Here is my data
[![enter image description here][1]][1]
my code
library(ggplot2)
library(reshape)
dt1 =read.csv("C:/Users/My DELL/Documents/R_data/machine learning/dt1.csv")
head(dt1)
dt1$month <- seq(nrow(dt1))
library(reshape2)
mm <- melt(subset(dt1,select=c(month,EgbeNa,UrejeNa,EroNa,RefNa,EgbeMg,UrejeMg,EroMg,RefMg
)),id.var="month")
head(mm)
library(lattice)
xyplot(value ~ month|variable,data=mm,type="l",
scales=list(y=list(relation="free")),
layout=c(1,8))
dt_repr = structure(list(Date = c("01-11-17", "01-12-17", "01-01-18", "01-02-18",
"01-03-18", "01-04-18", "01-05-18", "01-06-18", "01-07-18", "01-08-18",
"01-09-18", "01-10-18", "01-11-18", "01-12-18", "01-01-19", "01-02-19",
"01-03-19", "01-04-19", "01-05-19", "01-06-19", "01-07-19", "01-08-19",
"01-09-19", "01-10-19"), month = 1:24, EgbeNa = c(27.4, 29.25,
31.1, 20.4, 13.55, 14, 16.25, 18.5, 24.95, 16.2, 30.15, 28.6,
35.1, 36.5, 28.45, 31.5, 38.1, 28, 32.55, 30.5, 33.2, 30.8, 13,
24.3), UrejeNa = c(10.45, 9, 7.55, 13.35, 11.6, 12.475, 20.1625,
27.85, 21.5, 32.05, 17.65, 15.15, 25.7, 18.8, 26.85, 20.65, 23.5,
26.45, 30.2, 25.75, 28.3, 31.45, 44.4, 39.6), EroNa = c(44.45,
40.55, 36.65, 43, 39.825, 36.825, 44.1, 51.65, 44.2, 56.1, 61.3,
66.05, 15.75, 19.15, 13.05, 12.2, 21.7, 17.9, 14.6, 33.3, 21.2,
19.6, 32.7, 25.1), RefNa = c(10.55, 9.75, 12.35, 19.65, 10.6,
13.74, 22.62, 25.82, 20.4, 31.2, 16.95, 14.25, 15.03, 17.15,
12.75, 13.5, 20.45, 16.8, 15.5, 25.4, 19.5, 19.8, 26.7, 25.1),
EgbeMg = c(4.118, 4.7155, 5.313, 4.4865, 5.1535, 5.1295,
5.113, 5.103, 5.721, 5.285, 3.8575, 4.128, 5.4205, 6.2975,
5.134, 5.4605, 5.124, 4.203, 5.2635, 5.135, 6.092, 5.575,
4.139, 4.8645), UrejeMg = c(3.6655, 3.977, 4.288, 4.192,
4.676, 4.434, 4.7005, 4.966, 5.3895, 5.7165, 4.881, 4.1015,
3.743, 6.132, 6.0785, 6.1775, 6.3135, 6.028, 5.739, 6.126,
4.5155, 4.716, 5.2165, 5.678), EroMg = c(2.472, 2.31425,
2.1565, 2.2115, 2.184, 2.135, 4.135, 6.2005, 5.457, 5.981,
5.784, 5.885, 5.406, 5.248, 4.967, 4.449, 5.058, 5.1675,
5.667, 6.966, 5.17, 4.8965, 7.201, 6.538), RefMg = c(3.75,
3.87, 4.82, 4.132, 3.98, 4.23, 4.57, 5.01, 5.02, 4.67, 4.18,
4.51, 5.21, 5.18, 4.76, 4.29, 4.95, 5.07, 5.45, 5.86, 5.11,
4.79, 6.01, 5.24)), class = "data.frame", row.names = c(NA,
-24L)) #This data is reproducible
and the output
I want to use Date as my x-axis, the Date covers 24 months. It starts at 01-11-17 and ends at 01-10-19. Anyone can help please.
It is difficult to provide answers without using your data. You need to provide your data in a usable format as #r2evans says above. However, you can convert your Date row, which appears to be a string, to Date type and use that as your X-axis. You can format how the date should be displayed by adding the format in the scales list.
For example, in your case:
...
scales=list(
y=list(relation="free"),
x = list(format = "%m-%Y") # or whatever format you need
),
...
or whatever format you need.
