You can use a generator expression: >>> l = [1, 3, 7, 2, 6, 8, 10] >>> sum(1 for i in l if i % 4 == 3) 2 or even >>> sum(i % 4 == 3 for i in l) 2 which uses the fact that True == 1 and False == 0. Alternatively, … Read more
There’s not really a good way to do this, usually you see an extra item in the enum, i.e. enum foobar {foo, bar, baz, quz, FOOBAR_NR_ITEMS}; So then you can do: int fuz[FOOBAR_NR_ITEMS]; Still not very nice though. But of course you do realize that just the number of items in an enum is not … Read more
length(x) Get or set the length of vectors (including lists) and factors, and of any other R object for which a method has been defined. lengths(x) Get the length of each element of a list or atomic vector (is.atomic) as an integer or numeric vector.
The count() method returns the number of non-NaN values in each column: >>> df1.count() a 3 b 2 d 1 dtype: int64 Similarly, count(axis=1) returns the number of non-NaN values in each row.
THE way is :%s/pattern//gn
Query syntax var query = from p in context.People group p by p.name into g select new { name = g.Key, count = g.Count() }; Method syntax var query = context.People .GroupBy(p => p.name) .Select(g => new { name = g.Key, count = g.Count() });
Although the Django docs recommend using count rather than len: Note: Don’t use len() on QuerySets if all you want to do is determine the number of records in the set. It’s much more efficient to handle a count at the database level, using SQL’s SELECT COUNT(*), and Django provides a count() method for precisely … Read more
Ruby v2.7+ (latest) As of ruby v2.7.0 (released December 2019), the core language now includes Enumerable#tally – a new method, designed specifically for this problem: names = [“Jason”, “Jason”, “Teresa”, “Judah”, “Michelle”, “Judah”, “Judah”, “Allison”] names.tally #=> {“Jason”=>2, “Teresa”=>1, “Judah”=>3, “Michelle”=>1, “Allison”=>1} Ruby v2.4+ (currently supported, but older) The following code was not possible in … Read more
There’s a special function n() in dplyr to count rows (potentially within groups): library(dplyr) mtcars %>% group_by(cyl, gear) %>% summarise(n = n()) #Source: local data frame [8 x 3] #Groups: cyl [?] # # cyl gear n # (dbl) (dbl) (int) #1 4 3 1 #2 4 4 8 #3 4 5 2 #4 6 … Read more
The following should work $ad->getcodes()->distinct()->count(‘pid’);