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I am attempting to include a dummy regressor that notes the beginning of the pandemic and runs a regression with ARIMA errors.
My dataset revolves around breaking & entering's happening in Toronto from 2014 to 2021. The issue is that the trend takes a turn due to covid-19 around 2020.
Auto.arima provides me with a ARIMA(1,0,1) model as it is not taking into account the impact of covid-19 and is performing according to the implied return to the series average.
When trying to include a dummy regressor that notes the beginning of the pandemic and run a regression with ARIMA errors I get the following error:
In ifelse(time(BEDATA_GROUPEDtsssarima) >= yearmonth("2020-03"), :
Incompatible methods ("Ops.ts", ">=.vctrs_vctr") for ">="
Code:
# Create a binary time series that indicates the start of the pandemic
library(fpp3)
library(forecast)
library(zoo)
# Check if timeseries
class(BEDATA_GROUPED)
#Convert timeseries
BEDATA_GROUPEDtsssarima <- ts(BEDATA_GROUPED[,2], frequency = 12, start = c(2014, 1))
class(BEDATA_GROUPEDtsssarima)
#Plot
forecast::autoplot(BEDATA_GROUPEDtsssarima)
# Assume that the pandemic began in March 2020
pandemic_dummy <- ifelse(time(BEDATA_GROUPEDtsssarima) >= yearmonth("2020-03"), 1, 0)
# Use auto.arima() to fit an ARIMA model with the dummy variable as an exogenous variable
beddatamodel <- auto.arima(BEDATA_GROUPEDtsssarima, xreg = pandemic_dummy, ic="aic", trace = TRUE)
# Create a binary time series that indicates the start of the pandemic
# In this example, we will assume that the pandemic began in March 2020
pandemic_dummy <- ifelse(time(BEDATA_GROUPEDtsssarima) >= yearmonth("2020-03"), 1, 0)
# Use auto.arima() to fit an ARIMA model with the dummy variable as an exogenous variable
beddatamodel <- auto.arima(BEDATA_GROUPEDtsssarima, xreg = pandemic_dummy, ic="aic", trace = TRUE)
# Create a binary time series for the forecast period that includes the pandemic dummy variable
forecast_period <- time(BEDATA_GROUPEDtsssarima)["2022/01/01/":"2023/12/31/"]
pandemic_dummy_forecast <- ifelse(forecast_period >= yearmonth("2020-03"), 1, 0)
# Use the forecast()
forecast(pandemic_dummy_forecast)
Dataset:
structure(list(occurrence_yrmn = c("2014-January", "2014-February",
"2014-March", "2014-April", "2014-May", "2014-June", "2014-July",
"2014-August", "2014-September", "2014-October", "2014-November",
"2014-December", "2015-January", "2015-February", "2015-March",
"2015-April", "2015-May", "2015-June", "2015-July", "2015-August",
"2015-September", "2015-October", "2015-November", "2015-December",
"2016-January", "2016-February", "2016-March", "2016-April",
"2016-May", "2016-June", "2016-July", "2016-August", "2016-September",
"2016-October", "2016-November", "2016-December", "2017-January",
"2017-February", "2017-March", "2017-April", "2017-May", "2017-June",
"2017-July", "2017-August", "2017-September", "2017-October",
"2017-November", "2017-December", "2018-January", "2018-February",
"2018-March", "2018-April", "2018-May", "2018-June", "2018-July",
"2018-August", "2018-September", "2018-October", "2018-November",
"2018-December", "2019-January", "2019-February", "2019-March",
"2019-April", "2019-May", "2019-June", "2019-July", "2019-August",
"2019-September", "2019-October", "2019-November", "2019-December",
"2020-January", "2020-February", "2020-March", "2020-April",
"2020-May", "2020-June", "2020-July", "2020-August", "2020-September",
"2020-October", "2020-November", "2020-December", "2021-January",
"2021-February", "2021-March", "2021-April", "2021-May", "2021-June",
"2021-July", "2021-August", "2021-September", "2021-October",
"2021-November", "2021-December"), MCI = c(586, 482, 567, 626,
625, 610, 576, 634, 636, 663, 657, 556, 513, 415, 510, 542, 549,
618, 623, 666, 641, 632, 593, 617, 541, 523, 504, 536, 498, 552,
522, 519, 496, 541, 602, 570, 571, 492, 560, 525, 507, 523, 593,
623, 578, 657, 683, 588, 664, 582, 619, 512, 630, 644, 563, 654,
635, 732, 639, 748, 719, 567, 607, 746, 739, 686, 805, 762, 696,
777, 755, 675, 704, 617, 732, 609, 464, 487, 565, 609, 513, 533,
505, 578, 526, 418, 428, 421, 502, 452, 509, 492, 478, 469, 457,
457)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-96L))
I see you have used the fpp3 library, so I've had a go using the tidyverts tools. I've had a go at three models: a plain ARIMA, a plain regression using the pandemic dummy variable, and a dynamic model using both ARIMA and the dummy variable.
