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.
Related
I'm trying to flag rows that have values that match elements in 2 external vectors. I need these values to match, meaning it can't be any case, but something where both conditions are met (i.e. not using an %in% operator).
Here's what I've tried:
widths <- c(1.2, 1.7, 1.8, 1.8, 1.9)
species <- c(rep("versicolor", 3), rep("virginica", 2))
iris_flagged <- iris %>%
filter(Species != "setosa") %>%
mutate(flag = ifelse((Petal.Width == widths & Species == species), "check", "")) %>%
arrange(Species, Petal.Width) #to help with visualization
But when I run this some flags are missed. If you look at the data you'll see that rows 14, 50, 57, 59, 61:64, and 66:71 are missing flags.
A clunky, non-robust method could look like this, but I really want to make it robust and use these vectors to my advantage.
iris_flagged2 <- iris %>%
filter(Species != "setosa") %>%
mutate(flag = ifelse((Species == "versicolor" & Petal.Width == 1.2 |
Species == "versicolor" & Petal.Width == 1.7 |
Species == "versicolor" & Petal.Width == 1.8 |
Species == "virginica" & Petal.Width == 1.8 |
Species == "virginica" & Petal.Width == 1.9),
"check", "")) %>%
arrange(Species, Petal.Width)
Thanks for your help. I tried to search but wasn't sure how to phrase my question.
Here is one way using map2 to pass each pair of condition and reduce to combine them into one.
library(tidyverse)
iris2 <- iris %>% filter(Species != "setosa")
iris2 <- iris2 %>%
mutate(flag = ifelse(map2(species, widths,
~Species == .x & Petal.Width == .y) %>% reduce(`|`), 'check', ''))
iris2
Additional to the comment of #akrun, here is a possible solution for your task:
library(dplyr)
iris %>%
mutate(across(-Species, ~round(.,1)),
flag = ifelse(Petal.Width %in% widths &
Species %in% species, "check",""))
output:
Sepal.Length Sepal.Width Petal.Length Petal.Width Species flag
1 5.1 3.5 1.4 0.2 setosa
2 4.9 3.0 1.4 0.2 setosa
3 4.7 3.2 1.3 0.2 setosa
4 4.6 3.1 1.5 0.2 setosa
5 5.0 3.6 1.4 0.2 setosa
6 5.4 3.9 1.7 0.4 setosa
7 4.6 3.4 1.4 0.3 setosa
8 5.0 3.4 1.5 0.2 setosa
9 4.4 2.9 1.4 0.2 setosa
10 4.9 3.1 1.5 0.1 setosa
11 5.4 3.7 1.5 0.2 setosa
12 4.8 3.4 1.6 0.2 setosa
13 4.8 3.0 1.4 0.1 setosa
14 4.3 3.0 1.1 0.1 setosa
15 5.8 4.0 1.2 0.2 setosa
16 5.7 4.4 1.5 0.4 setosa
17 5.4 3.9 1.3 0.4 setosa
18 5.1 3.5 1.4 0.3 setosa
19 5.7 3.8 1.7 0.3 setosa
20 5.1 3.8 1.5 0.3 setosa
21 5.4 3.4 1.7 0.2 setosa
22 5.1 3.7 1.5 0.4 setosa
23 4.6 3.6 1.0 0.2 setosa
