绘图笔记
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print(sin(pi/2))
sink('tmpres01.txt', split=TRUE)
print(sin(pi/6))
print(cos(pi/6))
cat('t(10)的双侧0.05分位数(临界值)=', qt(1 - 0.05/2, 10), '\n')
sink()
x<-10000
for(i in 1:10)
{
y=x*(1.03)^i
print(y)
}
curve(x^2, -2, 2)
plot(x^2, -2, 2)######错误
######错误
plot(main = "example",xlab = "x",ylab = "y",xlim = c(0,100),ylim = c(0,100),col = "red",pch = 19)
###########散点图,红色的两个点
plot( c(0,100),c(0,100),main = "example",xlab = "x",ylab = "y",col = "red",pch = 19)
###########散点图,白色的两个点,所以不显示
plot( c(0,100),c(0,100),main = "example",xlab = "x",ylab = "y",col = "white",pch = 19)
x <- 1:10
beijing <- round(rnorm(10,mean = 20 , sd = 2),1)
shanghai <- round(rnorm(10,mean = 20 , sd = 3),1)
guangzhou <- round(rnorm(10,mean = 20 , sd = 1),1)
plot(x,beijing,type = 'l',ylim = c(16,30),lwd = 2,main = "北京上海和广州最近十天的气温变化趋势")
lines(x,shanghai,type = 'l',col = 'blue',lwd = 2)###type = 'l'代表可以画折线图
exp(1.059+0.04*x)/(1+exp(1.059+0.04))
}
# 利用curve函数进行绘制
curve(f1,from=-100,to=100)
# 编写函数:
phi(0.712+0.031*x)
# 知道标准正态分布的变形
# 标准正态分布是mean=0,sd=1
# 现在的分布,mean=0.712,sd=(0.031)^0.5
# 绘制从-1到2的图
# R产生正态分布的函数是dnorm
curve(dnorm(x,mean=0.712,sd=(0.031)^0.5),from=-1,to=2)
sink('tmpres01.txt', split=TRUE)
print(sin(pi/6))
print(cos(pi/6))
cat('t(10)的双侧0.05分位数(临界值)=', qt(1 - 0.05/2, 10), '\n')
sink()
x<-10000
for(i in 1:10)
{
y=x*(1.03)^i
print(y)
}
curve(x^2, -2, 2)
plot(x^2, -2, 2)######错误
######错误
plot(main = "example",xlab = "x",ylab = "y",xlim = c(0,100),ylim = c(0,100),col = "red",pch = 19)
###########散点图,红色的两个点
plot( c(0,100),c(0,100),main = "example",xlab = "x",ylab = "y",col = "red",pch = 19)
###########散点图,白色的两个点,所以不显示
plot( c(0,100),c(0,100),main = "example",xlab = "x",ylab = "y",col = "white",pch = 19)
x <- 1:10
beijing <- round(rnorm(10,mean = 20 , sd = 2),1)
shanghai <- round(rnorm(10,mean = 20 , sd = 3),1)
guangzhou <- round(rnorm(10,mean = 20 , sd = 1),1)
plot(x,beijing,type = 'l',ylim = c(16,30),lwd = 2,main = "北京上海和广州最近十天的气温变化趋势")
lines(x,shanghai,type = 'l',col = 'blue',lwd = 2)###type = 'l'代表可以画折线图
lines(x,guangzhou,type = 'l',col = 'red', lwd = 2)
函数的绘图
f1 <- function(x){exp(1.059+0.04*x)/(1+exp(1.059+0.04))
}
# 利用curve函数进行绘制
curve(f1,from=-100,to=100)
# 编写函数:
phi(0.712+0.031*x)
# 知道标准正态分布的变形
# 标准正态分布是mean=0,sd=1
# 现在的分布,mean=0.712,sd=(0.031)^0.5
# 绘制从-1到2的图
# R产生正态分布的函数是dnorm
curve(dnorm(x,mean=0.712,sd=(0.031)^0.5),from=-1,to=2)
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