seasonplot function in R is not a function, character or symbol - r

transport<- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12L), .Label = c("01.01.2001", "01.02.2001", "01.03.2001",
"01.04.2001", "01.05.2001", "01.06.2001", "01.07.2001", "01.08.2001",
"01.09.2001", "01.10.2001", "01.11.2001", "01.12.2001"), class = "factor"),
Market_82 = c(7000L, 7272L, 7668L, 7869L, 8057L, 8428L, 8587L,
8823L, 8922L, 9178L, 9306L, 9439L, 3725L, 4883L, 8186L, 7525L,
6335L, 4252L, 5642L, 1326L, 8605L, 3501L, 1944L, 7332L),
transport = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
), .Label = c("plane", "train"), class = "factor")), .Names = c("date",
"Market_82", "transport"), class = "data.frame", row.names = c(NA,
-24L))
let's create seasonalplot for each group(plane and train) separately
library(forecast)
par(mfrow = c(2, 1))
lapply(split(transport['Market_82'], transport$transport), seasonplot(ts(transport,frequency=12)))
then i get error
Error in match.fun(FUN) :
'seasonplot(ts(transport, frequency = 12))' is not a function, character or symbol
How to get seasonlap plot for two groups?


lapply wants a function, without the arguments in brackets. If you want to pass additional arguments to your function, list them after the function, e.g. lapply(func, arg1, arg2).
Also, seasonplot(ts(transport,frequency=12)) would plot both, plane and train data into one plot.
Since in your example you also want to build a time series object using ts, it is best to code it in a function you define within lapply:
Try:
lapply(split(transport['Market_82'], transport$transport), function(x)seasonplot(ts(x, frequency=12)))
Edit
To distinguish which group is for which plot, you could iterate over the names:
data = split(transport['Market_82'], transport$transport)
par(mfrow = c(2, 1))
lapply(names(data), function(x)seasonplot(ts(data[[x]], frequency=12), main=x))

Related

cld() output has a wrong order of factor levels

I am using R cld() function with emmeans, but the order of factor level in the output is different from what I set. Before calling cld(), the by.years output is also in the desired order (screenshot), but when I do cld(), the output is in the alphabetical order of Light - Moderate - No(screenshot). I also checked cld.years$Grazing.intensity, the levels are correct. Is there a way to specify the order of factor levels in the cld() output? Any help is appreciated.
# sample data
plants <- structure(list(Grazing.intensity = structure(c(3L, 2L, 3L, 3L, 3L, 1L, 3L, 2L, 2L, 2L, 1L, 2L, 3L, 3L, 3L), .Label = c("Light-grazing", "Moderate-grazing", "No-grazing"), class = "factor"), Grazing.intensity1 = structure(c(3L, 2L, 3L, 3L, 3L, 1L, 3L, 2L, 2L, 2L, 1L, 2L, 3L, 3L, 3L), .Label = c("LG", "MG", "NG"), class = "factor"), Years = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 2L), .Label = c("Dry-year", "Wet-year"), class = "factor"), Month = structure(c(2L, 2L, 2L, 1L, 3L, 3L, 1L, 1L, 3L, 1L, 3L, 3L, 2L, 2L, 3L), .Label = c("Aug.", "Jul.", "Sept."), class = "factor"), Plots = c(1L, 3L, 8L, 6L, 9L, 7L, 2L, 2L, 10L, 10L, 7L, 7L, 9L, 4L, 2L), Species.richness = c(8L, 6L, 10L, 11L, 9L, 5L, 7L, 13L, 10L, 6L, 5L, 5L, 14L, 8L, 10L)), class = "data.frame", row.names = c(NA, -15L))
# set the order of factor levels
plants$Grazing.intensity <- factor(plants$Grazing.intensity, levels =
c('No-grazing','Light-grazing','Moderate-grazing'))
attach(plants)
lmer.mod <- lmer(Species.richness ~ Grazing.intensity*Years + (1|Month), data = plants)
by.years <- emmeans(lmer.mod, specs = ~ Grazing.intensity:Years, by = 'Years', type = "response")
# display cld
cld.years <- cld(by.years, Letters = letters)
This is my first time posting sample data in StackOverflow, so it may be wrong.. I used dput().

I solved the issue. The order changed because the levels are displayed in the increasing order of emmean. I set sort = FALSE, and the result was displayed in the default order. I should have read the documentations more thoroughly.

