44  42. Practice Questions - TOPIC: Lists and Dataframes.

rm(list=ls())

# The following list contains the data for a chain of hardware stores
# From October 2021.
#
# The data for each store in the chain is stored in a separate list within
# the list of stores. 
#
# The first two items in the list are the year and month for the data.
# The other items in the list are lists of information for each store in 
# the chain. For simplicity, the data below only shows three stores. However, 
# there could be many stores in the chain, each with similarly structured data.
#
# Each store is allowed to set their own prices and salePrices. 
# Each store is also allowed to carry a somewhat different selection of 
# products than the other stores.
#
# Answer the questions below by referring to this data. You should 
# write the code so that it will continue to work even if the actual 
# values in the data change. 

stores = list(
  
  year=2021,
  
  month=10,
  
  list(
      address="123 main street",
      storeId="1000",
      manager="joe",
      products = data.frame(
        product =          c("hammer", "screwdriver", "hand saw", "wrench" ),
        price=             c(10.99,    8.00,          15.00,      5.00),
        salePrice=         c(8.99,     NA,            NA,         4.50),
        amountSold=        c(100,      200,           50,         75),
        amountInInventory= c(20,       12,            0,          5)
      )
  ),
  
  list(
    address="99 hickory place",
    storeId="1111",
    manager="sue",
    products = data.frame (
      product =          c("wrench","screwdriver", "screws", "nails","pliers" ),
      price=             c(11.99,    8.00,          3.50,    2.99,   4.99),
      salePrice=         c(NA,       7.00,          NA,      2.50,   4.50),
      amountSold=        c(80,       195,           200,     400,    80),
      amountInInventory= c(20,       15,            40,      15,     0)
    )
  ),
  
  list(
    address="450 broadway",
    storeId="2345",
    manager="carla",
    products = data.frame (
      product =          c("wrench","saw","hammer","screws", "nails","drill" ),
      price=             c(11.99,    8.00, 11.00,  3.50,    2.99,   44.99),
      salePrice=         c(NA,       7.00, 10.00,  2.50,    2.50,   NA),
      amountSold=        c(40,       100,  3,      100,     200,    10),
      amountInInventory= c(30,       5,    40,     0,       0,      3)
    )
  )
  
  # If there were more stores then the data for the other
  # stores would appear as separate lists here.
)

# NOTE - the following command may help you to understand the data better
str(stores)

List of 5
 $ year : num 2021
 $ month: num 10
 $      :List of 4
  ..$ address : chr "123 main street"
  ..$ storeId : chr "1000"
  ..$ manager : chr "joe"
  ..$ products:'data.frame':    4 obs. of  5 variables:
  .. ..$ product          : chr [1:4] "hammer" "screwdriver" "hand saw" "wrench"
  .. ..$ price            : num [1:4] 11 8 15 5
  .. ..$ salePrice        : num [1:4] 8.99 NA NA 4.5
  .. ..$ amountSold       : num [1:4] 100 200 50 75
  .. ..$ amountInInventory: num [1:4] 20 12 0 5
 $      :List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0
 $      :List of 4
  ..$ address : chr "450 broadway"
  ..$ storeId : chr "2345"
  ..$ manager : chr "carla"
  ..$ products:'data.frame':    6 obs. of  5 variables:
  .. ..$ product          : chr [1:6] "wrench" "saw" "hammer" "screws" ...
  .. ..$ price            : num [1:6] 11.99 8 11 3.5 2.99 ...
  .. ..$ salePrice        : num [1:6] NA 7 10 2.5 2.5 NA
  .. ..$ amountSold       : num [1:6] 40 100 3 100 200 10
  .. ..$ amountInInventory: num [1:6] 30 5 40 0 0 3

#----------------------------------------------------------------------------------
# QUESTION 1
# TOPIC: lists, paste
# 
# PART A
#
# Write a command to show the year and month of the data in the format
# month/year. For example if the year is 2021 and the month is 11
# then your command should show "11/2021". Remember that your command should
# continue to work even if the value of the year or month in the data changes.
#
# Use the $dollar-sign-notation for this answer.
#
# PART B
#
# Create a new variable, stores2, that has all the same values as the stores
# variable. However, instead of the first two values being year and month, 
# the first value in the stores2 variable should be a value named date that 
# contains the month and year in month/year format (as in part A). 
# The data for the first store should appear in the 2nd position in the 
# stores2 list.
#----------------------------------------------------------------------------------

part a - click here for answer

###################.
# Answer - part A
###################.

paste( stores$month , stores$year, sep="/")

[1] "10/2021"

part b - click here for answer

##################################.
# Answer - part B - ONE ANSWER
##################################.

