Overview

Teaching: 50 min
Exercises: 30 min

Questions

• How can I select specific rows and/or columns from a data frame?

• How can I combine multiple commands into a single command?

• How can create new columns or remove existing columns from a data frame?

• How can I reformat a dataframe to meet my needs?

Objectives

• Describe the purpose of the dplyr and tidyr packages.

• Select certain columns in a data frame with the dplyr function select.

• Select certain rows in a data frame according to filtering conditions with the dplyr function filter.

• Link the output of one dplyr function to the input of another function with the ‘pipe’ operator %>%.

• Add new columns to a data frame that are functions of existing columns with mutate.

• Use the split-apply-combine concept for data analysis.

• Use summarize, group_by, and count to split a data frame into groups of observations, apply a summary statistics for each group, and then combine the results.

• Describe the concept of a wide and a long table format and for which purpose those formats are useful.

• Describe what key-value pairs are.

• Reshape a data frame from long to wide format and back with the spread and gather commands from the tidyr package.

• Export a data frame to a csv file.

Data Manipulation using dplyr and tidyr

dplyr is a package for making tabular data manipulation easier by using a limited set of functions that can be combined to extract and summarize insights from your data. It pairs nicely with tidyr which enables you to swiftly convert between different data formats (long vs. wide) for plotting and analysis.

Similarly to readr, dplyr and tidyr are also part of the tidyverse. These packages were loaded in R’s memory when we called library(tidyverse) earlier.

What are dplyr and tidyr?

The package dplyr provides easy tools for the most common data manipulation tasks. It is built to work directly with data frames, with many common tasks optimized by being written in a compiled language (C++). An additional feature is the ability to work directly with data stored in an external database. The benefits of doing this are that the data can be managed natively in a relational database, queries can be conducted on that database, and only the results of the query are returned.

This addresses a common problem with R in that all operations are conducted in-memory and thus the amount of data you can work with is limited by available memory. The database connections essentially remove that limitation in that you can connect to a database of many hundreds of GB, conduct queries on it directly, and pull back into R only what you need for analysis.

The package tidyr addresses the common problem of wanting to reshape your data for plotting and use by different R functions. Sometimes we want data sets where we have one row per measurement. Sometimes we want a data frame where each measurement type has its own column, and rows are instead more aggregated groups. Moving back and forth between these formats is nontrivial, and tidyr gives you tools for this and more sophisticated data manipulation.

To learn more about dplyr and tidyr after the workshop, you may want to check out this handy data transformation with dplyr cheatsheet and this one about tidyr.

To make sure, everyone will use the same dataset for this lesson, we’ll read again the SAFI dataset that we downloaded earlier. As in the previous lesson, continue to type your code into your script file, SAFI_analysis.R and run it using keyboard shortcuts or the Run button. Dont forget to save, commit, and push your changes to your remote repository periodically!

## load the tidyverse
library(tidyverse)
library(lubridate)

interviews <- read_csv("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() and summarize(): create summary statistics on grouped data
• arrange(): sort results
• count(): count discrete values

Selecting columns and filtering rows

To select columns of a data frame, use select(). The first argument to this function is the data frame (surveys), and the subsequent arguments are the columns to keep.

select(interviews, village, no_membrs, years_liv)

To choose rows based on a specific criteria, use filter():

filter(interviews, village == "God")

# A tibble: 43 x 14
   key_ID village interview_date      no_membrs years_liv respondent_wall~
    <dbl> <chr>   <dttm>                  <dbl>     <dbl> <chr>           
 1      1 God     2016-11-17 00:00:00         3         4 muddaub         
 2      1 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     11 God     2016-11-21 00:00:00         6        20 sunbricks       
 9     12 God     2016-11-21 00:00:00         7        20 burntbricks     
10     13 God     2016-11-21 00:00:00         6         8 burntbricks     
# ... with 33 more rows, and 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 data frame and use that as input to the next function, like this:

interviews2 <- filter(interviews, village == "God")
interviews_god <- select(interviews2, no_membrs, years_liv)

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. Alternatively, you can create the final data frame by overwiting the intermediate result, reusing the same variable.

interviews_god <- filter(interviews, village == "God")
interviews_god <- select(interviews_god, no_membrs, years_liv)

This avoids cluttering your workspace but can lead to confusion. If only the first statement is executed, e.g. because the second one contained a typo, the result won’t be what you expcted.

