omg | Command-line bookmark manager

I used what you did and android 12 emulator also worked.I used this for versions less than android 12 I hope it works.

MenuItems.filter((item) => "Drinks") return always true

What you should be doing is comparing the category to drinks.

MenuItems.filter((item) => item.category === "Drinks")

You could use json module to parse content of script tags, which is accessible by .text field

Here is the example of parsing all script jsons and printing name:

Try this one, it has enough expressions in there to allow you to modify them based on how it is implemented:

We can use str.findall to find all the matching occurrences of regex pattern from each row, then map the occurrences to dict and create a new dataframe. This approach will avoid re-indexing the dataframe. Further you also have to fix your regex pattern to properly capture matching pairs.

I see a couple of issues. First, the code doesn't have a return statement for every logical branch. I mean that, for every condition in an if, then, else statement, the code should have a return statement or some sort of control-flow statement (e.g. a recursive call or a continue) that will eventually lead to a return statement.

The second issue is the while loop. Supposedly, right should be greater than left from the beginning of the function (unless the length of smallString is 0) because right is the length of smallerString minus 1. The while condition is left > right, so nothing inside of the while will be executed unless smallString has a negative length (which it doesn't).

By the way, if you want to check the entire bigString, you will need to iterate over bigString, not smallString. If you are only checking smallString.length characters, you won't be checking the entire bigString. Figuring out how to write this function is a good exercise, so I will leave the writing of it to you and refrain from providing an implementation myself. Keep up the good work!

I don't know how Breeze works.

But I think you could use flex in your favor.

If you want the link always on the right side, you can use something like this,

As was pointed out in comments, you used a positive sign in the computation of omg_n. There are different definitions of the DFT, so it isn't wrong by itself. However this would naturally lead to differences if you compare your results with an implementation that uses a negative sign, as is the case with numpy.fft.fft. Adjusting your implementation to also use a negative sign would cover all forward transform cases (leaving only small roundoff errors on the order of ~10^-16).

For the inverse transform cases, your implementation ends up scaling the result by 1/n at every stage, instead of only the final stage. To correct this, simply remove the scaling from the recursion, and normalize only on the final stage:

You could apply the flex utility class to your a element.