如何修复“数据将正确绘制绘图/数据更改”

时间:2019-04-22 00:25:01

标签: r modeling population

我创建了一个人口模型,该模型在某些时候确实按照我想要的去做。但是,随着我开始继续重新运行代码以产生持续正确的结果,以验证我是否正确执行了代码,数据将以错误的方式更改。本质上,数据将间隔开并且完全消失为全零。发生这种情况时,我没有对代码进行任何更改。

我尝试注释掉一些“ if语句”参数,将数据集从矩阵“ d1”和“ df2”相乘更改为“ d1”和“ pop_model”,然后将图从df2更改为pop_model。我什至把我的df2变量移到了我的论点之前,认为这是问题所在。

projection_matrix <- matrix(c(0,4.4,8.8,0.2,0,0,0,.4,.4),nrow=3,ncol=3,byrow=T)
projection_matrix

disease_matrix_1 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05),nrow=3,ncol=3,byrow=T)
disease_matrix_1
d1 <- disease_matrix_1

disease_matrix_2 <- matrix(c(0,.66,1.32,.15,0,0,0,.15,.15),nrow=3,ncol=3,byrow=T)
disease_matrix_2
d2 <- disease_matrix_2

disease_matrix_3 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05),nrow=3,ncol=3,byrow=T)
disease_matrix_3
d3 <- disease_matrix_3

disease_matrix_4 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_4
d4 <- disease_matrix_4

disease_matrix_5 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_5
d5 <- disease_matrix_5

disease_matrix_6 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_6
d6 <- disease_matrix_6

disease_matrix_7 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_7
d7 <- disease_matrix_7

disease_matrix_8 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_8
d8 <- disease_matrix_8

disease_matrix_9 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_9
d9 <- disease_matrix_9

disease_matrix_10 <- matrix(c(0,.22,.44,.05,0,0,0,.05,.05), nrow=3, ncol=3, byrow=T)
disease_matrix_10
d10 <- disease_matrix_10

abundance_yr0 <- c(0,0,12)
abundance_yr0

Year1 <- projection_matrix %*% abundance_yr0  # matrix multiplication!
Year1

Year2 <- projection_matrix %*% Year1 
Year2

t=200; N0=abundance_yr0; x=projection_matrix

-------------------------------------------------------

pop_model <- matrix(0,nrow=nrow(x),ncol=t+1)
pop_model[,1] <- N0

df2 <- replace(pop_model, pop_model > 2.00e+9, sample(2.00e+9:2.5e+9, 10000, replace=T))

for(i in 2:(t+1)){
  if(i == 135){
    pop_model[,i] <- d1 %*% df2[,i-1]
  }
  else{
    pop_model[,i] <- x %*% df2[,i-1]}
  if(i == 155){
    pop_model[,i] <- d2 %*% df2[,i-1]
  }
}
  if(i == 160){
    pop_model[,i] <- d3 %*% df2[,i-1]
  }
  if(i == 161){
    pop_model[,i] <- d4 %*% df2[,i-1]
  }
  if(i == 162){
    pop_model[,i] <- d5 %*% df2[,i-1]
  }
  if(i == 163){
    pop_model[,i] <- d6 %*% df2[,i-1]
  }
  if(i == 164){
    pop_model[,i] <- d7 %*% df2[,i-1]
  }
  if(i == 165){
    pop_model[,i] <- d8 %*% df2[,i-1]
  }
  if(i == 166){
    pop_model[,i] <- d9 %*% df2[,i-1]
  }
  if(i == 167){
    pop_model[,i] <- d10 %*% df2[,i-1]
  } 
}

plot(1,1,pch="",ylim=c(0,3.00e+9),xlim=c(0,t+1),xlab="Years",ylab="Abundance",xaxt="n")
cols <- topo.colors(ncol(x))

max(pop_model)

for(s in 1:ncol(x)){
  points(df2[s,],col=cols[s],type="l",lwd=2)
}
axis(1,at=seq(1,t+1),labels = seq(0,t))
legend("topleft",col=cols,lwd=rep(2,ncol(x)),legend=c("Infant", "Adolescent", "Adult"))

正确运行时,应生成总体模型图,该总体模型崩溃将在155-165附近,具体取决于您是否运行整个代码。

运行df2 [156]或df2 [3,156]之类的数字时,其数字应显着低于图表的其余部分,几乎接近于0。

0 个答案:

没有答案