I have a question about this simple code. I cannot understand why it produces two different plots.
boxplot(split(iris$Sepal.Length, iris$Species))
boxplot(iris[,1,1],iris[,1,2],iris[,1,3])
The answer to this issue can be seen by exploring the data for barplot:
The code split(iris$Sepal.Length, iris$Species) will produce these result:
$setosa
[1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0
[27] 5.0 5.2 5.2 4.7 4.8 5.4 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 5.3 5.0
$versicolor
[1] 7.0 6.4 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1 5.6 6.7 5.6 5.8 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6
[27] 6.8 6.7 6.0 5.7 5.5 5.5 5.8 6.0 5.4 6.0 6.7 6.3 5.6 5.5 5.5 6.1 5.8 5.0 5.6 5.7 5.7 6.2 5.1 5.7
$virginica
[1] 6.3 5.8 7.1 6.3 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4 6.8 5.7 5.8 6.4 6.5 7.7 7.7 6.0 6.9 5.6 7.7 6.3 6.7 7.2
[27] 6.2 6.1 6.4 7.2 7.4 7.9 6.4 6.3 6.1 7.7 6.3 6.4 6.0 6.9 6.7 6.9 5.8 6.8 6.7 6.7 6.3 6.5 6.2 5.9
Which are three different variables after using split() function. The plot is also different:
When splitting you create new variables according to Species.
For the second code: boxplot(iris[,1,1],iris[,1,2],iris[,1,3]) the output is the same variable for iris[,1,1],iris[,1,2],iris[,1,3]:
iris[,1,1]
[1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0
[27] 5.0 5.2 5.2 4.7 4.8 5.4 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 5.3 5.0 7.0 6.4
[53] 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1 5.6 6.7 5.6 5.8 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6 6.8 6.7
[79] 6.0 5.7 5.5 5.5 5.8 6.0 5.4 6.0 6.7 6.3 5.6 5.5 5.5 6.1 5.8 5.0 5.6 5.7 5.7 6.2 5.1 5.7 6.3 5.8 7.1 6.3
[105] 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4 6.8 5.7 5.8 6.4 6.5 7.7 7.7 6.0 6.9 5.6 7.7 6.3 6.7 7.2 6.2 6.1 6.4 7.2
[131] 7.4 7.9 6.4 6.3 6.1 7.7 6.3 6.4 6.0 6.9 6.7 6.9 5.8 6.8 6.7 6.7 6.3 6.5 6.2 5.9
iris[,1,2]
[1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0
[27] 5.0 5.2 5.2 4.7 4.8 5.4 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 5.3 5.0 7.0 6.4
[53] 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1 5.6 6.7 5.6 5.8 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6 6.8 6.7
[79] 6.0 5.7 5.5 5.5 5.8 6.0 5.4 6.0 6.7 6.3 5.6 5.5 5.5 6.1 5.8 5.0 5.6 5.7 5.7 6.2 5.1 5.7 6.3 5.8 7.1 6.3
[105] 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4 6.8 5.7 5.8 6.4 6.5 7.7 7.7 6.0 6.9 5.6 7.7 6.3 6.7 7.2 6.2 6.1 6.4 7.2
[131] 7.4 7.9 6.4 6.3 6.1 7.7 6.3 6.4 6.0 6.9 6.7 6.9 5.8 6.8 6.7 6.7 6.3 6.5 6.2 5.9
iris[,1,3]
[1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0
[27] 5.0 5.2 5.2 4.7 4.8 5.4 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 5.3 5.0 7.0 6.4
[53] 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1 5.6 6.7 5.6 5.8 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6 6.8 6.7
[79] 6.0 5.7 5.5 5.5 5.8 6.0 5.4 6.0 6.7 6.3 5.6 5.5 5.5 6.1 5.8 5.0 5.6 5.7 5.7 6.2 5.1 5.7 6.3 5.8 7.1 6.3
[105] 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4 6.8 5.7 5.8 6.4 6.5 7.7 7.7 6.0 6.9 5.6 7.7 6.3 6.7 7.2 6.2 6.1 6.4 7.2
[131] 7.4 7.9 6.4 6.3 6.1 7.7 6.3 6.4 6.0 6.9 6.7 6.9 5.8 6.8 6.7 6.7 6.3 6.5 6.2 5.9
That is why the output will be the same for the plots:
If you wnat to compare the first three variables in a boxplot you could use boxplot(iris[,1],iris[,2],iris[,3]) where iris[,1] is related to first variable in iris and so on.