You can also nest functions (i.e. one function inside of another), like this:

interviews_god <- select(filter(interviews, village == "God"), no_membrs, years_liv)

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. Pipes in R look like %>% and are made available via the magrittr package, installed automatically with dplyr. If you use RStudio, you can type the pipe with Ctrl

• Shift + M if you have a PC or Cmd + Shift + M if you have a Mac.

interviews %>%
    filter(village == "God") %>%
    select(no_membrs, years_liv)

# A tibble: 43 x 2
   no_membrs years_liv
       <dbl>     <dbl>
 1         3         4
 2         7         9
 3        10        15
 4         7         6
 5         7        40
 6         3         3
 7         6        38
 8         6        20
 9         7        20
10         6         8
# ... with 33 more rows

In the above code, we use the pipe to send the interviews dataset first through filter() to keep rows where village is “God”, then through select() to keep only the no_membrs and years_liv columns. 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 data frame 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 data frame interviews, then we filter for rows with village == "God", then we select columns no_membrs and years_liv. The dplyr functions by themselves are somewhat simple, but by combining them into linear workflows with the pipe, we can accomplish more complex manipulations of data frames.

If we want to create a new object with this smaller version of the data, we can assign it a new name:

interviews_god <- interviews %>%
    filter(village == "God") %>%
    select(no_membrs, years_liv)

interviews_god

# A tibble: 43 x 2
   no_membrs years_liv
       <dbl>     <dbl>
 1         3         4
 2         7         9
 3        10        15
 4         7         6
 5         7        40
 6         3         3
 7         6        38
 8         6        20
 9         7        20
10         6         8
# ... with 33 more rows

Note that the final data frame (interviews_god) 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.

Solution

interviews %>%
    filter(memb_assoc == "yes") %>%
    select(affect_conflicts, liv_count, no_meals)

# A tibble: 33 x 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
# ... with 23 more rows

This is a good place to save, commit, and push your new code!

Mutate

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):

interviews %>%
    mutate(people_per_room = no_membrs / rooms)

# A tibble: 131 x 15
   key_ID village interview_date      no_membrs years_liv respondent_wall~
    <dbl> <chr>   <dttm>                  <dbl>     <dbl> <chr>           
 1      1 God     2016-11-17 00:00:00         3         4 muddaub         
 2      1 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 Chirod~ 2016-11-16 00:00:00        12        70 burntbricks     
 9      9 Chirod~ 2016-11-16 00:00:00         8         6 burntbricks     
10     10 Chirod~ 2016-12-16 00:00:00        12        23 burntbricks     
# ... with 121 more rows, and 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:

interviews %>%
    filter(!is.na(memb_assoc)) %>%
    mutate(people_per_room = no_membrs / rooms)

# A tibble: 92 x 15
   key_ID village interview_date      no_membrs years_liv respondent_wall~
    <dbl> <chr>   <dttm>                  <dbl>     <dbl> <chr>           
 1      1 God     2016-11-17 00:00:00         7         9 muddaub         
 2      7 God     2016-11-17 00:00:00         6        38 muddaub         
 3      8 Chirod~ 2016-11-16 00:00:00        12        70 burntbricks     
 4      9 Chirod~ 2016-11-16 00:00:00         8         6 burntbricks     
 5     10 Chirod~ 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     
# ... with 82 more rows, and 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, so we’re asking for every row where memb_assoc is not missing..

Exercise

Create a new data frame 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 data frame.

Hint: think about how the commands should be ordered to produce this data frame!

Solution

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:

interviews %>%
    group_by(village) %>%
    summarize(mean_no_membrs = mean(no_membrs))

# A tibble: 3 x 2
  village  mean_no_membrs
  <chr>             <dbl>
1 Chirodzo           7.08
2 God                6.86
3 Ruaca              7.57

You may also have noticed that the output from these calls doesn’t run off the screen anymore. It’s one of the advantages of tbl_df over data frame.

