小编Bio*_*eek的帖子

我有以下四个方程(a,b,c,d),有几个不同的变量(x,t,v,w,n,f).我的目标是尝试找到所有可以生成方程(a,b,c,d)的所有正(和非)数的变量值集.常规循环只会遍历生成的每个序列,并系统地检查它是否生成正值.我希望它从每个序列中获取随机数并将其与R中的其他序列进行测试.例如(x = 8,t = 2.1,v = 13,w = 1,n = 10,f = 1)是可能的一组组合.

请不要建议解析性地解决这些问题,然后找出价值观.这些只是我正在处理的方程式的表示.我所拥有的方程非常复杂,超过15个变量.

#Equations
a <- x * t - 2*x
b <- v - x^2 
c <- x - w*t - t*t 
d <- (n - f)/t

x <- seq(from = 0.0001, to = 1000, by = 0.1)
t <- seq(from = 0.0001, to = 1000, by = 0.1)
v <- seq(from = 0.0001, to = 1000, by = 0.1)
w <- seq(from = 0.0001, to = 1000, …

我有这种格式的数据集(df)

index <- runif(n = 100,min = 0, max = 1)
type1 <- rep("low", 50)
type2 <- rep("high", 50)
type <- c(type1,type2)

level1 <- rep("single", 25)
level2 <- rep("multiple", 25)
level3 <- rep("single", 25)
level4 <- rep("multiple", 25)
level <- c(level1,level2,level3,level4)

block <- rep(1:5, 10)
set <- rep(1:5, 10)

df <- data.frame("index" = index,"type" = type, "level" = level, "block" = block, "set" = set)
df$block <- as.factor(df$block)
df$set <- as.factor(df$set)

我想创建一个看起来像这样的模型

model <- lmer(index ~ type * level + …

在一项实验中，我试图找到第一胎出生的时间。id 和rep 给出了四种动物（A1、A2、B1、B2）、它们的年龄和婴儿。对于每个 id 和rep，我只想保留婴儿第一次出生的行

id <- c("A","A","A","A","A","A","B","B","B","B","B","B","B","B","B")
rep <- c(1,1,1,2,2,2,1,1,1,1,2,2,2,2,2)
age <- c(0,1,2,0,1,2,0,1,2,3,0,1,2,3,4)
babies <- c(0,0,1,0,1,0,0,0,0,1,0,0,0,1,1)

df <- data.frame(id,rep,age,babies)

所以在这里，最终的数据框应该是这样的

id <- c("A","A","B","B")
rep <- c(1,2,1,2)
age <- c(2,1,3,3)
babies <- c(1,1,1,1)

df <- data.frame(id,rep,age,babies)

我有以下混合效应模型的输出。我想谈谈模型解释了多少变化。随机效应下的方差是否对应于残差（注：这里的试验是随机效应）所解释的变异？即 58.6 % 或者有其他方法来推断这一点

REML criterion at convergence: 71.9\n\nScaled residuals: \n     Min       1Q   Median       3Q      Max \n-1.82579 -0.59620  0.04897  0.62629  1.54639 \n\nRandom effects:\n Groups   Name        Variance Std.Dev.\n trial     (Intercept) 0.06008  0.2451  \n Residual              0.58633  0.7974  \nNumber of obs: 60, groups:  trial, 30\n\nFixed effects:\n                        Estimate Std. Error      df t value Pr(>|t|)    \n(Intercept)               1.5522     0.2684 12.6610  13.233  0.09888 \ndrugantho                 0.8871     0.1753 14.0000   1.043  0.31601    \ninterventionadded         0.2513     0.2553 14.0000  -1.276  0.32436 **   \nsexmale                   3.0026     0.6466 15.0000   4.066  0.00021  \n---\nSignif. codes:  0 \xe2\x80\x98***\xe2\x80\x99 …

我需要求解一个由四个方程组成的系统并求解(Alg,Blg,C,D)..其他只是变量.

dAlg = (ab*D + a1*f1*D*Alg*Blg + a2*f2*D*Alg*C)/e1 - m1*Alg;
dBlg = (f11*D*Blg)/eB - f1*Alg*Blg - mp*Blg;
dC = (f22*D*C)/eC - f2*Alg*C - mp*C;
dD = (f*R/(h + R)) - r*D - ab*D - a1*f1*D*Alg*Blg - f11*D*Blg - 
   a2*f2*D*Alg*C-f22*D*C;

所以我想解决{dAlg = 0,dBlg = 0,dC = 0,dD = 0},然后找出所有可能的解决方案组合.