Can not use is.na() function in mutate_if funciton in r

I tried to use is.na() in mutate_if() but I get an error:
Error in is_logical(.p) : object 'n_day' not found
n_day indeed in my dataframe and I thought because of the argument set of is.na() that I can not use it in mutate_if() but I don't know how to solve it.
Here's the idea if the value in n_day is NA, replace it with the value in n_cum at the same day.
Any help will be highly appreciated!
My code like this:
library(tidyverse)
t <- structure(list(city = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("a", "b"), class = "factor"),
time = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L), .Label = c("2012/1/1", "2012/1/2",
"2012/1/3", "2012/1/4", "2012/2/1", "2012/2/2", "2012/2/3",
"2012/2/4"), class = "factor"), n_cum = c(1L, 2L, 3L, 4L,
5L, 6L, 7L, 8L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L)), class = "data.frame", row.names = c(NA,
-16L))
t
t2 <- t %>% group_by(city) %>%
mutate(n_day = n_cum - lag(n_cum))
t2 %>% mutate_if(is.na(n_day), n_day = n_cum)

mutate_if is used to do operations on multiple columns at once(See documentation), this is not what you are looking for here as you only want to change one column.
The question can be solved using mutate and if_else :
t2 %>% mutate(n_day = if_else(is.na(n_day),n_cum,n_day))

Use mutate_at + if condition instead,
t2 %>% mutate_at(vars(n_day), ~ ifelse(is.na(.), n_cum, .))
In the case of multiple variables selection, just add them respectively into vars helper.

perform acf plot for each type of group in R

Say, here the mydata (little part)
transport<- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12L), .Label = c("01.01.2001", "01.02.2001", "01.03.2001",
"01.04.2001", "01.05.2001", "01.06.2001", "01.07.2001", "01.08.2001",
"01.09.2001", "01.10.2001", "01.11.2001", "01.12.2001"), class = "factor"),
Market_82 = c(7000L, 7272L, 7668L, 7869L, 8057L, 8428L, 8587L,
8823L, 8922L, 9178L, 9306L, 9439L, 3725L, 4883L, 8186L, 7525L,
6335L, 4252L, 5642L, 1326L, 8605L, 3501L, 1944L, 7332L),
transport = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
), .Label = c("plane", "train"), class = "factor")), .Names = c("date",
"Market_82", "transport"), class = "data.frame", row.names = c(NA,
-24L))
group variable - Transport.
For each type of transport i must get acf plot of time series.
something like this
How perform acf plot for each transport?
I have a lot of groups. How to do that plots were in folder
C:/Users/admin/Documents/myplot

akrun's answer is spot on. Since you tagged the question with ggplot2 you could also use ggAcf from the forcast package.
The first step is to split your data.
transport_split <- split(transport, transport$transport)
If you want to include the respective element of column transport in the title, subtitle etc. try with Map
out <- Map(
f = function(x, y)
forecast::ggAcf(x$Market_82) + labs(title = y),
x = transport_split,
y = names(transport_split)
)
out$train

We can do this with Acf from forecast
library(forecast)
par(mfrow = c(2, 1))
lapply(split(transport['Market_82'], transport$transport), Acf)
If we also want the title, then
lst <- lapply(split(transport['Market_82'], transport$transport), acf, plot = FALSE)
par(mfrow = c(2, 1))
lapply(names(lst), function(x) plot(lst[[x]], main = x))