24 5.1 3.3 1.7 0.5 setosa
25 4.8 3.4 1.9 0.2 setosa
26 5.0 3.0 1.6 0.2 setosa
27 5.0 3.4 1.6 0.4 setosa
28 5.2 3.5 1.5 0.2 setosa
29 5.2 3.4 1.4 0.2 setosa
30 4.7 3.2 1.6 0.2 setosa
31 4.8 3.1 1.6 0.2 setosa
32 5.4 3.4 1.5 0.4 setosa
33 5.2 4.1 1.5 0.1 setosa
34 5.5 4.2 1.4 0.2 setosa
35 4.9 3.1 1.5 0.2 setosa
36 5.0 3.2 1.2 0.2 setosa
37 5.5 3.5 1.3 0.2 setosa
38 4.9 3.6 1.4 0.1 setosa
39 4.4 3.0 1.3 0.2 setosa
40 5.1 3.4 1.5 0.2 setosa
41 5.0 3.5 1.3 0.3 setosa
42 4.5 2.3 1.3 0.3 setosa
43 4.4 3.2 1.3 0.2 setosa
44 5.0 3.5 1.6 0.6 setosa
45 5.1 3.8 1.9 0.4 setosa
46 4.8 3.0 1.4 0.3 setosa
47 5.1 3.8 1.6 0.2 setosa
48 4.6 3.2 1.4 0.2 setosa
49 5.3 3.7 1.5 0.2 setosa
50 5.0 3.3 1.4 0.2 setosa
51 7.0 3.2 4.7 1.4 versicolor
52 6.4 3.2 4.5 1.5 versicolor
53 6.9 3.1 4.9 1.5 versicolor
54 5.5 2.3 4.0 1.3 versicolor
55 6.5 2.8 4.6 1.5 versicolor
56 5.7 2.8 4.5 1.3 versicolor
57 6.3 3.3 4.7 1.6 versicolor
58 4.9 2.4 3.3 1.0 versicolor
59 6.6 2.9 4.6 1.3 versicolor
60 5.2 2.7 3.9 1.4 versicolor
61 5.0 2.0 3.5 1.0 versicolor
62 5.9 3.0 4.2 1.5 versicolor
63 6.0 2.2 4.0 1.0 versicolor
64 6.1 2.9 4.7 1.4 versicolor
65 5.6 2.9 3.6 1.3 versicolor
66 6.7 3.1 4.4 1.4 versicolor
67 5.6 3.0 4.5 1.5 versicolor
68 5.8 2.7 4.1 1.0 versicolor
69 6.2 2.2 4.5 1.5 versicolor
70 5.6 2.5 3.9 1.1 versicolor
71 5.9 3.2 4.8 1.8 versicolor check
72 6.1 2.8 4.0 1.3 versicolor
73 6.3 2.5 4.9 1.5 versicolor
74 6.1 2.8 4.7 1.2 versicolor check
75 6.4 2.9 4.3 1.3 versicolor
76 6.6 3.0 4.4 1.4 versicolor
77 6.8 2.8 4.8 1.4 versicolor
78 6.7 3.0 5.0 1.7 versicolor check
79 6.0 2.9 4.5 1.5 versicolor
80 5.7 2.6 3.5 1.0 versicolor
81 5.5 2.4 3.8 1.1 versicolor
82 5.5 2.4 3.7 1.0 versicolor
83 5.8 2.7 3.9 1.2 versicolor check
84 6.0 2.7 5.1 1.6 versicolor
85 5.4 3.0 4.5 1.5 versicolor
86 6.0 3.4 4.5 1.6 versicolor
87 6.7 3.1 4.7 1.5 versicolor
88 6.3 2.3 4.4 1.3 versicolor
89 5.6 3.0 4.1 1.3 versicolor
90 5.5 2.5 4.0 1.3 versicolor
91 5.5 2.6 4.4 1.2 versicolor check
92 6.1 3.0 4.6 1.4 versicolor
93 5.8 2.6 4.0 1.2 versicolor check
94 5.0 2.3 3.3 1.0 versicolor
95 5.6 2.7 4.2 1.3 versicolor
96 5.7 3.0 4.2 1.2 versicolor check
97 5.7 2.9 4.2 1.3 versicolor
98 6.2 2.9 4.3 1.3 versicolor
99 5.1 2.5 3.0 1.1 versicolor
100 5.7 2.8 4.1 1.3 versicolor
101 6.3 3.3 6.0 2.5 virginica
102 5.8 2.7 5.1 1.9 virginica check
103 7.1 3.0 5.9 2.1 virginica
104 6.3 2.9 5.6 1.8 virginica check
105 6.5 3.0 5.8 2.2 virginica
106 7.6 3.0 6.6 2.1 virginica
107 4.9 2.5 4.5 1.7 virginica check
