Import and Tidy Data
MATH-7360 Data Analysis
Dr. Xiang Ji @ Tulane University
Sep 16, 2020
rm(list = ls()) # clean-up workspace
library("tidyverse")## ── Attaching packages ────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── tidyverse 1.3.0 ──## ✓ ggplot2 3.3.2 ✓ purrr 0.3.4
## ✓ tibble 3.0.3 ✓ dplyr 1.0.2
## ✓ tidyr 1.1.1 ✓ stringr 1.4.0
## ✓ readr 1.3.1 ✓ forcats 0.5.0## ── Conflicts ───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()Announcement
We will go over HW1 next Monday
Mid-term evaluation (voluntary, anonymous, ~ 10 min)
Acknowledgement
Dr. Hua Zhou’s slides
Tidy data | r4ds chapter 12
“Happy families are all alike; every unhappy family is unhappy in its own way.” –– Leo Tolstoy
“Tidy datasets are all alike, but every messy dataset is messy in its own way.” –– Hadley Wickham
Tidy data
There are three interrelated rules which make a dataset tidy:
Each variable must have its own column.
Each observation must have its own row.
Each value must have its own cell.
Example table1
table1## # A tibble: 6 x 4 ## country year cases population ## <chr> <int> <int> <int> ## 1 Afghanistan 1999 745 19987071 ## 2 Afghanistan 2000 2666 20595360 ## 3 Brazil 1999 37737 172006362 ## 4 Brazil 2000 80488 174504898 ## 5 China 1999 212258 1272915272 ## 6 China 2000 213766 1280428583is tidy.
Example table2
table2## # A tibble: 12 x 4 ## country year type count ## <chr> <int> <chr> <int> ## 1 Afghanistan 1999 cases 745 ## 2 Afghanistan 1999 population 19987071 ## 3 Afghanistan 2000 cases 2666 ## 4 Afghanistan 2000 population 20595360 ## 5 Brazil 1999 cases 37737 ## 6 Brazil 1999 population 172006362 ## 7 Brazil 2000 cases 80488 ## 8 Brazil 2000 population 174504898 ## 9 China 1999 cases 212258 ## 10 China 1999 population 1272915272 ## 11 China 2000 cases 213766 ## 12 China 2000 population 1280428583is not tidy.
Example table3
table3## # A tibble: 6 x 3 ## country year rate ## * <chr> <int> <chr> ## 1 Afghanistan 1999 745/19987071 ## 2 Afghanistan 2000 2666/20595360 ## 3 Brazil 1999 37737/172006362 ## 4 Brazil 2000 80488/174504898 ## 5 China 1999 212258/1272915272 ## 6 China 2000 213766/1280428583is not tidy.
Example table4a
table4a## # A tibble: 3 x 3 ## country `1999` `2000` ## * <chr> <int> <int> ## 1 Afghanistan 745 2666 ## 2 Brazil 37737 80488 ## 3 China 212258 213766is not tidy.
Example table4b
table4b## # A tibble: 3 x 3 ## country `1999` `2000` ## * <chr> <int> <int> ## 1 Afghanistan 19987071 20595360 ## 2 Brazil 172006362 174504898 ## 3 China 1272915272 1280428583is not tidy.
