I'm trying to manually order my intersections with UpSetR but I don't know if it's possible to do it. order.by only allows freq or degree and I don't see any other parameter that could produce what I want.
a <- list(one = c(1, 2, 3, 5,11,19),
two = c(1, 2, 4, 5, 11, 13),
three = c(1, 5, 6, 7, 11, 19),
four = c(1, 5, 6, 8, 13, 19))
upset(fromList(a), sets = c("one", "two", "three", "four"))
On the left is what I actually have, on the right is what I want to reproduce.
Does anyone knows if it is possible? Thanks.
I just found the solution with ComplexUpset package :
if(!require(devtools)) install.packages("devtools")
devtools::install_github("krassowski/complex-upset")
upset(a,colnames(a),intersections=list(c("one"),c("one","two"),c("one","two","three","four"),c("one","three","four"),c("one","two","three"),c("three"),c("two"),c("four"),c("three","four"),c("two","four")),sort_intersections=FALSE)
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I have two datasets which contains a distrbution of 90 data points into 2 and 4 groups/rows and I would like to determine which one out of the two has better distributed the data and plot the result to visually see which one has done this. Better distribution means which one has made it so each group has a similar/same number of data. For example, we can see that the result of Grouped 2 the second group contains larger values for each column compared to the first column so 1 of the 2 groups contains larger values which means its not well distributed among the 2 groups.
I quite new to R so I am unsure how I could go about doing this. Would appreciate any insight into what approach could be used.
R
Grouped into 4
Values <- matrix(c(1, 6, 3, 6, 6, 8,
3, 3, 5, 3, 3, 3,
6, 7, 6, 7, 5, 4,
9, 4, 4, 5, 5, 3), nrow = 4, ncol = 6, byrow = TRUE)
Grouped into 2
Values <- matrix(c(3, 6, 4, 3, 4, 6,
12, 9, 12, 12, 11, 9), nrow = 2, ncol = 6, byrow = TRUE)
You can do this with some basic statistics, using hypothesis testing i.e. testing whether the two groups are statistically different or not. The stats package in R has a lot of tests that you can try and use, each with its own assumptions. Here is one:
Making the matrix
values <- matrix(c(3, 6, 4, 3, 4, 6,
12, 9, 12, 12, 11, 9), nrow = 2, ncol = 6, byrow = TRUE)
Conducting t-test
t.test(values[1, ], values[2, ], paired = FALSE)
Will give you this:
Welch Two Sample t-test
data: values[1, ] and values[2, ]
t = -7.9279, df = 9.945, p-value = 1.318e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-8.328203 -4.671797
sample estimates:
mean of x mean of y
4.333333 10.833333
The means of values[1, ] is smaller than values[2, ], with a p-value of 1.3e-05.
I create the radarchart below and I would like to know if it is possible to exclude all the person with value=6 from the hoverinfo text.
library(radarchart)
# Using the data frame interface
chartJSRadar(scores=skills)
# Or using a list interface
labs <- c("Communicator", "Data Wangler", "Programmer", "Technologist", "Modeller", "Visualizer")
scores <- list("Rich" = c(9, 7, 4, 5, 3, 7),
"Andy" = c(7, 6, 6, 2, 6, 9),
"Aimee" = c(6, 5, 8, 4, 7, 6))
# Default settings
chartJSRadar(scores=scores, labs=labs)
I have a vector of calls made on each days of a certain month.
callsperDayforMonth <- c(3, 1, 2, 1, 1, 3, 9, 1, 4, 2, 6, 4, 9, 13, 15, 2, 5, 5, 2, 7, 3, 0, 1, 2, 7, 1, 8, 6, 9, 4)
I also have a vector of factors which spans the range of the "callsperDayforMonth" vector.
"0-2" "3-5" "6-8" "9-11" "12-14" "16+"
I need to create a histogram, with the factors on the horizontal axis.
How can this be done.
The hist command has an argument breaks that can be a vector of the breakpoints to be used. That should do what you want.
Or you could use table and cut to do the counts yourself and create a barplot from the result.
For example:
library(ggplot2)
cuts <- cut(callsperDayforMonth,
breaks = c(-Inf,2, 5, 8, 11, 14, 16, Inf),
labels = c("0-2", "3-5", "6-8", "9-11", "12-14", "15-16", "16+"))
df <- data.frame(cuts, callsperDayforMonth)
ggplot(df, aes(x=cuts)) + geom_bar(stat = "count")
I'm have to use R instead of Matlab and I'm new to it.
I have a large array of data repeating like 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10...
I need to find the locations where values equal to 1, 4, 7, 10 are found to create a sample using those locations.
In this case it will be position(=corresponding value) 1(=1) 4(=4) 7(=7) 10(=10) 11(=1) 14(=4) 17(=7) 20(=10) and so on.
in MatLab it would be y=find(ismember(x,[1, 4, 7, 10 ])),
Please, help! Thanks, Pavel
something like this?
foo <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
bar <- c(1, 4, 7, 10)
which(foo %in% bar)
#> [1] 1 4 7 10 11 14 17 20
#nicola, feel free to copy my answer and get the recognition for your answer, simply trying to close answered questions.
The %in% operator is what you want. For example,
# data in x
targets <- c(1, 4, 7, 10)
locations <- x %in% targets
# locations is a logical vector you can then use:
y <- x[locations]
There'll be an extra step or two if you wanted the row and column indices of the locations, but it's not clear if you do. (Note, the logicals will be in column order).
I am trying to fit a curved line segment to a dataset. While I can create the line it is always connected back to the starting point. I can't figure out how to get rid of this. I would really appreciate any help. Here is the code
mscF25=c(-12.94382785, -11.0281518, -9.186403952, -7.691576905, -6.470229134, -5.43000796, -4.559074508, -12.87271022, -10.0646268, -6.796208225, -4.433351598, -2.928135666, -1.979265556, -1.38936463, -11.05819006, -7.785838826, -5.297330858, -3.674159165, -2.64702678, -1.980973252, -1.533714976, -11.83971039, -9.168353808, -6.89192172, -5.23424594, -4.033326594, -3.148798626, -2.480469911)
bscF25=c(4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10)
df25 <- data.frame(bscF25,mscF25)
plot(mscF25 ~ bscF25, data = df25)
ls25 <- loess(mscF25 ~ bscF25, data = df25, span = 3)
lines(df25$bscF25, ls25$fitted)
You might try the scatter.smooth function: "Plot and add a smooth curve computed by loess to a scatter plot"
scatter.smooth(x = df25$bscF25, y = df25$mscF25, span = 3)