Related
I have data with multiple variables. I tried to covert my data to numeric (each variable). I have tried as.numeric, but I got an error:
Error: (list) object cannot be coerced to type 'double'.
I also tried as.matrix, and it is fine. But I still need to convert my data to numeric. So, I tried as.numeric(unlist(iris[,1:4])), but it is not what I want. Here is my try:
data("iris")
iris[,1:4]
> is.data.frame(iris[,1:4])
[1] TRUE
> is.numeric(iris[,1:4])
[1] FALSE
This is not what I want.
> as.numeric(unlist(iris[,1:4]))
[1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0 5.0 5.2 5.2 4.7
[31] 4.8 5.4 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 5.3 5.0 7.0 6.4 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2
[61] 5.0 5.9 6.0 6.1 5.6 6.7 5.6 5.8 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6 6.8 6.7 6.0 5.7 5.5 5.5 5.8 6.0 5.4 6.0 6.7 6.3 5.6 5.5
[91] 5.5 6.1 5.8 5.0 5.6 5.7 5.7 6.2 5.1 5.7 6.3 5.8 7.1 6.3 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4 6.8 5.7 5.8 6.4 6.5 7.7 7.7 6.0
I would like to have it like this (without the name of the variables):
[,1] [,2] [,3] [,4]
[1,] 5.1 3.5 1.4 0.2
[2,] 4.9 3.0 1.4 0.2
[3,] 4.7 3.2 1.3 0.2
[4,] 4.6 3.1 1.5 0.2
I need to remove the name of variables and make it as a matrix as above.
any help, please?
We can do:
res<-sapply(iris[,-5], as.numeric)
attr(res,"dimnames") <- NULL
Or as #markus suggests simply:
unname(as.matrix(iris[,1:4]))
Result:
[,1] [,2] [,3] [,4]
[1,] 5.1 3.5 1.4 0.2
[2,] 4.9 3.0 1.4 0.2
[3,] 4.7 3.2 1.3 0.2
[4,] 4.6 3.1 1.5 0.2
There are 5 columns in "iris", which are Sepal.Length, Sepal.Width, Petal.Length, Petal.Width & Species. I have make a few tries as follows:
The function of unique() in each column was workable.
The function of sapply() was also good when I used the FUN as mean. However, I got an Error when I try to use the FUN as unique.
sapply(iris,unique)
$Sepal.Length
[1] 5.1 4.9 4.7 4.6 5.0 5.4 4.4 4.8 4.3 5.8 5.7 5.2 5.5 4.5 5.3 7.0 6.4 6.9 6.5 6.3 6.6 5.9 6.0 6.1 5.6 6.7 6.2
[28] 6.8 7.1 7.6 7.3 7.2 7.7 7.4 7.9
$Sepal.Width
[1] 3.5 3.0 3.2 3.1 3.6 3.9 3.4 2.9 3.7 4.0 4.4 3.8 3.3 4.1 4.2 2.3 2.8 2.4 2.7 2.0 2.2 2.5 2.6
$Petal.Length
[1] 1.4 1.3 1.5 1.7 1.6 1.1 1.2 1.0 1.9 4.7 4.5 4.9 4.0 4.6 3.3 3.9 3.5 4.2 3.6 4.4 4.1 4.8 4.3 5.0 3.8 3.7 5.1
[28] 3.0 6.0 5.9 5.6 5.8 6.6 6.3 6.1 5.3 5.5 6.7 6.9 5.7 6.4 5.4 5.2
$Petal.Width
[1] 0.2 0.4 0.3 0.1 0.5 0.6 1.4 1.5 1.3 1.6 1.0 1.1 1.8 1.2 1.7 2.5 1.9 2.1 2.2 2.0 2.4 2.3
$Species
[1] setosa versicolor virginica
Error in if (n <= 1L || lenl[n] <= width) n else max(1L, which.max(lenl > :
missing value where TRUE/FALSE needed
It's seen that sapply() and unique() have already done their works, but why the Error was showed on the console? I have tried to use "option(error=recover)";however, I couldn't figure it out.... Is it because the class of the Species is factor? How can I make it work?
