somebody to steer me in the right direction. I have been searching on the web. I am not sure if I took the right approach. What I need is that each time a person hovers over a particular keyword, it should display an alert box. In this example the word is else. When I run the code it does not give any errors and does not display anything when mouse hovers on the word. function on_func2() { var searchString = 'else'; var elements = document.getElementById('paragraph2'); for (var i = 0; i < elements.length; i++) { if (elements[i].innerHTML.indexOf(searchString) !== -1) { alert('Match'); break; } } }

    Read the article

  • LINQ and ordering of the result set

    - by vik20000in
    After filtering and retrieving the records most of the time (if not always) we have to sort the record in certain order. The sort order is very important for displaying records or major calculations. In LINQ for sorting data the order keyword is used. With the help of the order keyword we can decide on the ordering of the result set that is retrieved after the query.  Below is an simple example of the order keyword in LINQ.     string[] words = { "cherry", "apple", "blueberry" };     var sortedWords =         from word in words         orderby word         select word; Here we are ordering the data retrieved based on the string ordering. If required the order can also be made on any of the property of the individual like the length of the string.     var sortedWords =         from word in words         orderby word.Length         select word; You can also make the order descending or ascending by adding the keyword after the parameter.     var sortedWords =         from word in words         orderby word descending         select word; But the best part of the order clause is that instead of just passing a field you can also pass the order clause an instance of any class that implements IComparer interface. The IComparer interface holds a method Compare that Has to be implemented. In that method we can write any logic whatsoever for the comparision. In the below example we are making a string comparison by ignoring the case. string[] words = { "aPPLE", "AbAcUs", "bRaNcH", "BlUeBeRrY", "cHeRry"}; var sortedWords = words.OrderBy(a => a, new CaseInsensitiveComparer());  public class CaseInsensitiveComparer : IComparer<string> {     public int Compare(string x, string y)     {         return string.Compare(x, y, StringComparison.OrdinalIgnoreCase);     } }  But while sorting the data many a times we want to provide more than one sort so that data is sorted based on more than one condition. This can be achieved by proving the next order followed by a comma.     var sortedWords =         from word in words         orderby word , word.length         select word; We can also use the reverse() method to reverse the full order of the result set.     var sortedWords =         from word in words         select word.Reverse();                                 Vikram

