42Practice 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 betterstr(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$year, sep="/")
[1] "10/2021"
part b - click here for answer
##################################.# Answer - part B - ONE ANSWER##################################.# copy the values from storesstores2 = stores # set the first value in stores2 to be the month/yearstores2[[1]] =paste( stores$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 stores2stores2[[2]] =NULL# remove the 2nd value from stores2str(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 liststores2 =list()# set the date stores2$date =paste( stores$month , stores$year, sep="/")# copy the data for all the stores from the stores variablestores2[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.#----------------------------------------------------------------------------------
#----------------------------------------------------------------------------------# QUESTION 5# TOPIC: lists## Write a command that calculates the number of stores in the chain.#----------------------------------------------------------------------------------
#----------------------------------------------------------------------------------# 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
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
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#----------------------------------------------------------------------------------
#----------------------------------------------------------------------------------# 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.#----------------------------------------------------------------------------------
#----------------------------------------------------------------------------------# 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]]$productssaleProducts = products[ !is.na(products$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
#----------------------------------------------------------------------------------# 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 questionnames(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
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 codestoresWithNames$store1$products [ storesWithNames$store1$products$price >=10 , c("product","price")]
product price
1 hammer 10.99
3 hand saw 15.00
# Alternative answer with more than one line of codestore1products = storesWithNames$store1$productsstore1products[ store1products$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)])#----------------------------------------------------------------------------------
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$address)
##################.# ANSWER - PART B##################.sapply(stores[3:length(stores)],function(store) store$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.#----------------------------------------------------------------------------------
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.#----------------------------------------------------------------------------------