You can also group by multiple columns:

interviews %>%
    group_by(village, memb_assoc) %>%
    summarize(mean_no_membrs = mean(no_membrs))

# A tibble: 9 x 3
# Groups:   village [3]
  village  memb_assoc mean_no_membrs
  <chr>    <chr>               <dbl>
1 Chirodzo <NA>                 5.08
2 Chirodzo no                   8.06
3 Chirodzo yes                  7.82
4 God      <NA>                 6   
5 God      no                   7.13
6 God      yes                  8   
7 Ruaca    <NA>                 6.22
8 Ruaca    no                   7.18
9 Ruaca    yes                  9.5 

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.

interviews %>%
    filter(!is.na(memb_assoc)) %>%
    group_by(village, memb_assoc) %>%
    summarize(mean_no_membrs = mean(no_membrs))

# A tibble: 6 x 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):

interviews %>%
    filter(!is.na(memb_assoc)) %>%
    group_by(village, memb_assoc) %>%
    summarize(mean_no_membrs = mean(no_membrs),
              min_membrs = min(no_membrs))

# A tibble: 6 x 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           5

It 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:

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)

# A tibble: 6 x 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           5

To 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:

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))

# A tibble: 6 x 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          2

Counting

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:

interviews %>%
    count(village)

# A tibble: 3 x 2
  village      n
  <chr>    <int>
1 Chirodzo    39
2 God         43
3 Ruaca       49

For convenience, count() provides the sort argument to get results in decreasing order:

interviews %>%
    count(village, sort = TRUE)

# A tibble: 3 x 2
  village      n
  <chr>    <int>
1 Ruaca       49
2 God         43
3 Chirodzo    39

Exercise

1. 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?

Solution

interviews %>%
   count(no_meals)

# A tibble: 2 x 2
  no_meals     n
     <dbl> <int>
1        2    52
2        3    79

1. 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).

Solution

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()
  )

# A tibble: 3 x 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    49

1. What was the largest household interviewed in each month?

Solution

# if not already included, add month, year, and day columns
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))

# A tibble: 5 x 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            15

Save, commit, and push your changes.

Reshaping with gather and spread

In the spreadsheet lesson, we discussed how to structure our data leading to the four rules defining a tidy dataset:

1. Each variable has its own column
2. Each observation has its own row
3. Each value must have its own cell
4. Each type of observational unit forms a table

Here we examine the fourth rule: Each type of observational unit forms a table.

In interviews, each row contains the values of variables associated with each record (the unit), values such as the number of household members or posessions associated with each record. What if instead of comparing records, we wanted to look at differences in households grouped by different types of housing construction materials?

We’d need to create a new table where each row (the unit) is comprised of values of variables associated with each housing material (e.g. for respondent_wall_type). In practical terms this means the values of the wall construction materials in respondent_wall_type would become the names of column variables and the cells would contain TRUE or FALSE.

Having created a new table, we can now explore the relationship within and between household types - for example we could compare the ratio of household members to sleeping rooms grouped by type of construction material. The key point here is that we are still following a tidy data structure, but we have reshaped the data according to the observations of interest.

The opposite transformation would be to transform column names into values of a variable.

We can do both these of transformations with two tidyr functions, spread() and gather().

Spreading

spread() takes three principal arguments:

1. the data
2. the key column variable whose values will become new column names.
3. the value column variable whose values will fill the new column variables.

Further arguments include fill which, if set, fills in missing values with the value provided.

Let’s use spread() to transform interviews to create new columns for each type of wall construction material. We use the pipe as before too. Because both the key and value parameters must come from column values, we will create a dummy column (we’ll name it wall_type_logical) to hold the value TRUE, which we will then place into the appropriate column that corresponds to the wall construction material for that respondent. When using mutate() if you give a single value, it will be used for all observations in the dataset. We will use fill = FALSE in spread() to fill the rest of the new columns for that row with FALSE.

interviews_spread <- interviews %>%
    mutate(wall_type_logical = TRUE) %>%
    spread(key = respondent_wall_type, value = wall_type_logical, fill = FALSE)

View the interviews_spread data frame and notice that there is no longer a column titled respondent_wall_type. This is because there is a default parameter in spread() that drops the original column.