    Read the article

  • Python - pyparsing unicode characters

    - by mgj
    Hi..:) I tried using w = Word(printables), but it isn't working. How should I give the spec for this. 'w' is meant to process Hindi characters (UTF-8) The code specifies the grammar and parses accordingly. 671.assess :: ????? ::2 x=number + "." + src + "::" + w + "::" + number + "." + number If there is only english characters it is working so the code is correct for the ascii format but the code is not working for the unicode format. I mean that the code works when we have something of the form 671.assess :: ahsaas ::2 i.e. it parses words in the english format, but I am not sure how to parse and then print characters in the unicode format. I need this for English Hindi word alignment for purpose. The python code looks like this: # -*- coding: utf-8 -*- from pyparsing import Literal, Word, Optional, nums, alphas, ZeroOrMore, printables , Group , alphas8bit , # grammar src = Word(printables) trans = Word(printables) number = Word(nums) x=number + "." + src + "::" + trans + "::" + number + "." + number #parsing for eng-dict efiledata = open('b1aop_or_not_word.txt').read() eresults = x.parseString(efiledata) edict1 = {} edict2 = {} counter=0 xx=list() for result in eresults: trans=""#translation string ew=""#english word xx=result[0] ew=xx[2] trans=xx[4] edict1 = { ew:trans } edict2.update(edict1) print len(edict2) #no of entries in the english dictionary print "edict2 has been created" print "english dictionary" , edict2 #parsing for hin-dict hfiledata = open('b1aop_or_not_word.txt').read() hresults = x.scanString(hfiledata) hdict1 = {} hdict2 = {} counter=0 for result in hresults: trans=""#translation string hw=""#hin word xx=result[0] hw=xx[2] trans=xx[4] #print trans hdict1 = { trans:hw } hdict2.update(hdict1) print len(hdict2) #no of entries in the hindi dictionary print"hdict2 has been created" print "hindi dictionary" , hdict2 ''' ####################################################################################################################### def translate(d, ow, hinlist): if ow in d.keys():#ow=old word d=dict print ow , "exists in the dictionary keys" transes = d[ow] transes = transes.split() print "possible transes for" , ow , " = ", transes for word in transes: if word in hinlist: print "trans for" , ow , " = ", word return word return None else: print ow , "absent" return None f = open('bidir','w') #lines = ["'\ #5# 10 # and better performance in business in turn benefits consumers . # 0 0 0 0 0 0 0 0 0 0 \ #5# 11 # vHyaapaar mEmn bEhtr kaam upbhOkHtaaomn kE lIe laabhpHrdd hOtaa hAI . # 0 0 0 0 0 0 0 0 0 0 0 \ #'"] data=open('bi_full_2','rb').read() lines = data.split('!@#$%') loc=0 for line in lines: eng, hin = [subline.split(' # ') for subline in line.strip('\n').split('\n')] for transdict, source, dest in [(edict2, eng, hin), (hdict2, hin, eng)]: sourcethings = source[2].split() for word in source[1].split(): tl = dest[1].split() otherword = translate(transdict, word, tl) loc = source[1].split().index(word) if otherword is not None: otherword = otherword.strip() print word, ' <-> ', otherword, 'meaning=good' if otherword in dest[1].split(): print word, ' <-> ', otherword, 'trans=good' sourcethings[loc] = str( dest[1].split().index(otherword) + 1) source[2] = ' '.join(sourcethings) eng = ' # '.join(eng) hin = ' # '.join(hin) f.write(eng+'\n'+hin+'\n\n\n') f.close() ''' if an example input sentence for the source file is: 1# 5 # modern markets : confident consumers # 0 0 0 0 0 1# 6 # AddhUnIk baajaar : AshHvsHt upbhOkHtaa . # 0 0 0 0 0 0 !@#$% the ouptut would look like this :- 1# 5 # modern markets : confident consumers # 1 2 3 4 5 1# 6 # AddhUnIk baajaar : AshHvsHt upbhOkHtaa . # 1 2 3 4 5 0 !@#$% Output Explanation:- This achieves bidirectional alignment. It means the first word of english 'modern' maps to the first word of hindi 'AddhUnIk' and vice versa. Here even characters are take as words as they also are an integral part of bidirectional mapping. Thus if you observe the hindi WORD '.' has a null alignment and it maps to nothing with respect to the English sentence as it doesn't have a full stop. The 3rd line int the output basically represents a delimiter when we are working for a number of sentences for which your trying to achieve bidirectional mapping. What modification should i make for it to work if the I have the hindi sentences in Unicode(UTF-8) format.

