Numbers

Jeff Stevens

2025-03-05

Review

Group data wrangling challenge

Using the penguins data set from the {palmerpenguins} package, recreate this data frame.

# A tibble: 4 × 4
  species   island     male female
  <fct>     <fct>     <dbl>  <dbl>
1 Adelie    Torgersen 4035.  3396.
2 Adelie    Biscoe    4050   3369.
3 Adelie    Dream     4046.  3344.
4 Chinstrap Dream     3939.  3527.

Introduction

Mental model of data analysis

The problem

What’s different between these data sets?

What is needed to create data2 from data1?

data1
# A tibble: 12 × 3
    val1   val2    val3
   <dbl>  <dbl>   <dbl>
 1 0.372 0.129  0.00394
 2 0.338 0.898  0.00925
 3 0.889 0.324  0.00762
 4 0.736 0.477  0.00902
 5 0.493 0.928  0.00934
 6 0.909 0.200  0.00142
 7 0.650 0.846  0.00258
 8 0.535 0.0300 0.00172
 9 0.202 0.490  0.00329
10 0.613 0.649  0.00457
11 0.666 0.684  0.00325
12 0.510 0.356  0.00217
data2
# A tibble: 12 × 3
   percent_val1 log_val2 val3   
          <dbl>    <dbl> <chr>  
 1         37.2  -2.05   3.9e-03
 2         33.8  -0.107  9.3e-03
 3         88.9  -1.13   7.6e-03
 4         73.6  -0.740  9.0e-03
 5         49.3  -0.0747 9.3e-03
 6         90.9  -1.61   1.4e-03
 7         65.0  -0.167  2.6e-03
 8         53.5  -3.51   1.7e-03
 9         20.2  -0.713  3.3e-03
10         61.3  -0.432  4.6e-03
11         66.6  -0.379  3.3e-03
12         51    -1.03   2.2e-03

Set-up

Comparing and counting

Comparing numbers

x <- 1
y <- 1.00000000000001
z <- 1.001
x == y
[1] FALSE
all.equal(x, y)
[1] TRUE
all.equal(x, z)
[1] "Mean relative difference: 0.001"
all.equal(x, z, tolerance = 1e-2)
[1] TRUE
sqrt(2) ^ 2 == 2
[1] FALSE
all.equal(sqrt(2) ^ 2, 2)
[1] TRUE
dplyr::near(sqrt(2) ^ 2, 2)
[1] TRUE

Counts

As a reminder, we’ve already seen how to use dplyr::count()

count(flights, carrier)
# A tibble: 16 × 2
   carrier     n
   <chr>   <int>
 1 9E      18460
 2 AA      32729
 3 AS        714
 4 B6      54635
 5 DL      48110
 6 EV      54173
 7 F9        685
 8 FL       3260
 9 HA        342
10 MQ      26397
11 OO         32
12 UA      58665
13 US      20536
14 VX       5162
15 WN      12275
16 YV        601

Counts

We can also automatically sort by count.

count(flights, carrier, sort = TRUE)
# A tibble: 16 × 2
   carrier     n
   <chr>   <int>
 1 UA      58665
 2 B6      54635
 3 EV      54173
 4 DL      48110
 5 AA      32729
 6 MQ      26397
 7 US      20536
 8 9E      18460
 9 WN      12275
10 VX       5162
11 FL       3260
12 AS        714
13 F9        685
14 YV        601
15 HA        342
16 OO         32

And sum up totals instead of just count

count(flights, carrier, wt = distance)
# A tibble: 16 × 2
   carrier        n
   <chr>      <dbl>
 1 9E       9788152
 2 AA      43864584
 3 AS       1715028
 4 B6      58384137
 5 DL      59507317
 6 EV      30498951
 7 F9       1109700
 8 FL       2167344
 9 HA       1704186
10 MQ      15033955
11 OO         16026
12 UA      89705524
13 US      11365778
14 VX      12902327
15 WN      12229203
16 YV        225395

Counts

Remember n() counts inside a summarise()

flights |> 
  group_by(carrier) |> 
  summarise(n = n())
# A tibble: 16 × 2
   carrier     n
   <chr>   <int>
 1 9E      18460
 2 AA      32729
 3 AS        714
 4 B6      54635
 5 DL      48110
 6 EV      54173
 7 F9        685
 8 FL       3260
 9 HA        342
10 MQ      26397
11 OO         32
12 UA      58665
13 US      20536
14 VX       5162
15 WN      12275
16 YV        601

n_distinct() counts instances within a group

flights |> 
  group_by(dest) |> 
  summarise(carriers = 
              n_distinct(carrier))
# A tibble: 105 × 2
   dest  carriers
   <chr>    <int>
 1 ABQ          1
 2 ACK          1
 3 ALB          1
 4 ANC          1
 5 ATL          7
 6 AUS          6
 7 AVL          2
 8 BDL          2
 9 BGR          2
10 BHM          1
# ℹ 95 more rows

