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))))),”@”)),”|”))
Suppose you have this simplified input and you want to expand columns sequentially n times: A = [1 4 2 5 3 6]; szA = size(A); n = 3; There are few ways to do that: Replicate, then reshape: reshape(repmat(A,n,1),szA(1),n*szA(2)) Kronecker product: kron(A,ones(1,n)) Using FEX: expand(): expand(A,[1 n]) Since R2015a, repelem(): repelem(A,1,n) All yield the … Read more
numpy’s repeat function gets the job done: >>> import numpy as np >>> x = [2, 3, 4] >>> y = [1, 2, 3] >>> np.repeat(x, y) array([2, 3, 3, 4, 4, 4])
You could do this with a little Linq: if (testArray.Length != testArray.Distinct().Count()) { Console.WriteLine(“Contains duplicates”); } The Distinct extension method removes any duplicates, and Count gets the size of the result set. If they differ at all, then there are some duplicates in the list. Alternatively, here’s more complicated query, but it may be a … Read more
An ugly hack could be inserting multiple times —statically— the same image (using multiple backgrounds): E.g. To repeat horizontally an image (80×80) three times: background-image: url(‘bg_texture.png’), url(‘bg_texture.png’), url(‘bg_texture.png’); background-repeat: no-repeat, no-repeat, no-repeat; background-position: 0 top, 80px top, 160px top; Beware: not working on IE under 9 version (source).
Without splitting and joining and other string manipulations, =ARRAYFORMULA(FLATTEN(IF(SEQUENCE(ROWS(A2:A4),5),A2:A4))) A2:A4 The range to repeat 5 Number of times to repeat SEQUENCE to create a 2D array of numbers IF to transpose a 1D array A2:A4 to a 2D array of size equal to the SEQUENCE array created FLATTEN the 2D to 1D array. A1 Name … Read more
@Override public View getView(int pos, View convertView, ViewGroup parent) { if(convertView == null){ convertView = mLayoutInflater.inflate(zeilen_Layout, null); //Assignment code } return convertView; } The reason this is happening is because of how you’ve defined this method (which is getView()). It only modifies the convertView object if it is null, but after the first page of … Read more
You’re fortunate because in fact .NET regex does this (which I think is quite unique). Essentially in every Match, each Group stores every Captures that was made. So you can count how many times a repeatable pattern matched an input by: Making it a capturing group Counting how many captures were made by that group … Read more
Here’s one way I like to accomplish this: >> index = zeros(1,sum(rep)); >> index(cumsum([1 rep(1:end-1)])) = 1; index = 1 0 1 0 0 1 1 0 0 0 0 >> index = cumsum(index) index = 1 1 2 2 2 3 4 4 4 4 4 >> vv = v(index) vv = 3 3 … 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