    Read the article

  • Convert Java program to C

    - by imicrothinking
    I need a bit of guidance with writing a C program...a bit of quick background as to my level, I've programmed in Java previously, but this is my first time programming in C, and we've been tasked to translate a word count program from Java to C that consists of the following: Read a file from memory Count the words in the file For each occurrence of a unique word, keep a word counter variable Print out the top ten most frequent words and their corresponding occurrences Here's the source program in Java: package lab0; import java.io.File; import java.io.FileReader; import java.util.ArrayList; import java.util.Calendar; import java.util.Collections; public class WordCount { private ArrayList<WordCountNode> outputlist = null; public WordCount(){ this.outputlist = new ArrayList<WordCountNode>(); } /** * Read the file into memory. * * @param filename name of the file. * @return content of the file. * @throws Exception if the file is too large or other file related exception. */ public char[] readFile(String filename) throws Exception{ char [] result = null; File file = new File(filename); long size = file.length(); if (size > Integer.MAX_VALUE){ throw new Exception("File is too large"); } result = new char[(int)size]; FileReader reader = new FileReader(file); int len, offset = 0, size2read = (int)size; while(size2read > 0){ len = reader.read(result, offset, size2read); if(len == -1) break; size2read -= len; offset += len; } return result; } /** * Make article word by word. * * @param article the content of file to be counted. * @return string contains only letters and "'". */ private enum SPLIT_STATE {IN_WORD, NOT_IN_WORD}; /** * Go through article, find all the words and add to output list * with their count. * * @param article the content of the file to be counted. * @return words in the file and their counts. */ public ArrayList<WordCountNode> countWords(char[] article){ SPLIT_STATE state = SPLIT_STATE.NOT_IN_WORD; if(null == article) return null; char curr_ltr; int curr_start = 0; for(int i = 0; i < article.length; i++){ curr_ltr = Character.toUpperCase( article[i]); if(state == SPLIT_STATE.IN_WORD){ article[i] = curr_ltr; if ((curr_ltr < 'A' || curr_ltr > 'Z') && curr_ltr != '\'') { article[i] = ' '; //printf("\nthe word is %s\n\n",curr_start); if(i - curr_start < 0){ System.out.println("i = " + i + " curr_start = " + curr_start); } addWord(new String(article, curr_start, i-curr_start)); state = SPLIT_STATE.NOT_IN_WORD; } }else{ if (curr_ltr >= 'A' && curr_ltr <= 'Z') { curr_start = i; article[i] = curr_ltr; state = SPLIT_STATE.IN_WORD; } } } return outputlist; } /** * Add the word to output list. */ public void addWord(String word){ int pos = dobsearch(word); if(pos >= outputlist.size()){ outputlist.add(new WordCountNode(1L, word)); }else{ WordCountNode tmp = outputlist.get(pos); if(tmp.getWord().compareTo(word) == 0){ tmp.setCount(tmp.getCount() + 1); }else{ outputlist.add(pos, new WordCountNode(1L, word)); } } } /** * Search the output list and return the position to put word. * @param word is the word to be put into output list. * @return position in the output list to insert the word. */ public int dobsearch(String word){ int cmp, high = outputlist.size(), low = -1, next; // Binary search the array to find the key while (high - low > 1) { next = (high + low) / 2; // all in upper case cmp = word.compareTo((outputlist.get(next)).getWord()); if (cmp == 0) return next; else if (cmp < 0) high = next; else low = next; } return high; } public static void main(String args[]){ // handle input if (args.length == 0){ System.out.println("USAGE: WordCount <filename> [Top # of results to display]\n"); System.exit(1); } String filename = args[0]; int dispnum; try{ dispnum = Integer.parseInt(args[1]); }catch(Exception e){ dispnum = 10; } long start_time = Calendar.getInstance().getTimeInMillis(); WordCount wordcount = new WordCount(); System.out.println("Wordcount: Running..."); // read file char[] input = null; try { input = wordcount.readFile(filename); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); System.exit(1); } // count all word ArrayList<WordCountNode> result = wordcount.countWords(input); long end_time = Calendar.getInstance().getTimeInMillis(); System.out.println("wordcount: completed " + (end_time - start_time)/1000000 + "." + (end_time - start_time)%1000000 + "(s)"); System.out.println("wordsort: running ..."); start_time = Calendar.getInstance().getTimeInMillis(); Collections.sort(result); end_time = Calendar.getInstance().getTimeInMillis(); System.out.println("wordsort: completed " + (end_time - start_time)/1000000 + "." + (end_time - start_time)%1000000 + "(s)"); Collections.reverse(result); System.out.println("\nresults (TOP "+ dispnum +" from "+ result.size() +"):\n" ); // print out result String str ; for (int i = 0; i < result.size() && i < dispnum; i++){ if(result.get(i).getWord().length() > 15) str = result.get(i).getWord().substring(0, 14); else str = result.get(i).getWord(); System.out.println(str + " - " + result.get(i).getCount()); } } public class WordCountNode implements Comparable{ private String word; private long count; public WordCountNode(long count, String word){ this.count = count; this.word = word; } public String getWord() { return word; } public void setWord(String word) { this.word = word; } public long getCount() { return count; } public void setCount(long count) { this.count = count; } public int compareTo(Object arg0) { // TODO Auto-generated method stub WordCountNode obj = (WordCountNode)arg0; if( count - obj.getCount() < 0) return -1; else if( count - obj.getCount() == 0) return 0; else return 1; } } } Here's my attempt (so far) in C: #include <stdio.h> #include <stdlib.h> #include <stdbool.h> #include <string.h> // Read in a file FILE *readFile (char filename[]) { FILE *inputFile; inputFile = fopen (filename, "r"); if (inputFile == NULL) { printf ("File could not be opened.\n"); exit (EXIT_FAILURE); } return inputFile; } // Return number of words in an array int wordCount (FILE *filePointer, char filename[]) {//, char *words[]) { // count words int count = 0; char temp; while ((temp = getc(filePointer)) != EOF) { //printf ("%c", temp); if ((temp == ' ' || temp == '\n') && (temp != '\'')) count++; } count += 1; // counting method uses space AFTER last character in word - the last space // of the last character isn't counted - off by one error // close file fclose (filePointer); return count; } // Print out the frequencies of the 10 most frequent words in the console int main (int argc, char *argv[]) { /* Step 1: Read in file and check for errors */ FILE *filePointer; filePointer = readFile (argv[1]); /* Step 2: Do a word count to prep for array size */ int count = wordCount (filePointer, argv[1]); printf ("Number of words is: %i\n", count); /* Step 3: Create a 2D array to store words in the file */ // open file to reset marker to beginning of file filePointer = fopen (argv[1], "r"); // store words in character array (each element in array = consecutive word) char allWords[count][100]; // 100 is an arbitrary size - max length of word int i,j; char temp; for (i = 0; i < count; i++) { for (j = 0; j < 100; j++) { // labels are used with goto statements, not loops in C temp = getc(filePointer); if ((temp == ' ' || temp == '\n' || temp == EOF) && (temp != '\'') ) { allWords[i][j] = '\0'; break; } else { allWords[i][j] = temp; } printf ("%c", allWords[i][j]); } printf ("\n"); } // close file fclose (filePointer); /* Step 4: Use a simple selection sort algorithm to sort 2D char array */ // PStep 1: Compare two char arrays, and if // (a) c1 > c2, return 2 // (b) c1 == c2, return 1 // (c) c1 < c2, return 0 qsort(allWords, count, sizeof(char[][]), pstrcmp); /* int k = 0, l = 0, m = 0; char currentMax, comparedElement; int max; // the largest element in the current 2D array int elementToSort = 0; // elementToSort determines the element to swap with starting from the left // Outer a iterates through number of swaps needed for (k = 0; k < count - 1; k++) { // times of swaps max = k; // max element set to k // Inner b iterates through successive elements to fish out the largest element for (m = k + 1; m < count - k; m++) { currentMax = allWords[k][l]; comparedElement = allWords[m][l]; // Inner c iterates through successive chars to set the max vars to the largest for (l = 0; (currentMax != '\0' || comparedElement != '\0'); l++) { if (currentMax > comparedElement) break; else if (currentMax < comparedElement) { max = m; currentMax = allWords[m][l]; break; } else if (currentMax == comparedElement) continue; } } // After max (count and string) is determined, perform swap with temp variable char swapTemp[1][20]; int y = 0; do { swapTemp[0][y] = allWords[elementToSort][y]; allWords[elementToSort][y] = allWords[max][y]; allWords[max][y] = swapTemp[0][y]; } while (swapTemp[0][y++] != '\0'); elementToSort++; } */ int a, b; for (a = 0; a < count; a++) { for (b = 0; (temp = allWords[a][b]) != '\0'; b++) { printf ("%c", temp); } printf ("\n"); } // Copy rows to different array and print results /* char arrayCopy [count][20]; int ac, ad; char tempa; for (ac = 0; ac < count; ac++) { for (ad = 0; (tempa = allWords[ac][ad]) != '\0'; ad++) { arrayCopy[ac][ad] = tempa; printf("%c", arrayCopy[ac][ad]); } printf("\n"); } */ /* Step 5: Create two additional arrays: (a) One in which each element contains unique words from char array (b) One which holds the count for the corresponding word in the other array */ /* Step 6: Sort the count array in decreasing order, and print the corresponding array element as well as word count in the console */ return 0; } // Perform housekeeping tasks like freeing up memory and closing file I'm really stuck on the selection sort algorithm. I'm currently using 2D arrays to represent strings, and that worked out fine, but when it came to sorting, using three level nested loops didn't seem to work, I tried to use qsort instead, but I don't fully understand that function as well. Constructive feedback and criticism greatly welcome (...and needed)!

