residue | Real-time centralized logging server

I have been doing a script that takes two files to extract a specific part of the data to make a new file. If you want to see the complete file, here's a GitHub link: enter link description here

File one (report file) is a type of file that reports me when a value is >=0.5 (column N°6 is the value that interests me). This file is something like this (this is only a part):

how can I assign to a list from another method? I created WPF Application.

I tried:

I am plotting 100 data points for 9 different groups. One group's data points are much larger than all the other groups so when I make a box graph using pandas only that group is shown, while all other groups are smashed to the bottom. Here is what it looks like now: smushed box plot

I would like the Y axis to be more spaced out so that I can see the other groups' box graphs. Here is similar data in a scatter plot that has the spacing I am looking for: well spaced scatter plot

Here is my code at the moment:

A residue function (res) does the sum of y values for y>thr (threshold). It returns the residue between this sum and the target.

In this example, a target is calculated for y>70 and I would like to find, after minimization, y=70 starting from y=60.

I have a file that looks like this:

I have what I think is a interesting question, about google sheets and some Maths, here is the scenario:

4 numbers as follows:

64.20 | 107 | 535 | 1070

A reference number in which the previous numbers needs to fit leaving the minimum possible residue while setting the number of times each of them fitted in the reference number for example we could say the reference number is the following:

806.45

So here is the problem:

I'm calculating how many times those 4 numbers can fit in the reference number by starting from the higher to the lower number like this:

| 1070 | => =IF(E12/((I15+J15)+IF(H17,K17,0)+IF(H19,K19,0)) > 0,ROUNDDOWN(E12/((I15+J15)+IF(H17,K17,0)+IF(H19,K19,0))),0)

| 535 | => =IF(H15>0,ROUNDDOWN((E12-K15-IF(H17,K17,0)-IF(H19,K19,0))/(I14+J14)),ROUNDDOWN(E12/((I14+J14)+IF(H17,K17,0)+IF(H19,K19,0))))

| 107 | => =IF(OR(H15>0,H14>0),ROUNDDOWN((E12-K15-K14-IF(H17,K17,0)-IF(H19,K19,0))/(I13+J13)),ROUNDDOWN((E12-IF(H17,K17,0)-IF(H19,K19,0))/(I13+J13)))

| 64.20 | => =IF(OR(H15>0,H14>0,H13>0),ROUNDDOWN((E12-K15-K14-K13-IF(H17,K17,0)-IF(H19,K19,0))/(I12+J12)),ROUNDDOWN((E12-IF(H17,K17,0)-IF(H19,K19,0))/(I12+J12)))

As you can notice, I'm checking if the higher values has a concurrence, so I can substract the amount from the original number and calculate again how many times can fit the lower number in the result of that subtraction , you can also see that I'm including some checkboxes to the formula in order to add a fixed number to the main number.

This actually works, and as you can see in the example, the result is:

| 1070 | -> Fits 0 times

| 535 | -> Fits 1 time

| 107 | -> Fits 2 times

| 64.20 | -> Fits 0 times

The residue of 806.45 in this example is: 57.45

But each number that needs to fit on the main number needs to take in consideration others; IF you solve this exercise manually, you could get something much better.. like this:

| 1070 | -> Fits 0 times

| 535 | -> Fits 1 time

| 107 | -> Fits 0 times

| 64.20 | -> Fits 4 times

The residue of 806.45 in this example is: 14.65

When I’m talking about residue I mean the result when subtracting, I’m sorry if this is not clear, it’s hard to me to explain maths in English, since is not my native language, please see the spreadsheet and make a copy to better understand what I’m trying to do, or suggest me a way to explain it better if possible.

So what would you do to make it work more efficient and "smart" with the minimum possible residue after the calculation?

Here is the Google's spreadsheet for reference and practice, please make a copy so others can try their own solutions:

LINK TO SPREADSHEET

Thanks in advance for any help or hints.

I have a homework from my intro to programming class where they ask me to separate an int value into its different numbers so i can then multiply them.

For example, the number is 860007386.

The verification number is calculated by doing the steps below:

Multiply each digit by the numbers 41, 37, 29, 23, 19, 17, 13, 7 and 3 in that order. For example, In this case it should go like this: 8x41, 6x37, 0x29, 0x23, 0x19, 7x17, 3x13, 8x7, 6x3.

then add the product of all the operations above.

calculate the residue that you get from dividing the sum by 11.

If the residue is 1 or 0 then then that is the verification code.

If not then the verification code is given by substracting the residue of 11.

I am running a simple benzene simulation in GROMOS54a7. I want to calculate the RDF of the center of masses of each benzene molecule, using MDAnalysis 1.0.0.

Is this possible? I have create the rdf for the C molecules g_cc(r) using the following code in a Jupyter Notebook:

In the following Main method why isn't the last word (clapping) removed?