Here is one way how you could achieve your task:
library(tidyverse)
library(lubridate)
library(lattice)
df <- dt_repr %>%
pivot_longer(
cols = c(-Date, -month),
names_to = "names",
values_to = "values"
) %>%
mutate(Date = dmy(Date))
xyplot(values ~ Date|names,data=df,type="l",
scales=list(y=list(relation="free")),
layout=c(1,8))
I got the solution using this set of instruction:
#From Painless way to install a new version of R?
Run in the old version of R (or via RStudio)
setwd("C:/Temp/")
packages <- installed.packages()[,"Package"]
save(packages, file="Rpackages")
# INSTALL NEW R VERSION
if(!require(installr)) { install.packages("installr"); require(installr)} #load / install+load installr
# See here for more on installr: https://www.r-statistics.com/2013/03/updating-r-from-r-on-windows-using-the-installr-package/
# step by step functions:
check.for.updates.R() # tells you if there is a new version of R or not.
install.R() # download and run the latest R installer
# Install library - run in the new version of R. This calls package names and installs them from repos, thus all packages should be correct to the most recent version
setwd("C:/Temp/")
load("Rpackages")
for (p in setdiff(packages, installed.packages()[,"Package"]))
install.packages(p)
# Installr includes a package migration tool but this simply copies packages, it does not update them
copy.packages.between.libraries() # copy your packages to the newest R installation from the one version before it (if ask=T, it will ask you between which two versions to perform the copying)
Then all the error messages are gone, the missing packages tidyverse and ggplot2 came back and I have my desired plot with expected x axis
Related
I have date for 8 years. Sample of my data:
structure(list(Data = c("1/1/2015", "1/2/2015", "1/3/2015", "1/4/2015",
"1/5/2015", "1/6/2015", "1/7/2015", "1/8/2015", "1/9/2015", "1/10/2015",
"1/11/2015", "1/12/2015", "1/13/2015", "1/14/2015", "1/15/2015",
"1/16/2015", "1/17/2015", "1/18/2015", "1/19/2015", "1/20/2015",
"1/21/2015", "1/22/2015", "1/23/2015", "1/24/2015", "1/25/2015",
"1/26/2015", "1/27/2015", "1/28/2015", "1/29/2015", "1/30/2015",
"1/31/2015"), no2 = c(3.56, 11.13, 11.84, 4.88, 6.16, 12.56,
18.99, 24.74, 10.81, 12.7, 6.08, 7.34, 16.88, 16.65, 15.81, 20.78,
15.03, 11.82, 15.18, 17, 15.21, 13.86, 10.28, 8.34, 11.89, 7.22,
15.44, 10.55, 8.19, 5.04, 14.65), ws = c(10.84, 3.71, 2.08, 4.59,
6.18, 2.97, 2.13, 1.22, 1.92, 2.07, 3.09, 4.75, 2.12, 1.8, 1.9,
1.79, 1.58, 1.86, 1.58, 1.47, 1.7, 2.6, 2.67, 3.21, 1.78, 4.58,
1.79, 3.1, 3.49, 6.15, 2.59), wd = c(90, 112.5, 112.5, 270, 90,
135, 112.5, 112.5, 270, 315, 270, 112.5, 112.5, 135, 135, 112.5,
292.5, 135, 270, 135, 112.5, 112.5, 270, 112.5, 112.5, 112.5,
112.5, 112.5, 270, 270, 270)), class = "data.frame", row.names = c(NA,
-31L))
library(openair)
windRose(nitrogen,
key = list(header="Wind Rose Acri", footer="wind speed",
plot.style = c("ticks", "border"),
fit = "all", height = 1,
space = "top"))
pollutionRose(nitrogen, pollutant = "no2")
I want to show how the wind rose varies by month. The same problem (Wind rose with ggplot (R)?) but tried realised by function from Openair package.
You could convert your Data column to a name called date with date format and specify type argument with "month". type according to documenation:
type determines how the data are split i.e. conditioned, and then
plotted. The default is will produce a single plot using the entire
data. Type can be one of the built-in types as detailed in cutData
e.g. “season”, “year”, “weekday” and so on. For example, type =
"season" will produce four plots --- one for each season.
