比较R中两个数据集的散点图

我想用散点图比较两个相同长度的数据集，然后根据指定p值的基因列表给差异表达的基因上色。更具体地说，它如下所示：

---

将x轴替换为数据A，y轴替换为数据B和有效p值根据原始图中的轴范围为[0,1]这一事实，我认为您希望绘制两种条件和两组测试下p值的散点图

以下是打印本身的操作方式：

# Generate an artificial dataset
library(MASS)
set.seed(1)

# Suitably chosen covariance matrix
covariancematrix <- matrix(c(0.02,0.019,0.019,0.02), nrow=2)
#> print(cov2cor(covariancematrix))
#     [,1] [,2]
#[1,] 1.00 0.95
#[2,] 0.95 1.00

# Randomize 20,000 observations and constraint them to a p-value like range
pvalues <- mvrnorm(20000, mu=c(0.5,0.5), Sigma=covariancematrix)
colnames(pvalues) <- c("p value condition 1", "p value condition 2")
rownames(pvalues) <- paste("Probe/gene id", 1:20000)
# p-values should be within range [0,1]
pvalues[pvalues<0] <- 0
pvalues[pvalues>1] <- 1

#> str(pvalues)
# num [1:20000, 1:2] 0.461 0.54 0.52 0.518 0.61 ...
# - attr(*, "dimnames")=List of 2
#  ..$ : chr [1:20000] "Probe/gene id 1" "Probe/gene id 2" "Probe/gene id 3" "Probe/gene id 4" ...
#  ..$ : chr [1:2] "p value condition 1" "p value condition 2"
#> head(pvalues)
#                p value condition 1 p value condition 2
#Probe/gene id 1           0.4614707           0.4767356
#Probe/gene id 2           0.5398754           0.5583752
#Probe/gene id 3           0.5196980           0.5251556
#Probe/gene id 4           0.5182167           0.4471374
#Probe/gene id 5           0.6097540           0.5745387
#Probe/gene id 6           0.4652409           0.3940416

# The plotting call itself
plot(   # If this is a 2-column matrix, then first column pvalues[,1] will be by default the x-axis and second column pvalue[,2] the y-axis
    # Can be a matrix with 2 columns or a data.frame with 2 columns
    x = pvalues, 
    # Analogous if you have two separated vectors instead of two columns, change to something like:
    # x = pvalues[,1], # x-axis observation values
    # y = pvalues[,2], # y-axis observation values
    # x- and y-axis ranges [0,1]
    xlim=c(0,1), 
    ylim=c(0,1), 
    # Select filled dots as the symbols
    pch=16, 
    # Conditional color vector based on where the observation is located
    col=apply(pvalues, MARGIN=1, FUN=function(x) { ifelse(1*x[1] +0.05 > x[2] & 1*x[1] -0.05 < x[2], 
        # Color for dots inside the area (semi-transparent black)
        "#00000075", 
        # Color for dots outside the area (semi-transparent blue)
        "#0000FF75") 
        } ),
    # Axis labels
    xlab="p values in differences condition 1", 
    ylab="p values in differences condition 2"
    )
# Draw the lines, formula:
# y = 1*x + 0.05
abline(a=0.05, b=1, col="red", lwd=2)
# y = 1*x - 0.05
abline(a=-0.05, b=1, col="red", lwd=2)

#生成一个人工数据集
图书馆（弥撒）
种子（1）
#适当选择协方差矩阵
协方差矩阵打印（cov2cor（协方差矩阵））
#     [,1] [,2]
#[1,] 1.00 0.95
#[2,] 0.95 1.00
#随机化20000个观察值，并将其限制在类似p值的范围内
p值从原始图中的轴范围为[0,1]这一事实推断，我认为您希望在两种条件和两组测试下绘制p值的散点图

以下是打印本身的操作方式：

# Generate an artificial dataset
library(MASS)
set.seed(1)

# Suitably chosen covariance matrix
covariancematrix <- matrix(c(0.02,0.019,0.019,0.02), nrow=2)
#> print(cov2cor(covariancematrix))
#     [,1] [,2]
#[1,] 1.00 0.95
#[2,] 0.95 1.00

# Randomize 20,000 observations and constraint them to a p-value like range
pvalues <- mvrnorm(20000, mu=c(0.5,0.5), Sigma=covariancematrix)
colnames(pvalues) <- c("p value condition 1", "p value condition 2")
rownames(pvalues) <- paste("Probe/gene id", 1:20000)
# p-values should be within range [0,1]
pvalues[pvalues<0] <- 0
pvalues[pvalues>1] <- 1

#> str(pvalues)
# num [1:20000, 1:2] 0.461 0.54 0.52 0.518 0.61 ...
# - attr(*, "dimnames")=List of 2
#  ..$ : chr [1:20000] "Probe/gene id 1" "Probe/gene id 2" "Probe/gene id 3" "Probe/gene id 4" ...
#  ..$ : chr [1:2] "p value condition 1" "p value condition 2"
#> head(pvalues)
#                p value condition 1 p value condition 2
#Probe/gene id 1           0.4614707           0.4767356
#Probe/gene id 2           0.5398754           0.5583752
#Probe/gene id 3           0.5196980           0.5251556
#Probe/gene id 4           0.5182167           0.4471374
#Probe/gene id 5           0.6097540           0.5745387
#Probe/gene id 6           0.4652409           0.3940416

