Transforming contingency tables into frequency tables in R

Problem

We want to tranform a contingency table into a frequency table in R.

# Contingency table
tbl <- table (mtcars[, c("am", "gear")])

   gear
am   3  4  5
  0 15  4  0
  1  0  8  5

Frequency tables

We transform the contingency table into a data frame.

df <- as.data.frame(tbl)
df

  am gear Freq
1  0    3   15
2  1    3    0
3  0    4    4
4  1    4    8
5  0    5    0
6  1    5    5

Transforming frequency tables in contingency tables

To transform a frequency table back to a contingency table.

ftable(xtabs(Freq ~ am + gear, data = df)) 

   gear  3  4  5
am              
0       15  4  0
1        0  8  5

It is the equivalent of:

ftable(mtcars[, c("am", "gear")])

References

• Contingency tables in R
• Proportion tables in R

Proportion tables in R

Problem

We want to create proportion tables for one or multiple variables.

Solution

• One variable

tabla <- table(mtcars$am)
prop.table(tabla)

      0       1 
0.59375 0.40625

• Two variables

tabla <- table(mtcars[, c("am", "gear")])
prop.table(tabla)

   gear
am        3       4       5
  0 0.46875 0.12500 0.00000
  1 0.00000 0.25000 0.15625

The prop.table function has two arguments:
• x, table created with the function table
• margin, with three possible values:
  Null - x/sum(x) default like in the previous example.
  1 - proportion calculated by rows.
  2 - proportion calculated by columns.


# By row
prop.table(tabla, 1)

   gear
am          3         4         5
  0 0.7894737 0.2105263 0.0000000
  1 0.0000000 0.6153846 0.3846154

# By column
prop.table(tabla, 2)

   gear
am          3         4         5
  0 1.0000000 0.3333333 0.0000000
  1 0.0000000 0.6666667 1.0000000

• Three variables

tabla <- table(mtcars[, c("am", "gear", "cyl")])
prop.table(tabla)

, , cyl = 4

   gear
am        3       4       5
  0 0.03125 0.06250 0.00000
  1 0.00000 0.18750 0.06250

, , cyl = 6

   gear
am        3       4       5
  0 0.06250 0.06250 0.00000
  1 0.00000 0.06250 0.03125

, , cyl = 8

   gear
am        3       4       5
  0 0.37500 0.00000 0.00000
  1 0.00000 0.00000 0.06250

• Flat Contingency Table

In the previous example, a better approach would be to create a flat contingency table..

tabla <- ftable(mtcars[, c("am", "gear", "cyl")])
prop.table(tabla)

        cyl       4       6       8
am gear                            
0  3        0.03125 0.06250 0.37500
   4        0.06250 0.06250 0.00000
   5        0.00000 0.00000 0.00000
1  3        0.00000 0.00000 0.00000
   4        0.18750 0.06250 0.00000
   5        0.06250 0.03125 0.06250

• Percentage table

We can use the function round.

round(prop.table(tabla)*100, 2)

         cyl     4     6     8
am gear                      
0  3         3.12  6.25 37.50
   4         6.25  6.25  0.00
   5         0.00  0.00  0.00
1  3         0.00  0.00  0.00
   4        18.75  6.25  0.00
   5         6.25  3.12  6.25

round(prop.table(tabla, 1)*100, 2) # By row, am y gear.

        cyl     4     6     8
am gear                      
0  3         6.67 13.33 80.00
   4        50.00 50.00  0.00
   5          NaN   NaN   NaN
1  3          NaN   NaN   NaN
   4        75.00 25.00  0.00
   5        40.00 20.00 40.00

round(prop.table(tabla, 2)*100, 2) # By column, cyl

        cyl     4     6     8
am gear                      
0  3         9.09 28.57 85.71
   4        18.18 28.57  0.00
   5         0.00  0.00  0.00
1  3         0.00  0.00  0.00
   4        54.55 28.57  0.00
   5        18.18 14.29 14.29

References

• Contingency tables in R

Contingency tables in R

Problem

We want to create a contingency table for one or multiple variables.

Solution

• One variable

table(mtcars$am)

 0  1 
19 13 

• Two variables

table(mtcars$am, mtcars$gear)

     3  4  5
  0 15  4  0
  1  0  8  5

If we want to include the names of the variables:

table(mtcars[, c("am", "gear")]) 
tabla <- table(mtcars[, 9:10])
# or the argument dnn:
table(mtcars$am, mtcars$gear, dnn = c("am", "gear"))

   gear
am   3  4  5
  0 15  4  0
  1  0  8  5

• Three variables

table(mtcars[, c("am", "gear", "cyl")])

, , cyl = 4

   gear
am   3  4  5
  0  1  2  0
  1  0  6  2

, , cyl = 6

   gear
am   3  4  5
  0  2  2  0
  1  0  2  1

, , cyl = 8

   gear
am   3  4  5
  0 12  0  0
  1  0  0  2

• Flat contingency tables

In the previous example, a better approach would be to create a flat contingency table.

ftable(mtcars[, c("am", "gear", "cyl")])

        cyl  4  6  8
am gear             
0  3         1  2 12
   4         2  2  0
   5         0  0  0
1  3         0  0  0
   4         6  2  0
   5         2  1  2

We use the arguments row.vars and col.vars to provide the numbers or names of the variables to be used for the rows and columns of the flat contingency table. If neither of these two is given, the last variable is used for the columns. In our example the variable cyl.

ftable(mtcars[, c("am", "gear", "cyl")], col.vars = c(1, 2))

     am    0        1      
    gear  3  4  5  3  4  5
cyl                       
4         1  2  0  0  6  2
6         2  2  0  0  2  1
8        12  0  0  0  0  2

Alternative

The function xtabs creates contingency tables using a formula interface, each variable separated by +.