It is also possible to choose type as another variable in the data
frame. If that variable is numeric, then the data will be split into
four quantiles (if possible) and labelled accordingly. If type is an
existing character or factor variable, then those categories/levels
will be used directly. This offers great flexibility for understanding
the variation of different variables and how they depend on one
another.
Type can be up length two e.g. type = c("season", "weekday") will
produce a 2x2 plot split by season and day of the week. Note, when two
types are provided the first forms the columns and the second the
rows.
Please note you only provided one month:
library(openair)
# add month column
nitrogen$date <- as.POSIXct(nitrogen$Data, format = '%m/%d/%Y')
windRose(nitrogen,
key = list(header="Wind Rose Acri", footer="wind speed",
plot.style = c("ticks", "border"),
fit = "all", height = 1,
space = "top"),
type = 'month')
Created on 2022-12-13 with reprex v2.0.2
Here is an example with build-in data with type = 'month':
library(openair)
windRose(mydata, type = "month")
Created on 2022-12-13 with reprex v2.0.2
I would like to implement an (easy) automatic lane / band detection for thin layer chromatography in R. Below I have the rawdata and an image for a not-so-clean square wave signal that represent several bands.
The following image shows the wave and (introduced by hand) start (blue) and stop (red) of a lane.
I would like to automatically determine:
How many lanes are there? (in this example: 9)
How broad are they?
what is the distance between lanes?
also: what is the center of each lane would/could be helpful
Any strategy on how to achieve this in R would be highly welcome. A "rough" estimation of the values for the questions above would already help, as the "precise" values could later be manually adjusted. But the automatically determined values should be somewhat near the actual values, of course.
So far I tried a peak detection using the pracma-package, but this wasn't really useful as I have a square-wave-like signal, not a sharp peak... But maybe I missed something?
Here is the original raw data:
a1 <-c(305.91, 219.13, 117.2, 35.92, -4.89, -9.72, -0.34, 0.67, -15.81,
-42.09, -61.73, -62.25, -43.29, -15.69, 6.4, 14.45, 9.44, -0.57,
-6.75, -5.25, 0.96, 4.55, -1.1, -17.24, -38.05, -52.97, -52.16,
-32.31, 0.65, 34.12, 55.7, 60.34, 53.11, 45.13, 45.36, 53.58,
60.06, 52.48, 25.47, -14.03, -49.77, -65.91, -56.74, -29.88,
-0.87, 16.9, 19.89, 14.68, 11.42, 15.44, 23.25, 25.29, 13.3,
-13.08, -44.98, -68.97, -74.62, -60.26, -33.6, -7.01, 9.42, 13.02,
8.98, 5.86, 9.19, 17.13, 21.35, 12.71, -11.49, -43.9, -69.95,
-76.04, -58.01, -24.17, 9.41, 28.69, 29.92, 20.83, 14.06, 17.41,
27.93, 34.07, 24.37, -3.49, -39.75, -67.82, -74.25, -56.8, -25.3,
4.69, 21.34, 22.66, 16, 11.65, 15.07, 23.04, 25.92, 14.53, -12.82,
-47.78, -75.72, -83.84, -68.64, -38.22, -7.6, 10.16, 11.43, 3.18,
-2.93, 0.37, 10.2, 15.29, 4.32, -24.87, -61.99, -89.58, -93.53,
-71.9, -35.99, -3.03, 14.36, 14.91, 7.51, 3.53, 8.38, 18.02,
22.33, 12.73, -11.41, -41.52, -64.3, -69.08, -53.5, -24.72, 4.7,
23.57, 27.96, 22.7, 17.69, 20.8, 31.89, 42.05, 39.24, 17.06,
-19.32, -54.78, -73.28, -68.27, -47.01, -24.84, -13, -8.96, 2.88,
39.23, 102.66, 174.58, 222.62, 219)
I have a series of x and y values that I've used to build a linear model.
I can use predict() to find a value of y from a known value of x, but I'm struggling to calculate x from a known value of y. I've seen a few posts that talk about using the approx() function, but I can't figure out how to implement it for my use case. The idea is to write a function that takes a numerical value of y as an input and returns the expected value of x that it would correspond to, ideally with a prediction interval, eg "The expected value of x is 38.90, plus or minus 0.7", or something like that.
