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R - 像素矩阵的图像？

发布于 2024-11-01 13:35:45 字数 133 浏览 1 评论 0原文

如何在 R 中从矩阵生成图像？

矩阵值将对应于图像上的像素强度（尽管我现在只对 0,1 值白色或黑色感兴趣。），而列数和行数对应于图像上的垂直和水平位置。

我所说的制作图像是指将其显示在屏幕上并将其另存为 jpg 格式。

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谜泪 2024-11-08 13:35:45

您可以使用“图像”最简单地将其显示在屏幕上：

m = matrix(runif(100),10,10)
par(mar=c(0, 0, 0, 0))
image(m, useRaster=TRUE, axes=FALSE)

您还可以查看光栅包...

You can display it on the screen easiest using 'image':

m = matrix(runif(100),10,10)
par(mar=c(0, 0, 0, 0))
image(m, useRaster=TRUE, axes=FALSE)

You can also have a look at the raster package...

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醉梦枕江山 2024-11-08 13:35:45

设置没有边距的绘图：

par(mar = rep(0, 4))

用灰度对矩阵进行成像，就像 spacedman 的答案一样，但完全填充设备：

m = matrix(runif(100),10,10)
image(m, axes = FALSE, col = grey(seq(0, 1, length = 256)))

将其包装在对 png() 的调用中以创建文件：

png("simpleIm.png")
par(mar = rep(0, 4))
image(m, axes = FALSE, col = grey(seq(0, 1, length = 256)))
dev.off()

如果需要使用空间轴执行此操作（默认为 [ 0,1] 对于 X 和 Y），然后使用 image.default(x, y, z, ...) 形式，其中 x 和 y 给出 z 中像素的中心位置。 x 和 y 的长度可以是 dim(z) + 1，以给出该约定的角坐标。

像素中心（这是图像的默认值）：

x <- seq(0, 1, length = nrow(m))
y <- seq(0, 1, length = ncol(m))
image(x, y, m, col = grey(seq(0, 1, length = 256)))

像素角（需要 1 个额外的 x 和 y，0 现在是左下角）：

x <- seq(0, 1, length = nrow(m) + 1)
y <- seq(0, 1, length = ncol(m) + 1)
image(x, y, m, col = grey(seq(0, 1, length = 256)))

请注意，从 R 2.13 image.default 获得一个参数 useRaster< /code> 它使用非常高效的新图形函数 rasterImage 而不是旧的 image ，后者实际上是在引擎盖下多次调用 rect 进行绘制每个像素都是一个多边形。

Set up a plot with no margin:

par(mar = rep(0, 4))

Image the matrix with greyscale, like spacedman's answer but completely filling the device:

m = matrix(runif(100),10,10)
image(m, axes = FALSE, col = grey(seq(0, 1, length = 256)))

Wrap that in a call to png() to create the file:

png("simpleIm.png")
par(mar = rep(0, 4))
image(m, axes = FALSE, col = grey(seq(0, 1, length = 256)))
dev.off()

If you need to do this with spatial axes (defaults to [0,1] for X and Y) then use the image.default(x, y, z, ...) form where x and y give the central positions of the pixels in z. x and y can be of length dim(z) + 1 to give corner coordinates for that convention.

Centres of pixels (this is the default for image):

x <- seq(0, 1, length = nrow(m))
y <- seq(0, 1, length = ncol(m))
image(x, y, m, col = grey(seq(0, 1, length = 256)))

Corners of pixels (need 1 extra x and y, and 0 is now the very bottom left corner):

x <- seq(0, 1, length = nrow(m) + 1)
y <- seq(0, 1, length = ncol(m) + 1)
image(x, y, m, col = grey(seq(0, 1, length = 256)))

Note that from R 2.13 image.default gains an argument useRaster which uses the very efficient newish graphics function rasterImage rather than the old image which is effectively multiple calls to rect under the hood to draw every pixel as a polygon.