# copy the values from stores
stores2 = stores             

# set the first value in stores2 to be the month/year
stores2[[1]] = paste( stores$month , stores$year, sep="/")

# change the name of the first value to "date"
names(stores2)[1] = "date"

# remove the 2nd value (i.e. the month) from stores2
stores2[[2]] = NULL            # remove the 2nd value from stores2

str(stores2)

List of 4
 $ date: chr "10/2021"
 $     :List of 4
  ..$ address : chr "123 main street"
  ..$ storeId : chr "1000"
  ..$ manager : chr "joe"
  ..$ products:'data.frame':    4 obs. of  5 variables:
  .. ..$ product          : chr [1:4] "hammer" "screwdriver" "hand saw" "wrench"
  .. ..$ price            : num [1:4] 11 8 15 5
  .. ..$ salePrice        : num [1:4] 8.99 NA NA 4.5
  .. ..$ amountSold       : num [1:4] 100 200 50 75
  .. ..$ amountInInventory: num [1:4] 20 12 0 5
 $     :List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0
 $     :List of 4
  ..$ address : chr "450 broadway"
  ..$ storeId : chr "2345"
  ..$ manager : chr "carla"
  ..$ products:'data.frame':    6 obs. of  5 variables:
  .. ..$ product          : chr [1:6] "wrench" "saw" "hammer" "screws" ...
  .. ..$ price            : num [1:6] 11.99 8 11 3.5 2.99 ...
  .. ..$ salePrice        : num [1:6] NA 7 10 2.5 2.5 NA
  .. ..$ amountSold       : num [1:6] 40 100 3 100 200 10
  .. ..$ amountInInventory: num [1:6] 30 5 40 0 0 3

##################################.
# Answer - part B - ANOTHER ANSWER
##################################.

# create an empty list
stores2 = list()

# set the date 
stores2$date = paste( stores$month , stores$year, sep="/")

# copy the data for all the stores from the stores variable
stores2[2:(length(stores)-1)] = stores[3:length(stores)]

str(stores2)

List of 4
 $ date: chr "10/2021"
 $     :List of 4
  ..$ address : chr "123 main street"
  ..$ storeId : chr "1000"
  ..$ manager : chr "joe"
  ..$ products:'data.frame':    4 obs. of  5 variables:
  .. ..$ product          : chr [1:4] "hammer" "screwdriver" "hand saw" "wrench"
  .. ..$ price            : num [1:4] 11 8 15 5
  .. ..$ salePrice        : num [1:4] 8.99 NA NA 4.5
  .. ..$ amountSold       : num [1:4] 100 200 50 75
  .. ..$ amountInInventory: num [1:4] 20 12 0 5
 $     :List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0
 $     :List of 4
  ..$ address : chr "450 broadway"
  ..$ storeId : chr "2345"
  ..$ manager : chr "carla"
  ..$ products:'data.frame':    6 obs. of  5 variables:
  .. ..$ product          : chr [1:6] "wrench" "saw" "hammer" "screws" ...
  .. ..$ price            : num [1:6] 11.99 8 11 3.5 2.99 ...
  .. ..$ salePrice        : num [1:6] NA 7 10 2.5 2.5 NA
  .. ..$ amountSold       : num [1:6] 40 100 3 100 200 10
  .. ..$ amountInInventory: num [1:6] 30 5 40 0 0 3

#----------------------------------------------------------------------------------
# QUESTION 2
# TOPIC: lists, paste
# 
# Same as the previous question but DO NOT use the $dollar-sign-notation
# for accessing the list.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

# The [[double-bracket]] notation is equivalent to the $dollar-sign notation.
# NOTE - do NOT use the [single-bracket] notation.
#
# stores[[1]] in the data above is "2021"
#
# However, stores[1] in the data above is a LIST that contains "2021"

paste( stores[[2]] , stores[[1]], sep="/")

[1] "10/2021"

#----------------------------------------------------------------------------------
# QUESTION 3
# TOPIC: lists, c
# 
# Part A
# Write a single command that returns a vector that contains
# the addresses of the first and second stores.
# The answer should look as follows: 
  #   [1] "123 main street"  "99 hickory place"
#
# Part B
# Do this again, but this time use a different notation to access the list.
#----------------------------------------------------------------------------------

part a - click here for answer

##################.
# ANSWER - PART A
##################.

