R 以图方式固定一个节点坐标并保留一个节点的灵活坐标-解网

问：

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我真的很想找到一个解决方案！

假设我使用 plotly 绘制桑基图：

library(plotly)
library(dplyr)

input <- tibble(label = c("A1", "A2", "B1", "B2", "C1", "C2"),
                color = c("blue", "blue", "red", "blue", "blue", "blue"),
                source = c(0,1,0,2,3,0),
                target = c(2,3,3,4,4,5),
                
                x = c(0.1,0.3,0.5,0.7,1,0.9),
                y = c(0.3,0.2,0.6,0.2,0.3,0.7),
                
                value =  c(8,4,2,8,4,2))

(fig <- plot_ly(
  type = "sankey",
  orientation = "h",
  
  node = list(
    label = input$label,
    color = input$color,
    pad = 150, # distance between nodes
    x = input$x, 
    y = input$y
  ),
  
  link = list(
    source = input$source, 
    target = input$target, 
    value = input$value
  )
))

在此示例中，我修复了坐标和坐标。但是假设我真的只关心两者中的一个（在这种情况下），并且想情节地选择第二个坐标。xyx

所以实际上我想做这样的事情：

input <- tibble(label = c("A1", "A2", "B1", "B2", "C1", "C2"),
                color = c("blue", "blue", "red", "blue", "blue", "blue"),
                source = c(0,1,0,2,3,0),
                target = c(2,3,3,4,4,5),
                
                x = c(0.1,0.3,0.5,0.7,1,0.9),
                y = NA,
                
                value =  c(8,4,2,8,4,2))

然后以图形方式忽略 x 和 y 坐标 - 事实上，我得到的结果与我只是删除节点列表中的 x 和 y 参数（注意例如 B2 的位置）相同。

所以它等价于这个：

(fig <- plot_ly(
  type = "sankey",
  orientation = "h",
  
  node = list(
    label = input$label,
    color = input$color,
    pad = 150 # distance between nodes
  ),
  
  link = list(
    source = input$source, 
    target = input$target, 
    value = input$value
  )
))

关于如何修复其中一个坐标的任何想法？

r 图地

library(tibble)
library(plotly)
library(dplyr)

input <- tibble(label = c("A1", "A2", "B1", "B2", "C1", "C2"),
                color = c("blue", "blue", "red", "blue", "blue", "blue"),
                source = c(0,1,0,2,3,0),
                target = c(2,3,3,4,4,5),
                
                x = c(0.1,0.3,0.5,0.7,1,0.9),
                y = NA,
                
                value =  c(8,4,2,8,4,2))

(fig <- plot_ly(
  type = "sankey",
  orientation = "h",
  
  node = list(
    label = input$label,
    color = input$color,
    pad = 150, # distance between nodes
    x = input$x, 
    y = input$y
  ),
  
  link = list(
    source = input$source, 
    target = input$target, 
    value = input$value
  )
))

your_fig <- plotly_build(fig)
your_fig$x$data[[1]]$node$y = c(NA, NA, NA, 1, NA, NA, NA)
your_fig

^{创建于 2023-11-14 with reprex v2.0.2}

正如你所看到的，节点 B2 现在有一个不同的位置。而其他的仍然是一样的，无需手动更改。

R 以图方式固定一个节点坐标并保留一个节点的灵活坐标

R plotly fix one coordinate of nodes and leave one flexible

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