Related
Using a dataframe with missing values:
structure(list(id = c("id1", "test", "rew", "ewt"), total_frq_1 = c(54, 87, 10, 36), total_frq_2 = c(45, 24, 202, 43), total_frq_3 = c(24, NA, 25, 8), total_frq_4 = c(36, NA, 104, NA)), row.names = c(NA, 4L), class = "data.frame")
How is is possible to create a bar plot with the mean for every column, excluding the id column, but without filling the missing values with 0 but leaving out the row with missing values example for total_frq_3 24+25+8 = 57/3 = 19
You can use colMeans function and pass it the appropriate argument to ignore NA.
library(ggplot2)
xy <- structure(list(id = c("id1", "test", "rew", "ewt"),
total_frq_1 = c(54, 87, 10, 36), total_frq_2 = c(45, 24, 202, 43), total_frq_3 = c(24, NA, 25, 8),
total_frq_4 = c(36, NA, 104, NA)),
row.names = c(NA, 4L),
class = "data.frame")
xy.means <- colMeans(x = xy[, 2:ncol(xy)], na.rm = TRUE)
xy.means <- as.data.frame(xy.means)
xy.means$total <- rownames(xy.means)
ggplot(xy.means, aes(x = total, y = xy.means)) +
theme_bw() +
geom_col()
Or just use base image graphic
barplot(height = colMeans(x = xy[, 2:ncol(xy)], na.rm = TRUE))
I'm still a debutant with R so this is my issue.
I got a dataframe named dataset3.
structure(list(Varietes = c("Georgia", "Georgia", "Georgia", "Georgia", "Georgia", "Georgia", "Georgia", "Ruby", "Ruby", "Ruby", "Ruby", "Ruby", "Ruby", "Ruby", "Abelastone", "Abelastone", "Abelastone", "Abelastone", "Abelastone"), ligne.rep = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), Pied = c(1, 2, 3, 4, 5, 7, 8, 1, 3, 4, 5, 6, 7, 8, 1, 2, 3, 6, 7), Date.floraison.mâle = structure(c(1627689600, 1627430400, 1627689600, 1627344000,1627516800,1627603200, 1627689600, 1627516800, 1627776000, 1627603200, 1627344000,1627516800, 1627689600, 1627776000, 1627516800, 1627862400, 1627776000, 1627948800, 1628035200), tzone = "UTC", class = c("POSIXct", "POSIXt")), Date.floraison.femelle = structure(c(1627776000, 1627948800, 1627862400, 1627948800, 1628121600, 1628035200, 1627862400, 1627862400, 1628035200,1628121600, 1627862400, 1627948800, 1628380800, 1628121600, 1628553600, 1629158400, 1628640000, 1629849600, 1629158400), tzone = "UTC", class = c("POSIXct", "POSIXt")), ASIi = c(1, 6, 2, 7, 7, 5, 2, 4, 3, 6, 6, 5, 8, 4, 12, 15, 10, 22, 13), Hauteur.des.pieds = c(200, 210, 205, 215, 217, 207, 205, 208, 206, 215, 216, 212, 220, 215, 230, 228, 226, 240, 233), Hauteur.des.soies = c(100, 104, 102, 110, 108, 103, 102, 105, 105, 110, 112, 108, 113, 110, 123, 118, 116, 124, 122)), row.names = c(NA, -19L), class = c("tbl_df", "tbl", "data.frame"), na.action = structure(c(`6` = 6L, `10` = 10L, `20` = 20L, `21` = 21L, `24` = 24L), class = "omit"))
This is false data. I'm trying now to write a script that will be really helpful in the futur.
I would like to interroge my data to the simpliest way possible to get correltions, like for example "ASIi" and "Hauteur des pieds", but for each varieties of my column "Varietes". In my dataset 3 i got 3 varieties : Georgia, Abelastone, Ruby.
Because in my real dataset i will have 50 varieties and this will not easy if i have to isolate every variety.
Is there an easier way to do this?
Thank you very much
To get the correlation of the same variable, but for each variety
We can loop trough every combination of two varieties:
comb.var = combn(unique(dataset3$Varietes), 2) #Get the combinations
correlations = list() #Create empty list to store values
cols = c("Pied", "ASIi", "Hauteur.des.pieds", "Hauteur.des.soies") #Define the columns that you want to get correlations for
for(i in 1:ncol(comb.var)){
name = paste(comb.var[,i], collapse="/")
correlations[[name]] = numeric() #Create a empty vector for the combination
for(j in cols){
correlations[[name]][j] = cor(dataset3[dataset3$Varietes==comb.var[1,i],j],
dataset3[dataset3$Varietes==comb.var[2,i],j])}} #Fill the vector with each correlation
OBS: your example data had less observations for Abelastone, if in your real dataset every variety has the same amount of obs this won't be a problem, otherwise i can generalize the code a bit more. Output:
> correlations
$`Georgia/Ruby`
Pied ASIi Hauteur.des.pieds Hauteur.des.soies
0.9787781 0.1828599 0.1065409 0.3239701
$`Georgia/Abelastone`
Pied ASIi Hauteur.des.pieds Hauteur.des.soies
0.9048705 0.6451144 0.5370488 0.3387600
$`Ruby/Abelastone`
Pied ASIi Hauteur.des.pieds Hauteur.des.soies
0.93982306 0.05192741 0.44198418 0.21908902
To get the correlation of different variables, for each variety
We just reverse the logic of the loop:
comb.col = combn(cols, 2)
for(i in 1:ncol(comb.col)){
name = paste(comb.col[,i], collapse="/")
correlations[[name]] = numeric()
for(j in unique(dataset3$Varietes)){
correlations[[name]][j] = cor(dataset3[[comb.col[1,i]]][dataset3$Varietes==j],
dataset3[[comb.col[2,i]]][dataset3$Varietes==j])}}
Output:
> correlations
$`ASIi/Hauteur.des.pieds`
Georgia Ruby Abelastone
0.9340590 0.8795120 0.8174805
$`ASIi/Hauteur.des.soies`
Georgia Ruby Abelastone
0.8849296 0.8489775 0.5125693
$`Hauteur.des.pieds/Hauteur.des.soies`
Georgia Ruby Abelastone
0.9519358 0.9789882 0.8293315
I merged nine plots together and I would like to group them based on different characteristics (A,B,C). Is there a simple way to add labels or annotations at the bottom of plots? When using cowplot or GridExtra i receive the following error:
In as_grob.default(plot) :
Cannot convert object of class list into a grob.
Sample data
list(list(stats = structure(c(43, 96.5, 297.5, 707.5, 778), .Dim = c(5L,
1L)), n = 36, conf = structure(c(136.603333333333, 458.396666666667
), .Dim = 2:1), out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(2, 10.5, 55.5, 102, 128), .Dim = c(5L,
1L)), n = 36, conf = structure(c(31.405, 79.595), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(1, 3, 5.5, 77, 88), .Dim = c(5L,
1L)), n = 36, conf = structure(c(-13.9866666666667, 24.9866666666667
), .Dim = 2:1), out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(531, 632.5, 701, 726.5, 786), .Dim = c(5L,
1L)), n = 36, conf = structure(c(676.246666666667, 725.753333333333
), .Dim = 2:1), out = c(485, 464, 446), group = c(1, 1, 1
), names = ""), list(stats = structure(c(104,
109.5, 113.5, 121, 125), .Dim = c(5L, 1L)), n = 36, conf = structure(c(110.471666666667,
116.528333333333), .Dim = 2:1), out = c(91, 91, 88, 84, 84,
79), group = c(1, 1, 1, 1, 1, 1), names = ""),
list(stats = structure(c(28, 53.5, 83.5, 88, 91), .Dim = c(5L,
1L)), n = 36, conf = structure(c(74.415, 92.585), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(80, 89, 102.5, 153, 236), .Dim = c(5L,
1L)), n = 36, conf = structure(c(85.6466666666667, 119.353333333333
), .Dim = 2:1), out = c(343, 318, 299, 257), group = c(1,
1, 1, 1), names = """"), list(stats = structure(c(7,
12, 22.5, 44, 72), .Dim = c(5L, 1L)), n = 36, conf = structure(c(14.0733333333333,
30.9266666666667), .Dim = 2:1), out = numeric(0), group = numeric(0),
names = ""), list(stats = structure(c(5,
5, 6, 12.5, 21), .Dim = c(5L, 1L)), n = 36, conf = structure(c(4.025,
7.975), .Dim = 2:1), out = numeric(0), group = numeric(0),
names = ""))
Many thanks
I agree with the idea of using ggplot2 graphics with facets, but given your plot objects, you could do something like this (to get you started). I used ggplotify instead of cowplot because I ran into trouble with the figure margins, but you might be able to fix that by changing the null device (not tested).
Edit:
Added individual labels and y axis labels, as well as outer margins. You might have to adjust some of that depending on the output size of your composite plot. This may show you how you could adjust those settings for individual plots. Still, using ggplot2 to generate the plots would make things quite a bit easier.
library(grid)
library(gridExtra)
library(ggplotify)
sdt <- list(list(stats = structure(c(43, 96.5, 297.5, 707.5, 778), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(136.603333333333, 458.396666666667), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(2, 10.5, 55.5, 102, 128), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(31.405, 79.595), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(1, 3, 5.5, 77, 88), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(-13.9866666666667, 24.9866666666667), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(531, 632.5, 701, 726.5, 786), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(676.246666666667, 725.753333333333), .Dim = 2:1),
out = c(485, 464, 446), group = c(1, 1, 1), names = ""),
list(stats = structure(c(104, 109.5, 113.5, 121, 125), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(110.471666666667, 116.528333333333), .Dim = 2:1),
out = c(91, 91, 88, 84, 84, 79), group = c(1, 1, 1, 1, 1, 1), names = ""),
list(stats = structure(c(28, 53.5, 83.5, 88, 91), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(74.415, 92.585), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(80, 89, 102.5, 153, 236), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(85.6466666666667, 119.353333333333), .Dim = 2:1),
out = c(343, 318, 299, 257), group = c(1,1, 1, 1), names = ""),
list(stats = structure(c(7, 12, 22.5, 44, 72), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(14.0733333333333, 30.9266666666667), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""),
list(stats = structure(c(5, 5, 6, 12.5, 21), .Dim = c(5L, 1L)),
n = 36, conf = structure(c(4.025, 7.975), .Dim = 2:1),
out = numeric(0), group = numeric(0), names = ""))
sublabels <- paste0(rep(LETTERS[1:3], each=3), 1:3)
gplts <- lapply(1:9, function(x) as.grob(function(y=sdt[[x]]) {
par(oma=c(0,3,0,3))
bxp(y, ylab="values", main=sublabels[x])}))
grid.arrange(rectGrob(gp=gpar(col="red")), rectGrob(gp=gpar(col="green")),
rectGrob(gp=gpar(col="yellow")), nrow=1, newpage =T)
vp <- viewport(.33/2,0.45, gp = gpar(col="red"))
grid.text("Group A",
y = .1, just = c("center", "bottom"),
gp = gpar(fontsize=20), vp = vp)
vp <- viewport(.5,.45, gp = gpar(col="green"))
grid.text("Group B",
y = .1, just = c("center", "bottom"),
gp = gpar(fontsize=20), vp = vp)
vp <- viewport(1-(.33/2),.45, gp = gpar(col="yellow"))
grid.text("Group C",
y = .1, just = c("center", "bottom"),
gp = gpar(fontsize=20), vp = vp)
grid.arrange(grobs=gplts, nrow=1, newpage=F)
Created on 2021-03-25 by the reprex package (v1.0.0)
I am trying to develop a shiny dashboard app that is able to produce a bar graph for different outcome variables that can be selected by the user. To do so, I need to subset my data reactively to generate aggregate data frames. I am able to have the code below successfully filter my data reactively, but I am running into trouble when I try to use dplyr::summarise() reactively.
Here is my data
dput(head(df))
structure(
list(
geoid = c(
"01001020200",
"01001020300",
"01001020700",
"01001020802",
"01001021000",
"01001021100"
),
state = c(
"Alabama",
"Alabama",
"Alabama",
"Alabama",
"Alabama",
"Alabama"
),
county = c(
"Autauga County",
"Autauga County",
"Autauga County",
"Autauga County",
"Autauga County",
"Autauga County"
),
ozzone = structure(
c(1L, 1L, 2L, 1L, 1L, 1L),
.Label = c("non.oz", "oz"),
class = "factor"
),
tract_type = c(
"LICs",
"Contiguous",
"LICs",
"Contiguous",
"Contiguous",
"LICs"
),
investment_score_1_low_10_high = c(4,
6, 9, 10, 5, 6),
socioeconomic_change_flag_1_yes_blank_no = c(0,
0, 0, 0, 0, 0),
fips_county = c("01001", "01001", "01001", "01001",
"01001", "01001"),
total_empl = c(51809L, 51809L, 51809L, 51809L,
51809L, 51809L),
total_payroll = c(338395L, 338395L, 338395L,