# Look at the answer carefully.
# NOTE that stores[[3]] is the list for the first store
# Therefore stores[[3]]$address is the address of the first store.
#
# HOWEVER - be careful
# stores[3] is a LIST that contains the list for the first store.
# Therefore stores[3]$address is wrong since the list stores[3] only 
# contains a single item, i.e. another list, and does NOT contain an $address item.
# 
# You can see this more clearly if you use the str function. 
# Compare the output of the following 3 commands:
# str(stores)
# str(stores[[3]])
# str(stores[3])

c(stores[[3]]$address, stores[[4]]$address)

[1] "123 main street"  "99 hickory place"

part b - click here for answer

##################.
# ANSWER - PART B
##################.

c(stores[[3]][[1]], stores[[4]][[1]])

[1] "123 main street"  "99 hickory place"

#----------------------------------------------------------------------------------
# QUESTION 4
# TOPIC: lists
# 
# Write a command to display the number of different products
# that are carried by the first store.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

nrow(stores[[3]]$products)

[1] 4

#----------------------------------------------------------------------------------
# QUESTION 5
# TOPIC: lists
#
# Write a command that calculates the number of stores in the chain.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

length( stores ) - 2

[1] 3

#----------------------------------------------------------------------------------
# QUESTION 6
# TOPIC: lists
#
# Write a command to retrieve a list that contains 
# the year, month and just the data for the 2nd store.
# Store this new list in a variable called store2.
# The structure of the data should be exactly the same.
#
# Remember that the data above only contains three stores, but the command you write
# should work even if there were more stores (e.g. 100 stores) in the data.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

store2 = stores[c(1,2,4)]

# The following commands are NOT part of the answer but may help
# you to understand the results.
store2   # see the answer

$year
[1] 2021

$month
[1] 10

[[3]]
[[3]]$address
[1] "99 hickory place"

[[3]]$storeId
[1] "1111"

[[3]]$manager
[1] "sue"

[[3]]$products
      product price salePrice amountSold amountInInventory
1      wrench 11.99        NA         80                20
2 screwdriver  8.00       7.0        195                15
3      screws  3.50        NA        200                40
4       nails  2.99       2.5        400                15
5      pliers  4.99       4.5         80                 0

str(store2) # understand the structure of the answer

List of 3
 $ year : num 2021
 $ month: num 10
 $      :List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0

#----------------------------------------------------------------------------------
# QUESTION 7
# TOPIC: lists
#
# The first store in the data actually went out of business in 9/2021
# However, it was mistakenly left in the data for 10/2021.
# Write a command to create a variable named updatedStores that contains
# all of the data from the stores list but without the data for the first store.
#
# Remember that the data above only contains three stores, but the command you write
# should work even if there were more stores (e.g. 100 stores) in the data.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

# [single-brackets] since you are getting back a list of items
# The index -3 means get back everything except for the third item
# in the list, i.e. everything except for the data for the first store.
updatedStores = stores[-3] 

# The following commands are NOT part of the answer but may help
# you to understand the results.
updatedStores   # see the answer

$year
[1] 2021

$month
[1] 10

[[3]]
[[3]]$address
[1] "99 hickory place"

[[3]]$storeId
[1] "1111"

[[3]]$manager
[1] "sue"

[[3]]$products
      product price salePrice amountSold amountInInventory
1      wrench 11.99        NA         80                20
2 screwdriver  8.00       7.0        195                15
3      screws  3.50        NA        200                40
4       nails  2.99       2.5        400                15
5      pliers  4.99       4.5         80                 0


[[4]]
[[4]]$address
[1] "450 broadway"

[[4]]$storeId
[1] "2345"

[[4]]$manager
[1] "carla"