ggplot2 loop graph with conditional subsets

Data description:
I have a data set that is in long format with multiple different grouping variables (in data example: StandID and simID)
What I am trying to do:
I need to create simple scatter plots (x=predicted, y=observed) from this dataset for multiple columns based on a unique grouping variable.
An example of what I am trying to do using just standard plot is
obs=subset(example,simID=="OBS_OBS_OBS")
csfnw=example[example$simID== "CS_F_NW",]
plot(obs$X1HR,csfnw$X1HR)
I would need to do this for all simID and columns 9-14. (12 graphs total from data example)
What I have tried:
The problem I am running into is the y axis needs to remain the same, while cycling through the different subsets for the x axis.
I will admit up front, I have no idea what would be the best approach for this... I thought this would be easy for a split second because the data is already in long format and I would just be pointing to a subset of the data.
1) My original approach was to try and just splice up the data so that each simID had its own data frame, and compare it against the observation dataframe but I don't know how I would then pass it to ggplot.
2) My second idea was to make some kind of makeGraph function containing all the aesthetics I wanted essentially and use some kind of apply on it to pass everything through the function, but I could get neither to work.
makePlot=function(dat,x,y) {
ggplot(data=dat,aes(x=x,y=y))+geom_point(shape=Treat)+theme_bw()
}
What I could get to work was just breaking down the dataframe into the vectors of the variables I would then pass to some kind of loop/apply
sims=levels(example$simID)
sims2=sims[sims != "OBS_OBS_OBS"]
fuel_classes=colnames(example)[9:14]
Thank you
Data example:
example=structure(list(Year = structure(c(7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L), .Label = c("2001", "2002", "2003", "2004", "2005",
"2013", "2014", "2015"), class = "factor"), StandID = structure(c(10L,
2L, 6L, 22L, 14L, 18L, 34L, 26L, 30L, 10L, 2L, 6L, 22L, 14L,
18L, 34L, 26L, 30L, 10L, 2L, 6L, 22L, 14L, 18L, 34L, 26L, 30L
), .Label = c("1NB", "1NC", "1NT", "1NTB", "1RB", "1RC", "1RT",
"1RTB", "1SB", "1SC", "1ST", "1STB", "2NB", "2NC", "2NT", "2NTB",
"2RB", "2RC", "2RT", "2RTB", "2SB", "2SC", "2ST", "2STB", "3NB",
"3NC", "3NT", "3NTB", "3RB", "3RC", "3RT", "3RTB", "3SB", "3SC",
"3ST", "3STB"), class = "factor"), Block = structure(c(1L, 1L,
1L, 2L, 2L, 2L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L,
1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("1", "2", "3"
), class = "factor"), Aspect = structure(c(3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L), .Label = c("N", "R", "S"), class = "factor"),
Treat = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L), .Label = c("B", "C", "T", "TB"), class = "factor"),
Variant = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = c("CS", "OBS", "SN"), class = "factor"),
Fuels = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L), .Label = c("F", "NF", "OBS"), class = "factor"),
Weather = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = c("NW", "OBS", "W"), class = "factor"),
X1HR = c(0.321666667, 0.177777778, 0.216111111, 0.280555556,
0.255555556, 0.251666667, 0.296666667, 0.231111111, 0.22,
0.27556628, 0.298042506, 0.440185249, 0.36150676, 0.398630172,
0.367523015, 0.345717251, 0.349305987, 0.412227929, 0.242860824,
0.258737177, 0.394024998, 0.287317872, 0.321927488, 0.281322986,
0.313588411, 0.303123146, 0.383658946), X10HR = c(0.440555556,
0.32, 0.266666667, 0.292222222, 0.496666667, 0.334444444,
0.564444444, 0.424444444, 0.432777778, 0.775042951, 0.832148314,
1.08174026, 1.023838878, 0.976997674, 0.844206274, 0.929837704,
1.0527215, 1.089246511, 0.88642776, 0.920596302, 1.209707737,
1.083737493, 1.077612877, 0.92481339, 1.041637182, 1.149550319,
1.229776621), X100HR = c(0.953888889, 1.379444444, 0.881666667,
1.640555556, 2.321666667, 1.122222222, 1.907777778, 1.633888889,
1.208333333, 1.832724094, 2.149356842, 2.364475727, 2.493232965,
2.262988567, 1.903909683, 2.135747433, 2.256677628, 2.288722038,
1.997704744, 2.087135553, 2.524872541, 2.34671092, 2.338253498,
2.06796217, 2.176314831, 2.580271006, 2.857197046), X1000HR = c(4.766666667,
8.342222222, 3.803333333, 8.057777778, 10.11444444, 6.931111111,
6.980555556, 13.20611111, 1.853333333, 3.389177084, 4.915714741,
2.795267582, 2.48227787, 2.218413353, 1.64684248, 2.716156483,
2.913746119, 2.238629341, 3.449863434, 3.432626724, 3.617531776,
3.641639471, 3.453454971, 3.176793337, 3.459602833, 3.871166945,
2.683447838), LITTER = c(2.4, 2.219444444, 2.772222222, 2.596666667,
2.693888889, 2.226111111, 2.552222222, 3.109444444, 2.963333333,
2.882233381, 3.025934696, 3.174396992, 3.291081667, 2.897673607,
2.737119675, 2.987895727, 3.679605484, 2.769756079, 2.882241249,
3.02594161, 3.174404144, 3.291091681, 2.897681713, 2.737129688,
2.987901449, 3.679611444, 2.769766569), DUFF = c(1.483333333,
1.723888889, 0.901666667, 1.520555556, 1.49, 1.366111111,
0.551666667, 1.056111111, 0.786111111, 2.034614563, 2.349547148,
1.685223818, 2.301301956, 2.609308243, 2.21895647, 2.043699026,
2.142618418, 0.953421116, 4.968493462, 4.990526676, 5.012362003,
5.023665905, 4.974074364, 4.947199821, 4.976779461, 5.082509995,
3.55211544), simID = structure(c(5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L), .Label = c("CS_F_NW", "CS_F_W",
"CS_NF_NW", "CS_NF_W", "OBS_OBS_OBS", "SN_F_NW", "SN_F_W",
"SN_NF_NW", "SN_NF_W"), class = "factor")), .Names = c("Year",
"StandID", "Block", "Aspect", "Treat", "Variant", "Fuels", "Weather",
"X1HR", "X10HR", "X100HR", "X1000HR", "LITTER", "DUFF", "simID"
), row.names = c(37L, 38L, 39L, 40L, 41L, 42L, 43L, 44L, 45L,
82L, 83L, 84L, 85L, 86L, 87L, 88L, 89L, 90L, 127L, 128L, 129L,
130L, 131L, 132L, 133L, 134L, 135L), class = "data.frame")