Here's my data:
> dput(x)
c(4.66, 5.53, 5.62, 5.85, 6.26, 6.91, 7.04, 7.32, 7.43, 7.85,
8.1, 8.3, 8.34, 8.53, 8.69, 8.7, 8.73, 8.76, 8.96, 9.06, 9.42,
9.78, 10.3, 10.82, 10.98, 11.07, 11.09, 11.32, 11.75, 12.1, 12.46,
12.5, 12.99, 13.02, 13.28, 13.43, 13.96, 14, 14.07, 14.29, 14.57,
14.66, 15.21, 15.56, 15.97, 16.44, 16.8, 17.95, 18.33, 18.62,
18.92, 19.49, 19.9, 19.92, 20.14, 20.18, 21.19, 22.7, 23.25,
23.48, 23.49, 23.58, 23.7, 23.83, 23.83, 23.97, 24.05, 24.14,
24.15, 24.19, 24.32, 24.62, 24.9, 24.92, 25, 25.06, 25.31, 25.36,
25.86, 25.9, 25.95, 25.99, 26.08, 26.2, 26.27, 26.39, 26.5, 26.51,
26.68, 26.78, 26.82, 26.92, 26.92, 27.05, 27.05, 27.07, 27.32,
27.6, 27.77, 27.8, 27.91, 27.96, 27.97, 28.04, 28.05, 28.15,
28.2, 28.28, 28.37, 28.51, 28.53, 28.53, 28.66, 28.68, 28.72,
28.74, 28.82, 28.83, 28.83, 28.86, 28.89, 28.91, 29.04, 29.2,
29.35, 29.4, 29.42, 29.48, 29.53, 29.65, 29.67, 29.69, 29.7,
29.72, 29.93, 29.97, 30.03, 30.08, 30.09, 30.11, 30.18, 30.62,
30.66, 30.78, 31, 31.32, 31.43, 31.47, 31.69, 31.96, 32.33, 32.5,
32.5, 32.58, 32.7, 32.92, 33.2, 33.6, 33.72, 33.77, 33.95, 34.02,
34.08, 34.42, 34.79, 34.91, 34.99, 35.08, 35.15, 35.49, 35.6,
35.6, 35.74, 35.8, 36.05, 36.17, 36.3, 36.37, 36.84, 37.31, 37.95,
38.75, 38.78, 38.81, 38.9, 39.21, 39.31, 39.5, 42.68, 43.92,
43.95, 44.64, 45.7, 45.95, 46.25, 46.8, 49.08, 50.33, 51.23,
52.76, 53.06, 62)
> dput(y)
c(11.91, 13.491, 13.708, 13.984, 14.624, 15.688, 15.823, 16.105,
16.387, 17.004, 17.239, 17.498, 17.686, 17.844, 17.997, 18.044,
18.003, 18.191, 18.332, 18.25, 18.778, 19.237, 19.693, 20.177,
20.441, 20.876, 20.512, 20.894, 21.493, 21.539, 21.951, 21.763,
22.498, 22.451, 22.744, 22.785, 23.409, 23.314, 23.408, 23.567,
23.849, 23.978, 24.472, 24.678, 25.236, 25.547, 25.676, 26.81,
26.83, 27.275, 27.331, 27.844, 28.009, 28.244, 28.497, 28.555,
29.067, 30.412, 30.788, 30.965, 31.058, 31.423, 31.346, 31.118,
31.252, 31.258, 31.399, 31.605, 31.552, 31.881, 31.822, 31.91,
32.333, 32.174, 32.222, 32.704, 32.445, 32.557, 32.993, 32.845,
32.997, 32.909, 32.911, 33.121, 33.191, 33.156, 33.426, 33.332,
33.52, 33.526, 33.697, 33.379, 33.849, 33.726, 33.538, 33.885,
33.961, 34.284, 34.208, 33.896, 34.278, 34.355, 34.276, 34.267,
34.399, 34.507, 34.492, 34.531, 34.695, 34.642, 34.872, 34.772,
34.813, 34.942, 34.883, 34.948, 34.719, 34.983, 34.99, 35.136,
35.007, 34.026, 35.148, 35.201, 35.459, 35.418, 35.236, 35.411,
35.459, 35.5, 35.665, 35.724, 35.636, 35.667, 35.747, 35.788,
35.882, 35.9, 35.83, 36.106, 36.029, 36.364, 36.358, 36.517,
37.005, 36.74, 36.963, 36.634, 37.04, 37.48, 37.581, 37.78, 37.686,
38.262, 37.998, 37.986, 38.498, 39.296, 38.467, 38.779, 38.885,
38.72, 39.038, 38.932, 39.719, 39.654, 39.367, 40.072, 39.707,
39.742, 39.919, 40.054, 40.189, 40.197, 40.154, 40.383, 42.146,
40.595, 40.971, 41.441, 41.964, 42.328, 42.463, 42.627, 42.633,
42.721, 42.786, 42.857, 45.318, 45.665, 46.406, 46.335, 47.663,
47.181, 48.074, 48.109, 49.931, 50.377, 51.053, 52.451, 53.004,
65.889)
> model <- lm(y ~ poly(x,3,raw=TRUE))
> model
Call:
lm(formula = y ~ poly(x, 3, raw = TRUE))
Coefficients:
(Intercept) poly(x, 3, raw = TRUE)1 poly(x, 3, raw = TRUE)2 poly(x, 3, raw = TRUE)3
6.6096981 1.4736619 -0.0238935 0.0002445
Since you have fitted a low order polynomial in ordinary form (raw = TRUE), you can use polyroot to directly find x given y.