# The plotting call itself
plot(   # If this is a 2-column matrix, then first column pvalues[,1] will be by default the x-axis and second column pvalue[,2] the y-axis
    # Can be a matrix with 2 columns or a data.frame with 2 columns
    x = pvalues, 
    # Analogous if you have two separated vectors instead of two columns, change to something like:
    # x = pvalues[,1], # x-axis observation values
    # y = pvalues[,2], # y-axis observation values
    # x- and y-axis ranges [0,1]
    xlim=c(0,1), 
    ylim=c(0,1), 
    # Select filled dots as the symbols
    pch=16, 
    # Conditional color vector based on where the observation is located
    col=apply(pvalues, MARGIN=1, FUN=function(x) { ifelse(1*x[1] +0.05 > x[2] & 1*x[1] -0.05 < x[2], 
        # Color for dots inside the area (semi-transparent black)
        "#00000075", 
        # Color for dots outside the area (semi-transparent blue)
        "#0000FF75") 
        } ),
    # Axis labels
    xlab="p values in differences condition 1", 
    ylab="p values in differences condition 2"
    )
# Draw the lines, formula:
# y = 1*x + 0.05
abline(a=0.05, b=1, col="red", lwd=2)
# y = 1*x - 0.05
abline(a=-0.05, b=1, col="red", lwd=2)

#生成一个人工数据集
图书馆（弥撒）
种子（1）
#适当选择协方差矩阵
协方差矩阵打印（cov2cor（协方差矩阵））
#     [,1] [,2]
#[1,] 1.00 0.95
#[2,] 0.95 1.00
#随机化20000个观察值，并将其限制在类似p值的范围内
pvalues您尝试过什么？请给我们一个例子。@sgibb：我试过使用plot（x~y）；abline（lm（x~y）），但这也是不可复制的。@sgibb：如果你是指数据的dput，它相当大，不想把它挤到这里。我尝试使用的代码与上面的代码类似，但我只想知道它是如何完成的。p值的黑色/蓝色看起来不错。你在坚持什么？另外，在生成随机数时，请尝试使用
set.seed（1）
（或任何其他数字），以使结果完全可重复。您尝试过什么？请给我们一个例子。@sgibb：我试过使用plot（x~y）；abline（lm（x~y）），但这也是不可复制的。@sgibb：如果你是指数据的dput，它相当大，不想把它挤到这里。我尝试使用的代码与上面的代码类似，但我只想知道它是如何完成的。p值的黑色/蓝色看起来不错。你在坚持什么？另外，在生成随机数时，请尝试使用
set.seed（1）
（或任何其他数字），以使结果完全可重复。
# Generate an artificial dataset
library(MASS)
set.seed(1)

# Suitably chosen covariance matrix
covariancematrix <- matrix(c(0.02,0.019,0.019,0.02), nrow=2)
#> print(cov2cor(covariancematrix))
#     [,1] [,2]
#[1,] 1.00 0.95
#[2,] 0.95 1.00

# Randomize 20,000 observations and constraint them to a p-value like range
pvalues <- mvrnorm(20000, mu=c(0.5,0.5), Sigma=covariancematrix)
colnames(pvalues) <- c("p value condition 1", "p value condition 2")
rownames(pvalues) <- paste("Probe/gene id", 1:20000)
# p-values should be within range [0,1]
pvalues[pvalues<0] <- 0
pvalues[pvalues>1] <- 1

#> str(pvalues)
# num [1:20000, 1:2] 0.461 0.54 0.52 0.518 0.61 ...
# - attr(*, "dimnames")=List of 2
#  ..$ : chr [1:20000] "Probe/gene id 1" "Probe/gene id 2" "Probe/gene id 3" "Probe/gene id 4" ...
#  ..$ : chr [1:2] "p value condition 1" "p value condition 2"
#> head(pvalues)
#                p value condition 1 p value condition 2
#Probe/gene id 1           0.4614707           0.4767356
#Probe/gene id 2           0.5398754           0.5583752
#Probe/gene id 3           0.5196980           0.5251556
#Probe/gene id 4           0.5182167           0.4471374
#Probe/gene id 5           0.6097540           0.5745387
#Probe/gene id 6           0.4652409           0.3940416

# The plotting call itself
plot(   # If this is a 2-column matrix, then first column pvalues[,1] will be by default the x-axis and second column pvalue[,2] the y-axis
    # Can be a matrix with 2 columns or a data.frame with 2 columns
    x = pvalues, 
    # Analogous if you have two separated vectors instead of two columns, change to something like:
    # x = pvalues[,1], # x-axis observation values
    # y = pvalues[,2], # y-axis observation values
    # x- and y-axis ranges [0,1]
    xlim=c(0,1), 
    ylim=c(0,1), 
    # Select filled dots as the symbols
    pch=16, 
    # Conditional color vector based on where the observation is located
    col=apply(pvalues, MARGIN=1, FUN=function(x) { ifelse(1*x[1] +0.05 > x[2] & 1*x[1] -0.05 < x[2], 
        # Color for dots inside the area (semi-transparent black)
        "#00000075", 
        # Color for dots outside the area (semi-transparent blue)
        "#0000FF75") 
        } ),
    # Axis labels
    xlab="p values in differences condition 1", 
    ylab="p values in differences condition 2"
    )
# Draw the lines, formula:
# y = 1*x + 0.05
abline(a=0.05, b=1, col="red", lwd=2)
# y = 1*x - 0.05
abline(a=-0.05, b=1, col="red", lwd=2)