brook | Brook 一键安装脚本 - | Proxy library

You seem to have overcomplicated pushing an object to an Array

All you want to do is push to users2

It's all about spread operator, you should spread object inside array correctly, below example works fine.

The function feof will only tell you whether a previous input operation has already encountered end-of-file. It won't tell you whether you have now reached the end of file, so that the next input operation will fail. That function function is unable to predict whether the next input operation to fscanf or fgets will fail. Therefore, it should generally not be used as a loop condition. See this question for further information: Why is “while ( !feof (file) )” always wrong?

In your case, feof may return false and the subsequent function call to fscanf may return EOF due to encountering end-of-file. In that case, your posted code will ignore the return value of fscanf and behave as if fscanf had succeeded, and your posted code will attempt to process the non-existant input. This is likely to result in a bug.

Therefore, instead of using the function feof to determine whether the loop should be continued, you should check the return value of the input function.

You could rewrite your loop like this:

I'd create a named vector and use outer to build a matrix of differences. Calling your data frame df:

If you didn't have any condition on employee ID at all you'd end up with records where a self-match had occurred, e.g. the results would show "Gracie Gardner was hired on the same day as Gracie Gardner"

We could then put ON e1.employee_id <> e2.employee_id - this would prevent Gracie matching with Gracie, but you'd then find "Gracie Gardner was hired on the same day as Summer Payne" and "Summer Payne was hired on the same day as Gracie Gardner" - i.e. you'd get "duplicate records" in terms of "person paired with person", each name being mentioned both ways round

Using greater than prevents this, and effectively means that any given pair of names only appears once. Because Gracie's ID is less than Summer's, you'll get Gracie in e1 paired with Summer in e2 but you won't get Summer in e1 paired with Gracie in e2

Another way of visualizing it is with a square/matrix

try:

Locating element by link text "Job Details" gives you the a element while you need to get the h3 element text.
Try this instead:

I'm sure RegEx is the key here, as it's a pattern matching process similar to SQL LIKE, I think.

I wrote this on the assumption that if the tags don't have any of the "tag1 ... tag5 or tag_wrong1", then "Tag_after" is supposed to be the whole "Tag" value. E.g Matt's row tags are "tag8,tag9" and since it doesn't match any of the given tag patterns, Matt gets the tag_after value the same as his tags.

Input:

Here is the benchmarking for the existing to OP's question (borrow test data from @Marek Fiołka but with n <- 10000)