## pc: polynomial coefficients in increasing order
solvePC <- function (pc, y) {
pc[1] <- pc[1] - y
## all roots, including complex ones
roots <- polyroot(pc)
## keep real roots
Re(roots)[abs(Im(roots)) / Mod(roots) < 1e-10]
}
y0 <- 38.9 ## example y-value
x0 <- solvePC(coef(model), y0)
#[1] 34.28348
plot(x, y, col = 8)
lines(x, model$fitted, lwd = 2)
abline(h = y0)
abline(v = x0)
To get an interval estimate, we can use sampling methods.
## polyfit: an ordinary polynomial regression model fitted by lm()
rootCI <- function (polyfit, y, nSamples = 1000, level = 0.05) {
## sample regression coefficients from their joint distribution
pc <- MASS::mvrnorm(nSamples, coef(polyfit), vcov(polyfit))
## for each row (a sample), call solvePC()
roots <- apply(pc, 1, solvePC, y)
## confidence interval
quantile(roots, prob = c(0.5 * level, 1 - 0.5 * level))
}
## 95% confidence interval
rootCI(model, y = y0)
# 2.5% 97.5%
#34.17981 34.38828
You can use optim:
Predict the y values given x:
pred_y <- function(x)predict(model, data.frame(x))
pred_y(x = 10)
[1] 19.20145
Now to predict x given y, we do:
pred_x <- function(y) optim(1, \(x) (y-pred_y(x))^2, method='BFGS')[[1]]
pred_x(19.20145)
[1] 10
The uniroot function is intended for this type of problem.
#coefficients for the model
coeff <- c(6.6096981, 1.4736619, -0.0238935, 0.0002445)
#define the equation which one needs the root of
modely <- function(x, y) {
# could use the predict function here
my<-coeff[1] + coeff[2]*x + coeff[3]*x**2 + coeff[4]*x**3
y-my
}
#use the uniroot functiion
#In this example y=10
uniroot(modely, lower=-100, upper=100, y=10)
$root
[1] 2.391022
$f.root
[1] -1.208443e-08
$iter
[1] 10
$init.it
[1] NA
$estim.prec
[1] 6.103516e-05
In this case for y=10, x = 2.391022
I'm able to convert the following df to tsibble object and plot using gg_season():
library(tsibble)
library(feasts)
library(tidyr)
library(dplyr)
df <- structure(list(date = structure(c(18292, 18321, 18352, 18382,
18413, 18443, 18474, 18505, 18535, 18566, 18596, 18627, 18658,
18686, 18717, 18747, 18778, 18808, 18839, 18870, 18900, 18931,
18961, 18992), class = "Date"), value1 = c(-2.94, -40.61, -6.89,
3.04, -3.5, 0.18, 6.79, 9.08, 9.35, 10.92, 20.53, 18.04, 24.6,
154.6, 30.4, 32.1, 27.7, 32.1, 19.2, 25.4, 28, 26.9, 21.7, 20.9
), value2 = c(-12.66, 7.56, -1.36, -14.39, -16.18, 3.29, -0.69,
-1.6, 13.47, 4.83, 4.56, 7.58, 28.7, 18.9, 39.1, 44, 52, 37.1,
28.2, 32.7, 17.2, 20.4, 31.4, 19.5)), class = "data.frame", row.names = c(NA,
-24L))
# Convert to tsibble object and plot using gg_season()
df %>%
pivot_longer(value1:value2) %>%
mutate(date = yearmonth(date)) %>%
mutate(year = year(date)) %>%
as_tsibble(index = date, key = name) %>%
gg_season(value) +
geom_point() # +
# scale_color_manual(values = c('2020' = 'blue', '2021' = 'red'))
Now I try to customize colors based on different years, ie., setting blue for 2020, and red for 2021. I've added scale_color_manual(values = c('2020' = 'blue', '2021' = 'red')), but I didn't succeed yet, how could I do that correctly? Thanks.