我不知道如何使用求解包在R上进行此操作(我发现的示例似乎更简单的方程式)...但是当我使用另一个包时,我得到以下答案.我想在R中实现同样的目标

    {{Alg -> ((eC f11 - eB f22) mp)/(-eC f11 f2 + eB f1 f22), 
  Blg -> -((
    eC f f11 f2 R - eB f f1 f22 R + …

对于主y轴和x轴,我有通用标题,如"Tank's Ratio"和"Counts".我想要第二行标签,我指定比率和计数.例如.就在"Tank's Ratio"之下,我希望"#in water /#in sand"中的字体较小,但是沿着y轴.类似地,对于x轴.这是基本代码

data <- data.frame(set = c(1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 4, 4), density = c(1, 3, 3, 1, 3, 1, 1, 1, 3, 3, 1, 3), counts = c(100, 2, 3, 76, 33, 12, 44, 13, 54, 36, 65, 1), ratio = c(1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 90, 1))

library(ggplot2)

ggplot(data, aes(x = counts, y = ratio)) + 
  geom_point() + 
  ylab("Tank's Ratio") + 
  xlab("Counts") 

我有一个 2 x 2 阶乘数据集，我使用mean_cl_boot函数绘制了置信区间。我想使用适当的函数在 R 中计算它。我怎样才能做到这一点？

我的数据集示例如下：

df <- data.frame(
      fertilizer = c("N","N","N","N","N","N","N","N","N","N","N","N","P","P","P","P","P","P","P","P","P","P","P","P","N","N","N","N","N","N","N","N","N","N","N","N","P","P","P","P","P","P","P","P","P","P","P","P"), 
      level = c("low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","high","low","low","high","low"), 
      repro = c(0,90,2,4,0,80,1,90,2,33,56,0,99,100,66,80,1,0,2,33,0,0,1,2,90,5,2,2,5,8,0,1,90,2,4,66,0,0,0,0,1,2,90,5,2,5,8,55)
    )    

我知道有从图表中提取CI点的方式，但我并不想这样做。我想使用计算这个的函数。

这是方程a <- x * t - 2 * x。我想为 解这个方程t。所以基本上，设置a = 0并求解t。我是R求解方程包的新手。我需要解决复杂根的包。我使用的原始方程有实根和虚根。我只是在寻找代数解决方案，而不是数字解决方案。

我试过：

a <- x * t - 2 * x
solve(a,t)

我遇到了一个错误：

a <- x * t - 2 * x
solve(a,t)

我有一个 ggplot 存储在 say 中p，希望添加一个poly_stat_eq图层来显示曲线方程。我想更改文本的字体大小，但找不到有关如何实现它的文档

p + 
  stat_poly_eq(formula = y ~ x, 
               aes(label = paste(..eq.label.., ..rr.label.., sep = "~~~")),
               label.x = 4, label.y = -5, parse = TRUE) 

我有一些计数数据。我想用计数数据绘制直方图，并添加负二项式、正态密度函数和泊松密度函数，但将函数拟合到计数数据。

我尝试按照这个示例进行操作，但是（a）我无法拟合负二项式和泊松函数（b）无法将其缩放到计数数据级别（c）不知道如何将所有三个函数拟合到同一个图表上，并为每个函数添加图例(d)行另外，我怎样才能获得每个适合的基本统计数据？例如，负二项式拟合将生成参数 k。我怎样才能让它出现在情节上

set.seed(111)
counts <- rbinom(500,100,0.1) 
df <- data.frame(counts)

ggplot(df, aes(x = counts)) + 
  geom_histogram(aes(y=..density..),colour = "black", fill = "white") +
  stat_function(fun=dnorm,args=fitdistr(df$counts,"normal")$estimate)

ggplot(df, aes(x = counts)) + 
  geom_histogram(aes(y=..density..),colour = "black", fill = "white") +
  stat_function(fun=poisson,args=fitdistr(df$counts,"poisson")$estimate)

ggplot(df, aes(x = counts)) + 
  geom_histogram(aes(y=..density..),colour = "black", fill = "white") +
  stat_function(fun=dnbinom,args=fitdistr(df$counts,"dnbinom")$estimate)