Data Wrangling with dplyr
Last updated on 2024-12-03 | Edit this page
Overview
Questions
- How can I select specific rows and/or columns from a dataframe?
- How can I combine multiple commands into a single command?
- How can I create new columns or remove existing columns from a dataframe?
Objectives
- Describe the purpose of an R package and the
dplyrpackage. - Select certain columns in a dataframe with the
dplyrfunctionselect. - Select certain rows in a dataframe according to filtering conditions
with the
dplyrfunctionfilter. - Link the output of one
dplyrfunction to the input of another function with the ‘pipe’ operator%>%. - Add new columns to a dataframe that are functions of existing
columns with
mutate. - Use the split-apply-combine concept for data analysis.
- Use
summarize,group_by, andcountto split a dataframe into groups of observations, apply a summary statistics for each group, and then combine the results.
dplyr is a package for making tabular
data wrangling easier by using a limited set of functions that can be
combined to extract and summarize insights from your data.
Like readr,
dplyr is a part of the tidyverse. These
packages were loaded in R’s memory when we called
library(tidyverse) earlier.
Note
The packages in the tidyverse, namely
dplyr, tidyr
and ggplot2 accept both the British
(e.g. summarise) and American (e.g. summarize)
spelling variants of different function and option names. For this
lesson, we utilize the American spellings of different functions;
however, feel free to use the regional variant for where you are
teaching.
What is an R package?
The package dplyr provides easy tools
for the most common data wrangling tasks. It is built to work directly
with dataframes, with many common tasks optimized by being written in a
compiled language (C++) (not all R packages are written in R!).
There are also packages available for a wide range of tasks including
building plots (ggplot2, which we’ll see
later), downloading data from the NCBI database, or performing
statistical analysis on your data set. Many packages such as these are
housed on, and downloadable from, the Comprehensive
R Archive Network
(CRAN) using install.packages. This function makes the
package accessible by your R installation with the command
library(), as you did with tidyverse
earlier.
To easily access the documentation for a package within R or RStudio,
use help(package = "package_name").
To learn more about dplyr after the
workshop, you may want to check out this handy
data transformation with dplyr
cheatsheet.
Note
There are alternatives to the tidyverse packages for
data wrangling, including the package data.table.
See this comparison
for example to get a sense of the differences between using
base, tidyverse, and
data.table.
Learning dplyr
To make sure everyone will use the same dataset for this lesson, we’ll read again the SAFI dataset that we downloaded earlier.
R
## load the tidyverse
library(tidyverse)
library(here)
interviews <- read_csv(here("data", "SAFI_clean.csv"), na = "NULL")
## inspect the data
interviews
## preview the data
# view(interviews)
We’re going to learn some of the most common
dplyr functions:
-
select(): subset columns -
filter(): subset rows on conditions -
mutate(): create new columns by using information from other columns -
group_by()andsummarize(): create summary statistics on grouped data -
arrange(): sort results -
count(): count discrete values
Selecting columns and filtering rows
To select columns of a dataframe, use select(). The
first argument to this function is the dataframe
(interviews), and the subsequent arguments are the columns
to keep, separated by commas. Alternatively, if you are selecting
columns adjacent to each other, you can use a : to select a
range of columns, read as “select columns from ___ to ___.” You may have
done something similar in the past using subsetting.
select() is essentially doing the same thing as subsetting,
using a package (dplyr) instead of R’s base functions.
