.Rhistory

# Sıcaklık değerleri temp 'e aktarılır.
temp <- c(35, 88, 42, 84, 81, 30)
# Şehir isimleri city 'e aktarılır.
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
# Şehir isimleri ve sıcaklık değerlerini içeren dataframe oluşturulur.
city_temps <- data.frame(name = city, temperature = temp)
states <- murders$state
ranks <- rank(murders$population)
# Murder içerisinde ki state ve rank değerlerini içeren ayrı bir dataframe oluşturulur.
my_df <- data.frame(states = states, ranks = ranks)
# --Data Frames, Ranks ve Orders--
# his exercise is somewhat more challenging. We are going to repeat the previous exercise
# but this time order my_df so that the states are ordered from least populous to most.
# Create variables states and ranks to store the state names and ranks by population size
# respectively.
# Create an object ind that stores the indexes needed to order the population values,
# using the order command.
# For example we could define o <- order(murders$population)
# Create a data frame with both variables following the correct order.
# Use the bracket operator [ to re-order each column in the data frame.
# For example, states[o] orders the abbreviations based by population size.
# The columns of the data frame must be in the specific
# order and have the specific names: states, ranks.
# Define a variable states to be the state names from the murders data frame
states <- murders$state
# Define a variable ranks to determine the population size ranks
ranks <- rank(murders$population)
# Define a variable ind to store the indexes needed to order the population values
ind <- order(murders$population)
# Create a data frame my_df with the state name and its rank and ordered from least populous to most
my_df <- data.frame(states = states[ind], ranks = ranks[ind])
# Define a variable states to be the state names from the murders data frame
states <- murders$state
library(murders)
dslabs(murders)
library(murders)
dslabs(murders)
library(murders)
dslabs(murders)
library(murders)
dslabs(murders)
# --Ranks--
#data.frame fonksiyonu ile dataframe oluşturulabilinir.
# Sıcaklık değerleri temp 'e aktarılır.
temp <- c(35, 88, 42, 84, 81, 30)
# Şehir isimleri city 'e aktarılır.
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
# Şehir isimleri ve sıcaklık değerlerini içeren dataframe oluşturulur.
city_temps <- data.frame(name = city, temperature = temp)
states <- murders$state
ranks <- rank(murders$population)
# Murder içerisinde ki state ve rank değerlerini içeren ayrı bir dataframe oluşturulur.
my_df <- data.frame(states = states, ranks = ranks)
# --Data Frames, Ranks ve Orders--
# his exercise is somewhat more challenging. We are going to repeat the previous exercise
# but this time order my_df so that the states are ordered from least populous to most.
# Create variables states and ranks to store the state names and ranks by population size
# respectively.
# Create an object ind that stores the indexes needed to order the population values,
# using the order command.
# For example we could define o <- order(murders$population)
# Create a data frame with both variables following the correct order.
# Use the bracket operator [ to re-order each column in the data frame.
# For example, states[o] orders the abbreviations based by population size.
# The columns of the data frame must be in the specific
# order and have the specific names: states, ranks.
# Define a variable states to be the state names from the murders data frame
states <- murders$state
# Define a variable ranks to determine the population size ranks
ranks <- rank(murders$population)
# Define a variable ind to store the indexes needed to order the population values
ind <- order(murders$population)
# Create a data frame my_df with the state name and its rank and ordered from least populous to most
my_df <- data.frame(states = states[ind], ranks = ranks[ind])
library(murders)
dslabs(murders)
dslabs(murders)
library(murders)
dslabs(murders)
# --Ranks--
#data.frame fonksiyonu ile dataframe oluşturulabilinir.