Actually, I meet the same problem when I take the lesson of swirl. It has stocked me for few days...Could anyone help me to solve the problem? I will appreciate for your help. Thanks.
I'm trying to bootstrap the standard error of the mean using the purrr::rerun() function in R. For example, here I'm trying to find the bootsrapped standard error for the Sepal.Length variable
sample_the_mean <- function(x) {
the_sample <- sample(x, replace = TRUE)
mean(the_sample)
}
sample_the_mean(iris$Sepal.Length)
#> [1] 5.894667
Seemed to work fine when used one time. Here's with purrr::rerun(); I'll just show the first list element of the output (but the list has an element for every iteration, so 10 total elements):
out <- purrr::rerun(10, sample_the_mean, x = iris$Sepal.Length)
out[[1]]
#> [[1]]
#> function (x)
#> {
#> the_sample <- sample(x, replace = TRUE)
#> mean(the_sample)
#> }
#>
#> $x
#> [1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7 5.4
#> [18] 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0 5.0 5.2 5.2 4.7 4.8 5.4 5.2 5.5
#> [35] 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6 5.3 5.0 7.0
#> [52] 6.4 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1 5.6 6.7 5.6 5.8
#> [69] 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6 6.8 6.7 6.0 5.7 5.5 5.5 5.8 6.0 5.4
#> [86] 6.0 6.7 6.3 5.6 5.5 5.5 6.1 5.8 5.0 5.6 5.7 5.7 6.2 5.1 5.7 6.3 5.8
#> [103] 7.1 6.3 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4 6.8 5.7 5.8 6.4 6.5 7.7 7.7
#> [120] 6.0 6.9 5.6 7.7 6.3 6.7 7.2 6.2 6.1 6.4 7.2 7.4 7.9 6.4 6.3 6.1 7.7
#> [137] 6.3 6.4 6.0 6.9 6.7 6.9 5.8 6.8 6.7 6.7 6.3 6.5 6.2 5.9
As you can see, not the mean, but rather the sample itself is returned. Any thoughts on why this is? How can I do this differently? I'd prefer to not use a package (other than purrr, in this particular case).
The following works:
set.seed(2018)
purrr::rerun(10, sample_the_mean(iris$Sepal.Length))
#[[1]]
#[1] 5.73
#
#[[2]]
#[1] 5.810667
#
#[[3]]
#[1] 5.868667
#
#[[4]]
#[1] 5.902
#
#[[5]]
#[1] 5.844
#
#[[6]]
#[1] 5.746667
#
#[[7]]
#[1] 5.877333
#
#[[8]]
#[1] 5.853333
#
#[[9]]
#[1] 5.821333
#
#[[10]]
#[1] 5.768
You can see from ?rerun that ... refers to the expressions to be re-run. So in your case you need to specify a single expression as sample_the_mean(iris$Sepal.Length), which will get captured as a quosure and then evaluated. Perhaps type rerun into an R terminal to see what's going on under the hood.
data("iris")
len<-split(iris$Petal.Length,iris$Species)
len[5]
len data:
$setosa
[1] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 1.5 1.6 1.4 1.1 1.2 1.5 1.3 1.4 1.7 1.5 1.7 1.5 1.0 1.7 1.9 1.6 1.6 1.5 1.4 1.6
[31] 1.6 1.5 1.5 1.4 1.5 1.2 1.3 1.4 1.3 1.5 1.3 1.3 1.3 1.6 1.9 1.4 1.6 1.4 1.5 1.4
$versicolor
[1] 4.7 4.5 4.9 4.0 4.6 4.5 4.7 3.3 4.6 3.9 3.5 4.2 4.0 4.7 3.6 4.4 4.5 4.1 4.5 3.9 4.8 4.0 4.9 4.7 4.3 4.4 4.8 5.0 4.5 3.5
[31] 3.8 3.7 3.9 5.1 4.5 4.5 4.7 4.4 4.1 4.0 4.4 4.6 4.0 3.3 4.2 4.2 4.2 4.3 3.0 4.1
$virginica
[1] 6.0 5.1 5.9 5.6 5.8 6.6 4.5 6.3 5.8 6.1 5.1 5.3 5.5 5.0 5.1 5.3 5.5 6.7 6.9 5.0 5.7 4.9 6.7 4.9 5.7 6.0 4.8 4.9 5.6 5.8
[31] 6.1 6.4 5.6 5.1 5.6 6.1 5.6 5.5 4.8 5.4 5.6 5.1 5.1 5.9 5.7 5.2 5.0 5.2 5.4 5.1
Error:NULL
We need to loop through the list (lapply- returns a list, while sapply returns a vector) and get the 5th element
sapply(len, '[', 5)
#setosa versicolor virginica
# 1.4 4.6 5.8
You can also use purrr's map function within the tidyverse.