You were actually on the right track. If all plots are the same, just make one function and then use loops to loop over the subsets. For your example this can be done like this:
library(ggplot2)
# the plot function
plotFun = function(dat, title) {
ggplot(data=dat) +
geom_point(aes(x = x, y = y), shape=18) +
ggtitle(title) +
theme_bw()
}
# columns of interest
colIdx = 9:14
# split on all values of simID
dfList = split(example, example$simID)
# simID has never appearing factors. These are removed
dfList = dfList[lapply(dfList, nrow) != 0]
# make empty array for saving plots
plotList = array(list(), dim = c(length(dfList), length(dfList), length(colIdx)),
dimnames = list(names(dfList), names(dfList), names(example)[colIdx]))
# the first two loops loop over all unique combinations of dfList
for (i in 2:length(dfList)) {
for (j in 1:(i-1)) {
# loop over target variables
for (k in seq_along(colIdx)) {
# store variables to plot in a temporary dataframe
tempDf = data.frame(x = dfList[[i]][, colIdx[k]],
y = dfList[[j]][, colIdx[k]])
# add a title so we can see in the plot what is plotted vs what
title = paste0(names(dfList)[i], ":", names(dfList[[i]])[colIdx[k]], " VS ",
names(dfList)[j], ":", names(dfList[[j]])[colIdx[k]])
# make and save plot
plotList[[i, j, k]] = plotFun(tempDf, title)
}
}
}
# call the plots like this
plotList[[2, 1, 4]]
# Note that we only filled the lower triangle of combinations
# therefore indexing with [[1, 1, 1]] just returns NULL
plotList[, , 1]
This process can probably be more optimized, but when creating graphs I would go for clarity above speed since speed usually isn't an issue.

How to display value labels in R outputs?

I am using the Survey Package in R to do some initial descriptive stats for my dissertation. With the outputs for both the svymean and the barplot, I would like the value labels to be displayed for the variable-it would make the descriptive statistics much easier to interpret. Instead of the output labels of: F1RTRCC1, F1RTRCC2-I would like to see the value labels of "academic" and "occupational" to be displayed.
How do I go about making this happen?
I am including a minimal reproducible example with a small subset of my actual data:
#Calling Survey Package
library (survey)
#Using dput to display a subset of my actual data
MRE1 <- structure(list(ï..STU_ID = c(101101L, 101102L, 101104L, 101105L, 101106L, 101107L),
PSU = c(1L, 1L, 1L, 1L, 1L, 1L), STRAT_ID = c(101L,
101L, 101L, 101L, 101L, 101L), BYSCTRL = c(1L, 1L, 1L, 1L, 1L,
1L), G10COHRT = c(1L, 1L, 1L, 1L, 1L, 1L), F1RTRCC = c(2L, 1L,
4L, 2L, 2L, 4L), F1SEX = c(2L, 2L, 2L, 2L, 2L, 1L), F1RACE = c(5L,
2L, 7L, 3L, 4L, 4L), F1PARED = c(5L, 5L, 2L, 2L, 1L, 2L), F1SES2QU = c(2L,
4L, 1L, 1L, 1L, 1L), F1HIMATH = c(5L, 6L, 6L, 4L, 5L, 4L), F1RGPP2 = c(2L,
4L, 4L, 4L, 4L, 1L), F3ATTAINMENT = c(3L, 10L, 6L, 4L, 4L, 3L
), F3EDSTAT = c(5L, 5L, 5L, 2L, 2L, 5L)), .Names = c("ï..STU_ID",
"PSU", "STRAT_ID", "BYSCTRL", "G10COHRT", "F1RTRCC", "F1SEX",
"F1RACE", "F1PARED", "F1SES2QU", "F1HIMATH", "F1RGPP2", "F3ATTAINMENT",
"F3EDSTAT"), row.names = c(NA, 6L), class = "data.frame")
#Svymean of Variable F1RTRCC
#There is an error coming here: Error in UseMethod("svymean", design) :
#no applicable method for 'svymean' applied to an object of class "data.frame"-I
#think this is likely related to my dput subset of my data, as when I am using
#this function with my full dataset, this error does not display.
CC <- svymean(~F1RTRCC, MRE1, na.rm=T)
CC
#Barplot for F1RTRCC
barplot(CC)
With the outputs for both the svymean and the barplot, I would like the value labels to be displayed.
Instead of F1RTRCC1, F1RTRCC2-I would like to see the value labels of "academic" and "occupational" to be displayed.
How do I go about making this happen?

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