# Sıcaklık değerleri temp 'e aktarılır.
temp <- c(35, 88, 42, 84, 81, 30)
# Şehir isimleri city 'e aktarılır.
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
# Şehir isimleri ve sıcaklık değerlerini içeren dataframe oluşturulur.
city_temps <- data.frame(name = city, temperature = temp)
states <- murders$state
ranks <- rank(murders$population)
# Murder içerisinde ki state ve rank değerlerini içeren ayrı bir dataframe oluşturulur.
my_df <- data.frame(states = states, ranks = ranks)
# --Data Frames, Ranks ve Orders--
# his exercise is somewhat more challenging. We are going to repeat the previous exercise
# but this time order my_df so that the states are ordered from least populous to most.
# Create variables states and ranks to store the state names and ranks by population size
# respectively.
# Create an object ind that stores the indexes needed to order the population values,
# using the order command.
# For example we could define o <- order(murders$population)
# Create a data frame with both variables following the correct order.
# Use the bracket operator [ to re-order each column in the data frame.
# For example, states[o] orders the abbreviations based by population size.
# The columns of the data frame must be in the specific
# order and have the specific names: states, ranks.
# Define a variable states to be the state names from the murders data frame
states <- murders$state
# Define a variable ranks to determine the population size ranks
ranks <- rank(murders$population)
# Define a variable ind to store the indexes needed to order the population values
ind <- order(murders$population)
# Create a data frame my_df with the state name and its rank and ordered from least populous to most
my_df <- data.frame(states = states[ind], ranks = ranks[ind])
dslabs(murder)
library(murders)
library(murder)
library(dslabs)
dslabs(murders)
# --Ranks--
#data.frame fonksiyonu ile dataframe oluşturulabilinir.
# Sıcaklık değerleri temp 'e aktarılır.
temp <- c(35, 88, 42, 84, 81, 30)
# Şehir isimleri city 'e aktarılır.
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
# Şehir isimleri ve sıcaklık değerlerini içeren dataframe oluşturulur.
city_temps <- data.frame(name = city, temperature = temp)
states <- murders$state
ranks <- rank(murders$population)
# Murder içerisinde ki state ve rank değerlerini içeren ayrı bir dataframe oluşturulur.
my_df <- data.frame(states = states, ranks = ranks)
# --Data Frames, Ranks ve Orders--
# his exercise is somewhat more challenging. We are going to repeat the previous exercise
# but this time order my_df so that the states are ordered from least populous to most.
# Create variables states and ranks to store the state names and ranks by population size
# respectively.
# Create an object ind that stores the indexes needed to order the population values,
# using the order command.
# For example we could define o <- order(murders$population)
# Create a data frame with both variables following the correct order.
# Use the bracket operator [ to re-order each column in the data frame.
# For example, states[o] orders the abbreviations based by population size.
# The columns of the data frame must be in the specific
# order and have the specific names: states, ranks.
# Define a variable states to be the state names from the murders data frame
states <- murders$state
# Define a variable ranks to determine the population size ranks
ranks <- rank(murders$population)
# Define a variable ind to store the indexes needed to order the population values
ind <- order(murders$population)
# Create a data frame my_df with the state name and its rank and ordered from least populous to most
my_df <- data.frame(states = states[ind], ranks = ranks[ind])
my_df <- data.frame(states = states[ind], ranks = ranks[ind])
my_df
library(dslabs)
data(na_example)  #NA içeren bir vektör çağırmak için na_example kullanılabilir
# Checking the structure
str(na_example)
na_example
mean(na_example)
# Use is.na to create a logical index ind that tells which entries are NA
ind <- is.na(na_example)
ind
# Determine how many NA ind has using the sum function
sum(ind)
x <- c(1, 2, 3)
ind <- c(FALSE, TRUE, FALSE)
x[!ind]
ind2 <- is.na(na_example)
mean(na_example[!ind2]
mean(na_example[!ind2])
mean(na_example[!ind2]
data(na_example)  #NA içeren bir vektör çağırmak için na_example kullanılabilir
ind2 <- is.na(na_example)
ind
ind2
mean(na_example[!ind2]
mean(na_example[!ind2]
mean(na_example[!ind2]
mean(na_example[!ind2]
mean(na_example[!ind2])
# --Factors & Tables--
library(murders)
dslabs(murders)