R
# to select columns throughout the dataframe
select(interviews, village, no_membrs, months_lack_food)
# to do the same thing with subsetting
interviews[c("village","no_membrs","months_lack_food")]
# to select a series of connected columns
select(interviews, village:respondent_wall_type)
To choose rows based on specific criteria, we can use the
filter() function. The argument after the dataframe is the
condition we want our final dataframe to adhere to (e.g. village name is
Chirodzo):
R
# filters observations where village name is "Chirodzo"
filter(interviews, village == "Chirodzo")
OUTPUT
# A tibble: 39 × 14
key_ID village interview_date no_membrs years_liv respondent_wall_type
<dbl> <chr> <dttm> <dbl> <dbl> <chr>
1 8 Chirodzo 2016-11-16 00:00:00 12 70 burntbricks
2 9 Chirodzo 2016-11-16 00:00:00 8 6 burntbricks
3 10 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
4 34 Chirodzo 2016-11-17 00:00:00 8 18 burntbricks
5 35 Chirodzo 2016-11-17 00:00:00 5 45 muddaub
6 36 Chirodzo 2016-11-17 00:00:00 6 23 sunbricks
7 37 Chirodzo 2016-11-17 00:00:00 3 8 burntbricks
8 43 Chirodzo 2016-11-17 00:00:00 7 29 muddaub
9 44 Chirodzo 2016-11-17 00:00:00 2 6 muddaub
10 45 Chirodzo 2016-11-17 00:00:00 9 7 muddaub
# ℹ 29 more rows
# ℹ 8 more variables: rooms <dbl>, memb_assoc <chr>, affect_conflicts <chr>,
# liv_count <dbl>, items_owned <chr>, no_meals <dbl>, months_lack_food <chr>,
# instanceID <chr>You may also have noticed that the output from these call doesn’t run
off the screen anymore. It’s one of the advantages of
tbl_df (also called tibble), the central data class in the
tidyverse, compared to normal dataframes in R.
We can also specify multiple conditions within the
filter() function. We can combine conditions using either
“and” or “or” statements. In an “and” statement, an observation (row)
must meet every criteria to be included in the
resulting dataframe. To form “and” statements within dplyr, we can pass
our desired conditions as arguments in the filter()
function, separated by commas:
R
# filters observations with "and" operator (comma)
# output dataframe satisfies ALL specified conditions
filter(interviews, village == "Chirodzo",
rooms > 1,
no_meals > 2)
OUTPUT
# A tibble: 10 × 14
key_ID village interview_date no_membrs years_liv respondent_wall_type
<dbl> <chr> <dttm> <dbl> <dbl> <chr>
1 10 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
2 49 Chirodzo 2016-11-16 00:00:00 6 26 burntbricks
3 52 Chirodzo 2016-11-16 00:00:00 11 15 burntbricks
4 56 Chirodzo 2016-11-16 00:00:00 12 23 burntbricks
5 65 Chirodzo 2016-11-16 00:00:00 8 20 burntbricks
6 66 Chirodzo 2016-11-16 00:00:00 10 37 burntbricks
7 67 Chirodzo 2016-11-16 00:00:00 5 31 burntbricks
8 68 Chirodzo 2016-11-16 00:00:00 8 52 burntbricks
9 199 Chirodzo 2017-06-04 00:00:00 7 17 burntbricks
10 200 Chirodzo 2017-06-04 00:00:00 8 20 burntbricks
# ℹ 8 more variables: rooms <dbl>, memb_assoc <chr>, affect_conflicts <chr>,
# liv_count <dbl>, items_owned <chr>, no_meals <dbl>, months_lack_food <chr>,
# instanceID <chr>We can also form “and” statements with the &
operator instead of commas:
R
# filters observations with "&" logical operator
# output dataframe satisfies ALL specified conditions
filter(interviews, village == "Chirodzo" &
rooms > 1 &
no_meals > 2)
OUTPUT
# A tibble: 10 × 14
key_ID village interview_date no_membrs years_liv respondent_wall_type
<dbl> <chr> <dttm> <dbl> <dbl> <chr>
1 10 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