Pivoting
Typical issues:
One variable might be spread across multiple columns
One variable might be scattered across multiple rows
Longer
pivot_longer()columns into a new pair of variables.table4a %>% pivot_longer(c(`1999`, `2000`), names_to = "year", values_to = "cases")## # A tibble: 6 x 3 ## country year cases ## <chr> <chr> <int> ## 1 Afghanistan 1999 745 ## 2 Afghanistan 2000 2666 ## 3 Brazil 1999 37737 ## 4 Brazil 2000 80488 ## 5 China 1999 212258 ## 6 China 2000 213766
We can pivot table4b longer too and then join them
tidy4a <- table4a %>% pivot_longer(c(`1999`, `2000`), names_to = "year", values_to = "cases") tidy4b <- table4b %>% pivot_longer(c(`1999`, `2000`), names_to = "year", values_to = "population") left_join(tidy4a, tidy4b)## Joining, by = c("country", "year")## # A tibble: 6 x 4 ## country year cases population ## <chr> <chr> <int> <int> ## 1 Afghanistan 1999 745 19987071 ## 2 Afghanistan 2000 2666 20595360 ## 3 Brazil 1999 37737 172006362 ## 4 Brazil 2000 80488 174504898 ## 5 China 1999 212258 1272915272 ## 6 China 2000 213766 1280428583
Wider
pivot_wider()is the opposite ofpivot_longer().table2## # A tibble: 12 x 4 ## country year type count ## <chr> <int> <chr> <int> ## 1 Afghanistan 1999 cases 745 ## 2 Afghanistan 1999 population 19987071 ## 3 Afghanistan 2000 cases 2666 ## 4 Afghanistan 2000 population 20595360 ## 5 Brazil 1999 cases 37737 ## 6 Brazil 1999 population 172006362 ## 7 Brazil 2000 cases 80488 ## 8 Brazil 2000 population 174504898 ## 9 China 1999 cases 212258 ## 10 China 1999 population 1272915272 ## 11 China 2000 cases 213766 ## 12 China 2000 population 1280428583table2 %>% pivot_wider(names_from = type, values_from = count)## # A tibble: 6 x 4 ## country year cases population ## <chr> <int> <int> <int> ## 1 Afghanistan 1999 745 19987071 ## 2 Afghanistan 2000 2666 20595360 ## 3 Brazil 1999 37737 172006362 ## 4 Brazil 2000 80488 174504898 ## 5 China 1999 212258 1272915272 ## 6 China 2000 213766 1280428583
Separating
table3 %>% separate(rate, into = c("cases", "population"))## # A tibble: 6 x 4 ## country year cases population ## <chr> <int> <chr> <chr> ## 1 Afghanistan 1999 745 19987071 ## 2 Afghanistan 2000 2666 20595360 ## 3 Brazil 1999 37737 172006362 ## 4 Brazil 2000 80488 174504898 ## 5 China 1999 212258 1272915272 ## 6 China 2000 213766 1280428583
Seperate into numeric values:
table3 %>% separate(rate, into = c("cases", "population"), convert = TRUE)## # A tibble: 6 x 4 ## country year cases population ## <chr> <int> <int> <int> ## 1 Afghanistan 1999 745 19987071 ## 2 Afghanistan 2000 2666 20595360 ## 3 Brazil 1999 37737 172006362 ## 4 Brazil 2000 80488 174504898 ## 5 China 1999 212258 1272915272 ## 6 China 2000 213766 1280428583
Separate at a fixed position:
table3 %>% separate(year, into = c("century", "year"), sep = 2)## # A tibble: 6 x 4 ## country century year rate ## <chr> <chr> <chr> <chr> ## 1 Afghanistan 19 99 745/19987071 ## 2 Afghanistan 20 00 2666/20595360 ## 3 Brazil 19 99 37737/172006362 ## 4 Brazil 20 00 80488/174504898 ## 5 China 19 99 212258/1272915272 ## 6 China 20 00 213766/1280428583
Unite
table5## # A tibble: 6 x 4 ## country century year rate ## * <chr> <chr> <chr> <chr> ## 1 Afghanistan 19 99 745/19987071 ## 2 Afghanistan 20 00 2666/20595360 ## 3 Brazil 19 99 37737/172006362 ## 4 Brazil 20 00 80488/174504898 ## 5 China 19 99 212258/1272915272 ## 6 China 20 00 213766/1280428583
unite()is the inverse ofseparate().table5 %>% unite(new, century, year, sep = "")## # A tibble: 6 x 3 ## country new rate ## <chr> <chr> <chr> ## 1 Afghanistan 1999 745/19987071 ## 2 Afghanistan 2000 2666/20595360 ## 3 Brazil 1999 37737/172006362 ## 4 Brazil 2000 80488/174504898 ## 5 China 1999 212258/1272915272 ## 6 China 2000 213766/1280428583