[[4]]$products
  product price salePrice amountSold amountInInventory
1  wrench 11.99        NA         40                30
2     saw  8.00       7.0        100                 5
3  hammer 11.00      10.0          3                40
4  screws  3.50       2.5        100                 0
5   nails  2.99       2.5        200                 0
6   drill 44.99        NA         10                 3

str(updatedStores) # understand the structure of the answer

List of 4
 $ year : num 2021
 $ month: num 10
 $      :List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0
 $      :List of 4
  ..$ address : chr "450 broadway"
  ..$ storeId : chr "2345"
  ..$ manager : chr "carla"
  ..$ products:'data.frame':    6 obs. of  5 variables:
  .. ..$ product          : chr [1:6] "wrench" "saw" "hammer" "screws" ...
  .. ..$ price            : num [1:6] 11.99 8 11 3.5 2.99 ...
  .. ..$ salePrice        : num [1:6] NA 7 10 2.5 2.5 NA
  .. ..$ amountSold       : num [1:6] 40 100 3 100 200 10
  .. ..$ amountInInventory: num [1:6] 30 5 40 0 0 3

#----------------------------------------------------------------------------------
# QUESTION 8
# TOPIC: lists, %in%
#
# The stores are allowed to carry different product lines.
# Write a command that shows those products that are carried by the 
# second store that are not carried by the first store.
# 
# PART A - write this answer using the %in% operator
# PART B - write this answer without using the %in% operator
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

# PART A
stores[[4]]$product [ ! ( stores[[4]]$product %in% stores[[3]]$product ) ]

      product price salePrice amountSold amountInInventory
1      wrench 11.99        NA         80                20
2 screwdriver  8.00       7.0        195                15
3      screws  3.50        NA        200                40
4       nails  2.99       2.5        400                15
5      pliers  4.99       4.5         80                 0

# PART B
setdiff(stores[[4]]$product, stores[[3]]$product)

[[1]]
[1] "wrench"      "screwdriver" "screws"      "nails"       "pliers"     

[[2]]
[1] 11.99  8.00  3.50  2.99  4.99

[[3]]
[1]  NA 7.0  NA 2.5 4.5

[[4]]
[1]  80 195 200 400  80

[[5]]
[1] 20 15 40 15  0

#----------------------------------------------------------------------------------
# QUESTION 9
# TOPIC: lists, is.na
#
# The products that are on sale have a number for the sale price.
# The products that are NOT on sale have NA for the sale price.
# Write a command to return a VECTOR of the names of the products
# that are on sale in the first store.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

stores[[3]]$product [ !is.na(stores[[3]]$salePrice) ]

data frame with 0 columns and 4 rows

#----------------------------------------------------------------------------------
# QUESTION 10
# TOPIC: lists, is.na, parallel vectors
#
# See the previous question.
#
# For this question create a variable named saleProducts that
# contains a dataframe with just those rows from the first store's products
# which are on sale. (You may answer using several lines of code).
#
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

products = stores[[3]]$products
saleProducts = products[ !is.na(products$salePrice) , ]

# The following commands are NOT part of the answer but may help
# you to understand the results.
products   

      product price salePrice amountSold amountInInventory
1      hammer 10.99      8.99        100                20
2 screwdriver  8.00        NA        200                12
3    hand saw 15.00        NA         50                 0
4      wrench  5.00      4.50         75                 5

saleProducts

  product price salePrice amountSold amountInInventory
1  hammer 10.99      8.99        100                20
4  wrench  5.00      4.50         75                 5

#----------------------------------------------------------------------------------
# QUESTION 11
# TOPIC: lists, names
#
# The list, stores, contains names for the year and month entries,
# but does not contain names for the two stores. 
# 
# PART A
#
# Copy the data for stores into a variable named storesWithNames
# i.e. storesWithNames = stores
# Then modify storesWithNames so that
#  - the list of data for the 1st store has the name "store1" in the main list
#  - the list of data for the 2nd store has the name "store2" in the main list
#  - etc ... for all of the stores (you code should work even if there were more stores)
#
#
# PART B
# Now use the storesWithNames variable to display just the product names and prices 
# in the 1st store, i.e. store1, whose prices are $10 or more.
# Use the new names that you just created in PART A as part of your answer.
#----------------------------------------------------------------------------------

part a - click here for answer

####################.
# ANSWER - Part A
####################.

storesWithNames = stores   # as directed to do in the question
names(storesWithNames)[3:length(storesWithNames)] = paste0("store", 1:(length(storesWithNames)-2))

# The following commands are NOT part of the answer but may help
# you to understand the results.
storesWithNames   # see the answer

$year
[1] 2021

$month
[1] 10

$store1
$store1$address
[1] "123 main street"

$store1$storeId
[1] "1000"

$store1$manager
[1] "joe"

$store1$products
      product price salePrice amountSold amountInInventory
1      hammer 10.99      8.99        100                20
2 screwdriver  8.00        NA        200                12
3    hand saw 15.00        NA         50                 0
4      wrench  5.00      4.50         75                 5