# --Factors--
# Using the str() command, we saw that the region column stores a factor.
# You can corroborate this by using the class command on the region column.
str(murders)
# --Factors & Tables--
library(dslabs)
dslabs(murders)
# --Factors--
# Using the str() command, we saw that the region column stores a factor.
# You can corroborate this by using the class command on the region column.
str(murders)
# The function levels shows us the categories for the factor.
# We can see the class of the region variable using class
class(murders$region)
# Determine the number of regions included in this variable
a <- levels(murders$region)
length(a)
# --Table--
# The function table takes a vector as input and returns the frequency of
# each unique element in the vector.
# We will use the table function to answer this question.
# Use the table function in one line of code to
# create a table showing the number of states per region.
# The "c" in `c()` is short for "concatenate," which is the action of connecting items into a chain
# The function `c()` connects all of the strings within it into a single vector, which we can assign to `x`
x <- c("a", "a", "b", "b", "b", "c")
# Here is an example of what the table function does
table(x)
# Write one line of code to show the number of states per region
murder_regiontable <- table(murders$region)
#Shown Table "murder_regiontable"
murder_regiontable
# --Factors & Tables--
library(dslabs)
dslabs(murders)
# --Factors--
# Using the str() command, we saw that the region column stores a factor.
# You can corroborate this by using the class command on the region column.
str(murders)
# The function levels shows us the categories for the factor.
levels(murders)
# The function levels shows us the categories for the factor.
levels(murders$region)
a <- levels(murders$region)
length(a)
x <- c("a", "a", "b", "b", "b", "c")
# Here is an example of what the table function does
table(x)
# Write one line of code to show the number of states per region
murder_regiontable <- table(murders$region)
murder_regiontable
library(murders)
dslabs(murders)
# -- Sort --
states <- murders$state
# State değerlerini alfabetik olarak sırala
states <- sort(states)
# 1. değeri göster
states[1]
pop <- murders$pop
# Sort the object and save it in the same object
pop <- sort(pop)
# Report the smallest population size
pop[1]
library(dslabs)
dslabs(murders)
# -- Sort --
states <- murders$state
# State değerlerini alfabetik olarak sırala
states <- sort(states)
# 1. değeri göster
states[1]
pop <- murders$pop
# Sort the object and save it in the same object
pop <- sort(pop)
# Report the smallest population size
pop[1]
states[1]
pop <- murders$pop
pop
# if you go down the rows of murders.
# Access population from the dataset and store it in pop
pop2 <- murders$population
ord <- order(pop2)
ord
# --Sequences--
# 32 'den 99 'a kadar olan sayıları içeren vektörü oluşturmak için;
m <- 32:99
m
# Create a vector with the multiples of 7, smaller than 50.
seq(7, 49, 7)
# by the same amount but are of the prespecified length.
#For example, this line of code
a <- seq(0, 100, length.out = 5)    #produces the numbers 0, 25, 50, 75, 100.
lenght(a)
# by the same amount but are of the prespecified length.
#For example, this line of code
a <- seq(0, 100, length.out = 5)    #produces the numbers 0, 25, 50, 75, 100.
lenght(a)
#For example, this line of code
a <- seq(0, 100, length.out = 5)    #produces the numbers 0, 25, 50, 75, 100.
length(a)
a
cost <- c(50, 75, 90, 100, 150)
# --Character Vectors--
food <- c("pizza", "burgers", "salads", "cheese", "pasta")
# --Connecting Numeric and Character Vectors--
# We have successfully assigned the temperatures as numeric values to temp and the city
# names as character values to city. But can we associate the temperature to its related city?
# Yes! We can do so using a code we already know - names.
# We assign names to the numeric values.
# Associate the cost values with its corresponding food item
names(cost) <- food
cost
# Numeric/Character Vectors and Connecting Numeric and Character Vectors
# --Numeric Vectors--
cost <- c(50, 75, 90, 100, 150)
# --Character Vectors--
food <- c("pizza", "burgers", "salads", "cheese", "pasta")
# --Connecting Numeric and Character Vectors--
#Numeric ve Character vektörlerini birleştirme.