    Read the article

  • Python sort strings started with digits

    - by vlad
    I have the next list: a = ['1th Word', 'Another Word', '10th Word'] print a.sort() >>> ['10th Word', '1th Word', 'Another Word'] But I need: ['1th Word', '10th Word','Another Word'] Is there an easy way to do this? I tried: r = re.compile(r'(\d+)') def sort_by_number(s): m = r.match(s) return m.group(0) x.sort(key=sort_by_number) But some strings do not have numbers and this leads to an errors. Thanks.

    Read the article

  • Auto-Suggest via &lsquo;Trie&rsquo; (Pre-fix Tree)

    - by Strenium
    Auto-Suggest (Auto-Complete) “thing” has been around for a few years. Here’s my little snippet on the subject. For one of my projects, I had to deal with a non-trivial set of items to be pulled via auto-suggest used by multiple concurrent users. Simple, dumb iteration through a list in local cache or back-end access didn’t quite cut it. Enter a nifty little structure, perfectly suited for storing and matching verbal data: “Trie” (http://tinyurl.com/db56g) also known as a Pre-fix Tree: “Unlike a binary search tree, no node in the tree stores the key associated with that node; instead, its position in the tree defines the key with which it is associated. All the descendants of a node have a common prefix of the string associated with that node, and the root is associated with the empty string. Values are normally not associated with every node, only with leaves and some inner nodes that correspond to keys of interest.” This is a very scalable, performing structure. Though, as usual, something ‘fast’ comes at a cost of ‘size’; fortunately RAM is more plentiful today so I can live with that. I won’t bore you with the detailed algorithmic performance here - Google can do a better job of such. So, here’s C# implementation of all this. Let’s start with individual node: Trie Node /// <summary> /// Contains datum of a single trie node. /// </summary> public class AutoSuggestTrieNode {     public char Value { get; set; }       /// <summary>     /// Gets a value indicating whether this instance is leaf node.     /// </summary>     /// <value>     ///     <c>true</c> if this instance is leaf node; otherwise, a prefix node <c>false</c>.     /// </value>     public bool IsLeafNode { get; private set; }       public List<AutoSuggestTrieNode> DescendantNodes { get; private set; }         /// <summary>     /// Initializes a new instance of the <see cref="AutoSuggestTrieNode"/> class.     /// </summary>     /// <param name="value">The phonetic value.</param>     /// <param name="isLeafNode">if set to <c>true</c> [is leaf node].</param>     public AutoSuggestTrieNode(char value = ' ', bool isLeafNode = false)     {         Value = value;         IsLeafNode = isLeafNode;           DescendantNodes = new List<AutoSuggestTrieNode>();     }       /// <summary>     /// Gets the descendants of the pre-fix node, if any.     /// </summary>     /// <param name="descendantValue">The descendant value.</param>     /// <returns></returns>     public AutoSuggestTrieNode GetDescendant(char descendantValue)     {         return DescendantNodes.FirstOrDefault(descendant => descendant.Value == descendantValue);     } }   Quite self-explanatory, imho. A node is either a “Pre-fix” or a “Leaf” node. “Leaf” contains the full “word”, while the “Pre-fix” nodes act as indices used for matching the results.   Ok, now the Trie: Trie Structure /// <summary> /// Contains structure and functionality of an AutoSuggest Trie (Pre-fix Tree) /// </summary> public class AutoSuggestTrie {     private readonly AutoSuggestTrieNode _root = new AutoSuggestTrieNode();       /// <summary>     /// Adds the word to the trie by breaking it up to pre-fix nodes + leaf node.     /// </summary>     /// <param name="word">Phonetic value.</param>     public void AddWord(string word)     {         var currentNode = _root;         word = word.Trim().ToLower();           for (int i = 0; i < word.Length; i++)         {             var child = currentNode.GetDescendant(word[i]);               if (child == null) /* this character hasn't yet been indexed in the trie */             {                 var newNode = new AutoSuggestTrieNode(word[i], word.Count() - 1 == i);                   currentNode.DescendantNodes.Add(newNode);                 currentNode = newNode;             }             else                 currentNode = child; /* this character is already indexed, move down the trie */         }     }         /// <summary>     /// Gets the suggested matches.     /// </summary>     /// <param name="word">The phonetic search value.