I hereby claim:
- I am bestdan on github.
- I am dpegan271 (https://keybase.io/dpegan271) on keybase.
- I have a public key ASBCTT8RIBBl4GrD_mXa50Qkb8gxy2XmNULJld4SBNfB2wo
To claim this, I am signing this object:
| #! /usr/bin/env bash | |
| # Create a git commit message using claude | |
| gcaim() { | |
| local staged_diff | |
| staged_diff=$(git diff --cached 2>/dev/null) | |
| if [[ -n "$staged_diff" ]]; then | |
| local prompt | |
| read -r -d '' prompt <<'EOF' | |
| Generate a commit message from the git diff provided via stdin. |
| #!/usr/bin/env sh | |
| # # These functions make it easy to look at the output of a dart | |
| # # test with large non-matching objects. | |
| # | |
| # # To use: | |
| # # One and done: | |
| # auditDartTest test/example_test.dart 'this test fails' | |
| # |
| #' @title S&P 500 heatmap | |
| #' @author Daniel Egan | |
| #' @description Creates real and nominal triangle heatmaps based on the S&P 500. | |
| #' @details Last update 2012-06-01 | |
| ### ---- | |
| library(quantmod) | |
| library(RColorBrewer) | |
| #Some useful matrix functions |
| #' @title Log-means versus Medians | |
| #' @author Daniel Egan | |
| #' @description When data has a power law or extremely skewed distribution, | |
| #' using a log-mean usually results in more stable and useful central estimates | |
| #' compared to a mean or a median. | |
| #' https://towardsdatascience.com/on-average-youre-using-the-wrong-average-geometric-harmonic-means-in-data-analysis-2a703e21ea0 | |
| library(dplyr) | |
| library(tidyr) | |
| library(ggplot2) |
| library(ggplot2) | |
| library(tidyr) | |
| library(dplyr) | |
| # Normalize | |
| data(mtcars) | |
| mtcars_norm <- as.data.frame(apply(mtcars, 2, scale)) | |
| # Make long |
| # Version 1: LOC incentive --> inflated code | |
| add_up = function(inputs){ | |
| total = inputs[1] | |
| for i in 2:length(inputs){ | |
| total = total + inputs[i] | |
| } | |
| return(total) | |
| } |
| def cs_service_bot(): | |
| welcome_message = """Hello! Welcome to the DNS Cable Company's Service Portal. Are you a new or existing customer? | |
| \n[1] New Customer | |
| \n[2] Existing Customer | |
| \n""" | |
| response = input(welcome_message) | |
| if response == "1": | |
| new_customer() |
I hereby claim:
To claim this, I am signing this object:
| rm(list=ls()) | |
| library(quantmod) | |
| library(lubridate) | |
| library(PerformanceAnalytics) | |
| #' Required 'drawdown' size | |
| #' Default: -6% return | |
| drawdownHurdle<- -0.06 | |
| # Grab data ----------------------------------------------------------------------------------- |
| effFrontier = function (data, nports = 20, shorts=T,wmin=0, wmax=1) | |
| { | |
| rcov<-cov(data) | |
| averet<-colMeans(data) | |
| mxret = max((colMeans(data))) | |
| mnret = 0#-mxret Long only... | |
| n.assets = ncol(data) | |
| reshigh = rep(wmax,n.assets) | |
| if( shorts ) | |
| { |