338395L, 338395L, 338395L),
total_establishments = c(5090L, 5090L,
5090L, 5090L, 5090L, 5090L),
largest_employer = c(72L, 72L, 72L,
72L, 72L, 72L),
largest_employer_bypayroll = c(44L, 44L, 44L,
44L, 44L, 44L),
trend_employee_change = c(
2735.60000000046,
2735.60000000046,
2735.60000000046,
2735.60000000046,
2735.60000000046,
2735.60000000046
),
trend_payroll_change = c(
23074.8000000037,
23074.8000000037,
23074.8000000037,
23074.8000000037,
23074.8000000037,
23074.8000000037
),
trend_establishment_change = c(
53.4000000000084,
53.4000000000084,
53.4000000000084,
53.4000000000084,
53.4000000000084,
53.4000000000084
),
damage_cost_weather_total = c(20000, 20000, 20000, 20000,
20000, 20000),
deaths_weather_total = c(0L, 0L, 0L, 0L, 0L, 0L),
medianrent = c(537, 633, 525, 680, 409, 303),
vacancyrate = c(
0.108200455580866,
0.113652113652114,
0.0436681222707424,
0.0512166859791425,
0.229962546816479,
0.21030303030303
),
total_pop = c(503, 827, 900, 2989, 740, 813),
undertwo_percent = c(
0.391650099403579,
0.351874244256348,
0.397777777777778,
0.17096018735363,
0.301351351351351,
0.263222632226322
),
mobility_rate = c(
0.133702166897188,
0.0737753882915173,
0.196514423076923,
0.172716680111141,
0.0641304347826087,
0.0681084570690769
),
unemploy_rate = c(
0.0176991150442478,
0.0273203592814371,
0.109881724532621,
0.0127906976744186,
0.0344982078853047,
0.0281910728269381
),
median_income = c(41287, 46806, 41250, 64439,
46607, 36450),
renter_percent = c(
0.337653478854025,
0.310596310596311,
0.331877729257642,
0.268110942458949,
0.328686327077748,
0.365986394557823
),
blackaa_percent = c(
0.5451197053407,
0.264697193500739,
0.145906432748538,
0.152916262243007,
0.258583690987124,
0.530922930542341
),
hispanic_percent = c(
0.0105893186003683,
0.0803545051698671,
0.0400584795321637,
0.0137651107385511,
0.00822603719599428,
0.00666032350142721
),
transit_score_mean = c(0, 0, 0, 0, 0, 0),
life_expectancy = c(75.67, 75.67, 75.67, 75.67, 75.67, 75.67),
trend_life_expectancy = c(5.1, 5.1, 5.1, 5.1, 5.1, 5.1),
median_monthly_housing_costs = c(885,
885, 885, 885, 885, 885),
pestilence_2018 = c(2, 2, 2, 2, 2,
2),
total_pop_county = c(6772, 6772, 6772, 6772, 6772, 6772),
deaths_weather_pop = c(0, 0, 0, 0, 0, 0),
cost_weather_pop = c(
2.95333727111636,
2.95333727111636,
2.95333727111636,
2.95333727111636,
2.95333727111636,
2.95333727111636
),
Male_HSgrad = c(75, 68, 211, 189, 97,
42),
Male_SomeCollege = c(28, 18, 51, 111, 74, 38),
Male_AssocDeg = c(4,
6, 0, 63, 0, 21),
Male_BachDeg = c(7, 9, 0, 11, 0, 9),
Male_GradDeg = c(0,
0, 0, 29, 6, 0),
MaleEduAboveHS = c(114, 101, 262, 403, 177,
110),
Total_Male18.24 = c(145, 123, 285, 455, 202, 110),
MaleEduHSAbove_pop = c(
0.786206896551724,
0.821138211382114,
0.919298245614035,
0.885714285714286,
0.876237623762376,
1
),
Female_HSgrad = c(11, 60, 87, 156, 23, 83),
Female_SomeCollege = c(22,
25, 13, 47, 54, 65),
Female_AssocDeg = c(0, 0, 20, 82, 0,
0),
Female_BachDeg = c(5, 26, 0, 19, 0, 11),
Female_GradDeg = c(5,
16, 0, 0, 0, 0),
FemaleEduAboveHS = c(43, 127, 120, 304,
77, 159),
Total_Female18.24 = c(53, 127, 192, 581, 92, 198),
FemaleEduHSAbove_pop = c(
0.811320754716981,
1,
0.625,
0.523235800344234,
0.83695652173913,
0.803030303030303
)
),
row.names = c(NA,
6L),
class = "data.frame"
)
Here is my code
#List of potential outcome variables to be plotted
variables <- c("total_empl", "total_payroll", "total_establishments", "largest_employer", "largest_employer_bypayroll", "trend_employee_change", "trend_payroll_change", "trend_establishment_change", "damage_cost_weather_total", "deaths_weather_total", "medianrent", "vacancyrate", "total_pop", "undertwo_percent", "mobility_rate", "unemploy_rate", "median_income", "renter_percent", "blackaa_percent", "hispanic_percent", "median_monthly_housing_costs", "MaleEduAboveHS_pop", "FemaleEduHSAbove_pop")
# Define inputs
selectInput('state_name', label = 'Select a state', choices = lookup)
selectInput('DV', label = 'Outcome Measure', choices = variables)
#Filter data based on the State and outcome measure the user would like to investigate.