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宁愿没拥抱 2024-11-08 13:35:45

我用两种方法之一制作一个矩阵（其中垂直轴向下增加）。下面是使用 heatmap.2() 的第一种方法。它可以更好地控制图中数值的格式（请参阅下面的 formatC 语句），但在更改布局时处理起来有点困难。

 library(gplots)

 #Build the matrix data to look like a correlation matrix
 x <- matrix(rnorm(64), nrow=8)
 x <- (x - min(x))/(max(x) - min(x)) #Scale the data to be between 0 and 1
 for (i in 1:8) x[i, i] <- 1.0 #Make the diagonal all 1's

 #Format the data for the plot
 xval <- formatC(x, format="f", digits=2)
 pal <- colorRampPalette(c(rgb(0.96,0.96,1), rgb(0.1,0.1,0.9)), space = "rgb")

 #Plot the matrix
 x_hm <- heatmap.2(x, Rowv=FALSE, Colv=FALSE, dendrogram="none", main="8 X 8 Matrix Using Heatmap.2", xlab="Columns", ylab="Rows", col=pal, tracecol="#303030", trace="none", cellnote=xval, notecol="black", notecex=0.8, keysize = 1.5, margins=c(5, 5))

在此处输入图像描述

I do a matrix (where the vertical axis increases going down) one of two ways. Below is the first way using heatmap.2(). It has more control over how the numeric values are formatted in the plot (see the formatC statement below), but is a little harder to deal with when changing the layout.

 library(gplots)

 #Build the matrix data to look like a correlation matrix
 x <- matrix(rnorm(64), nrow=8)
 x <- (x - min(x))/(max(x) - min(x)) #Scale the data to be between 0 and 1
 for (i in 1:8) x[i, i] <- 1.0 #Make the diagonal all 1's

 #Format the data for the plot
 xval <- formatC(x, format="f", digits=2)
 pal <- colorRampPalette(c(rgb(0.96,0.96,1), rgb(0.1,0.1,0.9)), space = "rgb")

 #Plot the matrix
 x_hm <- heatmap.2(x, Rowv=FALSE, Colv=FALSE, dendrogram="none", main="8 X 8 Matrix Using Heatmap.2", xlab="Columns", ylab="Rows", col=pal, tracecol="#303030", trace="none", cellnote=xval, notecol="black", notecex=0.8, keysize = 1.5, margins=c(5, 5))

enter image description here

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清音悠歌 2024-11-08 13:35:45

您可以创建矩阵的热图。

library(pheatmap)

# Create a 10x10 matrix of random numbers
m = matrix(runif(100), 10, 10)

# Save output to jpeg
jpeg("heatmap.jpg")

pheatmap(m, cluster_row = FALSE, cluster_col = FALSE, color=gray.colors(2,start=1,end=0))

dev.off()

请参阅 ?pheatmap 了解更多选项。

You can create a heatmap of the matrix.

library(pheatmap)

# Create a 10x10 matrix of random numbers
m = matrix(runif(100), 10, 10)

# Save output to jpeg
jpeg("heatmap.jpg")

pheatmap(m, cluster_row = FALSE, cluster_col = FALSE, color=gray.colors(2,start=1,end=0))

dev.off()

See ?pheatmap for more options.

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雾里花 2024-11-08 13:35:45

尝试水平图：

library(lattice)
levelplot(matrix)

Try levelplot:

library(lattice)
levelplot(matrix)

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却一份温柔 2024-11-08 13:35:45

这是第二种方法（同样，垂直轴向下增加）。此方法更易于布局，但对图中显示的数值格式的控制较少。

 library(plotrix)

 #Build the matrix data to look like a correlation matrix
 n <- 8
 x <- matrix(runif(n*n), nrow=n)
 xmin <- 0
 xmax <- 1
 for (i in 1:n) x[i, i] <- 1.0 #Make the diagonal all 1's

 #Generate the palette for the matrix and the legend.  Generate labels for the legend
 palmat <- color.scale(x, c(1, 0.4), c(1, 0.4), c(0.96, 1))
 palleg <- color.gradient(c(1, 0.4), c(1, 0.4), c(0.96, 1), nslices=100)
 lableg <- c(formatC(xmin, format="f", digits=2), formatC(1*(xmax-xmin)/4, format="f", digits=2), formatC(2*(xmax-xmin)/4, format="f", digits=2), formatC(3*(xmax-xmin)/4, format="f", digits=2), formatC(xmax, format="f", digits=2))

