readr | A modern web reading app | GraphQL library

You can use the read.csv() function. But there would be some warning messages (or use suppressWarnings() to wrap around the read.csv() function). If you wish to avoid warning messages, using the scan() method in the next section.

You can't pipe your data frame directly into parse_number. You would need to pipe into a mutate:

I found the issue. On a hunch, I investigated the whitespace.

There are too many issues in your code. We need to work from scratch. In general, here are some major concerns:

1. Don't do expensive operations so many times. Things like pivot_* and *_join are not cheap since they change the structure of the entire dataset. Don't use them so freely as if they come with no cost.

2. Do not repeat yourself. I saw filter(Id == idd, Category == ...) several times in your function. The rows that are filtered out won't come back. This is just a waste of computational power and makes your code unreadable.

3. Think carefully before you code. It seems that you want the regression results for multiple idd, date2 and Category. Then, should the function be designed to only take scalar inputs so that we can run it many times each involving several expensive data operations on a relatively large dataset, or should it be designed to take vector inputs, do fewer operations, and return them all at once? The answer to this question should be clear.

Now I will show you how I would approach this problem. The steps are

1. Find the relevant subset for each group of idd, dmda and CategoryChosse at once. We can use one or two joins to find the corresponding subset. Since we also need to calculate the median for each Week group, we would also want to find the corresponding dates that are in the same Week group for each dmda.

2. Pivot the data from wide to long, once and for all. Use row id to preserve row relationships. Call the column containing those "DRMXX" day and the column containing values value.

3. Find if trailing zeros exist for each row id. Use rev(cumsum(rev(x)) != 0) instead of a long and inefficient pipeline.

4. Compute the median-adjusted values by each group of "Id", "Category", ..., "day", and "Week". Doing things by group is natural and efficient in a long data format.

5. Aggregate the Week group. This follows directly from your code, while we will also filter out days that are smaller than the difference between each dmda and the corresponding date1 for each group.

6. Run lm for each group of Id, Category and dmda identified.

7. Use data.table for greater efficiency.

8. (Optional) Use a different median function rewritten in c++ since the one in base R (stats::median) is a bit slow (stats::median is a generic method considering various input types but we only need it to take numerics in this case). The median function is adapted from here.

Below shows the code that demonstrates the steps

You may use maxsum=. Example:

This is fairly convoluted, but I think a couple of minor adjustments might be helpful here. First, you might want to double check what inputs you include in req. Second, you may want to filter your data based on input$idd when you provide code options. Let me know if this is helpful.

I think you can use the following solution:

pivot_longer checks the column types and by specifying -Category in cols, it will select all the remaining columns. But, in the OP's dataset, there is a character column 'DTT' in addition to other numeric columns ('DR0'). An option is to either remove the 'DTT' (by %>% select(-DTT) %>% pivot_longer(..) and use the OP's code or use cols = starts_with("DR0")

The way your events.http is configured looks wrong. Try replacing it with: