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I want to plot a boxplot with the regular /boxplot function of R - not ggplot.
Y and X axis are continously numeric varibales (x-axis 6 forces: 1.0, 1.3, 1.6, 2.0, 2.5, 3.1 [N])
On Y-axis the participants ratings (1 to 7).
I would like to plot it, with quantified spacing on x-axis, and also later add regression line into the plot. Cant find anything for the regular /boxplot function.
Code so far:
kraft_ou <- data.frame(VR1_100$ou_kraft,
VR1_125$ou_kraft,
VR1_160$ou_kraft,
VR1_200$ou_kraft,
VR1_250$ou_kraft,
VR1_310$ou_kraft)
colnames(kraft_ou) <- c("kraft_100", "kraft_125", "kraft_160", "kraft_200", "kraft_250", "kraft_310")
kraft_ou
boxplot(kraft_ou,
names=c("1,0 [N]", "1,3 [N]","1,6 [N]","2,0 [N]","2,5 [N]","3,1 [N]"),
col = "bisque",
ylim = c(1, 7))
points(1:6, meanskraftou, pch=4)
text(1:6, meanskraftou + 0.24, labels = meanskraftou)
abline(h=4)
data (n=30 ratings from 1 to 7 for each of the 6 forces):
dput(kraft_ou)
structure(list(kraft_100 = c(4, 3, 5, 5, 3, 4, 2, 4, 4, 5, 4,
5, 5, 4, 4, 3, 4, 4, 5, 4, 6, 5, 4, 5, 5, 5, 4, 4, 4, 4), kraft_125 = c(4,
4, 5, 6, 4, 3, 4, 4, 4, 5, 4, 5, 4, 5, 4, 3, 4, 4, 4, 6, 6, 4,
4, 5, 3, 5, 4, 4, 4, 5), kraft_160 = c(5, 6, 6, 6, 6, 4, 6, 5,
6, 5, 4, 3, 6, 6, 6, 5, 5, 5, 5, 6, 6, 6, 5, 5, 4, 6, 4, 5, 5,
5), kraft_200 = c(6, 5, 6, 6, 5, 4, 5, 5, 6, 7, 5, 3, 5, 5, 5,
4, 7, 6, 5, 5, 7, 6, 5, 6, 6, 6, 5, 4, 5, 3), kraft_250 = c(5,
6, 6, 7, 6, 6, 6, 6, 7, 7, 6, 5, 7, 7, 5, 5, 6, 6, 7, 7, 6, 6,
5, 5, 5, 7, 4, 6, 6, 5), kraft_310 = c(7, 7, 7, 7, 6, 5, 6, 6,
6, 7, 4, 5, 7, 6, 5, 5, 7, 6, 5, 6, 6, 6, 5, 6, 5, 6, 5, 6, 6,
6)), class = "data.frame", row.names = c(NA, -30L))
You can use the at argument to specify x locations for your boxplots, though to get them narrow enough to avoid overplotting, you need to add an invisible box and set the relative widths of the visible boxes to a smaller value:
boxplot(cbind(kraft_ou, n = rep(NA, nrow(kraft_ou))),
names=c("1,0 [N]", "1,3 [N]","1,6 [N]","2,0 [N]","2,5 [N]","3,1 [N]",
" "),
col = "bisque",
ylim = c(1, 7), width = c(0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 1),
at = c(1, 1.3, 1.6, 2.0, 2.5, 3.1, 3.1))
abline(h = 4)
To add a regression line, you would need to have all your data frame values in a single y variable, and a vector of their corresponding x axis positions:
abline(lm(unlist(kraft_ou) ~ rep(c(1, 1.3, 1.6, 2.0, 2.5, 3.1), each = 30)))
I have a problem similar to what is found here. I have a loop which runs through some modelling for different pairs of variables. Probably should not have used loops to go through them, but right now that is too late. Then I want to create a plot for each run. At first nothing showed before looking at that post. Looking at the post and implementing the best answer i could at least print the plots, but they still were not stored. The idea is to generate the plots, and then use grid.arrange to plot them together. Could someone show how to fix it? Here is some random data and the loop from example:
col1 <- c(2, 4, 1, 2, 5, 1, 2, 0, 1, 4, 4, 3, 5, 2, 4, 3, 3, 6, 5, 3, 6, 4, 3, 4, 4, 3, 4,
2, 4, 3, 3, 5, 3, 5, 5, 0, 0, 3, 3, 6, 5, 4, 4, 1, 3, 3, 2, 0, 5, 3, 6, 6, 2, 3)
col2 <- c(2, 4, 4, 0, 4, 4, 4, 4, 1, 4, 4, 3, 5, 0, 4, 5, 3, 6, 5, 3, 6, 4, 4, 2, 4, 4, 4,
1, 1, 2, 2, 3, 3, 5, 0, 3, 4, 2, 4, 5, 5, 4, 4, 2, 3, 5, 2, 6, 5, 2, 4, 6, 3, 3)
col3 <- c(2, 5, 4, 1, 4, 2, 3, 0, 1, 3, 4, 2, 5, 1, 4, 3, 4, 6, 3, 4, 6, 4, 1, 3, 5, 4, 3,
2, 1, 3, 2, 2, 2, 4, 0, 1, 4, 4, 3, 5, 3, 2, 5, 2, 3, 3, 4, 2, 4, 2, 4, 5, 1, 3)
data2 <- data.frame(col1,col2,col3)
data2[,1:3] <- lapply(data2[,1:3], as.factor)
colnames(data2)<- c("A","B","C")
myplots <- vector('list', ncol(data2))
for (i in seq_along(data2)) {
message(i)
myplots[[i]] <- local({
i <- i
p1 <- ggplot(data2, aes(x = data2[[i]])) +
geom_histogram(fill = "lightgreen") +
xlab(colnames(data2)[i])
print(p1)
})
}
I tried to change print to return, but to no avail. I get the plots printed in the View window in Rstudio, but the plots are not stored at all.
You can use the following code -
library(ggplot2)
myplots <- vector('list', ncol(data2))
for (i in seq_along(data2)) {
myplots[[i]] <- ggplot(data2, aes(x = .data[[colnames(data2)[i]]])) +
geom_histogram(fill = "lightgreen")
}
However, using lapply would be easier.
myplots <- lapply(names(data2), function(x)
ggplot(data2, aes(x = .data[[x]])) + geom_histogram(fill = "lightgreen"))
Plot the list of plots with grid.arrange.
gridExtra::grid.arrange(grobs = myplots)
data
A <- c(2, 4, 1, 2, 5, 1, 2, 0, 1, 4, 4, 3, 5, 2, 4, 3, 3, 6, 5, 3, 6, 4, 3, 4, 4, 3, 4,
2, 4, 3, 3, 5, 3, 5, 5, 0, 0, 3, 3, 6, 5, 4, 4, 1, 3, 3, 2, 0, 5, 3, 6, 6, 2, 3)
B <- c(2, 4, 4, 0, 4, 4, 4, 4, 1, 4, 4, 3, 5, 0, 4, 5, 3, 6, 5, 3, 6, 4, 4, 2, 4, 4, 4,
1, 1, 2, 2, 3, 3, 5, 0, 3, 4, 2, 4, 5, 5, 4, 4, 2, 3, 5, 2, 6, 5, 2, 4, 6, 3, 3)
C <- c(2, 5, 4, 1, 4, 2, 3, 0, 1, 3, 4, 2, 5, 1, 4, 3, 4, 6, 3, 4, 6, 4, 1, 3, 5, 4, 3,
2, 1, 3, 2, 2, 2, 4, 0, 1, 4, 4, 3, 5, 3, 2, 5, 2, 3, 3, 4, 2, 4, 2, 4, 5, 1, 3)
data2 <- data.frame(A,B,C)
Does this work for you?, With patchwork and purrr::reduce we can club these graphs to stack(horizontal or vertical) with each other. You can also use slashes(/) instead of plus(+) in reduce to make it appended vertically instead of horizontally. If you want to plot histogram you should have continuous data , In case you do want to plot counts for discrete data you should try geom_bar. If you do want to check for geom_bar then you need to convert the columns into factors. I am not so sure what plot you want to carry out, I am assuming that you have continuous data and you want to carry out histogram here. Please let me know if it doesn't work in your scenario.
library(tidyverse)
library(patchwork)
data2 <- data.frame(col1, col2, col3) ## No conversion of factors
nm <- names(data2)
g1 <- reduce(map2(data2,nm, ~ggplot(data2,aes(x =.x )) + geom_histogram(fill = "yellow4") + labs(x=.y, y = 'count')), `+`)
print(g1)
Or with slashes:
g2 <- reduce(map2(data2,nm, ~ggplot(data2,aes(x =.x )) + geom_histogram(fill = "yellow4") + labs(x=.y, y = 'count')), `/`)
print(g2)
Or if you want to have for loops then probably you can do this as well, you already have intialised myplots so not adding it here:
for (i in seq_along(data2)) {
myplots[[i]] <-
ggplot(data2, aes(x = data2[[i]])) +
geom_histogram(fill = "lightgreen") +
xlab(colnames(data2)[i])
}
Explanation:
Now you can use reduce with your myplots to arrange them, Note here myplots should be containing your 3 plots :
reduce(myplots, `+`)
for arranging it.
The map2 and reduce is similar solution, with map2 you are getting 3 plots saved into a list, so 3 objects are returned from below code:
plots <- map2(data2,nm, ~ggplot(data2,aes(x =.x )) + geom_histogram(fill = "yellow4") + labs(x=.y, y = 'count'))
To add them (arrange) them all you have to do is to use patchwork like below:
plots[[1]] + plots[[2]] + plots[[3]], but then its quite cumbersome, so we use reduce to make it happen like below:
reduce(plots, `+`)
Also like I mentioned earlier you can use slash instead of plus to make the arrangement vertical than horizontal. with plot_layout option in patchwork, you can create more flexible plots. You can check here .
with gridExtra : gridExtra::grid.arrange(grobs = (myplots)), again instead of myplots, it can be any list that contain ggplot objects.
I am wanting to plot graph clusters that I define by myself. I am using the simplified undirected enron data.
library(igraphdata)
data("enron")
g <- as.undirected(enron)
g <- simplify(g)
rm("enron")
member <- c(1, 8, 9, 9, 10, 10, 8, 7, 4, 1, 2, 6, 3, 1, 2, 8, 7, 2, 1, 5,
1, 7, 6, 4, 8, 4, 8, 10, 3, 6, 1, 4, 7, 4, 3, 7, 9, 10, 3, 8, 1,
9, 8, 2, 7, 2, 9, 5, 1, 2, 6, 10, 3, 3, 2, 1, 9, 10, 3, 5, 6, 5,
5, 3, 7, 6, 9, 10, 8, 10, 8, 8, 10, 10, 10, 8, 7, 7, 9, 1, 9, 2, 9,
7, 2, 7, 7, 3, 2, 5, 2, 1, 6, 5, 10, 4, 3, 2, 4, 6, 4, 9, 5, 4,
1, 10, 2, 3, 4, 3, 6, 3, 6, 4, 6, 8, 2, 4, 5, 1, 5, 1, 4, 10, 4, 7,
5, 9, 10, 1, 2, 1, 5, 7, 5, 3, 5, 8, 7, 9, 5, 8, 1, 5, 3, 3, 3, 10,
1, 7, 8, 4, 1, 10, 9, 6, 9, 9, 4, 2, 6, 4, 6, 3, 5, 6, 9, 7, 6, 6,
4, 8, 6, 8, 8, 2, 5, 4, 3, 2, 9, 10, 2, 7)
I have tried many ways but none looks good. The best I can make is
edges_data_frame <- get.data.frame(g, what = "edges")
w.mem <- rep(0, length(E(g)))
for (i in 1:length(E(g))){
w.mem[i] <- ifelse(member[edges_data_frame$from[i]] == member[edges_data_frame$to[i]], 500, 1)
}
mem <- make_clusters(g,member)
E(g)$weight <- w.mem
colors <- rainbow(max(membership(mem)))
layout <- layout.fruchterman.reingold(g, weights=w.mem)
set.seed(1234)
plot(g, vertex.color=colors[mem$membership],
mark.groups=communities(mem),
vertex.label = NA,
edge.width = 1, edge.color = "lightgray", vertex.size = 5)
my first trial
I found that the "deleting edges plot" looks much cleaner
coGrph <- delete_edges(g, E(g)[crossing(mem, g)])
col_vector <- c('#e6194b', '#3cb44b', '#ffe119', '#4363d8', '#f58231', '#911eb4', '#46f0f0', '#f032e6', '#bcf60c', '#fabebe', '#008080', '#e6beff', '#9a6324', '#fffac8', '#800000', '#aaffc3', '#808000', '#ffd8b1', '#000075', '#808080', '#ffffff', '#000000')
temp <- sapply(1:length(V(g)), FUN = function(i) {col_vector[member[i]]})
V(coGrph)$color <- temp
plot(coGrph, vertex.label = NA, vertex.size = 5)
my second trial
However, this plot has some missing edges and does not reflect the true connection of the plot. I want to use this plot and add the deleted edges back to the plot without changing the positions I have right now. Is it possible?
Thank you very much I really appreciate your help.
Yes. Use your coGrph to create a layout, but then plot the original graph.