$store2
$store2$address
[1] "99 hickory place"

$store2$storeId
[1] "1111"

$store2$manager
[1] "sue"

$store2$products
      product price salePrice amountSold amountInInventory
1      wrench 11.99        NA         80                20
2 screwdriver  8.00       7.0        195                15
3      screws  3.50        NA        200                40
4       nails  2.99       2.5        400                15
5      pliers  4.99       4.5         80                 0


$store3
$store3$address
[1] "450 broadway"

$store3$storeId
[1] "2345"

$store3$manager
[1] "carla"

$store3$products
  product price salePrice amountSold amountInInventory
1  wrench 11.99        NA         40                30
2     saw  8.00       7.0        100                 5
3  hammer 11.00      10.0          3                40
4  screws  3.50       2.5        100                 0
5   nails  2.99       2.5        200                 0
6   drill 44.99        NA         10                 3

str(storesWithNames) # understand the structure of the answer

List of 5
 $ year  : num 2021
 $ month : num 10
 $ store1:List of 4
  ..$ address : chr "123 main street"
  ..$ storeId : chr "1000"
  ..$ manager : chr "joe"
  ..$ products:'data.frame':    4 obs. of  5 variables:
  .. ..$ product          : chr [1:4] "hammer" "screwdriver" "hand saw" "wrench"
  .. ..$ price            : num [1:4] 11 8 15 5
  .. ..$ salePrice        : num [1:4] 8.99 NA NA 4.5
  .. ..$ amountSold       : num [1:4] 100 200 50 75
  .. ..$ amountInInventory: num [1:4] 20 12 0 5
 $ store2:List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0
 $ store3:List of 4
  ..$ address : chr "450 broadway"
  ..$ storeId : chr "2345"
  ..$ manager : chr "carla"
  ..$ products:'data.frame':    6 obs. of  5 variables:
  .. ..$ product          : chr [1:6] "wrench" "saw" "hammer" "screws" ...
  .. ..$ price            : num [1:6] 11.99 8 11 3.5 2.99 ...
  .. ..$ salePrice        : num [1:6] NA 7 10 2.5 2.5 NA
  .. ..$ amountSold       : num [1:6] 40 100 3 100 200 10
  .. ..$ amountInInventory: num [1:6] 30 5 40 0 0 3

part b - click here for answer

####################.
# ANSWER - Part B
####################.

# Answer with one line of code
storesWithNames$store1$products [ storesWithNames$store1$products$price >= 10 , c("product","price")]

   product price
1   hammer 10.99
3 hand saw 15.00

  # Alternative answer with more than one line of code
store1products = storesWithNames$store1$products
store1products[ store1products$price >= 10, c("product", "price")]

   product price
1   hammer 10.99
3 hand saw 15.00

#----------------------------------------------------------------------------------
# QUESTION 12
# TOPIC: lists
#
# Write the following function: 
#    addStore(stores, address, storeId, manager, products)
#
# Arguments:
#    stores - the complete stores list
#    address - the address of the new store
#    storeId - the id of the new store
#    manager - the manager of the new store
#    products - the products dataframe of the new store
#
# Returns:
#    A new stores vector that includes an entry for the new store.
#
# Your code should add the new store as at the end of the list of stores.
# Right now the stores list only contains 3 stores so your code should 
# add the new store as the 4th store - however, your code should work no
# matter how many stores are in the list.
#
# Example: 
#    YOUR CODE TO DEFINE THE FUNCTION GOES HERE
#
#    stores = addStore(stores, "4 apple drive", "4000", "dana",
#                      data.frame(product = c("hammer","saw"),
#                                 price = c(10,15),
#                                 salePrice=c(NA,NA),
#                                 amountSold = c(100,200),
#                                 amountInInventory = c(20,30)))
#
#    stores = addStore(stores, "5 orange grove", "5000", "jan",
#                      data.frame(product = c("wrench","hammer"),
#                                 price = c(12,9.50),
#                                 salePrice=c(NA,NA),
#                                 amountSold = c(90,80),
#                                 amountInInventory = c(10,20)))
#
#    str(stores[c(6,7)])
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

addStore = function(stores, address, storeId, manager, products){
  stores[[length(stores)+1]] = list(address = address,
                                    storeId = storeId,
                                    manager = manager, 
                                    products=products)
  return(stores)
}

