I’ve came with a solution. It gets an approximation of the spending factor because there is no direct formula. Getting the exact spending factor is too CPU intensive. So, I’m calculating a near approximation first. This value will always be above the solution. I then decrease that number until I’m getting below the total paid … Read more
The code given looks OK, but if you want the sum without including the last number, as is generally the case you should change the for loop like this for(; index < endno; index ++)
While @SamGrondahl is on the right track, unfortunately, if you enter a negative number it gives unexpected results. This should work properly with negative numbers: int sumdigits(int number) { int sign = number < 0 ? -1 : 1; number = sign * number; // abs the number if (number < 10) return sign * … Read more
I think you may be confusing type variables which are used in type annotations, and the actual values being passed to the function. In your definition of mygcd: mygcd c = a + b Unless a and b are defined elsewhere, this will be an error. In other words, these a, b, and c are … Read more
Here is one way to accomplish this: #make a data frame df <- data.frame(a = c(1,2,3), b = c(1,NA,3), c = c(1,2,3)) #use rowSums on a dataframe made with each of your columns and specify na.rm = TRUE df$mySum <- rowSums(cbind(df$a,df$b), na.rm = TRUE)
You could just use zip like, >>> list1 [[1, 2, 3], [4, 5, 6]] >>> list2 [[10, 2, 3], [11, 5, 6]] >>> [[x+y for x,y in zip(l1, l2)] for l1,l2 in zip(list1,list2)] [[11, 4, 6], [15, 10, 12]] or if you are not sure, if both list will be of same length, then you … Read more
The fizzbuzz numbers which are less or equal to N are integers between 0 and N that are divisible by 15. Their sum is sum(15*i for i = 1 to N/15) That’s equal to 15*sum(i for i i=1 to N/15), or 15*(N/15)*(1+N/15)/2. (Note, here / means rounding down integer division). The sum of fizzbuzz numbers … Read more
This is a lot like the APL Reduction operator, which has a cool feature of letting you reduce over any dimension (e.g. +/tab versus +/[1]tab). Over the first dimension, LINQ has an equivalent already, the Aggregate function. With some other extension functions, you can use it to do what you want: public static class Ext … Read more
Because $product refers to each iteration of $output[“PriceInformation”][“PriceDetails”][“Price”], you can’t sum the entire array like this. The best way to do it would be to add it to a variable as you go: $your_sum = 0; foreach($output as $value) { $your_sum += $value[‘PriceInformation’][‘PriceDetails’][‘Price’]; } echo ‘Sum: ‘ . $your_sum;
In base R (for sum) there’s xtabs: > xtabs(sales ~ Date + area, mydf) area Date beijing shanghai 201204 41 23 201205 17 71 To get it as a data.frame, wrap it in as.data.frame.matrix. To update this with the approach that is making the rounds these days, you can also use a combination of “dplyr” … Read more