names(cost) <- food
cost
temp <- c(35, 88, 42, 84, 81, 30)
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
names(temp) <- city
temp
# --Subsetting Vectors--
# If we want to display only selected values from the object, R can help us do that easily.
# For example, if we want to see the cost of the last 3 items in our food list, we would type:
# cost of the last 3 items in our food list:
cost[3:5]
# Note here, that we could also type cost[c(3,4,5)] and get the same result.
# The : operator helps us condense the code and get consecutive values.
# temperatures of the first three cities in the list:
temp[1:3]
# In the previous question, we accessed the temperature for consecutive cities (1st three).
# But what if we want to access the temperatures for any 2 specific cities?
# An example: To access the cost of pizza (1st) and pasta (5th food item) in our list,
# the code would be:
# Access the cost of pizza and pasta from our food list
cost_new <- cost[c(1,5)]
# Access the temperatures of Paris and San Juan
temp_new <- temp[c(3,5)]
cost[3:5]
#cost vektöründe 3. elemandan 5. elemana kadar göster.
cost[3:5]
# Access the cost of pizza and pasta from our food list
cost_new <- cost[c(1,5)]
cost
cost_new
a <- murders$abb
b <- murders[["abb"]]
identical(a,b)
library(dslabs)
data(murders)
# murders içerisinde bulunan population 'a ulaşmak için "$" veya "[["population"]]" kullanılabilir.
p <- murders$population
o <- murders[["population"]]
# 2 değer birbirine eşit ise TRUE döndürür.
identical(o, p)
a <- murders$abb
b <- murders[["abb"]]
identical(a,b)
# R has another type of vector we have not described, the integer class.
#You can create an integer by adding the letter L after a whole number. If you type
class(3L)
c
'den 10 'a kadar 0.5 artaran değerlerin vektörünü oluştur.
class(seq(1, 10, 0.5))
# Sayıların sonuna "L" ekleyerek integer oluşturulabilir.
class(3L)
3L - 3
#
class(1)
# Define the vector mx
mx <- c(1, 3, 5,"a")
# Note that the x is character vector
mx
mx <- as.numeric(mx)
mx
x <- 1:100
# Compute the sum
sum(1/x^2)
pi/6
pi
sum(x)
100*101/2
sum(1/x^2)
x
murder_rate <- murders$total/murders$population*100000
# Compute the average murder rate using `mean` and store it in object named `avg`
avg <- mean(murder_rate)
# --Logical Vectors--
library(dslabs)
data(murders)
murder_rate <- murders$total / murders$population * 100000
low <-  murder_rate < 1
which(murder_rate<1)
murders$state[low]
ind <- low & murders$region=="Northeast"
# murder_rate < 1 ve murders region = Northeast olan değer
murders$state[ind]
# --Exaple--
# In a previous exercise we computed the murder rate for each state and the average of these numbers.
# How many states are below the average?
murder_rate <- murders$total/murders$population*100000
# Compute the average murder rate using `mean` and store it in object named `avg`
avg <- mean(murder_rate)
# How many states have murder rates below avg ? Check using sum
x <- murder_rate < avg
sum(x)
avg <- mean(murder_rate)
avg
abbs <- c("AK", "MI", "IA")
# Match the abbs to the murders$abb and store in ind
ind <- match(abbs, murders$abb)
# Print state names from ind
murders$state[ind]
abbs <- c("AK", "MI", "IA")
ind <- match(abbs, murders$abb)
library(dslabs)
dslabs(murders)
# --Match--
abbs <- c("AK", "MI", "IA")
# Oluşturulan abbs vektörü içerisinde murders$abb ile eşleşen değerlerin indexleri
ind <- match(abbs, murders$abb)
murders$state[ind]
# --%in%--
# If rather than an index we want a logical that tells us whether or not
# each element of a first vector is in a second, we can use the function %in%.
# For example:
# x <- c(2, 3, 5)
# y <- c(1, 2, 3, 4)
# x%in%y
#Gives us two TRUE followed by a FALSE because 2 and 3 are in y but 5 is not.