Hope this helps! :-)
library(tsibble)
library(fable)
library(fabletools)
library(feasts)
library(dplyr)
Create a tsibble:
BEDATA_GROUPED <- BEDATA_GROUPED |>
mutate(Month = yearmonth(occurrence_yrmn)) |>
as_tsibble(index = Month)
autoplot(BEDATA_GROUPED)
Assume that the pandemic began in March 2020
and create a dummy variable:
pandemic_start <- yearmonth("2020-03-01")
BEDATA_GROUPED <- BEDATA_GROUPED |>
mutate(pandemic_dummy = ifelse(Month >= pandemic_start, 1, 0))
Work up a plain ARIMA:
BEDATA_GROUPED_arima <- BEDATA_GROUPED |>
model(ARIMA(MCI, stepwise = FALSE))
BEDATA_GROUPED_arima |>
gg_tsresiduals()
BEDATA_GROUPED_arima |>
forecast(h = 5) |>
autoplot()
Work up a plain regression:
BEDATA_GROUPED_TSLM <- BEDATA_GROUPED |>
model(TSLM(MCI ~ pandemic_dummy)) |>
report()
BEDATA_GROUPED_TSLM |>
gg_tsresiduals()
Make a data set to predict on:
new_data <- structure(list(Month = structure(c(18993, 19024, 19052, 19083,
19113), class = c("yearmonth", "vctrs_vctr")), pandemic_dummy = c(1,
1, 1, 1, 1)), class = c("tbl_ts", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -5L), key = structure(list(.rows = structure(list(
1:5), ptype = integer(0), class = c("vctrs_list_of", "vctrs_vctr",
"list"))), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-1L)), index = structure("Month", ordered = TRUE), index2 = "Month", interval = structure(list(
year = 0, quarter = 0, month = 1, week = 0, day = 0, hour = 0,
minute = 0, second = 0, millisecond = 0, microsecond = 0,
nanosecond = 0, unit = 0), .regular = TRUE, class = c("interval",
"vctrs_rcrd", "vctrs_vctr")))
Forecast plain regression:
BEDATA_GROUPED_TSLM |>
forecast(new_data = new_data) |>
autoplot()
Work up a dynamic regression, with ARIMA and the pandemic dummy variablee:
BEDATA_GROUPED_dyn_ARIMA <- BEDATA_GROUPED |>
model(ARIMA(MCI ~ pandemic_dummy)) |>
report()
BEDATA_GROUPED_dyn_ARIMA |>
gg_tsresiduals()
BEDATA_GROUPED_dyn_ARIMA |>
forecast(new_data = new_data) |>
autoplot()
These are my sets of four mean values:
meanf1hindi = c(253, 297, 377, 426, 476, 518, 560, 620, 657, 697)
meanf2hindi = c(850, 887, 1017, 1080, 1197, 1342, 1694, 1820, 2265)
meanf1tamil = c(260, 304, 390, 435, 483, 527, 563, 628, 670, 704)
meanf2tamil = c(891, 826, 1018, 1068, 1188, 1355, 1709, 1834, 1976, 2303)
I would like to make a linear graph of meanf1hindi and meanf2hindi together, and do the same with meanf1tamil and meanf2tamil.
This is what I did so far, and don't know how to proceed further:
plot(meanf1hindi, meanf2hindi)
Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ
You get the error because the length differs for your vectors. What you can do is make the two vectors' length the same by removing one value for the longer vector in this case remove one value of meanf1hindi by doing this:
> length(meanf1hindi)
[1] 10
> length(meanf2hindi)
[1] 9
plot(meanf1hindi[-1], meanf2hindi)
Output:
I'm trying to cut credit scores to ranges in order to assign their rating.
