You can just use table(): > a <- table(numbers) > a numbers 4 5 23 34 43 54 56 65 67 324 435 453 456 567 657 2 1 2 2 1 1 2 1 2 1 3 1 1 1 1 Then you can subset it: > a[names(a)==435] 435 3 Or convert it into … Read more
To do this in any ES5-compatible environment, such as Node.js, Chrome, Internet Explorer 9+, Firefox 4+, or Safari 5+: Object.keys(obj).length Browser compatibility Object.keys documentation (includes a method you can add to non-ES5 browsers)
Using data.table: library(data.table) dt = as.data.table(df) # or coerce to data.table by reference: # setDT(df) dt[ , count := .N, by = .(name, type)] For pre-data.table 1.8.2 alternative, see edit history. Using dplyr: library(dplyr) df %>% group_by(name, type) %>% mutate(count = n()) Or simply: add_count(df, name, type) Using plyr: plyr::ddply(df, .(name, type), transform, count = … Read more
str.count(sub[, start[, end]]) Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation. >>> sentence=”Mary had a little lamb” >>> sentence.count(‘a’) 4
# To sort the list in place… ut.sort(key=lambda x: x.count, reverse=True) # To return a new list, use the sorted() built-in function… newlist = sorted(ut, key=lambda x: x.count, reverse=True) More on sorting by keys.
Counting the occurrences / frequency of array elements
If you only want a single item’s count, use the count method: >>> [1, 2, 3, 4, 1, 4, 1].count(1) 3 Important: this is very slow if you are counting multiple different items Each count call goes over the entire list of n elements. Calling count in a loop n times means n * n … Read more
Your resource is not a list but an object. Better implementation would be something like this. [HttpGet(“{id}”)] public IActionResult Get(string id) { using (var unitOfWork = new UnitOfWork(_db)) { var r = unitOfWork.Resources.Get(id); if (r == null) { return NotFound(); } return Ok(ConvertResourceModel(r)); } }
import java.io.File; import java.io.FilenameFilter; public class FileCountFilter { public File[] findFilesInDir(String dirName){ File dir = new File(dirName); return dir.listFiles(new FilenameFilter() { public boolean accept(File dir, String filename) { return filename.toLowerCase().endsWith(“.txt”); } } ); } } The Length of File[] returned by findFilesInDir method would be what you are looking for.
Its sum value by each order u will need. According to your original question orderamount <100 then add 5 WITH cte AS ( SELECT OrderID ,OrderDate ,Customer ,ProductAmount ,sum(orderamount) orderamount FROM mytable GROUP BY OrderID ,OrderDate ,Customer ,ProductAmount ) SELECT OrderID ,OrderDate ,Customer ,ProductAmount ,CASE WHEN orderamount < 100 THEN orderamount + 5 ELSE orderamount … Read more