quantitative | quantitative - Quantitative finance back testing library | Data Manipulation library

The data is incoming from an elasticsearch url and has the following form :

I am able to label a bar chart created in Vega Lite, with either its percentage of the total or its absolute number, like so:

And here's the spec:

In first steps in algorithmic design and analysis I am following the book Algorithm Design by Kleinberg and Tardos I came across the Question/Case you can find down the page the solution is indeed f(n) = sqrt(n). My concerns are :

1- Why I still find log(n) more acceptable .I still could not grasp the plus value from sqrt(n) even when it is said that we will use more jars / trials .

2- from where did we get the sqrt(n) ?. Using k jars (trials) I could think of n/k incrementation but then lim n→∞ f(n) /n toward infinity is 1/k which is not 0. I got the feeling that the '2' in n^1/2 is tightly related to k = 2 , if yes how.

Thank you.

Case: Assume that a factory is doing some stress-testing on various models of glass jars to determine the height from which they can be dropped and still not break. The setup for this experiment, on a particular type of jar, is as follows. You have a ladder with n rungs, and you want to find the highest rung from which you can drop a copy of the jar and not have it break. We call this the highest safe rung. It might be natural to try binary search: drop a jar from the middle rung, see if it breaks, and then recursively try from rung n/4 or 3n/4 depending on the outcome. But this has the drawback that you could break a lot of jars in finding the answer. If your primary goal were to conserve jars, on the other hand, you could try the following strategy. Start by dropping a jar from the first rung, then the second rung, and so forth, climbing one higher each time until the jar breaks. In this way, you only need a single jar—at the moment it breaks, you have the correct answer—but you may have to drop it n times. So here is the trade-off: it seems you can perform fewer drops if you’re willing to break more jars. To understand better how this trade- off works at a quantitative level, let’s consider how to run this experiment given a fixed “budget” of k ≥ 1 jars. In other words, you have to determine the correct answer—the highest safe rung—and can use at most k jars in doing so. Now, please solve these two questions:

1. Suppose you are given a budget of k = 2 jars. Describe a strategy for finding the highest safe rung that requires you to drop a jar at most f(n) times, for some function f(n) that grows slower than linearly. (In other words, it should be the case that lim n→∞ f(n)/n = 0.)

In total I have 21k observations. In order to do a cluster analysis, I would like to group the 21K observations by the column "Neighborhood" (there are a total of 140 neighborhoods). So I would like to group by "neighborhood" and get the mean of the quantitative variables for each neighborhood (e.g. "buy price") and the mode for the qualitative variables (e.g. "energy certificate", "has parking", etc). So that the dataset is simply 140 rows (neighborhoods) and their means and modes depending on the variable concerned. I hope someone can help me. Thank you very much in advance.

I did a quantitative proteomics experiment to measure the differential expression of proteins in cells between two conditions. The output is a list of peptides, the protein they map to, and the their abundance for the experimental and control condition. Each protein has several detected peptides, and I need to pull out the median peptide abundance per protein, per condition into a new data frame. A simple version is as follows below:

Is there a way to write code for this in R? Note that I have about 6000 proteins, and about 60,000 detected peptides. Not all peptides were detected in both condition 1 and 2, but I would still need to take the median of all peptides per protein for each condition separately.

The goal is to do statistical analysis between the median peptide abundance for each protein so I can see if the values are significantly different.

Thanks in advance!

I'm trying to use altair to visualize data, but I struggle to use it in the way I would like to use it: by not embedding the data inside the generated .html chart, but by referencing the local .csv file that contains the data. Otherwise, it would result in duplicating the data and therefore doubling the required storage.

Here's what I tried:

I am sort of puzzled with the outcome of the code here below. The data frame I called aux (the data) contains a factor and a quantitative variable. I want to plot mean values of the quantitative variable according to levels of the factor.

The code creates also a second data frame containing those grouped mean values.

Then there are two plots. The first one is fine by me: it plots the right values in two different ways, that is using stat_summary() on the original aux data frame or geom_point() on the aux.grouped data frame.

However, when I try to plot the log10 values of the quantitative variable, stat_summary() does not plot what I would have expected. I get that the use of log10 under aes on the ggplot mapping line may at the origin of this issue. What I do not get is what is stat_summary() plotting instead and why does not it plot, if it comes to an unmatched mapping issue, the non-log10 values instead.

Thanks a lot for your help.

Best,

David

I have the below script that returns data in a list format per quote of (i). I set up an empty list, and then query with the API function get_kline_data, and pass each output into my klines_list with the .extend function

So I was revising what this guy asked: How do I "fill down"/expand observations with respect to a time variable?

I need the same thing for my dataset:

So they send him to check this:Complete column with group_by and complete (i tried to replicate the answers codes, but they didn't worked)

So my dataset looks like this (I present a simplification, in the real dataset there are more variables, and the real dimensions are 631230 obs. of 21 variables)

df