# try it

newStores = stores
newStores = addStore(newStores, "4 apple drive", "4000", "dana",
                  data.frame(product = c("hammer","saw"),
                             price = c(10,15),
                             salePrice=c(NA,NA),
                             amountSold = c(100,200),
                             amountInInventory = c(20,30)))

newStores = addStore(newStores, "5 orange grove", "5000", "jan",
                  data.frame(product = c("wrench","hammer"),
                             price = c(12,9.50),
                             salePrice=c(NA,NA),
                             amountSold = c(90,80),
                             amountInInventory = c(10,20)))

str(newStores[c(6,7)])

List of 2
 $ :List of 4
  ..$ address : chr "4 apple drive"
  ..$ storeId : chr "4000"
  ..$ manager : chr "dana"
  ..$ products:'data.frame':    2 obs. of  5 variables:
  .. ..$ product          : chr [1:2] "hammer" "saw"
  .. ..$ price            : num [1:2] 10 15
  .. ..$ salePrice        : logi [1:2] NA NA
  .. ..$ amountSold       : num [1:2] 100 200
  .. ..$ amountInInventory: num [1:2] 20 30
 $ :List of 4
  ..$ address : chr "5 orange grove"
  ..$ storeId : chr "5000"
  ..$ manager : chr "jan"
  ..$ products:'data.frame':    2 obs. of  5 variables:
  .. ..$ product          : chr [1:2] "wrench" "hammer"
  .. ..$ price            : num [1:2] 12 9.5
  .. ..$ salePrice        : logi [1:2] NA NA
  .. ..$ amountSold       : num [1:2] 90 80
  .. ..$ amountInInventory: num [1:2] 10 20

str(newStores)

List of 7
 $ year : num 2021
 $ month: num 10
 $      :List of 4
  ..$ address : chr "123 main street"
  ..$ storeId : chr "1000"
  ..$ manager : chr "joe"
  ..$ products:'data.frame':    4 obs. of  5 variables:
  .. ..$ product          : chr [1:4] "hammer" "screwdriver" "hand saw" "wrench"
  .. ..$ price            : num [1:4] 11 8 15 5
  .. ..$ salePrice        : num [1:4] 8.99 NA NA 4.5
  .. ..$ amountSold       : num [1:4] 100 200 50 75
  .. ..$ amountInInventory: num [1:4] 20 12 0 5
 $      :List of 4
  ..$ address : chr "99 hickory place"
  ..$ storeId : chr "1111"
  ..$ manager : chr "sue"
  ..$ products:'data.frame':    5 obs. of  5 variables:
  .. ..$ product          : chr [1:5] "wrench" "screwdriver" "screws" "nails" ...
  .. ..$ price            : num [1:5] 11.99 8 3.5 2.99 4.99
  .. ..$ salePrice        : num [1:5] NA 7 NA 2.5 4.5
  .. ..$ amountSold       : num [1:5] 80 195 200 400 80
  .. ..$ amountInInventory: num [1:5] 20 15 40 15 0
 $      :List of 4
  ..$ address : chr "450 broadway"
  ..$ storeId : chr "2345"
  ..$ manager : chr "carla"
  ..$ products:'data.frame':    6 obs. of  5 variables:
  .. ..$ product          : chr [1:6] "wrench" "saw" "hammer" "screws" ...
  .. ..$ price            : num [1:6] 11.99 8 11 3.5 2.99 ...
  .. ..$ salePrice        : num [1:6] NA 7 10 2.5 2.5 NA
  .. ..$ amountSold       : num [1:6] 40 100 3 100 200 10
  .. ..$ amountInInventory: num [1:6] 30 5 40 0 0 3
 $      :List of 4
  ..$ address : chr "4 apple drive"
  ..$ storeId : chr "4000"
  ..$ manager : chr "dana"
  ..$ products:'data.frame':    2 obs. of  5 variables:
  .. ..$ product          : chr [1:2] "hammer" "saw"
  .. ..$ price            : num [1:2] 10 15
  .. ..$ salePrice        : logi [1:2] NA NA
  .. ..$ amountSold       : num [1:2] 100 200
  .. ..$ amountInInventory: num [1:2] 20 30
 $      :List of 4
  ..$ address : chr "5 orange grove"
  ..$ storeId : chr "5000"
  ..$ manager : chr "jan"
  ..$ products:'data.frame':    2 obs. of  5 variables:
  .. ..$ product          : chr [1:2] "wrench" "hammer"
  .. ..$ price            : num [1:2] 12 9.5
  .. ..$ salePrice        : logi [1:2] NA NA
  .. ..$ amountSold       : num [1:2] 90 80
  .. ..$ amountInInventory: num [1:2] 10 20