bar <- reactive({
st <- df %>%
filter(state == input$state_name)
bp <- st %>%
group_by(tract_type) %>%
summarise(Outcome = mean(st[,input$DV]))
return(bp)
})
bar
UPDATE
Right now, this code successfully filters the data by the input$state_name, but there is an issue with the calculation of means. The result is this:
# A tibble: 2 x 2
tract_type Outcome
<chr> <dbl>
1 Contiguous 468296.
2 LICs 468296.
As you can see, the means that are calculated are identical. In fact, these values correspond to the grand average mean for whichever variable is chosen for input$DV. Therefore, the filtered st data is not being successfully grouped into the two levels of tract_type.
I see what you are trying to do. The difference is that in your reactive part you try to calculate the mean of a string, which won't work. What you want to do is summarise one of the columns in df by providing the name
In the following example, I specify the summarising variable manually. Note that investment_score_1_low_10_high does not have quotes. investment_score_1_low_10_high is what is called a symbol in R.
st <- df %>%
filter(state == "Alabama") %>%
group_by(tract_type) %>%
summarise(Outcome = mean(investment_score_1_low_10_high))
But I think this should work:
bar <- reactive({
# Create a symbol from string.
mean_variable <- sym(input$DV)
bp <- df %>%
filter(state == input$state_name) %>%
group_by(tract_type) %>%
summarise(Outcome = mean(!! mean_variable, na.rm = TRUE))
return(bp)
})
Extra information about the use of !! and what it does can be found here: Here
And even better with examples Here
Solution derived by #dylanvanw
bar <- reactive({
# Create a symbol from string.
mean_variable <- sym(input$DV)
bp <- df %>%
filter(state == input$state_name) %>%
group_by(tract_type) %>%
summarise(Outcome = mean(!! mean_variable, na.rm = TRUE))
return(bp)
})
I am using function msm2surv in msm package and trying to convert longitudinal data to the format flexsurve package likes. Following is my sample called tmp.
tmp <- structure(list(id = c(89, 90, 90, 91, 91, 91, 92, 92, 93, 93,
94, 94, 94, 95, 95, 96), days = c(9157, 0, 9156, 0, 8394, 9156,
0, 9156, 0, 8079, 0, 8933, 9003, 0, 8430, 0), event = c(1, 1,
1, 1, 2, 2, 1, 1, 1, 5, 1, 3, 6, 1, 4, 1)), row.names = c(NA,
-16L), class = c("grouped_df", "tbl_df", "tbl", "data.frame"), vars = "id", drop = TRUE, indices = list(
0L, 1:2, 3:5, 6:7, 8:9, 10:12, 13:14, 15L), group_sizes = c(1L,
2L, 3L, 2L, 2L, 3L, 2L, 1L), biggest_group_size = 3L, labels = structure(list(
id = c(89, 90, 91, 92, 93, 94, 95, 96)), row.names = c(NA,
-8L), class = "data.frame", vars = "id", drop = TRUE, .Names = "id"), .Names = c("id",
"days", "event"))
running codes:
library(msm)
Q <- matrix(c(
0,1,1,1,1,0,
0,0,1,1,1,1,
0,0,0,1,1,1,
0,0,0,0,1,1,
0,0,0,0,0,0,
0,0,0,0,0,0
), nrow=6, ncol=6,
byrow=TRUE,
dimnames=list(from=1:6,to=1:6))
dat <- msm2Surv(data=tmp, subject="id", time="days", state="event", Q=Q)
It gives me the error: Error in rowSums(Qf) : 'x' must be an array of at least two dimensions.
I checked the dimension of the dataframe, it sounds ok to me. But the error is there. Does everyone know how to resolve the problem/error?
Many thanks!
We can convert the tbl_df to data.frame and it should work
out <- msm2Surv(data=as.data.frame(tmp), subject="id",
time="days", state="event", Q=Q)
dim(out)
#[1] 31 8
The "tmp" dataset is a grouped tbl_df and has lots of attributes. By converting to data.frame, we remove those attributes.