# Store the 5 abbreviations in abbs. (remember that they are character vectors)
abbs <- c("MA", "ME", "MI", "MO", "MU")
# Use the %in% command to check if the entries of abbs are abbreviations in the the murders data frame
abbs%in%murders$abb
# --Example2--
#We are again working with the characters abbs <- c("MA", "ME", "MI", "MO", "MU")
#In a previous exercise we computed the index abbs%in%murders$abb. Based on that,
#and using the which function and the ! operator, get the index of the entries of
#abbs that are not abbreviations.
#Show the entries of abbs that are not actual abbreviations.
# Store the 5 abbreviations in abbs. (remember that they are character vectors)
abbs <- c("MA", "ME", "MI", "MO", "MU")
# Use the `which` command and `!` operator to find out which index abbreviations are not actually part of the dataset and store in `ind`
ind <- which(!(abbs%in%murders$abb))
# Names of abbreviations in `ind`
abbs[ind]
ind <- match(abbs, murders$abb)
ind
murders$state[ind]
x <- c(2, 3, 5)
y <- c(1, 2, 3, 4)
x%in%y
abbs <- c("MA", "ME", "MI", "MO", "MU")
ind <- which(!(abbs%in%murders$abb))
abbs[ind]
x <- c(88, 100, 83, 92, 94)
rank(-x)
library(dplyr)
# --dplyr--
library(dplyr)
library(dslabs)
data(murders)
# --mutate--
# rank(x) x vektörü(n elemanlı) içerisindeki en küçük değere "1" en düşük değere ise "n" değeri yazar
# rank(-x) yazarak en yüksek değere "1" en düşük değere "n" yazdırılabilir.
x <- c(88, 100, 83, 92, 94)
rank(-x)
rate <-  murders$total/ murders$population * 100000
rank_rate <- rank(-rate)
murders <- mutate(murders, rank = rank_rate)
murders
# -- select--
# With dplyr we can use select to show only certain columns.
# For example with this code we would only show the states and population sizes:
# select(murders, state, population)
# Load dplyr
library(dplyr)
library(dslabs)
dslabs(murders)
# Use select to show the state names and abbreviations in murders.
select(murders, state, abb)
# --filter--
# The dplyr function filter is used to choose specific rows of the data frame to keep.
# Unlike select which is for columns, filter is for rows.
# For example you can show just the New York row like this:
filter(murders, state == "New York")
#You can use other logical vectors to filter rows.
# --Example--
# Use filter to show the top 5 states with the highest murder rates.
# After we add murder rate and rank, do not change the murders dataset,
# just show the result. Note that you can filter based on the rank column.
# Add the necessary columns
murders <- mutate(murders, rate = total/population * 100000, rank = rank(-rate))
# Filter to show the top 5 states with the highest murder rates
filter(murders, rank <=5)
# --filter with " != "--
# We can remove rows using the != operator.
# For example to remove Florida we would do this:
no_florida <- filter(murders, state != "Florida")
# --Example--
# Create a new data frame called no_south that removes states from the South region.
# How many states are in this category? You can use the function nrow for this.
# Use filter to create a new data frame no_south
no_south <- filter(murders, region != "South")
# Use nrow() to calculate the number of rows
nrow(no_south)
e select to show the state names and abbreviations in murders.
select(murders, state, abb)
# --filter--
#murders içerisinde sadece New York 'a ait değerlere ulaşmak için.
filter(murders, state == "New York")
# mutate() komutu ile var olan dataframe 'e
murders <- mutate(murders, rate = total/population * 100000, rank = rank(-rate))
murders
no_south <- filter(murders, region != "South")
nrow(no_south)
no_south
# any() komutunda herhangi bir değer TRUE ise
z <- c(TRUE, TRUE, FALSE)
any(z)
all(z)
help("any")
z <- c(FALSE, FALSE, FALSE)
any(z)