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View pdz
Scribble:
B7Z2Y1_HUMAN
ARHG7_HUMAN
ARHG8_HUMAN
KCNA5_HUMAN
VWCE_HUMAN
DNML1_HUMAN
PKP4_HUMAN
NXPE2_HUMAN
View inplace_comparison.R
library(Rcpp)
`%count<%` <- cppFunction('
size_t count_less(NumericVector x, NumericVector y) {
const size_t nx = x.size();
const size_t ny = y.size();
if (nx > 1 & ny > 1) stop("Only one parameter can be a vector!");
View gist:b1b6ad8d37c63b1e929d
> library(purrr)
> ll <- list(list(a=1:3, b=4), list(a=5:7, b=8))
> ll %>% map(lift_dl(c)) %>% map_call(rbind)
a1 a2 a3 b
[1,] 1 2 3 4
[2,] 5 6 7 8
> Reduce(rbind, ll)
a b
init Integer,3 4
Integer,3 8
View initial_param_vs_runtime.R
library(purrr)
runtimes <- get_elapsed_time(fit)[,2]
inits <- get_inits(fit)
## traditional conversion to a matrix
# m.init <- do.call(rbind, lapply(inits, function(l) do.call(c, l)))
## using purrr
View fitting_with_uncertainty.Rmd
---
title: "Fitting with uncertainty"
author: "Michael Kuhn"
date: "24 Aug 2015"
output: html_document
---
In this toy example, we assume that we've independently measured values $x$ and $y$ and want to find a linear relationship between them, accounting for measurement uncertainty. Each $x$ and $y$ value is assigned a different uncertainty, and the challenge is to take this information into account. A standard linear model will treat all points equally.
When we treat each measurement as a multivariate normal distribution, we can find a point along the proposed fitted line that is maximizing the probability density function (i.e. that has a maximum likelihood). Thus, given a slope and intercept, we virtually move all measurements to their most likely point along the line, and use the likelihood at this point.
View code.R
library(dplyr)
a <- data.frame(foo = 1:10, bar = "bar")
b <- tbl_df(a)
a[,1]
b[,1]
paste0(a[,1], "!")
paste0(b[,1], "!")
View gist:9ad4da2e039e5c6741e2
library(ggplot2)
library(gtable)
# create example data
set.seed(42)
dataset_names <- c("Human", "Mouse", "Fly", "Worm")
datasets <- data.frame(name = factor(dataset_names, levels=dataset_names), parity = factor(c(0, 0, 1, 0)), v50 = runif(4, max=0.5), y=1:4)
data <- data.frame( dataset1 = rep(datasets$name, 4), dataset2 = rep(datasets$name, each = 4), z = runif(16,min = 0, max = 0.5) )
pal <- c("#dddddd", "#aaaaaa")
View InChi.js
// International Chemical Identifier Regex, by lo sauer - lsauer.com
// Morphine InchI:
var x="InChI=1S/C17H19NO3/c1-18-7-6-17-10-3-5-13(20)16(17)21-15-12(19)4-2-9(14(15)17)8-11(10)18/h2-5,10-11,13,16,19-20H,6-8H2,1H3/t10-,11+,13-,16-,17-/m0/s1"
// applying an organic character-subset
// we could check for the length property, but in case of 0 matches 'null' is returned -> hence !!.. \ generally equal to Boolean(..)
!!x.trim().match(/^((InChI=)?[^J][0-9BCOHNSOPrIFla+\-\(\)\\\/,pqbtmsih]{6,})$/ig)
>true
//generic:
x.trim().match(/^((InChI=)?[^J][0-9a-z+\-\(\)\\\/,]+)$/ig)
View plot with densities.R
library(ggplot2)
library(gridExtra)
mtcars$cyl <- ordered(mtcars$cyl)
p <- ggplot(mtcars, aes(mpg, hp, colour = cyl)) + geom_point()
p1 <- p + theme(legend.position = "none")
p2 <- ggplot(mtcars, aes(x=mpg, group=cyl, colour=cyl))
p2 <- p2 + stat_density(fill = NA, position="dodge")
View ggplot2_order.R
library(ggplot2)
df <- data.frame( n=c("a","a","b","b","c","c"), x = rep(c(1,2), 3), y = rep(c(1), 6), l = as.factor(c(1,1,0,0,0,0)))
# Contents of df:
# n x y l
# 1 a 1 1 1
# 2 a 2 1 1
# 3 b 1 1 0
# 4 b 2 1 0