 #Set up the plot area and plot the matrix
 par(mar=c(5, 5, 5, 8))
 color2D.matplot(x, cellcolors=palmat, main=paste(n, " X ", n, " Matrix Using Color2D.matplot", sep=""), show.values=2, vcol=rgb(0,0,0), axes=FALSE, vcex=0.7)
 axis(1, at=seq(1, n, 1)-0.5, labels=seq(1, n, 1), tck=-0.01, padj=-1)

 #In the axis() statement below, note that the labels are decreasing.  This is because
 #the above color2D.matplot() statement has "axes=FALSE" and a normal axis()
 #statement was used.
 axis(2, at=seq(1, n, 1)-0.5, labels=seq(n, 1, -1), tck=-0.01, padj=0.7)

 #Plot the legend
 pardat <- par()
 color.legend(pardat$usr[2]+0.5, 0, pardat$usr[2]+1, pardat$usr[2], paste(" ", lableg, sep=""), palleg, align="rb", gradient="y", cex=0.7)

在此处输入图像描述

Here's the second way (again, where the vertical axis increases going down). This method is easier to layout, but has less control over the format of the numeric values displayed in the plot.

 library(plotrix)

 #Build the matrix data to look like a correlation matrix
 n <- 8
 x <- matrix(runif(n*n), nrow=n)
 xmin <- 0
 xmax <- 1
 for (i in 1:n) x[i, i] <- 1.0 #Make the diagonal all 1's

 #Generate the palette for the matrix and the legend.  Generate labels for the legend
 palmat <- color.scale(x, c(1, 0.4), c(1, 0.4), c(0.96, 1))
 palleg <- color.gradient(c(1, 0.4), c(1, 0.4), c(0.96, 1), nslices=100)
 lableg <- c(formatC(xmin, format="f", digits=2), formatC(1*(xmax-xmin)/4, format="f", digits=2), formatC(2*(xmax-xmin)/4, format="f", digits=2), formatC(3*(xmax-xmin)/4, format="f", digits=2), formatC(xmax, format="f", digits=2))

 #Set up the plot area and plot the matrix
 par(mar=c(5, 5, 5, 8))
 color2D.matplot(x, cellcolors=palmat, main=paste(n, " X ", n, " Matrix Using Color2D.matplot", sep=""), show.values=2, vcol=rgb(0,0,0), axes=FALSE, vcex=0.7)
 axis(1, at=seq(1, n, 1)-0.5, labels=seq(1, n, 1), tck=-0.01, padj=-1)

 #In the axis() statement below, note that the labels are decreasing.  This is because
 #the above color2D.matplot() statement has "axes=FALSE" and a normal axis()
 #statement was used.
 axis(2, at=seq(1, n, 1)-0.5, labels=seq(n, 1, -1), tck=-0.01, padj=0.7)

 #Plot the legend
 pardat <- par()
 color.legend(pardat$usr[2]+0.5, 0, pardat$usr[2]+1, pardat$usr[2], paste(" ", lableg, sep=""), palleg, align="rb", gradient="y", cex=0.7)

enter image description here

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我喜欢麦丽素 2024-11-08 13:35:45

使用ggplot2：

library(tidyverse)
n <- 12
m <- matrix(rnorm(n*n),n,n)
rownames(m) <- colnames(m) <- 1:n
df <- as.data.frame(m) %>% gather(key='y', value='val')
df$y <- as.integer(df$y)
df$x <- rep(1:n, n)
ggplot(df, aes(x, y, fill= val)) + 
   geom_tile() +
   geom_text(aes(x, y, label=round(val,2))) +
   scale_fill_gradient(low = "white", high = "red") + 
   theme_bw()

With ggplot2:

library(tidyverse)
n <- 12
m <- matrix(rnorm(n*n),n,n)
rownames(m) <- colnames(m) <- 1:n
df <- as.data.frame(m) %>% gather(key='y', value='val')
df$y <- as.integer(df$y)
df$x <- rep(1:n, n)
ggplot(df, aes(x, y, fill= val)) + 
   geom_tile() +
   geom_text(aes(x, y, label=round(val,2))) +
   scale_fill_gradient(low = "white", high = "red") + 
   theme_bw()