ابتدا پکیج MASS
را نصب و فراخوانی کنید
این داده ها مربوط به نقاشان کلاسیک قرن 18 میباشند که در رسته های ترکیب بندی ،رنگ بندی،... دسته بندی شده اند
با این دستور میتوانیم مشاهده کنیم که سطح جایی که اموزش دیدن چگونه است
که در 8 سطح میباشد
@R_Experts
را نصب و فراخوانی کنید
> library(MASS) # load the MASS package
> painters
Composition Drawing Colour Expression School
Da Udine 10 8 16 3 A
Da Vinci 15 16 4 14 A
Del Piombo 8 13 16 7 A
Del Sarto 12 16 9 8 A
Fr. Penni 0 15 8 0 A
Guilio Romano 15 16 4 14 A
این داده ها مربوط به نقاشان کلاسیک قرن 18 میباشند که در رسته های ترکیب بندی ،رنگ بندی،... دسته بندی شده اند
> painters$School
[1] A A A A A A A A A A B B B B B B C C C C C C D D D D
[27] D D D D D D E E E E E E E F F F F G G G G G G G H H
[53] H H
Levels: A B C D E F G H
با این دستور میتوانیم مشاهده کنیم که سطح جایی که اموزش دیدن چگونه است
که در 8 سطح میباشد
@R_Experts
> school = painters$School # the painter schools
> school.freq = table(school) # apply the table function
> school.freq
school
A B C D E F G H
10 6 6 10 7 4 7 4
> cbind(school.freq)
school.freq
A 10
B 6
C 6
D 10
E 7
F 4
G 7
H 4
همان طور که مشاهده میکنید جدول فراوانی بین تعداد نقاشان و سطح جایی که اموزش دیدن رو اوردیم و به یک جدول تبدیل کردیم
و این جدول رو سپس ستونی نمودیم
مثلا
A 10
نشان دهنده ی این است که در سطح اموزش A
10 نفر اموزش دیدند
@R_Experts
> school.relfreq = school.freq / nrow(painters)
> school.relfreq
school
A B C D E F
0.185185 0.111111 0.111111 0.185185 0.129630 0.074074
G H
0.129630 0.074074
> barplot(school.freq)
> colors = c("red", "yellow", "green", "violet", + "orange", "blue", "pink", "cyan")
> barplot(school.freq, # apply the barplot function
+ col=colors) # set the color palette
ابتدا فراوانی نسبی این داده ها را محاسبه و سپس نمودار هیستوگرام انها را رسم کرد هایم که نمودار ها را در پایین مشاهده میکنید 👇👇👇
@R_Experts
> pie(school.freq)
> colors = c("red", "yellow", "green", "violet",
+ "orange", "blue", "pink", "cyan")
> pie(school.freq, # apply the pie function
+ col=colors) # set the color palette
نمودار دایره ای مربوطه👇👇👇👇
@R_Experts
#Solution_Practice_3
> cc<-c(1,2,3,4,5,2,1,2,3,4,3,2,1,2,3,4,3,2,1,2,5,4,3,2,1)
> A<-matrix(cc,5,5)
> A
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[2,] 2 1 2 3 4
[3,] 3 2 1 2 3
[4,] 4 3 2 1 2
[5,] 5 4 3 2 1
> b<-matrix(c(7,-1,-3,5,17),5,1)
> b
[,1]
[1,] 7
[2,] -1
[3,] -3
[4,] 5
[5,] 17
> x<-solve(A)%*%b
> x
[,1]
[1,] -2
[2,] 3
[3,] 5
[4,] 2
[5,] -4
#factor_function
فرم کلی این تابع به صورت :
factor(x = character(), levels, labels = levels,
exclude = NA, ordered = is.ordered(x))
که از یک رشته شی محتويات انها را خارج و در
Levels
قرار میدهد و میتوان به انها برچسب نیز نسبت داد ، و ordered ترتیب را برای ما مشخص میکند
ترتیب اعضای فاکتور را مشخص میکند
@R_Experts
فرم کلی این تابع به صورت :
factor(x = character(), levels, labels = levels,
exclude = NA, ordered = is.ordered(x))
که از یک رشته شی محتويات انها را خارج و در
Levels
قرار میدهد و میتوان به انها برچسب نیز نسبت داد ، و ordered ترتیب را برای ما مشخص میکند
> mons = c("March","April","January","November","January",
+ "September","October","September","November","August",
+ "January","November","November","February","May","August",
+ "July","December","August","August","September","November",
+ "February","April")
> mons = factor(mons)
> table(mons)
mons
April August December February January July
2 4 1 2 3 1
March May November October September
1 1 5 1 3
ordered=TRUEترتیب اعضای فاکتور را مشخص میکند
> fert = c(10,20,20,50,10,20,10,50,20)
> fert = factor(fert,levels=c(10,20,50),ordered=TRUE)
> fert
[1] 10 20 20 50 10 20 10 50 20
Levels: 10 < 20 < 50
@R_Experts
#abs
x: Numeric value, array or vector
@R_Experts
abs(x)
x: Numeric value, array or vector
> abs(-1)
[1] 1
> abs(20)
[1] 20
> abs(0)
[1] 0
> x <- c(-2,4,0,45,9,-4)
> abs(x)
[1] 2 4 0 45 9 4
> x <- matrix(c(-3,5,-7,1,-9,4),nrow=3,ncol=2,byrow=TRUE)
> abs(x[1,])
[1] 3 5
> abs(x[,1])
[1] 3 7 9
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#atan
atan() function returns the radian arctangent of number data.
x: Numeric value, array or vector
@R_Experts
atan() function returns the radian arctangent of number data.
atan(x)
x: Numeric value, array or vector
> atan(1)
[1] 0.7853982
> atan(0)
[1] 0
> atan(0.5)
[1] 0.4636476
> x <- c(1, 0, 0.5)
> atan(x)
[1] 0.7853982 0.0000000 0.4636476
@R_Experts
#beta
beta() function return the beta function and the natural logarithm of the beta function.
