In newer versions of MATLAB (R2015a and later) the easiest way to do this is using the repelem function: B = repelem(A, 2, 2); For older versions, a short alternative to the other (largely) indexing-based solutions is to use the functions kron and ones: >> A = [2 6; 8 4]; >> B = kron(A, … Read more
Just simply use below snippet private final Handler handler = new Handler(); private Runnable runnable = new Runnable() { public void run() { // // Do the stuff // handler.postDelayed(this, 1000); } }; runnable.run(); To stop it use handler.removeCallbacks(runnable); Should do the trick.
$values = array_count_values($array); arsort($values); $popular = array_slice(array_keys($values), 0, 5, true); array_count_values() gets the count of the number of times each item appears in an array arsort() sorts the array by number of occurrences in reverse order array_keys() gets the actual value which is the array key in the results from array_count_values() array_slice() gives us the … Read more
You can use a simple JOIN to get the desired result as below: SELECT t1.*, t2.number + 1 RepeatNumber FROM TableA t1 JOIN master.dbo.spt_values t2 ON t2.type=”P” AND t2.number < t1.Quantity The above query repeats each record by the specified number in Quantity column. Note for master.dbo.spt_values on type=”P”: This table is used for getting … Read more
You can use the concat function: In [13]: pd.concat([x]*5) Out[13]: a b 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 If you only want to repeat the values and not the index, you can do: In [14]: pd.concat([x]*5, ignore_index=True) Out[14]: a b 0 1 2 1 1 2 … Read more
=ARRAYFORMULA(TRIM(TRANSPOSE(SPLIT(QUERY( REPT(A1:A3&”♠”, 5), ,999^99), “♠”)))) =SORT(TRIM(TRANSPOSE(SPLIT(QUERY(ARRAYFORMULA( REPT(A1:A3&”♠”, 5)), ,999^99), “♠”))), 1, 0)
a = [‘1’, ‘1’, ‘1’, ‘1’, ‘1’, ‘1’, ‘2’, ‘2’, ‘2’, ‘2’, ‘7’, ‘7’, ‘7’, ’10’, ’10’] print a.count(“1”) It’s probably optimized heavily at the C level. Edit: I randomly generated a large list. In [8]: len(a) Out[8]: 6339347 In [9]: %timeit a.count(“1”) 10 loops, best of 3: 86.4 ms per loop Edit edit: This … Read more
Solutions: Use np.repeat: Version 1: Try using np.repeat: newdf = pd.DataFrame(np.repeat(df.values, 3, axis=0)) newdf.columns = df.columns print(newdf) The above code will output: Person ID ZipCode Gender 0 12345 882 38182 Female 1 12345 882 38182 Female 2 12345 882 38182 Female 3 32917 271 88172 Male 4 32917 271 88172 Male 5 32917 271 88172 … Read more
Give a try on below formula- =INDEX(SPLIT(FLATTEN(SPLIT(JOIN(“”,INDEX(REPT(BYROW(A2:D3,LAMBDA(x,TEXTJOIN(“|”,0,x)))&”@”,E2:E3))),”@”)),”|”)) To make it dynamic spill array, use- =INDEX(SPLIT(FLATTEN(SPLIT(JOIN(“”,INDEX(REPT(BYROW(A2:INDEX(D2:D,MATCH(“zzz”,D2:D)),LAMBDA(x,TEXTJOIN(“|”,0,x)))&”@”,E2:INDEX(E2:E,MATCH(9^9,E2:E))))),”@”)),”|”))