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countfis or sum if array to work out if given time is between the value of two columns


countfis or sum if array to work out if given time is between the value of two columns

By : Konstantinos Daras
Date : November 22 2020, 10:56 AM
I hope this helps . If I understand correctly, you want a 0 or 1 for each of the times in row 10 that you can use a conditional format on to show red or nothing. In A10 try,
=COUNTIFS($A$2:$A$5,"<"&A$10, $B$2:$B$5,">"&A$10)
code :


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Swap rows or columns of a 2d array in O(1) time in c++

Swap rows or columns of a 2d array in O(1) time in c++


By : Thomas Wyrick
Date : March 29 2020, 07:55 AM
I hope this helps you . You can use an indirect array on both rows and columns. In other words to access an element i,j you use
code :
data[rowix[i]][colix[j]]
data[i][j]
template<int ROWS, int COLS, typename T>
struct Mat2d {
    T data[ROWS][COLS];
    int colix[COLS], rowix[ROWS];
    Mat2d() {
        for (int i=0; i<ROWS; i++) {
            for (int j=0; j<COLS; j++) {
                data[i][j] = T();
            }
        }
        for (int i=0; i<ROWS; i++) rowix[i] = i;
        for (int j=0; j<COLS; j++) colix[j] = j;
    }
    T& operator()(int i, int j) { return data[rowix[i]][colix[j]]; }
    T operator()(int i, int j) const { return data[rowix[i]][colix[j]]; }
    void swapRows(int i1, int i2) { std::swap(rowix[i1], rowix[i2]); }
    void swapCols(int j1, int j2) { std::swap(colix[j1], colix[j2]); }
};
OpenCL: 2-10x run time summing columns rather that rows of square array?

OpenCL: 2-10x run time summing columns rather that rows of square array?


By : patafix
Date : March 29 2020, 07:55 AM
With these it helps This is a clear case of example of coalescence. It is not about how the index is used (in the rows or columns), but how the memory is accessed in the HW. Just a matter of following step by step how the real accesses are performed and in which order.
Lets analyze it properly:
code :
WI_0 will read 0, Gs, 2Gs, 3Gs, .... (k-1)Gs
WI_1 will read 1, Gs+1, 2Gs+1, 3Gs+1, .... (k-1)Gs+1
...
First iteration: 0, 1, 2, 3 ... N-1  -> Groups into few memory access
Second iteration: Gs, Gs+1, Gs+2, ... Gs+N-1  ->  Groups into few memory access
...
WI_0 will read 0, 1, 2, 3, .... (k-1)
WI_1 will read Gs, Gs+1, Gs+2, Gs+3, .... Gs+(k-1)   
...
First iteration: 0, Gs, 2Gs, 3Gs -> Scattered read, no grouping
Second iteration: 1, Gs+1, 2Gs+1, 3Gs+1 ->Scattered read, no grouping
...
How to modify N columns of numpy array at the same time?

How to modify N columns of numpy array at the same time?


By : dopppler
Date : March 29 2020, 07:55 AM
I hope this helps . Use indexing:
Here is an example:
code :
>>> P[:, [0, 1, 3]] += 10
>>> 
>>> P
array([[11, 12,  3, 18,  6],
       [14, 15,  6, 14,  5],
       [10,  8,  5, 13,  0]])
how to compare 2 columns of a 2d array at a time with columns of another array in python

how to compare 2 columns of a 2d array at a time with columns of another array in python


By : Fukuda
Date : March 29 2020, 07:55 AM
will help you EDIT
if array1[:2] == array2[:2]: compares all items from the index 0 to 2(2 is not included), and comes up with the same result as if array1[0] == array2[0] and array1[1] == array2[1]:. Also, it is simpler.( to Wyatt for comment)
code :
def compare_columns(array1, array2):
    if len(array1) != len(array2):
        return False # If row numbers are not same, return false

    for row_number in range(len(array1)):
        if array1[row_number][:2] != array2[row_number][:2]:
            return False # If the content is not equal, return false

    return True # All of the content is equal, the return true

# For example, these are 2-dimensional arrays
array1 = [["1.1", "1.2", "Lord of the Day of Judgment!"],
          ["2.1", "2.2", "Lord of the Day of Judgment!"]]
array2 = [["1.1", "1.2", "مَالِكِ يَوْمِ الدِّينِ"],
          ["2.1", "2.2", "مَالِكِ يَوْمِ الدِّينِ"]]

array3 = []       
if compare_columns(array1, array2):
       array3.append('matches: {!r}'.format(array1))
print(array3)
["matches: [['1.1', '1.2', 'Lord of the Day of Judgment!'], ['2.1', '2.2', 'Lord of the Day of Judgment!']]"]
array1 = ["1stcolumn", "2ndColumn", "1-3rdColumn"]
array2 = ["1stcolumn", "2ndColumn", "2-3rdColumn"]
array3 = []
if array1[0] == array2[0] and array1[1] == array2[1]:
       array3.append('matches: {!r}'.format(array1))
print(array3)
["matches: ['1stcolumn', '2ndColumn', '1-3rdColumn']"]
Sort a range or array based on two columns that contain the date and time

Sort a range or array based on two columns that contain the date and time


By : RAB Guild
Date : March 29 2020, 07:55 AM
will be helpful for those in need Given your comment - you want to sort the written chunk - you have two methods available. One is to sort written data after writing, by using the Spreadsheet service's Range#sort(sortObject) method. The other is to sort the data before writing, using the JavaScript Array#sort(sortFunction()) method.
Currently, your sort code //regWeek.sort([{ column: 1, ascending: true }]); is attempting to sort a JavaScript array, using the sorting object expected by the Spreadsheet service. Thus, you can simply chain this .sort(...) call to your write call, as Range#setValues() returns the same Range, allowing repeated Range method calling (e.g. to set values, then apply formatting, etc.).
code :
ss.getRange(ss.getLastRow() + 1, 2, regWeek.length, regWeek[0].length)
   .setValues(regWeek)
   /* other "chainable" Range methods you want to apply to
       the cells you just wrote to. */
   .sort([{column: 1, ascending: true}, ...]);
regWeek.sort(function (r1, r2) {
  // r1 and r2 are elements in the regWeek array, i.e.
  // they are each a row array if regWeek is an array of arrays:
  // Sort ascending on the first column, which is index 0:
  // if r1[0] = 1, r2[0] = 2, then 1 - 2 is -1, so r1 sorts before r2
  return r1[0] - r2[0];
});
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