Cómo representar funciones en R

Problema

Queremos representar funciones en R.

Solución

En nuestro ejemplo queremos representar la siguiente ecuación de segundo grado: 2x² + 20x + 3 = 0

• Opción 1: base package

f <- function(x) {
  x ^ 2 + 20 * x + 3
}
curve(expr = f, from = -100, to = 100)


• Opción 2: ggplot2

library(ggplot2)
ggplot(data.frame(x = c(-100, 100)), aes(x)) +
  stat_function(
    fun = function(x) {
      x ^ 2 + 20 * x + 3
    }
  ) +
  geom_hline(yintercept = 0) +
  geom_vline(xintercept = 0)

 

Notas

Si queremos añadir más funciones:

• Opción 1: Base package

f1 <- function(x) {
  x ^ 2 + 20 * x + 3
}
f2 <- function(x) {
  x ^ 2 + 50 * x + 100
}
curve(expr = f1, from = -100, to = 100)
curve(expr = f2, from = -100, to = 100, col  = 2, add = TRUE)


• Opción 2: ggplot2

ggplot(data.frame(x = c(-100, 100)), aes(x)) +
  stat_function(
    fun = function(x) {
      x ^ 2 + 20 * x + 3
    }
  ) +
  stat_function(
    fun = function(x) {
      x ^ 2 + 50 * x + 100
    },
    colour = "red"
  ) +
  geom_hline(yintercept = 0) +
  geom_vline(xintercept = 0)
 

Entradas relacionadas

• Resolver ecuaciones de segundo grado en R

Referencias

• ggplot2 reference

How to plot functions in R

Problem

We would like to plot functions in R.

Solution

In our example we will plot the following quadratic equation: 2x² + 20x + 3 = 0

• Option 1: base package

f <- function(x) {
  x ^ 2 + 20 * x + 3
}
curve(expr = f, from = -100, to = 100)


• Option 2: ggplot2

library(ggplot2)
ggplot(data.frame(x = c(-100, 100)), aes(x)) +
  stat_function(
    fun = function(x) {
      x ^ 2 + 20 * x + 3
    }
  ) +
  geom_hline(yintercept = 0) +
  geom_vline(xintercept = 0)

 

Notes

If we want to add more functions:

• Option 1: Base package

f1 <- function(x) {
  x ^ 2 + 20 * x + 3
}
f2 <- function(x) {
  x ^ 2 + 50 * x + 100
}
curve(expr = f1, from = -100, to = 100)
curve(expr = f2, from = -100, to = 100, col  = 2, add = TRUE)


• Option 2: ggplot2

ggplot(data.frame(x = c(-100, 100)), aes(x)) +
  stat_function(
    fun = function(x) {
      x ^ 2 + 20 * x + 3
    }
  ) +
  stat_function(
    fun = function(x) {
      x ^ 2 + 50 * x + 100
    },
    colour = "red"
  ) +
  geom_hline(yintercept = 0) +
  geom_vline(xintercept = 0)
 

References

• ggplot2 reference

SQL CASE or multiple if else statements in R

Problem

We would like to create in R a conditional statement equivalent to the CASE statement in SQL.

    idnat     idbp
1  french mainland
2  french   colony
3  french overseas
4 foreign  foreign

df <-structure(list(idnat = structure(c(2L, 2L, 2L, 1L), .Label = c("foreign", 
"french"), class = "factor"), idbp = structure(c(3L, 1L, 4L, 
2L), .Label = c("colony", "foreign", "mainland", "overseas"), class = "factor")), .Names = c("idnat", 
"idbp"), class = "data.frame", row.names = c(NA, -4L))

Solution

There are multiple alternatives. We will provide a couple of options using dplyr and sqldf.

