46

我能够绘制散点图并根据一个标准为点着色，即我可以将所有 >=3 的点着色为红色，其余点着色为黑色。我希望能够以这种方式着色点：

  1. =3 红色

  2. <=1 颜色蓝色
  3. 其余为黑色

我在下面的代码完成了第 1 步和第 3 步，但我不确定如何合并第 2 步的第二个参数

data<- read.table('sample_data.txtt', header=TRUE, row.name=1)
pos<- data$col_name1
cn<- data$col_name2
plot(pos,cn, ylim=c(0,5), col="blue")
plot(pos,cn, col=ifelse(cn>=3,"red","black"), ylim=c(0,10))

任何帮助都会很棒！！！提前致谢

在此处输入图像描述

4

4 回答 4

59

最好的办法是在数据对象中添加一列来表示点颜色。然后通过过滤更新它的部分。

data<- read.table('sample_data.txtt', header=TRUE, row.name=1)
# Create new column filled with default colour
data$Colour="black"
# Set new column values to appropriate colours
data$Colour[data$col_name2>=3]="red"
data$Colour[data$col_name2<=1]="blue"
# Plot all points at once, using newly generated colours
plot(data$col_name1,data$col_name2, ylim=c(0,5), col=data$Colour, ylim=c(0,10))

应该清楚如何使其适应具有更多颜色和条件的图。

于 2013-07-09T14:51:23.513 回答
20

也可以指定ifelse()两次：

plot(pos,cn, col= ifelse(cn >= 3, "red", ifelse(cn <= 1,"blue", "black")), ylim = c(0, 10))
于 2013-07-09T14:59:21.937 回答
3

最好使用 cut() 创建一个新的因子变量。我还使用 ggplot2 添加了一些选项。

df <- data.frame(
  X1=seq(0, 5, by=0.001),
  X2=rnorm(df$X1, mean = 3.5, sd = 1.5)
)

# Create new variable for plotting
df$Colour <- cut(df$X2, breaks = c(-Inf, 1, 3, +Inf), 
                 labels = c("low", "medium", "high"), 
                 right = FALSE)

### Base Graphics

plot(df$X1, df$X2, 
     col = df$Colour, ylim = c(0, 10), xlab = "POS", 
     ylab = "CS", main = "Plot Title", pch = 21)

plot(df$X1,df$X2, 
     col = df$Colour, ylim = c(0, 10), xlab = "POS", 
     ylab = "CS", main = "Plot Title", pch = 19, cex = 0.5)

# Using `with()` 

with(df, 
     plot(X1, X2, xlab="POS", ylab="CS", col = Colour, pch=21, cex=1.4)
     )

# Using ggplot2
library(ggplot2)

# qplot()
qplot(df$X1, df$X2, colour = df$Colour)

# ggplot()
p <- ggplot(df, aes(X1, X2, colour = Colour)) 
p <- p + geom_point() + xlab("POS") + ylab("CS")
p

p + facet_grid(Colour~., scales = "free")
于 2013-07-09T16:53:31.410 回答
1

这是一种使用阈值和相关颜色查找表将颜色映射到感兴趣变量的方法。 

 # make a grid 'Grd' of points and number points for side of square 'GrdD'
Grd <- expand.grid(seq(0.5,400.5,10),seq(0.5,400.5,10))
GrdD <- length(unique(Grd$Var1))

# Add z-values to the grid points
Grd$z <- rnorm(length(Grd$Var1), mean = 10, sd =2)

# Make a vector of thresholds 'Brks' to colour code z 
Brks <- c(seq(0,18,3),Inf)

# Make a vector of labels 'Lbls' for the colour threhsolds
Lbls <- Lbls <- c('0-3','3-6','6-9','9-12','12-15','15-18','>18')

# Make a vector of colours 'Clrs' for to match each range
Clrs <- c("grey50","dodgerblue","forestgreen","orange","red","purple","magenta")

# Make up lookup dataframe 'LkUp' of the lables and colours 
LkUp <- data.frame(cbind(Lbls,Clrs),stringsAsFactors = FALSE)

# Add a new variable 'Lbls' the grid dataframe mapping the labels based on z-value
Grd$Lbls <- as.character(cut(Grd$z, breaks = Brks, labels = Lbls))

# Add a new variable 'Clrs' to the grid dataframe based on the Lbls field in the grid and lookup table
Grd <- merge(Grd,LkUp, by.x = 'Lbls')

# Plot the grid using the 'Clrs' field for the colour of each point
plot(Grd$Var1,
     Grd$Var2,
     xlim = c(0,400),
     ylim = c(0,400),
     cex = 1.0,
     col = Grd$Clrs,
     pch = 20,
     xlab = 'mX',
     ylab = 'mY',
     main = 'My Grid',
     axes = FALSE,
     labels = FALSE,
     las = 1
)

axis(1,seq(0,400,100))
axis(2,seq(0,400,100),las = 1)
box(col = 'black')

legend("topleft", legend = Lbls, fill = Clrs, title = 'Z')
于 2019-06-09T05:30:59.630 回答