2 49 Chirodzo 2016-11-16 00:00:00 6 26 burntbricks
3 52 Chirodzo 2016-11-16 00:00:00 11 15 burntbricks
4 56 Chirodzo 2016-11-16 00:00:00 12 23 burntbricks
5 65 Chirodzo 2016-11-16 00:00:00 8 20 burntbricks
6 66 Chirodzo 2016-11-16 00:00:00 10 37 burntbricks
7 67 Chirodzo 2016-11-16 00:00:00 5 31 burntbricks
8 68 Chirodzo 2016-11-16 00:00:00 8 52 burntbricks
9 199 Chirodzo 2017-06-04 00:00:00 7 17 burntbricks
10 200 Chirodzo 2017-06-04 00:00:00 8 20 burntbricks
# ℹ 8 more variables: rooms <dbl>, memb_assoc <chr>, affect_conflicts <chr>,
# liv_count <dbl>, items_owned <chr>, no_meals <dbl>, months_lack_food <chr>,
# instanceID <chr>In an “or” statement, observations must meet at least one of the specified conditions. To form “or” statements we use the logical operator for “or,” which is the vertical bar (|):
R
# filters observations with "|" logical operator
# output dataframe satisfies AT LEAST ONE of the specified conditions
filter(interviews, village == "Chirodzo" | village == "Ruaca")
OUTPUT
# A tibble: 88 × 14
key_ID village interview_date no_membrs years_liv respondent_wall_type
<dbl> <chr> <dttm> <dbl> <dbl> <chr>
1 8 Chirodzo 2016-11-16 00:00:00 12 70 burntbricks
2 9 Chirodzo 2016-11-16 00:00:00 8 6 burntbricks
3 10 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
4 23 Ruaca 2016-11-21 00:00:00 10 20 burntbricks
5 24 Ruaca 2016-11-21 00:00:00 6 4 burntbricks
6 25 Ruaca 2016-11-21 00:00:00 11 6 burntbricks
7 26 Ruaca 2016-11-21 00:00:00 3 20 burntbricks
8 27 Ruaca 2016-11-21 00:00:00 7 36 burntbricks
9 28 Ruaca 2016-11-21 00:00:00 2 2 muddaub
10 29 Ruaca 2016-11-21 00:00:00 7 10 burntbricks
# ℹ 78 more rows
# ℹ 8 more variables: rooms <dbl>, memb_assoc <chr>, affect_conflicts <chr>,
# liv_count <dbl>, items_owned <chr>, no_meals <dbl>, months_lack_food <chr>,
# instanceID <chr>Pipes
What if you want to select and filter at the same time? There are three ways to do this: use intermediate steps, nested functions, or pipes.
With intermediate steps, you create a temporary dataframe and use that as input to the next function, like this:
R
interviews2 <- filter(interviews, village == "Chirodzo")
interviews_ch <- select(interviews2, village:respondent_wall_type)
This is readable, but can clutter up your workspace with lots of objects that you have to name individually. With multiple steps, that can be hard to keep track of.
You can also nest functions (i.e. one function inside of another), like this:
R
interviews_ch <- select(filter(interviews, village == "Chirodzo"),
village:respondent_wall_type)
This is handy, but can be difficult to read if too many functions are nested, as R evaluates the expression from the inside out (in this case, filtering, then selecting).
The last option, pipes, are a recent addition to R. Pipes
let you take the output of one function and send it directly to the
next, which is useful when you need to do many things to the same
dataset. There are two Pipes in R: 1) %>% (called
magrittr pipe; made available via the
magrittr package, installed automatically
with dplyr) or 2) |>
(called native R pipe and it comes preinstalled with R v4.1.0 onwards).
Both the pipes are, by and large, function similarly with a few
differences (For more information, check: https://www.tidyverse.org/blog/2023/04/base-vs-magrittr-pipe/).
The choice of which pipe to be used can be changed in the Global
settings in R studio and once that is done, you can type the pipe
with:
- Ctrl + Shift + M if you have a PC or Cmd + Shift + M if you have a Mac.