• Option 1: dplyr

library(dplyr)
df %>%
  mutate(idnat2 = case_when(
    idbp == 'mainland' ~ "mainland",
    idbp %in% c("colony", "overseas") ~ "overseas",
    TRUE ~ "foreign"
  ))

• Option 2: sqldf

library(sqldf)
sqldf(
  "SELECT idnat, idbp,
        CASE
          WHEN idbp IN ('colony', 'overseas') THEN 'overseas'
          ELSE idbp
        END AS idnat2
       FROM df"
)
 

Results

    idnat     idbp   idnat2
1  french mainland mainland
2  french   colony overseas
3  french overseas overseas
4 foreign  foreign  foreign

References

• A general vectorised if
• SQL CASE Statement
• stackoverflow

How to highlight specific data points in a scatter plot with ggplot2

Problem

We would like to highlight specific data points in a scatter plot created with ggplot2. In our example, we will use the mpg dataset (Fuel economy data from 1999 and 2008 for 38 popular models of car). We want to highlight in red and a larger point size those data points whose displ (engine displacement) are greater than 5 and hwy (highway miles per gallon) greater than 20.

Solution

Option 1

We add the conditions for both attributes, colour and size, inside geom_point. Then we control manually those using scale_colour_manual and scale_size_manual respectively. Finally, we remove the legend.

ggplot(data = mpg) + 
  geom_point(mapping = aes(x = displ, y = hwy, colour = displ > 5 & hwy > 20, size = displ > 5 & hwy > 20)) + 
  scale_colour_manual(values = c("black", "red")) + 
  scale_size_manual(values =c(1.5, 3))+
  theme(legend.position = "none")

Option 2

We create two layers of data points using geom_point. The first layer include all data points in black (by default but e). The second layer adds the points we would like to highlight in red with a larger point size.

ggplot(data = mpg) + 
  geom_point(mapping = aes(x = displ, y = hwy), colour= "black") +
  geom_point(data = subset(mpg, displ > 5 & hwy > 20), aes(x = displ, y = hwy), colour= "red", size = 3)
 

In ggplot2 the layers are added sequentially. If we change the order of the layers, we would get the following result.

ggplot(data = mpg) + 
   geom_point(data = subset(mpg, displ > 5 & hwy > 20), aes(x = displ, y = hwy), colour= "red", size = 3) +
   geom_point(mapping = aes(x = displ, y = hwy), colour= "black")
 

Cómo filtrar múltiples valores en una columna en R

Problema

Queremos filtrar múltiples valores en una columna en R. En nuestro ejemplo, queremos filtrar las filas que contengan la cadena de texto Tom o Lynn en la columna name.

Ejemplo

  days  name
1   88  Lynn
2   11   Tom
3    2 Chris
4    5  Lisa
5   22  Kyla
6    1   Tom
7  222  Lynn
8    2  Lynn

df <-
  data.frame(
    days = c(88, 11, 2, 5, 22, 1, 222, 2),
    name = c("Lynn", "Tom", "Chris", "Lisa", "Kyla", "Tom", "Lynn", "Lynn")
  ) 

Solución

Base package

df[df$name %in% c("Tom", "Lynn"), ] # or
target <- c("Tom", "Lynn")
df[df$name %in% target, ]

  days name
1   88 Lynn
2   11  Tom
6    1  Tom
7  222 Lynn
8    2 Lynn

dplyr

library(dplyr)
filter(df, name %in% c("Tom", "Lynn")) # or
target <- c("Tom", "Lynn")
filter(df, name %in% target)

data.table

library(data.table)
DT <- data.table(df)
DT[name %in% target]

sqldf

library(sqldf)
# Dos alternativas:
sqldf('SELECT *
      FROM df 
      WHERE name = "Tom" OR name = "Lynn"')
sqldf('SELECT *
      FROM df 
      WHERE name IN ("Tom", "Lynn")')

Entradas relacionadas

• Filtrar filas que contengan una cadena de texto en R

Referencias

• stackoverflow

How to filter multiple values on a column in R

Problem

We want to filter multiple values on a column in R. In our example, we want to subset the rows containing the string Tom or Lynn for the column name.

