What you need is calculating rolling total. The fastest way in SQL Server 2005 / 2008 / 2008 R2 I know is described here: https://stackoverflow.com/a/13744550/1744834. You have to have unique secuential column inside each category to use this method, like this: create table #t (no int, category int, value int, id int, primary key (category, … Read more
The Java compiler thinks that it is possible that loops won’t execute (For example it nums[0] is bigger than the length) . In that case your method won’t call a return. So you have to put a return at the end. public int[] twoSum(int[] nums, int target) { for(int i=nums[0];i<nums.length;i++){ for(int x:nums){ if (x!=i & … Read more
There is a function combinations in itertools and it can be used to generate combinations. import random import itertools # Generate a random list of 30 numbers mylist = random.sample(range(-50,50), 30) combins = itertools.combinations(mylist, 3) interested = list(filter(lambda combin: sum(combin) == 0, combins)) print(interested) Note that the results from filter() and itertools.combinations() are iterables and … Read more
One reasonably simple approach: n = 5 s = sum(range(n+1)) for i in range(n): s -= i print(s) 15 14 12 9 5 1 solved Sum from 1 to n, 2 to n, … n in python
@fia Maybe you should do your calculation elsewhere first for example on paper or in Excel to ensure you know exactly what you should get. It would also help you figure out the order of the calculation before writing it in SQL. According to the figures shown the values you’ve stated seems to be correct … Read more
d = [{‘A’: [3, 45, 34, 4, 2, 5, 94, 2139, 230345, 283047, 230847]}, {‘B’: [92374, 324, 345, 345, 45879, 34857987, 3457938457]}, {‘C’: [23874923874987, 2347]}] [{x.keys()[0]:sum(x.values()[0])} for x in d] solved I have a very big list of dictionaries and I want to sum the insides
The crucial point in OP’s approach is the staggered aggregation (see the related question row not consolidating duplicates in R when using multiple months in Date Filter). The OP wants to aggregate data across a number of files which apparently are too large to be loaded altogether and combined into a large data.table. Instead, each … Read more
Please try this query. I hope it’s will help you. SELECT catgory_id,COUNT(*) FROM tbl_name GROUP BY catgory_id; Thank you! solved MYSQL Select Query with SUM