Continuing your "second trial"
set.seed(1234)
LOcG = layout_nicely(coGrph)
V(g)$color <- temp
plot(g, layout=LOcG, vertex.label = NA, vertex.size = 5)
My goal is to find much simpler code, which can generalize, that shows the relationships between responses to two survey questions. In the MWE, one question asked respondents to rank eight marketing selections from 1 to 8 and the other asked them to rank nine attribute selections from 1 to 9. Higher rankings indicate the respondent favored the selection more. Here is the data frame.
structure(list(Email = c("a", "b", "c", "d", "e", "f", "g", "h",
"i"), Ads = c(2, 1, 1, 1, 1, 2, 1, 1, 1), Alumni = c(3, 2, 2,
3, 2, 3, 2, 2, 2), Articles = c(6, 4, 3, 2, 3, 4, 3, 3, 3), Referrals = c(4,
3, 4, 8, 7, 8, 8, 6, 4), Speeches = c(7, 7, 6, 7, 4, 7, 4, 5,
5), Updates = c(8, 6, 6, 5, 5, 5, 5, 7, 6), Visits = c(5, 8,
7, 6, 6, 6, 6, 4, 8), `Business Savvy` = c(10, 6, 10, 10, 4,
4, 6, 8, 9), Communication = c(4, 3, 8, 3, 3, 9, 7, 6, 7), Experience = c(7,
7, 7, 9, 2, 8, 5, 9, 5), Innovation = c(2, 1, 4, 2, 1, 2, 2,
1, 1), Nearby = c(3, 2, 2, 1, 5, 3, 3, 2, 2), Personal = c(8,
10, 6, 8, 6, 10, 4, 3, 3), Rates = c(9, 5, 9, 6, 9, 7, 10, 5,
4), `Staffing Model` = c(6, 8, 5, 5, 7, 5, 8, 7, 8), `Total Cost` = c(5,
4, 3, 7, 8, 6, 9, 4, 6)), row.names = c(NA, -9L), class = c("tbl_df",
"tbl", "data.frame"))
If numeric rankings cannot be used for my solution to calculating relationships (correlations), please correct me.
Hoping they can be used, I arrived at the following plodding code, which I hope calculates the correlation matrix of each method selection against each attribute selection.
library(psych)
dataframe2 <- psych::corr.test(dataframe[ , c(2, 9:17)])[[1]][1:10] # the first method vs all attributes
dataframe3 <- psych::corr.test(dataframe[ , c(3, 9:17)])[[1]][1:10] # the 2nd method vs all attributes and so on
dataframe4 <- psych::corr.test(dataframe[ , c(4, 9:17)])[[1]][1:10]
dataframe5 <- psych::corr.test(dataframe[ , c(5, 9:17)])[[1]][1:10]
dataframe6 <- psych::corr.test(dataframe[ , c(6, 9:17)])[[1]][1:10]
dataframe7 <- psych::corr.test(dataframe[ , c(7, 9:17)])[[1]][1:10]
dataframe8 <- psych::corr.test(dataframe[ , c(8, 9:17)])[[1]][1:10]
# create a dataframe from the rbinded rows
bind <- data.frame(rbind(dataframe2, dataframe3, dataframe4, dataframe5, dataframe6, dataframe7, dataframe8))
Rename rows and columns:
colnames(bind) <- c("Sel", colnames(dataframe[9:17]))
rownames(bind) <- colnames(dataframe[2:8])
How can I accomplish the above more efficiently?
By the way, the bind data frame also allows one to produce a heat map with the DataExplorer package.
library(DataExplorer)
DataExplorer::plot_correlation(bind)
[Summary]
In the scope of our discussion, there are two ways to get the correlation data.
Use stats::cor, i.e., cor(subset(dataframe, select = -Email))
Use psych::corr.test, i.e., corr.test(subset(dataframe, select = -Email))[[1]]
Then you may subset the correlation matrix with the desired rows and columns.
In order to use DataExplorer::plot_correlation, you can simply do plot_correlation(dataframe, type = "c"). Note: the output heatmap will include correlations for all columns, so you can just ignore columns that are not of interests.