Example

  days  name
1   88  Lynn
2   11   Tom
3    2 Chris
4    5  Lisa
5   22  Kyla
6    1   Tom
7  222  Lynn
8    2  Lynn

df <-
  data.frame(
    days = c(88, 11, 2, 5, 22, 1, 222, 2),
    name = c("Lynn", "Tom", "Chris", "Lisa", "Kyla", "Tom", "Lynn", "Lynn")
  ) 

Solution

Base package

df[df$name %in% c("Tom", "Lynn"), ] # or
target <- c("Tom", "Lynn")
df[df$name %in% target, ]

  days name
1   88 Lynn
2   11  Tom
6    1  Tom
7  222 Lynn
8    2 Lynn

dplyr

library(dplyr)
filter(df, name %in% c("Tom", "Lynn")) # or
target <- c("Tom", "Lynn")
filter(df, name %in% target)

data.table

library(data.table)
DT <- data.table(df)
DT[name %in% target]

sqldf

library(sqldf)
# Two alternatives:
sqldf('SELECT *
      FROM df 
      WHERE name = "Tom" OR name = "Lynn"')
sqldf('SELECT *
      FROM df 
      WHERE name IN ("Tom", "Lynn")')

Related posts

• Filter rows containing a certain string in R

References

• stackoverflow

How to draw a stratified sample in R

Problem

We want to draw a stratified sample from a data frame in R.

Solution

Let's look at two examples, with one or several groups.

One group

We extract 3 records from each of the species: setosa, versicolor y virginica.

Base package

set.seed(1)
iris1 <- lapply(split(iris, iris$Species), function(x) x[sample(nrow(x), 3), ])
do.call("rbind", iris1) 

              Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
setosa.14              4.3         3.0          1.1         0.1     setosa
setosa.19              5.7         3.8          1.7         0.3     setosa
setosa.28              5.2         3.5          1.5         0.2     setosa
versicolor.96          5.7         3.0          4.2         1.2 versicolor
versicolor.60          5.2         2.7          3.9         1.4 versicolor
versicolor.94          5.0         2.3          3.3         1.0 versicolor
virginica.148          6.5         3.0          5.2         2.0  virginica
virginica.133          6.4         2.8          5.6         2.2  virginica
virginica.131          7.4         2.8          6.1         1.9  virginica

dplyr

library(dplyr)
set.seed(1)
iris %>%
  group_by(Species) %>%
  sample_n(., 3)

 Source: local data frame [9 x 5]
Groups: Species

  Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
1          4.3         3.0          1.1         0.1     setosa
2          5.7         3.8          1.7         0.3     setosa
3          5.2         3.5          1.5         0.2     setosa
4          5.7         3.0          4.2         1.2 versicolor
5          5.2         2.7          3.9         1.4 versicolor
6          5.0         2.3          3.3         1.0 versicolor
7          6.5         3.0          5.2         2.0  virginica
8          6.4         2.8          5.6         2.2  virginica
9          7.4         2.8          6.1         1.9  virginica

Two groups

For each number of cylinders (4, 6 u 8) we will extract 2 records with automatic transmission (am = 0) and 2 with manual transmission (am = 1).

Base package

set.seed(1)
mtcars1 <- lapply(split(mtcars, list(mtcars$cyl, mtcars$am)), function(x) x[sample(nrow(x), 2), ])
do.call("rbind", mtcars1) 

                       mpg cyl  disp  hp drat    wt  qsec vs am gear carb
0.4.Merc 240D         24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
0.4.Toyota Corona     21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1
1.4.Fiat X1-9         27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
1.4.Lotus Europa      30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2
0.6.Hornet 4 Drive    21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
0.6.Merc 280          19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4
1.6.Ferrari Dino      19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6
1.6.Mazda RX4 Wag     21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
0.8.Chrysler Imperial 14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
0.8.Hornet Sportabout 18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
1.8.Ford Pantera L    15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4
1.8.Maserati Bora     15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8

dplyr

set.seed(1)
mtcars %>%
  group_by(cyl, am) %>%
  sample_n(., 2)

Source: local data frame [12 x 11]
Groups: cyl, am

    mpg cyl  disp  hp drat    wt  qsec vs am gear carb
1  24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
2  21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1
3  27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
4  30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2
5  21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
6  19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4
7  19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6
8  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
9  14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
10 18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
11 15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4
12 15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8

Related posts

• How to select a random sample in R

References

• stackoverflow