Gathering

The opposing situation could occur if we had been provided with data in the form of interviews_spread, where the building materials are column names, but we wish to treat them as values of a respondent_wall_type variable instead.

In this situation we are gathering the column names and turning them into a pair of new variables. One variable represents the column names as values, and the other variable contains the values previously associated with the column names. We will do this in two steps to make this process a bit clearer.

gather() takes four principal arguments:

1. the data
2. the key column variable we wish to create from column names.
3. the value column variable we wish to create and fill with values associated with the key.
4. the names of the columns we use to fill the key variable (or to drop).

To recreate our original data frame, we will use the following:

1. the data - interviews_spread
2. the key column will be “respondent_wall_type” (as a character string). This is the name of the new column we want to create.
3. the value column will be wall_type_logical. This will be either TRUE or FALSE.
4. the names of the columns we will use to fill the key variable are burntbricks:sunbricks (the column named “burntbricks” up to and including the column named “sunbricks” as they are ordered in the data frame).

interviews_gather <- interviews_spread %>%
    gather(key = respondent_wall_type, value = "wall_type_logical",
           burntbricks:sunbricks)

This creates a data frame with 524 rows (4 rows per interview respondent). The four rows for each respondent differ only in the value of the “respondent_wall_type” and “dummy” columns. View the data to see what this looks like.

Only one row for each interview respondent is informative - we know that if the house walls are made of “sunbrick” they aren’t made of any other the other materials. Therefore, we can get filter our dataset to only keep values where wall_type_logical is TRUE. Because, wall_type_logical is already either TRUE or FALSE, when passing the column name to filter(), it will automatically already only keep rows where this column has the value TRUE. We can then remove the wall_type_logical column. We do all of these steps together in the next chunk of code:

interviews_gather <- interviews_spread %>%
    gather(key = "respondent_wall_type", value = "wall_type_logical",
           burntbricks:sunbricks) %>%
    filter(wall_type_logical) %>%
    select(-wall_type_logical)

View both interviews_gather and interviews_spread and compare their structure. Notice that the rows have been reordered in interviews_gather such that all of the respondents with a particular wall type are grouped together.

Applying spread() to clean our data

Now that we’ve learned about gather() and spread() we’re going to put these functions to use to fix a problem with the way that our data is structured. In the spreadsheets lesson, we learned that it’s best practice to have only a single piece of information in each cell of your spreadsheet. In this dataset, we have several columns which contain multiple pieces of information. For example, the items_owned column contains information about whether our respondents owned a fridge, a television, etc. To make this data easier to analyze, we will split this column and create a new column for each item. Each cell in that column will either be TRUE or FALSE and will indicate whether that interview respondent owned that item.

interviews_items_owned <- interviews %>%
    mutate(split_items = str_split(items_owned, ";")) %>%
    unnest() %>%
    mutate(items_owned_logical = TRUE) %>%
    spread(key = split_items, value = items_owned_logical, fill = FALSE)
nrow(interviews_items_owned)

[1] 131

There are a couple of new concepts in this code chunk. Let’s walk through it line by line. First we create a new object (interviews_items_owned) based on the interviews dataframe.

interviews_items_owned <- interviews %>%

Then we use the new function str_split() to split the column items_owned based on the presence of semi-colons (;). This creates a new column split_items that contains each item as a list.

mutate(split_items = str_split(items_owned, ";")) %>%

Now that we have the items_owned column as a list, we can use the tidyr function unnest() to create a long format version of the dataset. In this long format version, there are 131 rows (one row for each unique item for each respondent).

unnest() %>%

Lastly, we use spread() to switch from long format to wide format. This creates a new column for each of the unique values in the split_items column and fills those columns with TRUE or FALSE.

mutate(items_owned_logical = TRUE) %>%
    spread(key = split_items, value = items_owned_logical, fill = FALSE)

View the interviews_items_owned data frame. It should have 131 rows (the same number of rows you had originally), but extra columns for each item.