#----------------------------------------------------------------------------------
# QUESTION 13
# TOPIC: lists
#
# PART A
# Use lapply to retrieve a list of just addresses from all of the
# stores in the stores list. Your code should work no matter how many
# stores are in the list.
#
# PART B
# Modify your answer to use sapply so that you get a vector instead
# of a list.
#----------------------------------------------------------------------------------

part a - click here for answer

##################.
# ANSWER - PART A
##################.

lapply(stores[3:length(stores)],
       function(store) store$address)

[[1]]
[1] "123 main street"

[[2]]
[1] "99 hickory place"

[[3]]
[1] "450 broadway"

part b - click here for answer

##################.
# ANSWER - PART B
##################.

sapply(stores[3:length(stores)],
       function(store) store$address)

                                                         
 "123 main street" "99 hickory place"     "450 broadway" 

#----------------------------------------------------------------------------------
# QUESTION 14
# TOPIC: lists, dataframes, rbind
#
# Create a single products dataframe that combines the products
# from the 1st and 2nd stores. Hint - use rbind.
#
# The combined dataframe should include a new column, location, that records
# the address of the store from which the data in a particular row comes from.
# Setup the new dataframe so that the location column is the first column.
#
# Store the new dataframe in a variable named, combinedProducts.
#
# You may use more than one line of code.
#----------------------------------------------------------------------------------

click here for answer

###########.
# ANSWER
###########.

df1 = stores[[3]]$products
df1$location = stores[[3]]$address

df2 = stores[[4]]$products
df2$location = stores[[4]]$address

combinedProducts = rbind ( df1, df2 )

combinedProducts = combinedProducts[ , c(ncol(combinedProducts), 1:(ncol(combinedProducts) - 1))]

combinedProducts

          location     product price salePrice amountSold amountInInventory
1  123 main street      hammer 10.99      8.99        100                20
2  123 main street screwdriver  8.00        NA        200                12
3  123 main street    hand saw 15.00        NA         50                 0
4  123 main street      wrench  5.00      4.50         75                 5
5 99 hickory place      wrench 11.99        NA         80                20
6 99 hickory place screwdriver  8.00      7.00        195                15
7 99 hickory place      screws  3.50        NA        200                40
8 99 hickory place       nails  2.99      2.50        400                15
9 99 hickory place      pliers  4.99      4.50         80                 0

#----------------------------------------------------------------------------------
# QUESTION 15
# TOPIC: lists
#
# PART A
# 
# Use lapply to create a list that contains just the products dataframes from 
# all of the stores. Each dataframe in the list of dataframes should
# include a location column that contains the address of the store
# for the products. Name the list productDataframes.
#
# PART B
# 
# Look up the documentation for the do.call function. 
# Use do.call to combine ALL of the product dataframes from PART A 
# into a single combined dataframe named allProducts.
#----------------------------------------------------------------------------------

part a - click here for answer

##################.
# ANSWER - PART A
##################.

productsWithLocation = function(store) {
  products = store$products
  products$location = store$address
  products = products[ c( ncol(products), 2:(ncol(products) - 1)) ]
  return(products)
}

productDataframes = lapply(stores[3:length(stores)], productsWithLocation)

part b - click here for answer

##################.
# ANSWER - PART B
##################.

allProducts = do.call ( rbind, productDataframes )
allProducts

           location price salePrice amountSold amountInInventory
1   123 main street 10.99      8.99        100                20
2   123 main street  8.00        NA        200                12
3   123 main street 15.00        NA         50                 0
4   123 main street  5.00      4.50         75                 5
5  99 hickory place 11.99        NA         80                20
6  99 hickory place  8.00      7.00        195                15
7  99 hickory place  3.50        NA        200                40
8  99 hickory place  2.99      2.50        400                15
9  99 hickory place  4.99      4.50         80                 0
10     450 broadway 11.99        NA         40                30
11     450 broadway  8.00      7.00        100                 5
12     450 broadway 11.00     10.00          3                40
13     450 broadway  3.50      2.50        100                 0
14     450 broadway  2.99      2.50        200                 0
15     450 broadway 44.99        NA         10                 3