</param>     /// <returns></returns>     public List<string> GetSuggestedMatches(string word)     {         var currentNode = _root;         word = word.Trim().ToLower();           var indexedNodesValues = new StringBuilder();         var resultBag = new ConcurrentBag<string>();           for (int i = 0; i < word.Trim().Length; i++)  /* traverse the trie collecting closest indexed parent (parent can't be leaf, obviously) */         {             var child = currentNode.GetDescendant(word[i]);               if (child == null || word.Count() - 1 == i)                 break; /* done looking, the rest of the characters aren't indexed in the trie */               indexedNodesValues.Append(word[i]);             currentNode = child;         }           Action<AutoSuggestTrieNode, string> collectAllMatches = null;         collectAllMatches = (node, aggregatedValue) => /* traverse the trie collecting matching leafNodes (i.e. "full words") */             {                 if (node.IsLeafNode) /* full word */                     resultBag.Add(aggregatedValue); /* thread-safe write */                   Parallel.ForEach(node.DescendantNodes, descendandNode => /* asynchronous recursive traversal */                 {                     collectAllMatches(descendandNode, String.Format("{0}{1}", aggregatedValue, descendandNode.Value));                 });             };           collectAllMatches(currentNode, indexedNodesValues.ToString());           return resultBag.OrderBy(o => o).ToList();     }         /// <summary>     /// Gets the total words (leafs) in the trie. Recursive traversal.     /// </summary>     public int TotalWords     {         get         {             int runningCount = 0;               Action<AutoSuggestTrieNode> traverseAllDecendants = null;             traverseAllDecendants = n => { runningCount += n.DescendantNodes.Count(o => o.IsLeafNode); n.DescendantNodes.ForEach(traverseAllDecendants); };             traverseAllDecendants(this._root);               return runningCount;         }     } }   Matching operations and Inserts involve traversing the nodes before the right “spot” is found. Inserts need be synchronous since ordering of data matters here. However, matching can be done in parallel traversal using recursion (line 64). Here’s sample usage:   [TestMethod] public void AutoSuggestTest() {     var autoSuggestCache = new AutoSuggestTrie();       var testInput = @"Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer nec odio. Praesent libero.                 Sed cursus ante dapibus diam. Sed nisi. Nulla quis sem at nibh elementum imperdiet. Duis sagittis ipsum. Praesent mauris.                 Fusce nec tellus sed augue semper porta. Mauris massa. Vestibulum lacinia arcu eget nulla. Class aptent taciti sociosqu ad                 litora torquent per conubia nostra, per inceptos himenaeos. Curabitur sodales ligula in libero. Sed dignissim lacinia nunc.                 Curabitur tortor. Pellentesque nibh. Aenean quam. In scelerisque sem at dolor. Maecenas mattis. Sed convallis tristique sem.                 Proin ut ligula vel nunc egestas porttitor. Morbi lectus risus, iaculis vel, suscipit quis, luctus non, massa. Fusce ac                 turpis quis ligula lacinia aliquet. Mauris ipsum. Nulla metus metus, ullamcorper vel, tincidunt sed, euismod in, nibh. Quisque                 volutpat condimentum velit. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Nam                 nec ante. Sed lacinia, urna non tincidunt mattis, tortor neque adipiscing diam, a cursus ipsum ante quis turpis. Nulla                 facilisi. Ut fringilla. Suspendisse potenti. Nunc feugiat mi a tellus consequat imperdiet. Vestibulum sapien. Proin quam. Etiam                 ultrices. Suspendisse in justo eu magna luctus suscipit. Sed lectus. Integer euismod lacus luctus magna. Quisque cursus, metus                 vitae pharetra auctor, sem massa mattis sem, at interdum magna augue eget diam. Vestibulum ante ipsum primis in faucibus orci                 luctus et ultrices posuere cubilia Curae; Morbi lacinia molestie dui. Praesent blandit dolor. Sed non quam. In vel mi sit amet                 augue congue elementum. Morbi in ipsum sit amet pede facilisis laoreet. Donec lacus nunc, viverra nec.";       testInput.Split(' ').ToList().ForEach(word => autoSuggestCache.AddWord(word));       var testMatches = autoSuggestCache.GetSuggestedMatches("le"); }   ..and the result: That’s it!

    Read the article

  • How can I force a Word plugin to be re-enabled after being automatically disabled?

    - by Gerald
    I have been developing a plugin for MS Word (2007) and recently it was causing some crashes on exit and I guess Word automatically disabled it. I went into Word Options under COM Add-ins and checked the box to re-enable it, but when I click okay and then open Word again it's still shown as disabled. I tried looking for something in the registry for this but I wasn't able to find anything. Is there some file or registry key that contains the enabled/disabled status of plugins?

    Read the article

  • Microsoft Office 2013 Issue

    - by Liz
    A few days ago I opened my microsoft office programs and discovered taht they are missing the editing icons at the top, some of them will appear if you scroll over them, but not all. Also, in PowerPoint the slides show in the side window with a red "x" I have tried to uninstall and reinstall office 2013, but I have had no luck. This issue is in every office program (excel, PP, word, access, outlook, etc). I also can't see the text when I type. Its there, I can see it when I print the document, but nothing on the screen. Does anyone have a solution for this issue??

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 106 107 108 109 110 111 112 113 114 115 116 117  | Next Page >