[Original Answer]
## Create data
dataframe <- structure(
list(
Email = c("a", "b", "c", "d", "e", "f", "g", "h", "i"),
Ads = c(2, 1, 1, 1, 1, 2, 1, 1, 1),
Alumni = c(3, 2, 2, 3, 2, 3, 2, 2, 2),
Articles = c(6, 4, 3, 2, 3, 4, 3, 3, 3),
Referrals = c(4, 3, 4, 8, 7, 8, 8, 6, 4),
Speeches = c(7, 7, 6, 7, 4, 7, 4, 5, 5),
Updates = c(8, 6, 6, 5, 5, 5, 5, 7, 6),
Visits = c(5, 8, 7, 6, 6, 6, 6, 4, 8),
`Business Savvy` = c(10, 6, 10, 10, 4, 4, 6, 8, 9),
Communication = c(4, 3, 8, 3, 3, 9, 7, 6, 7),
Experience = c(7, 7, 7, 9, 2, 8, 5, 9, 5),
Innovation = c(2, 1, 4, 2, 1, 2, 2, 1, 1),
Nearby = c(3, 2, 2, 1, 5, 3, 3, 2, 2),
Personal = c(8, 10, 6, 8, 6, 10, 4, 3, 3),
Rates = c(9, 5, 9, 6, 9, 7, 10, 5, 4),
`Staffing Model` = c(6, 8, 5, 5, 7, 5, 8, 7, 8),
`Total Cost` = c(5, 4, 3, 7, 8, 6, 9, 4, 6)
),
row.names = c(NA, -9L),
class = c("tbl_df", "tbl", "data.frame")
)
Following your example strictly, we can do the following:
## Calculate correlation
df2 <- subset(dataframe, select = -Email)
marketing_selections <- names(df2)[1:7]
attribute_selections <- names(df2)[8:16]
corr_matrix <- psych::corr.test(df2)[[1]]
bind <- subset(corr_matrix,
subset = rownames(corr_matrix) %in% marketing_selections,
select = attribute_selections)
DataExplorer::plot_correlation(bind)
WARNING
However, is this what you really want? psych::corr.test generates the correlation matrix, and DataExplorer::plot_correlation calculates the correlation again. It is like the correlation of the correlation.
For a sample dataframe:
df <- structure(list(antibiotic = c(0.828080341411847, 1.52002304506738,
1.31925434545302, 1.66681722567074, 1.17791610945551, 0.950096368502059,
1.10507733691997, 1.0568193215304, 1.03853131016669, 1.02313195567946,
0.868629787234043, 0.902126485349154, 1.12005679002801, 1.88261441540084,
0.137845900627507, 1.07040656448604, 1.41496470588235, 1.30978543173373,
1.16931780610558, 1.05894439450366, 1.24805122785724, 1.21318238007025,
0.497310305098053, 0.872362356327429, 0.902584749481137, 0.999731895498823,
0.907560340983954, 1.05930840957587, 1.40457554864091, 1.09747179272879,
0.944219456216072, 1.10363111431903, 0.974649273935516, 0.989983064420841,
1.14784471036171, 1.17232858907798, 1.44675812720393, 0.727078405331282,
1.36341361598635, 1.06120293299474, 1.06920290856811, 0.711007267992205,
1.39034247642439, 0.710873996527168, 1.30529753573398, 0.781191310196629,
0.921788181250106, 0.932214675722466, 0.752289683770589, 0.942392026874501
), year = c(3, 1, 4, 1, 2, 4, 1, 3, 4, 3, 4, 1, 2, 3, 4, 1, 1,
4, 1, 1, 1, 1, 4, 1, 3, 3, 1, 4, 1, 4, 2, 1, 1, 1, 3, 4, 3, 2,
2, 2, 3, 3, 1, 2, 3, 2, 3, 4, 4, 1), imd.decile = c(8, 2, 5,
5, 4, 3, 2, 8, 6, 4, 3, 6, 9, 2, 5, 3, 5, 6, 4, 2, 9, 11, 2,
8, 3, 5, 7, 8, 7, 4, 9, 7, 6, 4, 8, 10, 5, 6, 6, 11, 6, 4, 2,
4, 10, 8, 2, 8, 4, 3)), .Names = c("antibiotic", "year", "imd.decile"
), row.names = c(17510L, 6566L, 24396L, 2732L, 13684L, 28136L,
1113L, 15308L, 28909L, 21845L, 23440L, 1940L, 8475L, 22406L,
27617L, 4432L, 3411L, 27125L, 6891L, 6564L, 1950L, 5683L, 25240L,
5251L, 20058L, 18068L, 5117L, 29066L, 2807L, 24159L, 12309L,
6044L, 7629L, 2336L, 16583L, 23921L, 17465L, 14911L, 8879L, 13929L,
17409L, 19421L, 7239L, 11570L, 15283L, 8283L, 16246L, 27950L,
23723L, 4411L), class = "data.frame")
I am trying to graph imd.decile by antibiotic for each year
library(ggplot2)
p <- ggplot(df, aes(x = imd.decile, y = antibiotic, group = factor(year))) +
stat_summary(geom = "line", fun.y = mean)
p
How do I add the wave to colour the corresponding graph and add a legend (I can't seem to use the aes command correctly).