# One variable
xtabs(~ am, mtcars)
# Two variables
xtabs(~ am + gear, mtcars)
# Three variables
xtabs(~ am + gear + cyl, mtcars)
# Flat contingency table
ftable(xtabs(~ am + gear + cyl, mtcars))

Related posts

• Spanish version
• Proportion tables in R

How to draw square cells with geom_tile in ggplot2

Problem

In the following plot created using geom_tile we have rectangular cells. How can we draw squared cells instead?

set.seed(1)
df <- data.frame(val = rnorm(100), 
                 gene = rep(letters[1:20], 5), 
                 cell = c(sapply(LETTERS[1:5], 
                                 function(l) rep(l, 20))))
library(ggplot2)
ggplot(df, aes(y = gene, x = cell, fill = val)) +
  geom_tile(color = "white")

Soluction

We add coord_fixed() or coord_equal( ):

The default, ratio = 1, ensures that one unit on the x-axis is the same length as one unit on the y-axis.

ggplot(df, aes(y = gene, x = cell, fill = val)) +
  geom_tile(color = "white") +
  coord_fixed() # or coord_equal()

References

• stackoverflow
• coord_equal

Related posts

• Spanish version

Solving a system of linear equations in R

Problem

We want to solve a system of linear equations in R.

2x + 3y + 3z = 20

x + 4y + 3z = 15

5x + 3y + 4z = 30

Solution

We use the function solve.

a <- rbind(c(2, 3, 3), 
           c(1, 4, 3), 
           c(5, 3, 4))
b <- c(20, 15, 30)
solve(a, b)

[1] -1.25 -6.25 13.75

If we'd like the resulst in fractions, we use the function fractions from the package MASS.

library(MASS)
fractions(solve(a, b))

[1]  -5/4 -25/4  55/4

Related posts

Spanish version

Spacing between legend keys in ggplot2

Problem

We want to add some spacing between the elements of the legend to the following plot using ggplot2.

library(tidyverse)
mtcars %>%
  mutate(transmission = ifelse(am, "manual", "automatic")) %>%
  ggplot() +
  aes(x = transmission, fill = transmission) +
  geom_bar() +
  labs(fill = NULL) +
  theme(
    #legend.spacing.x = unit(.5, "char"), # adds spacing to the left too
    legend.position = "top",
    legend.justification = c(0, 0),
    legend.title = element_blank(),
    legend.margin = margin(c(5, 5, 5, 0))
  )

Solution

First option: spacing between the 4 elements

We use legend.spacing.x to add the spacing between the 4 elements the legend's key and text.

mtcars %>%
  mutate(transmission = ifelse(am, "manual", "automatic")) %>%
  ggplot() +
  aes(x = transmission, fill = transmission) +
  geom_bar() +
  labs(fill = NULL) +
  theme(
    legend.spacing.x = unit(.5, "char"),
    legend.position = "top",
    legend.justification = c(0, 0),
    legend.title = element_blank(),
    legend.margin = margin(c(5, 5, 5, 0)))

Second option: spacing between the 2 groups automatic and manual

We pass the argument margin to he element_text function to add spacing only between the 2 groups, the two types of transmissions automatic and manual, between the text automatic and the key (blue square) of manual

mtcars %>%
  mutate(transmission = ifelse(am, "manual", "automatic")) %>%
  ggplot() +
  aes(x = transmission, fill = transmission) +
  geom_bar() +
  labs(fill = NULL) +
  theme(legend.position = "top",
    legend.justification = c(0, 0),
    legend.title = element_blank(),
    legend.margin = margin(c(5, 5, 5, 0)),
    legend.text = element_text(margin = margin(r = 10, unit = "pt")))

Related posts

• Spanish version

References

• stackoverflow
• Modify components of a theme

How to calculate elapsed and remaining days between dates in R

Problem

We would like to calculate the number of elapsed and remaining days in a year.

Solution

From a starting date in the year

date<- as.Date("2019-03-29")
# Elapsed days
date - as.Date(ISOdate(format(Sys.Date(), "%Y"), 1, 1)) + 1

Time difference of 88 days

# Remaining days
as.Date(ISOdate(format(Sys.Date(), "%Y"), 12, 31)) - date

Time difference of 277 days

From today

# Elapsed days
Sys.Date() - as.Date(ISOdate(format(Sys.Date(), "%Y"), 1, 1)) + 1 

Time difference of 153 days

# Remaining days
as.Date(ISOdate(format(Sys.Date(), "%Y"), 12, 31)) - Sys.Date() 

Time difference of 212 days

Related entries

• Spanish version
• How to get week numbers from dates in R?