Reference:
how to change the color in geom_point or lines in ggplot
...
gg_season(value, pal = c("#3333FF", "#FF3333")) +
geom_point()
The year scale here is a continuous one (explaining why the scale_color_manual line produces "Error: Continuous value supplied to discrete scale"). But we can give gg_season a vector of color codes to use in its pal parameter.
I am writing a function which returns a plotly object. I managed to control the colors already. However I have trouble controlling the linetype. Currently I use something like:
plot_ly(colors=c(rep(c("#CD0C18","#1660A7"),each=3),'#9467bd'),linetypes = c(rep(c("dot","dash","solid"),2),"dot")) %>%
add_trace(data=long_data,x=~month,y=~temperature,color=~measure,linetype=~measure,type="scatter",mode="lines",line=list(width=4)) %>%
layout(title = "Average High and Low Temperatures in New York",
xaxis = list(title = "Months", categoryorder="array", categoryarray=month),
yaxis = list (title = "Temperature (degrees F)"))
which returns me a warning:
Warning message:
plotly.js only supports 6 different linetypes
The warning makes sense, since measure has seven levels. However I would like to control the linetype without getting a warning every time I have more than 6 traces to plot - is there a way?
My sample data:
month <- c('January', 'February', 'March', 'April', 'May', 'June', 'July',
'August', 'September', 'October', 'November', 'December')
high_2000 <- c(32.5, 37.6, 49.9, 53.0, 69.1, 75.4, 76.5, 76.6, 70.7, 60.6, 45.1, 29.3)
low_2000 <- c(13.8, 22.3, 32.5, 37.2, 49.9, 56.1, 57.7, 58.3, 51.2, 42.8, 31.6, 15.9)
mid_2000 <-apply(rbind(high_2000,low_2000),2,mean)
high_2007 <- c(36.5, 26.6, 43.6, 52.3, 71.5, 81.4, 80.5, 82.2, 76.0, 67.3, 46.1, 35.0)
low_2007 <- c(23.6, 14.0, 27.0, 36.8, 47.6, 57.7, 58.9, 61.2, 53.3, 48.5, 31.0, 23.6)
high_2014 <- c(28.8, 28.5, 37.0, 56.8, 69.7, 79.7, 78.5, 77.8, 74.1, 62.6, 45.3, 39.9)
low_2014 <- c(12.7, 14.3, 18.6, 35.5, 49.9, 58.0, 60.0, 58.6, 51.7, 45.2, 32.2, 29.1)
data <- data.frame(month, high_2000, low_2000,mid_2000, high_2007, low_2007, high_2014, low_2014)
long_data<-tidyr::gather(data,measure,temperature,-month)
As can be seen here, the warning arises in
validLinetypes <- as.character(Schema$traces$scatter$attributes$line$dash$values)
if (length(pal) > length(validLinetypes)) {
warning("plotly.js only supports 6 different linetypes", call. = FALSE)
}
So, if you want to disable this warning alone, there are only two things you can do: override the whole function or manually extend Schema$traces$scatter$attributes$line$dash$values. The latter is somewhat less intrusive and can be done with
tmp <- plotly:::Schema
tmp$traces$scatter$attributes$line$dash$values <- c(tmp$traces$scatter$attributes$line$dash$values, rep(NA, 100))
assignInNamespace("Schema", tmp, ns = "plotly")
Here we add NA 100 times so that up to 106 line types now wouldn't provoke a warning. The last line overrides the Schema variable with tmp in the plotly package environment.
The vector Schema$traces$scatter$attributes$line$dash$values only gets used (through validLinetypes) here four times, and looking at those it seems like this cheating doesn't have any likely side effects.