If you don’t want to modify the list in-place (e.g. for passing the list with an element removed to a function), you can use indexing: negative indices mean “don’t include this element”. x <- list(“a”, “b”, “c”, “d”, “e”); # example list x[-2]; # without 2nd element x[-c(2, 3)]; # without 2nd and 3rd Also, … Read more
Sure, you can use .get_loc(): In [45]: df = DataFrame({“pear”: [1,2,3], “apple”: [2,3,4], “orange”: [3,4,5]}) In [46]: df.columns Out[46]: Index([apple, orange, pear], dtype=object) In [47]: df.columns.get_loc(“pear”) Out[47]: 2 although to be honest I don’t often need this myself. Usually access by name does what I want it to (df[“pear”], df[[“apple”, “orange”]], or maybe df.columns.isin([“orange”, “pear”])), … Read more
It actually is possible to do what you want, but you have to use the functional form of the indexing operator. When you perform an indexing operation using (), you are actually making a call to the subsref function. So, even though you can’t do this: value = magic(5)(3, 3); You can do this: value … Read more
There are two “sys” catalog views you can consult: sys.indexes and sys.index_columns. Those will give you just about any info you could possibly want about indices and their columns. EDIT: This query’s getting pretty close to what you’re looking for: SELECT TableName = t.name, IndexName = ind.name, IndexId = ind.index_id, ColumnId = ic.index_column_id, ColumnName = … Read more
The function match works on vectors: x <- sample(1:10) x # [1] 4 5 9 3 8 1 6 10 7 2 match(c(4,8),x) # [1] 1 5 match only returns the first encounter of a match, as you requested. It returns the position in the second argument of the values in the first argument. For … 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
Basically an index on a table works like an index in a book (that’s where the name came from): Let’s say you have a book about databases and you want to find some information about, say, storage. Without an index (assuming no other aid, such as a table of contents) you’d have to go through … 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);
Provide the :name option to add_index, e.g.: add_index :studies, [“user_id”, “university_id”, “subject_name_id”, “subject_type_id”], unique: true, name: ‘my_index’ If using the :index option on references in a create_table block, it takes the same options hash as add_index as its value: t.references :long_name, index: { name: :my_index }
To select the ith row, use iloc: In [31]: df_test.iloc[0] Out[31]: ATime 1.2 X 2.0 Y 15.0 Z 2.0 Btime 1.2 C 12.0 D 25.0 E 12.0 Name: 0, dtype: float64 To select the ith value in the Btime column you could use: In [30]: df_test[‘Btime’].iloc[0] Out[30]: 1.2 There is a difference between df_test[‘Btime’].iloc[0] (recommended) … Read more