B(a,b) = Γ(a)Γ(b)/Γ(a+b)
beta(a, b)
a,b: non-negative numeric vectors
@R_Experts
beta() function return the beta function and the natural logarithm of the beta function.
B(a,b) = Γ(a)Γ(b)/Γ(a+b)
beta(a, b)
a,b: non-negative numeric vectors
> beta(4,9)
[1] 0.0005050505
> x <- c(3,6, 4)
> y <- c(7,4, 12)
> beta(x,y)
[1] 0.0039682540 0.0019841270 0.0001831502
@R_Experts
#choose()
choose(n,r)
n: n elements
r: r subset elements
...
nCr = n!/(r! * (n-r)!)
@R_Experts
choose(n,r)
n: n elements
r: r subset elements
...
nCr = n!/(r! * (n-r)!)
> choose(5,2)
[1] 10
> choose(2,1)
[1] 2
@R_Experts
#log( )
function computes natural logarithms (Ln) for a number or vector. log10 computes common logarithms (Lg).log2 computes binary logarithms (Log2). log(x,b) computes logarithms with base b.
@R_Experts
function computes natural logarithms (Ln) for a number or vector. log10 computes common logarithms (Lg).log2 computes binary logarithms (Log2). log(x,b) computes logarithms with base b.
>log(5) #ln5
[1] 1.609438
>log10(5) #lg5
[1] 0.69897
>log2(5) #log25
[1] 2.321928
>log(9,base=3) #log39 = 2
[1] 2
>x <- rep(1:12)
>x
[1] 1 2 3 4 5 6 7 8 9 10 11 12
>log(x)
[1] 0.0000000 0.6931472 1.0986123 1.3862944 1.6094379 1.7917595 1.9459101
[8] 2.0794415 2.1972246 2.3025851 2.3978953 2.4849066
>log(x,6)
[1] 0.0000000 0.3868528 0.6131472 0.7737056 0.8982444 1.0000000 1.0860331
[8] 1.1605584 1.2262944 1.2850972 1.3382908 1.3868528
@R_Experts
log10() function computes base 10 logarithm.
x: numeric vector
@R_Experts
log10(x)
x: numeric vector
> log10(100)
[1] 2
> x <- c(100,1000, 10000)
> log10(x)
[1] 2 3 4
@R_Experts
#exp
exp(x) function compute the exponential value of a number or number vector, ex.
expm1() function computes exp() minus 1:»>حاصل منهای 1
@R_Experts
exp(x) function compute the exponential value of a number or number vector, ex.
> x <- 5
> exp(x)
[1] 148.4132
> y <- rep(1:20)
> exp(y)
[,1] [,2] [,3] [,4] [,5]
[1,] 2.718282 20.08554 148.4132 1096.633 8103.084
[2,] 7.389056 54.59815 403.4288 2980.958 22026.466
expm1() function computes exp() minus 1:»>حاصل منهای 1
> expm1(5)
[1] 147.4132
> expm1(rep(1:20))
[,1] [,2] [,3] [,4] [,5]
[1,] 1.718282 19.08554 147.4132 1095.633 8102.084
[2,] 6.389056 53.59815 402.4288 2979.958 22025.466
@R_Experts
#factorial
factorial() function computes the factorial of a number.
factorial(x)
x: numeric vector
@R_Experts
factorial() function computes the factorial of a number.
factorial(x)
x: numeric vector
> factorial(2) #2 × 1
[1] 2
> factorial(1) #1 × 1
[1] 1
> factorial(3) #3 × 2 × 1
[1] 6
> factorial(4) #4 × 3 × 2 × 1
[1] 24
> factorial(c(4,3,2))
[1] 24 6 2
@R_Experts
#max_min
max() function computes the maximun value of a vector. min() function computes the minimum value of a vector.
max(x,na.rm=FALSE)
min(x,na.rm=FALSE)
• x: number vector
• na.rm: whether NA should be removed, if not, NA will be returned
...
Missing value affect the results:
After define na.rm=TRUE, result is meaningful:
Compare more than 1 vectors:
@R_Experts
max() function computes the maximun value of a vector. min() function computes the minimum value of a vector.
max(x,na.rm=FALSE)
min(x,na.rm=FALSE)
• x: number vector
• na.rm: whether NA should be removed, if not, NA will be returned
...
> x <- c(1,2.3,2,3,4,8,12,43,-4,-1)
> max(x)
[1] 43
> min(x)
[1] -4
Missing value affect the results:
> y<- c(x,NA)
> y
[1] 1.0 2.3 2.0 3.0 4.0 8.0 12.0 43.0 -4.0 -1.0 NA
> max(y)
[1] NA
> min(y)
[1] NA
After define na.rm=TRUE, result is meaningful:
> max(y,na.rm=TRUE)
[1] 43
Compare more than 1 vectors:
> x2 <- c(-100,-43,0,3,1,-3)
> min(x,x2)
[1] -100
@R_Experts