I tried this:
x <- sample(510:780, 50, replace=TRUE)
cut(x, breaks=c(300, 579, 669, 739, 799), include.lowest=TRUE)
result:
[300,579] (579,669] (669,739] (739,799]
but I want something like:
(300, 579] (580, 669] (670, 739] (740,799] (739,799]
something like the ranges here: https://www.experian.com/blogs/ask-experian/credit-education/score-basics/what-is-a-good-credit-score/
Maybe you can try the code below:
bks <- c(300, 579, 669, 739, 799)
bks <- sort(c(bks, bks[-c(1,length(bks))]+1))
r <- cut(x, breaks= bks, include.lowest=F)
f <- levels(r)[-seq(2,length(levels(r)),by = 2)]
r <- factor(r[r %in% f])
Such that you get:
> levels(r)
[1] "(300,579]" "(580,669]" "(670,739]" "(740,799]"
I don't think you need what you are showing here. Based on which group the data at the intervals should be part of you can use either findInterval or cut like :
findInterval(579, c(300, 579, 669, 739, 799), left.open = TRUE)
#[1] 1
findInterval(579, c(300, 579, 669, 739, 799))
#[1] 2
With cut
cut(579, c(300, 579, 669, 739, 799))
#[1] (300,579]
#Levels: (300,579] (579,669] (669,739] (739,799]
cut(579, c(300, 579, 669, 739, 799), right = FALSE)
#[1] [579,669)
#Levels: [300,579) [579,669) [669,739) [739,799)
Chemist here (so not very good with statistical analysis) and novice in R:
I have various sets of data where the yield of a reaction is monitored with time such as:
The data:
df <- structure(list(time = c(15, 30, 45, 60, 75, 90, 105, 120, 135,
150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300, 315, 330,
345, 360, 375, 390, 405, 420, 435, 450, 465, 480, 495, 510, 525,
540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720,
735, 750, 765, 780, 795, 810, 825, 840, 855, 870, 885, 900, 915,
930, 945, 960, 975, 990, 1005, 1020, 1035, 1050, 1065, 1080,
1095, 1110, 1125, 1140, 1155, 1170, 1185, 1200, 1215, 1230, 1245,
1260, 1275, 1290, 1305, 1320, 1335, 1350, 1365, 1380, 1395, 1410,
1425, 1440, 1455, 1470, 1485, 1500, 1515, 1530, 1545, 1560, 1575,
1590, 1605, 1620, 1635, 1650, 1665, 1680, 1695, 1710, 1725, 1740,
1755, 1770, 1785, 1800, 1815, 1830, 1845, 1860, 1875, 1890, 1905,
1920, 1935, 1950, 1965, 1980, 1995, 2010, 2025, 2040, 2055, 2070,
2085, 2100, 2115, 2130), yield = c(9.3411, 9.32582, 10.5475,
13.5358, 17.3376, 16.7444, 20.7234, 19.8374, 24.327, 27.4162,
27.38, 31.3926, 29.3289, 32.2556, 33.0025, 35.3358, 35.8986,
40.1859, 40.3886, 42.2828, 41.23, 43.8108, 43.9391, 43.9543,
48.0524, 47.8295, 48.674, 48.2456, 50.2641, 50.7147, 49.6828,
52.8877, 51.7906, 57.2553, 53.6175, 57.0186, 57.6598, 56.4049,
57.1446, 58.5464, 60.7213, 61.0584, 57.7481, 59.9151, 64.475,
61.2322, 63.5167, 64.6289, 64.4245, 62.0048, 65.5821, 65.8275,
65.7584, 68.0523, 65.4874, 68.401, 68.1503, 67.8713, 69.5478,
69.9774, 73.4199, 66.7266, 70.4732, 67.5119, 69.6107, 70.4911,
72.7592, 69.3821, 72.049, 70.2548, 71.6336, 70.6215, 70.8611,
72.0337, 72.2842, 76.0792, 75.2526, 72.7016, 73.6547, 75.6202,
76.5013, 74.2459, 76.033, 78.4803, 76.3058, 73.837, 74.795, 76.2126,
75.1816, 75.3594, 79.9158, 77.8157, 77.8152, 75.3712, 78.3249,
79.1198, 77.6184, 78.1244, 78.1741, 77.9305, 79.7576, 78.0261,
79.8136, 75.5314, 80.2177, 79.786, 81.078, 78.4183, 80.8013,
79.3855, 81.5268, 78.416, 78.9021, 79.9394, 80.8221, 81.241,
80.6111, 79.7504, 81.6001, 80.7021, 81.1008, 82.843, 82.2716,
83.024, 81.0381, 80.0248, 85.1418, 83.1229, 83.3334, 83.2149,
84.836, 79.5156, 81.909, 81.1477, 85.1715, 83.7502, 83.8336,
83.7595, 86.0062, 84.9572, 86.6709, 84.4124)), .Names = c("time",
"yield"), row.names = c(NA, -142L), class = "data.frame")
What i want to do to the data:
I need to smooth the data in order to plot the 1st derivative. In the paper the author mentioned that one can fit a high order polynomial and use that to do the processing which i think is wrong since we dont really know the true relationship between time and yield for the data and is definitely not polyonymic. I tried regardless and the plot of the derivative did not make any chemical sense as expected. Next i looked into loess using: loes<-loess(Yield~Time,data=df,span=0.9) which gave a much better fit. However, the best results so far was using :
spl <- smooth.spline(df$Time, y=df$Yield,cv=TRUE)
colnames(predspl)<-c('Time','Yield')
pred.der<-as.data.frame(predict(spl, deriv=1))
colnames(pred.der)<-c('Time', 'Yield')
which gave the best fit especially in the initial data points (by visual inspection).