R
# the following example is run using magrittr pipe but the output will be same with the native pipe
interviews %>%
filter(village == "Chirodzo") %>%
select(village:respondent_wall_type)
OUTPUT
# A tibble: 39 × 5
village interview_date no_membrs years_liv respondent_wall_type
<chr> <dttm> <dbl> <dbl> <chr>
1 Chirodzo 2016-11-16 00:00:00 12 70 burntbricks
2 Chirodzo 2016-11-16 00:00:00 8 6 burntbricks
3 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
4 Chirodzo 2016-11-17 00:00:00 8 18 burntbricks
5 Chirodzo 2016-11-17 00:00:00 5 45 muddaub
6 Chirodzo 2016-11-17 00:00:00 6 23 sunbricks
7 Chirodzo 2016-11-17 00:00:00 3 8 burntbricks
8 Chirodzo 2016-11-17 00:00:00 7 29 muddaub
9 Chirodzo 2016-11-17 00:00:00 2 6 muddaub
10 Chirodzo 2016-11-17 00:00:00 9 7 muddaub
# ℹ 29 more rowsR
#interviews |>
# filter(village == "Chirodzo") |>
# select(village:respondent_wall_type)
In the above code, we use the pipe to send the
interviews dataset first through filter() to
keep rows where village is “Chirodzo”, then through
select() to keep only the columns from village
to respondent_wall_type. Since %>% takes
the object on its left and passes it as the first argument to the
function on its right, we don’t need to explicitly include the dataframe
as an argument to the filter() and select()
functions any more.
Some may find it helpful to read the pipe like the word “then”. For
instance, in the above example, we take the dataframe
interviews, then we filter for rows
with village == "Chirodzo", then we
select columns village:respondent_wall_type.
The dplyr functions by themselves are
somewhat simple, but by combining them into linear workflows with the
pipe, we can accomplish more complex data wrangling operations.
If we want to create a new object with this smaller version of the data, we can assign it a new name:
R
interviews_ch <- interviews %>%
filter(village == "Chirodzo") %>%
select(village:respondent_wall_type)
interviews_ch
OUTPUT
# A tibble: 39 × 5
village interview_date no_membrs years_liv respondent_wall_type
<chr> <dttm> <dbl> <dbl> <chr>
1 Chirodzo 2016-11-16 00:00:00 12 70 burntbricks
2 Chirodzo 2016-11-16 00:00:00 8 6 burntbricks
3 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
4 Chirodzo 2016-11-17 00:00:00 8 18 burntbricks
5 Chirodzo 2016-11-17 00:00:00 5 45 muddaub
6 Chirodzo 2016-11-17 00:00:00 6 23 sunbricks
7 Chirodzo 2016-11-17 00:00:00 3 8 burntbricks
8 Chirodzo 2016-11-17 00:00:00 7 29 muddaub
9 Chirodzo 2016-11-17 00:00:00 2 6 muddaub
10 Chirodzo 2016-11-17 00:00:00 9 7 muddaub
# ℹ 29 more rowsNote that the final dataframe (interviews_ch) is the
leftmost part of this expression.
Exercise
Using pipes, subset the interviews data to include
interviews where respondents were members of an irrigation association
(memb_assoc) and retain only the columns
affect_conflicts, liv_count, and
no_meals.
R
interviews %>%
filter(memb_assoc == "yes") %>%
select(affect_conflicts, liv_count, no_meals)
OUTPUT
# A tibble: 33 × 3
affect_conflicts liv_count no_meals
<chr> <dbl> <dbl>
1 once 3 2
2 never 2 2
3 never 2 3
4 once 3 2
5 frequently 1 3
6 more_once 5 2
7 more_once 3 2
8 more_once 2 3
9 once 3 3
10 never 3 3
# ℹ 23 more rowsMutate
Frequently you’ll want to create new columns based on the values in
existing columns, for example to do unit conversions, or to find the
ratio of values in two columns. For this we’ll use
mutate().