You may notice that the last column in called \\``. This is because the respondents did not own any of the items that was in the interviewer's list. We can use the `rename()` function to change this name to something more meaningful:

interviews_items_owned <- interviews_items_owned %>%
    rename(no_listed_items = `<NA>`)

This format of the data allows us to do interesting things, like make a table showing the number of respondents in each village who owned a particular item:

interviews_items_owned %>%
    filter(bicycle) %>%
    group_by(village) %>%
    count(bicycle)

# A tibble: 3 x 3
# Groups:   village [3]
  village  bicycle     n
  <chr>    <lgl>   <int>
1 Chirodzo TRUE       17
2 God      TRUE       23
3 Ruaca    TRUE       20

Or calculate the average number of items from the list owned by respondents in each village:

interviews_items_owned %>%
    mutate(number_items = rowSums(select(., bicycle:television))) %>%
    group_by(village) %>%
    summarize(mean_items = mean(number_items))

# A tibble: 3 x 2
  village  mean_items
  <chr>         <dbl>
1 Chirodzo       4.54
2 God            3.98
3 Ruaca          5.57

Exercise

1. Create a new data frame (named interviews_months_no_food) that has one column for each month and records TRUE or FALSE for whether each interview respondent was lacking food in that month.

Solution

interviews_months_no_food <- interviews %>%
  mutate(split_months = str_split(months_lack_food, ";")) %>%
  unnest() %>%
  mutate(months_lack_food_logical  = TRUE) %>%
  spread(key = split_months, value = months_lack_food_logical, fill = FALSE)

1. How many months (on average) were respondents without food if they did belong to an irrigation association? What about if they didn’t?

Solution

interviews_months_no_food %>%
  mutate(number_months = rowSums(select(., Apr:Sept))) %>%
  group_by(memb_assoc) %>%
  summarize(mean_months = mean(number_months))

# A tibble: 3 x 2
  memb_assoc mean_months
  <chr>            <dbl>
1 <NA>              2.95
2 no                2.31
3 yes               2.64

Exporting data

Now that you have learned how to use dplyr to extract information from or summarize your raw data, you may want to export these new data sets to share them with your collaborators or for archival purposes.

Similar to the read_csv() function used for reading CSV files into R, there is a write_csv() function that generates CSV files from data frames.

Before using write_csv(), we are going to create a new folder, data_output, in our working directory that will store this generated dataset. We don’t want to write generated datasets in the same directory as our raw data. It’s good practice to keep them separate. The data folder should only contain the raw, unaltered data, and should be left alone to make sure we don’t delete or modify it. In contrast, our script will generate the contents of the data_output directory, so even if the files it contains are deleted, we can always re-generate them.

In preparation for our next lesson on plotting, we are going to create a version of the dataset where each of the columns includes only one data value. To do this, we will use spread to expand the months_lack_food and items_owned columns. We will also create a couple of summary columns.

interviews_plotting <- interviews %>%
    ## spread data by items_owned
    mutate(split_items = str_split(items_owned, ";")) %>%
    unnest() %>%
    mutate(items_owned_logical = TRUE) %>%
    spread(key = split_items, value = items_owned_logical, fill = FALSE) %>%
    rename(no_listed_items = `<NA>`) %>%
    ## spread data by months_lack_food
    mutate(split_months = str_split(months_lack_food, ";")) %>%
    unnest() %>%
    mutate(months_lack_food_logical = TRUE) %>%
    spread(key = split_months, value = months_lack_food_logical, fill = FALSE) %>%
    ## add some summary columns
    mutate(number_months_lack_food = rowSums(select(., Apr:Sept))) %>%
    mutate(number_items = rowSums(select(., bicycle:television)))

Now we can save this data frame to our data_output directory.

write_csv(interviews_plotting, path = "data_output/interviews_plotting.csv")

Save, commit, and push your changes if you haven’t done so already. Notice that you now have several files to commit, as we have just saved a new data output file. You can commit them individually or together, however, it is good practice to package commits in such a way that the changes are related. For example, if your new script saves your data output file, you can commit the script together with the new csv.

Key Points

• Use the dplyr package to manipulate dataframes.

• Use select() to choose variables from a dataframe.

• Use filter() to choose data based on values.

• Use group_by() and summarize() to work with subsets of data.

• Use mutate() to create new variables.

• Use the tidyr package to change the layout of dataframes.

• Use gather() to go from wide to long format.

• Use spread() to go from long to wide format.