An index is used to speed up searching in the database. MySQL have some good documentation on the subject (which is relevant for other SQL servers as well): http://dev.mysql.com/doc/refman/5.0/en/mysql-indexes.html An index can be used to efficiently find all rows matching some column in your query and then walk through only that subset of the table … Read more
This is an example of the greatest-n-per-group problem that has appeared regularly on StackOverflow. Here’s how I usually recommend solving it: SELECT c.*, p1.* FROM customer c JOIN purchase p1 ON (c.id = p1.customer_id) LEFT OUTER JOIN purchase p2 ON (c.id = p2.customer_id AND (p1.date < p2.date OR (p1.date = p2.date AND p1.id < p2.id))) … Read more
You can use operator.itemgetter: from operator import itemgetter a = [-2, 1, 5, 3, 8, 5, 6] b = [1, 2, 5] print(itemgetter(*b)(a)) # Result: (1, 5, 5) Or you can use numpy: import numpy as np a = np.array([-2, 1, 5, 3, 8, 5, 6]) b = [1, 2, 5] print(list(a[b])) # Result: [1, … Read more
Use the unaccent module for that – which is completely different from what you are linking to. unaccent is a text search dictionary that removes accents (diacritic signs) from lexemes. Install once per database with: CREATE EXTENSION unaccent; If you get an error like: ERROR: could not open extension control file “/usr/share/postgresql/<version>/extension/unaccent.control”: No such file … Read more
With a clustered index the rows are stored physically on the disk in the same order as the index. Therefore, there can be only one clustered index. With a non clustered index there is a second list that has pointers to the physical rows. You can have many non clustered indices, although each new index … Read more
Using C++ 11 lambdas: #include <iostream> #include <vector> #include <numeric> // std::iota #include <algorithm> // std::sort, std::stable_sort using namespace std; template <typename T> vector<size_t> sort_indexes(const vector<T> &v) { // initialize original index locations vector<size_t> idx(v.size()); iota(idx.begin(), idx.end(), 0); // sort indexes based on comparing values in v // using std::stable_sort instead of std::sort // to … Read more
String are immutable in Java. You can’t change them. You need to create a new string with the character replaced. String myName = “domanokz”; String newName = myName.substring(0,4)+’x’+myName.substring(5); Or you can use a StringBuilder: StringBuilder myName = new StringBuilder(“domanokz”); myName.setCharAt(4, ‘x’); System.out.println(myName);
either: df[‘index1’] = df.index or, .reset_index: df = df.reset_index(level=0) so, if you have a multi-index frame with 3 levels of index, like: >>> df val tick tag obs 2016-02-26 C 2 0.0139 2016-02-27 A 2 0.5577 2016-02-28 C 6 0.0303 and you want to convert the 1st (tick) and 3rd (obs) levels in the index … Read more
If you want to re-index starting to zero, simply do the following: $iZero = array_values($arr); If you need it to start at one, then use the following: $iOne = array_combine(range(1, count($arr)), array_values($arr)); Here are the manual pages for the functions used: array_values() array_combine() range()
You may want to try the following: SELECT l.position, l.username, l.score, @curRow := @curRow + 1 AS row_number FROM league_girl l JOIN (SELECT @curRow := 0) r; The JOIN (SELECT @curRow := 0) part allows the variable initialization without requiring a separate SET command. Test case: CREATE TABLE league_girl (position int, username varchar(10), score int); … Read more