108 7.3 2.9 6.3 1.8 virginica check
109 6.7 2.5 5.8 1.8 virginica check
110 7.2 3.6 6.1 2.5 virginica
111 6.5 3.2 5.1 2.0 virginica
112 6.4 2.7 5.3 1.9 virginica check
113 6.8 3.0 5.5 2.1 virginica
114 5.7 2.5 5.0 2.0 virginica
115 5.8 2.8 5.1 2.4 virginica
116 6.4 3.2 5.3 2.3 virginica
117 6.5 3.0 5.5 1.8 virginica check
118 7.7 3.8 6.7 2.2 virginica
119 7.7 2.6 6.9 2.3 virginica
120 6.0 2.2 5.0 1.5 virginica
121 6.9 3.2 5.7 2.3 virginica
122 5.6 2.8 4.9 2.0 virginica
123 7.7 2.8 6.7 2.0 virginica
124 6.3 2.7 4.9 1.8 virginica check
125 6.7 3.3 5.7 2.1 virginica
126 7.2 3.2 6.0 1.8 virginica check
127 6.2 2.8 4.8 1.8 virginica check
128 6.1 3.0 4.9 1.8 virginica check
129 6.4 2.8 5.6 2.1 virginica
130 7.2 3.0 5.8 1.6 virginica
131 7.4 2.8 6.1 1.9 virginica check
132 7.9 3.8 6.4 2.0 virginica
133 6.4 2.8 5.6 2.2 virginica
134 6.3 2.8 5.1 1.5 virginica
135 6.1 2.6 5.6 1.4 virginica
136 7.7 3.0 6.1 2.3 virginica
137 6.3 3.4 5.6 2.4 virginica
138 6.4 3.1 5.5 1.8 virginica check
139 6.0 3.0 4.8 1.8 virginica check
140 6.9 3.1 5.4 2.1 virginica
141 6.7 3.1 5.6 2.4 virginica
142 6.9 3.1 5.1 2.3 virginica
143 5.8 2.7 5.1 1.9 virginica check
144 6.8 3.2 5.9 2.3 virginica
145 6.7 3.3 5.7 2.5 virginica
146 6.7 3.0 5.2 2.3 virginica
147 6.3 2.5 5.0 1.9 virginica check
148 6.5 3.0 5.2 2.0 virginica
149 6.2 3.4 5.4 2.3 virginica
150 5.9 3.0 5.1 1.8 virginica check
My line of code is
df1<-rbind(df1,assign(paste(x,"_name_",_Date,sep=""),Result))
Basically
assign(paste(x,"_name_",_Date,sep=""),Result)
is the same as
df2
When i do
df1<-rbind(df1,df2)
it works but this needs to be dynamic and constantly changing as i do these updates weekly.
We need get to return the value from the object name string i.e. assign only assign it to an object and it doesn't return the value.
rbind(df1, {
nm1 <- paste(x,"_name_",_Date,sep="")
assign(nm1, Result)
get(nm1)})
Using a small reproducible example
rbind(head(iris), {
nm1 <- 'newobj'
assign(nm1, tail(iris))
get(nm1)})
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#1 5.1 3.5 1.4 0.2 setosa
#2 4.9 3.0 1.4 0.2 setosa
#3 4.7 3.2 1.3 0.2 setosa
#4 4.6 3.1 1.5 0.2 setosa
#5 5.0 3.6 1.4 0.2 setosa
#6 5.4 3.9 1.7 0.4 setosa
#145 6.7 3.3 5.7 2.5 virginica
#146 6.7 3.0 5.2 2.3 virginica
#147 6.3 2.5 5.0 1.9 virginica
#148 6.5 3.0 5.2 2.0 virginica
#149 6.2 3.4 5.4 2.3 virginica
#150 5.9 3.0 5.1 1.8 virginica
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
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