We might be interested in the ratio of number of household members to rooms used for sleeping (i.e. avg number of people per room):
R
interviews %>%
mutate(people_per_room = no_membrs / rooms)
OUTPUT
# A tibble: 131 × 15
key_ID village interview_date no_membrs years_liv respondent_wall_type
<dbl> <chr> <dttm> <dbl> <dbl> <chr>
1 1 God 2016-11-17 00:00:00 3 4 muddaub
2 2 God 2016-11-17 00:00:00 7 9 muddaub
3 3 God 2016-11-17 00:00:00 10 15 burntbricks
4 4 God 2016-11-17 00:00:00 7 6 burntbricks
5 5 God 2016-11-17 00:00:00 7 40 burntbricks
6 6 God 2016-11-17 00:00:00 3 3 muddaub
7 7 God 2016-11-17 00:00:00 6 38 muddaub
8 8 Chirodzo 2016-11-16 00:00:00 12 70 burntbricks
9 9 Chirodzo 2016-11-16 00:00:00 8 6 burntbricks
10 10 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
# ℹ 121 more rows
# ℹ 9 more variables: rooms <dbl>, memb_assoc <chr>, affect_conflicts <chr>,
# liv_count <dbl>, items_owned <chr>, no_meals <dbl>, months_lack_food <chr>,
# instanceID <chr>, people_per_room <dbl>We may be interested in investigating whether being a member of an irrigation association had any effect on the ratio of household members to rooms. To look at this relationship, we will first remove data from our dataset where the respondent didn’t answer the question of whether they were a member of an irrigation association. These cases are recorded as “NULL” in the dataset.
To remove these cases, we could insert a filter() in the
chain:
R
interviews %>%
filter(!is.na(memb_assoc)) %>%
mutate(people_per_room = no_membrs / rooms)
OUTPUT
# A tibble: 92 × 15
key_ID village interview_date no_membrs years_liv respondent_wall_type
<dbl> <chr> <dttm> <dbl> <dbl> <chr>
1 2 God 2016-11-17 00:00:00 7 9 muddaub
2 7 God 2016-11-17 00:00:00 6 38 muddaub
3 8 Chirodzo 2016-11-16 00:00:00 12 70 burntbricks
4 9 Chirodzo 2016-11-16 00:00:00 8 6 burntbricks
5 10 Chirodzo 2016-12-16 00:00:00 12 23 burntbricks
6 12 God 2016-11-21 00:00:00 7 20 burntbricks
7 13 God 2016-11-21 00:00:00 6 8 burntbricks
8 15 God 2016-11-21 00:00:00 5 30 sunbricks
9 21 God 2016-11-21 00:00:00 8 20 burntbricks
10 24 Ruaca 2016-11-21 00:00:00 6 4 burntbricks
# ℹ 82 more rows
# ℹ 9 more variables: rooms <dbl>, memb_assoc <chr>, affect_conflicts <chr>,
# liv_count <dbl>, items_owned <chr>, no_meals <dbl>, months_lack_food <chr>,
# instanceID <chr>, people_per_room <dbl>The ! symbol negates the result of the
is.na() function. Thus, if is.na() returns a
value of TRUE (because the memb_assoc is
missing), the ! symbol negates this and says we only want
values of FALSE, where memb_assoc is
not missing.
Exercise
Create a new dataframe from the interviews data that
meets the following criteria: contains only the village
column and a new column called total_meals containing a
value that is equal to the total number of meals served in the household
per day on average (no_membrs times no_meals).
Only the rows where total_meals is greater than 20 should
be shown in the final dataframe.
Hint: think about how the commands should be ordered to produce this data frame!
R
interviews_total_meals <- interviews %>%
mutate(total_meals = no_membrs * no_meals) %>%
filter(total_meals > 20) %>%
select(village, total_meals)
Split-apply-combine data analysis and the summarize() function
Many data analysis tasks can be approached using the
split-apply-combine paradigm: split the data into groups, apply
some analysis to each group, and then combine the results.
dplyr makes this very easy through the use
of the group_by() function.