library(tidyverse)
split(iris$Petal.Length,iris$Species) %>% map(5)
$setosa
[1] 1.4
$versicolor
[1] 4.6
$virginica
[1] 5.8
I am very new to R and I am currently playing around with it, I've run into an issue with plotting the following dataframe that I imported from a CSV.
studentname dateofbirth GSF3A3U FJÖ1UF05AU EÐLI2GR05BT FOR3D3U FOR3L3DU FOR4A3U ROB2B3U STÆR3FV05ET USA1012 WIN3B3DU userid
1 Ada Gauidóttir 13.8.1997 8.3 8.0 4.0 6.8 8.5 8.1 4.0 5.9 9.0 9.4 1
2 Gjaflaug Amildóttir 14.6.1998 6.0 6.6 6.2 8.9 4.7 9.4 8.5 8.1 4.3 5.3 2
3 Unndís Jónasdóttir 2.11.1998 8.7 7.8 6.9 10.0 7.0 10.0 9.3 5.4 7.2 5.8 3
4 Sigjón Elfráðurson 14.10.1996 9.3 8.9 6.2 8.1 9.7 5.5 6.8 9.0 6.9 4.2 4
5 Þórbjörg Rökkvidóttir 12.10.2000 4.9 6.9 5.2 6.9 5.3 5.5 5.6 4.8 8.9 9.2 5
6 Richard Hlérson 3.2.2000 9.4 7.7 8.4 6.1 6.4 9.6 4.9 7.2 9.3 7.0 6
7 Tala Arnalddóttir 18.8.1997 7.9 7.1 6.9 6.0 9.3 5.4 8.1 6.8 5.8 6.7 7
8 Petrína Estefandóttir 24.9.1994 9.6 4.9 5.0 8.4 7.9 8.7 5.5 10.0 4.0 9.5 8
9 Tanja Finnlaugurdóttir 11.7.1993 6.7 6.5 6.9 8.3 6.3 9.6 9.1 4.2 9.6 4.7 9
10 Elly Amosdóttir 6.7.2001 4.8 7.0 4.3 9.5 7.1 4.2 6.6 5.3 9.0 4.4 10
I am trying to plot this data so that the course names are on the X-axis and the rows for each course(the grades) are displayed on the Y-axis.
If I use the code below:
students[,3:12, drop=FALSE]
This result below is exactly what I'm looking for, but how do I scatterplot this?
GSF3A3U FJÖ1UF05AU EÐLI2GR05BT FOR3D3U FOR3L3DU FOR4A3U ROB2B3U STÆR3FV05ET USA1012 WIN3B3DU
1 8.3 8.0 4.0 6.8 8.5 8.1 4.0 5.9 9.0 9.4
2 6.0 6.6 6.2 8.9 4.7 9.4 8.5 8.1 4.3 5.3
3 8.7 7.8 6.9 10.0 7.0 10.0 9.3 5.4 7.2 5.8
4 9.3 8.9 6.2 8.1 9.7 5.5 6.8 9.0 6.9 4.2
5 4.9 6.9 5.2 6.9 5.3 5.5 5.6 4.8 8.9 9.2
6 9.4 7.7 8.4 6.1 6.4 9.6 4.9 7.2 9.3 7.0
7 7.9 7.1 6.9 6.0 9.3 5.4 8.1 6.8 5.8 6.7
8 9.6 4.9 5.0 8.4 7.9 8.7 5.5 10.0 4.0 9.5
9 6.7 6.5 6.9 8.3 6.3 9.6 9.1 4.2 9.6 4.7
10 4.8 7.0 4.3 9.5 7.1 4.2 6.6 5.3 9.0 4.4