The problem i have:
The issue however is that the derivative looks really good only up to t=500s and then it starts wiggling more and more towards the end. This shouldnt happen from a chemistry point of view and it is just a result of overfitting towards the end of the data due to the increase of the noise. I know this since for some experiments that i have performed 3 times and averaged the data (so the noise decreased) the wiggling is much smaller in the plot of the derivative.
What i have tried so far:
I tried different values of spar which although it smoothens correctly the later data it causes a poor fit in the initial data (which are the most important). I also tried to reduce the number of knots but i got a similar result with the one from changing the spar value. What i think i need is to have a larger amount of knots in the begining which will smoothly decrease to a small number of knots towards the end to avoid that overfitting.
The question:
Is my reasoning correct here? Does anyone know how can i have the above effect in order to get a smooth derivative without any wiggling? Do i need to try a different fit other than the spline maybe? I have attached a pic in the end where you can see the derivative from the smooth.spline vs time and a black line (drawn by hand) of what it should look like. Thank you for your help in advance.
I think you're on the right track on having more closely spaced knots for the spline at the start of the curve. You can specify knot locations for smooth.spline using all.knots (at least on R >= 3.4.3; I skimmed the release notes for R, but couldn't pinpoint the version where this became available).
Below is an example, and the resulting, smoother fit for the derivative after some manual work of trying out different knot positions:
with(df, {
kn <- c(0, c(50, 100, 200, 350, 500, 1500) / max(time), 1)
s <- smooth.spline(time, yield, cv = T)
s2 <- smooth.spline(time, yield, all.knots = kn)
ds <- predict(s, d = 1)
ds2 <- predict(s2, d = 1)
np <- list(mfrow = c(2, 1), mar = c(4, 4, 1, 2))
withr::with_par(np, {
plot(time, yield)
lines(s)
lines(s2, lty = 2, col = 'red')
plot(ds, type = 'l', ylim = c(0, 0.15))
lines(ds2, lty = 2, col = 'red')
})
})
You can probably fine tune the locations further, but I wouldn't be too concerned about it. The primary fits are already near enough indistinguishable, and I'd say you're asking quite a lot from these data in terms of identifying details about the derivative (this should be evident if you plot(time[-1], diff(yield) / diff(time)) which gives you an impression about the level of information your data carry about the derivative).
Created on 2018-02-15 by the reprex package (v0.2.0).
I've got a strange problem with plotting the fitted values of a glm.
My code is:
Data <- data.frame("Sp" = c(111.4, 185, 231, 272.5, 309, 342, 371, 399,
424, 447, 469, 489, 508, 527, 543, 560, 575, 589, 603, 616, 630, 642, 653,
664, 675, 685, 695, 705, 714, 725, 731, 740), "nrC" = 1:32)
modell <- glm(Sp ~ nrC, data = Data, family = Gamma)
pred <- predict(modell, newdata = data.frame("nrC" = 1:32), type = "response")
plot(Data$nrC, Data$Sp, xlim = c(0, 40), ylim = c(50, 1000))
lines(Data$nrC, pred, col = "blue")
The blue line representing the fitted values seems to be ok, apart from being horizontally mirrored.
I'm relatively new to this, so maybe I'm missing something obvious here, but I can't figure out what's wrong.
Doing the same with the data presented here works perfectly fine.
I'be grateful for any hints!
The gamma distribution isn't quite right for this data set. The data shown in the plot as you have it formulated shows a square root-ish looking function. Try specifying the model like this:
modell <- glm(Sp ~ sqrt(nrC), data = Data, family = gaussian)
pred <- predict(modell, newdata = data.frame("nrC" = 1:32), type = "response")
plot(Data$nrC, Data$Sp, xlim = c(0, 40), ylim = c(50, 1000))
lines(Data$nrC, pred, col = "blue")