The summarize() function
group_by() is often used together with
summarize(), which collapses each group into a single-row
summary of that group. group_by() takes as arguments the
column names that contain the categorical variables for
which you want to calculate the summary statistics. So to compute the
average household size by village:
R
interviews %>%
group_by(village) %>%
summarize(mean_no_membrs = mean(no_membrs))
OUTPUT
# A tibble: 3 × 2
village mean_no_membrs
<chr> <dbl>
1 Chirodzo 7.08
2 God 6.86
3 Ruaca 7.57You can also group by multiple columns:
R
interviews %>%
group_by(village, memb_assoc) %>%
summarize(mean_no_membrs = mean(no_membrs))
OUTPUT
`summarise()` has grouped output by 'village'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 9 × 3
# Groups: village [3]
village memb_assoc mean_no_membrs
<chr> <chr> <dbl>
1 Chirodzo no 8.06
2 Chirodzo yes 7.82
3 Chirodzo <NA> 5.08
4 God no 7.13
5 God yes 8
6 God <NA> 6
7 Ruaca no 7.18
8 Ruaca yes 9.5
9 Ruaca <NA> 6.22Note that the output is a grouped tibble of nine rows by three
columns which is indicated by the by two first lines with the
#. To obtain an ungrouped tibble, use the
ungroup function:
R
interviews %>%
group_by(village, memb_assoc) %>%
summarize(mean_no_membrs = mean(no_membrs)) %>%
ungroup()
OUTPUT
`summarise()` has grouped output by 'village'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 9 × 3
village memb_assoc mean_no_membrs
<chr> <chr> <dbl>
1 Chirodzo no 8.06
2 Chirodzo yes 7.82
3 Chirodzo <NA> 5.08
4 God no 7.13
5 God yes 8
6 God <NA> 6
7 Ruaca no 7.18
8 Ruaca yes 9.5
9 Ruaca <NA> 6.22Notice that the second line with the # that previously
indicated the grouping has disappeared and we now only have a 9x3-tibble
without grouping. When grouping both by village and
membr_assoc, we see rows in our table for respondents who
did not specify whether they were a member of an irrigation association.
We can exclude those data from our table using a filter step.
R
interviews %>%
filter(!is.na(memb_assoc)) %>%
group_by(village, memb_assoc) %>%
summarize(mean_no_membrs = mean(no_membrs))
OUTPUT
`summarise()` has grouped output by 'village'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 6 × 3
# Groups: village [3]
village memb_assoc mean_no_membrs
<chr> <chr> <dbl>
1 Chirodzo no 8.06
2 Chirodzo yes 7.82
3 God no 7.13
4 God yes 8
5 Ruaca no 7.18
6 Ruaca yes 9.5 Once the data are grouped, you can also summarize multiple variables at the same time (and not necessarily on the same variable). For instance, we could add a column indicating the minimum household size for each village for each group (members of an irrigation association vs not):
R
interviews %>%
filter(!is.na(memb_assoc)) %>%
group_by(village, memb_assoc) %>%
summarize(mean_no_membrs = mean(no_membrs),
min_membrs = min(no_membrs))
OUTPUT
`summarise()` has grouped output by 'village'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 6 × 4
# Groups: village [3]
village memb_assoc mean_no_membrs min_membrs
<chr> <chr> <dbl> <dbl>
1 Chirodzo no 8.06 4
2 Chirodzo yes 7.82 2
3 God no 7.13 3
4 God yes 8 5
5 Ruaca no 7.18 2
6 Ruaca yes 9.5 5It is sometimes useful to rearrange the result of a query to inspect
the values. For instance, we can sort on min_membrs to put
the group with the smallest household first:
R
interviews %>%