Counting NAs

To count NAs, you can sum() up TRUE responses to is.na()

flights |> 
  group_by(dest) |> 
  summarize(n_cancelled = sum(is.na(dep_time)))
# A tibble: 105 × 2
   dest  n_cancelled
   <chr>       <int>
 1 ABQ             0
 2 ACK             0
 3 ALB            20
 4 ANC             0
 5 ATL           317
 6 AUS            21
 7 AVL            12
 8 BDL            31
 9 BGR            15
10 BHM            25
# ℹ 95 more rows

Counting NAs

This trick can be used for any logical vector

sum(flights$month == 1)
[1] 27004
nrow(filter(flights, month == 1))
[1] 27004

Transforming numbers

Mathematical transformations

sqrt(seq(0, 100, 10))
 [1]  0.000000  3.162278  4.472136  5.477226  6.324555  7.071068  7.745967
 [8]  8.366600  8.944272  9.486833 10.000000
log(runif(10))
 [1] -1.65340685 -1.29070057 -1.07721774 -0.67318319 -0.16257538 -1.08017070
 [7] -0.54354157 -0.04112082 -0.58662299 -1.58314303
 [1] 0.6346852 1.0213321 0.8641012 0.3110083 1.1331782 1.0825151 1.2655997
 [8] 0.3102413 1.0683179 1.1974808

Rounding

Control significant digits with round()

round(123.456, 2)  # two digits
[1] 123.46
round(123.456, 1)  # one digit
[1] 123.5
round(123.456) # whole number
[1] 123
round(123.456, -1) # round to nearest ten
[1] 120
round(123.456, -2) # round to nearest hundred
[1] 100

Formatting numbers

Formatting

When numbers get too big, too small, or need other formatting, use format()

(x <- 0.0020)
[1] 0.002
format(x, scientific = TRUE)
[1] "2e-03"
format(x, nsmall = 4)
[1] "0.0020"
(y <- 12345678.9)
[1] 12345679
format(y, scientific = TRUE, 
       digits = 3)
[1] "1.23e+07"
format(y, big.mark = ",")
[1] "12,345,679"

Cutting numbers into ranges

If you need to bin numbers into ranges, use cut()

set.seed(1)
(x <- runif(12, min = 0, max = 100))
 [1] 26.550866 37.212390 57.285336 90.820779 20.168193 89.838968 94.467527
 [8] 66.079779 62.911404  6.178627 20.597457 17.655675
cut(x, breaks = c(0, 33, 66, 100))
 [1] (0,33]   (33,66]  (33,66]  (66,100] (0,33]   (66,100] (66,100] (66,100]
 [9] (33,66]  (0,33]   (0,33]   (0,33]  
Levels: (0,33] (33,66] (66,100]
cut(x, breaks = 3, labels = c("Low", "Medium", "High"))
 [1] Low    Medium Medium High   Low    High   High   High   Medium Low   
[11] Low    Low   
Levels: Low Medium High

Solving the problem

library(tidyverse)
nrows <- 12
set.seed(20250305)
data1 <- tibble(val1 = runif(nrows), 
                val2 = runif(nrows), 
                val3 = runif(nrows, 0.001, 0.01))

Solving the problem

data1
# A tibble: 12 × 3
    val1   val2    val3
   <dbl>  <dbl>   <dbl>
 1 0.372 0.129  0.00394
 2 0.338 0.898  0.00925
 3 0.889 0.324  0.00762
 4 0.736 0.477  0.00902
 5 0.493 0.928  0.00934
 6 0.909 0.200  0.00142
 7 0.650 0.846  0.00258
 8 0.535 0.0300 0.00172
 9 0.202 0.490  0.00329
10 0.613 0.649  0.00457
11 0.666 0.684  0.00325
12 0.510 0.356  0.00217
data2
# A tibble: 12 × 3
   percent_val1 log_val2 val3   
          <dbl>    <dbl> <chr>  
 1         37.2  -2.05   3.9e-03
 2         33.8  -0.107  9.3e-03
 3         88.9  -1.13   7.6e-03
 4         73.6  -0.740  9.0e-03
 5         49.3  -0.0747 9.3e-03
 6         90.9  -1.61   1.4e-03
 7         65.0  -0.167  2.6e-03
 8         53.5  -3.51   1.7e-03
 9         20.2  -0.713  3.3e-03
10         61.3  -0.432  4.6e-03
11         66.6  -0.379  3.3e-03
12         51    -1.03   2.2e-03

Let’s code!

Numbers