filter(!is.na(memb_assoc)) %>%
group_by(village, memb_assoc) %>%
summarize(mean_no_membrs = mean(no_membrs),
min_membrs = min(no_membrs)) %>%
arrange(min_membrs)
OUTPUT
`summarise()` has grouped output by 'village'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 6 × 4
# Groups: village [3]
village memb_assoc mean_no_membrs min_membrs
<chr> <chr> <dbl> <dbl>
1 Chirodzo yes 7.82 2
2 Ruaca no 7.18 2
3 God no 7.13 3
4 Chirodzo no 8.06 4
5 God yes 8 5
6 Ruaca yes 9.5 5To sort in descending order, we need to add the desc()
function. If we want to sort the results by decreasing order of minimum
household size:
R
interviews %>%
filter(!is.na(memb_assoc)) %>%
group_by(village, memb_assoc) %>%
summarize(mean_no_membrs = mean(no_membrs),
min_membrs = min(no_membrs)) %>%
arrange(desc(min_membrs))
OUTPUT
`summarise()` has grouped output by 'village'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 6 × 4
# Groups: village [3]
village memb_assoc mean_no_membrs min_membrs
<chr> <chr> <dbl> <dbl>
1 God yes 8 5
2 Ruaca yes 9.5 5
3 Chirodzo no 8.06 4
4 God no 7.13 3
5 Chirodzo yes 7.82 2
6 Ruaca no 7.18 2Counting
When working with data, we often want to know the number of
observations found for each factor or combination of factors. For this
task, dplyr provides count().
For example, if we wanted to count the number of rows of data for each
village, we would do:
R
interviews %>%
count(village)
OUTPUT
# A tibble: 3 × 2
village n
<chr> <int>
1 Chirodzo 39
2 God 43
3 Ruaca 49For convenience, count() provides the sort
argument to get results in decreasing order:
R
interviews %>%
count(village, sort = TRUE)
OUTPUT
# A tibble: 3 × 2
village n
<chr> <int>
1 Ruaca 49
2 God 43
3 Chirodzo 39Exercise
How many households in the survey have an average of two meals per day? Three meals per day? Are there any other numbers of meals represented?
R
interviews %>%
count(no_meals)
OUTPUT
# A tibble: 2 × 2
no_meals n
<dbl> <int>
1 2 52
2 3 79Exercise (continued)
Use group_by() and summarize() to find the
mean, min, and max number of household members for each village. Also
add the number of observations (hint: see ?n).
R
interviews %>%
group_by(village) %>%
summarize(
mean_no_membrs = mean(no_membrs),
min_no_membrs = min(no_membrs),
max_no_membrs = max(no_membrs),
n = n()
)
OUTPUT
# A tibble: 3 × 5
village mean_no_membrs min_no_membrs max_no_membrs n
<chr> <dbl> <dbl> <dbl> <int>
1 Chirodzo 7.08 2 12 39
2 God 6.86 3 15 43
3 Ruaca 7.57 2 19 49Exercise (continued)
What was the largest household interviewed in each month?
R
# if not already included, add month, year, and day columns
library(lubridate) # load lubridate if not already loaded
interviews %>%
mutate(month = month(interview_date),
day = day(interview_date),
year = year(interview_date)) %>%
group_by(year, month) %>%
summarize(max_no_membrs = max(no_membrs))
OUTPUT
`summarise()` has grouped output by 'year'. You can override using the
`.groups` argument.OUTPUT
# A tibble: 5 × 3
# Groups: year [2]
year month max_no_membrs
<dbl> <dbl> <dbl>
1 2016 11 19
2 2016 12 12
3 2017 4 17
4 2017 5 15
5 2017 6 15Key Points
- Use the
dplyrpackage to manipulate dataframes. - Use
select()to choose variables from a dataframe. - Use
filter()to choose data based on values. - Use
group_by()andsummarize()to work with subsets of data. - Use
mutate()to create new variables.