r - Plotly Sankey 微调；节点沿 x 轴对齐，下降

Question

下图与我要找的很接近，但是我想知道以下是否可行:

节点左对齐而不是
沿 x 轴对齐？，例如，只有 2 个节点的流
将在 x 轴的一半处完成，而不是在 x 最大处(在我的非玩具桑基图中，这是左对齐的，但是，我无法计算出差异)

仅删除节点上的悬停文本(而不是链接上)。我尝试了“标签”、“文本”、“值”、“百分比”、“名称”与“+”或“全部”或“无”或“跳过”的各种组合，但这些似乎都没有有区别。

例如，使用 NA 处理下车，我​​不想看到从 SA 到 Drop(蓝色节点)的链接，但确实希望看到 x=-1 处的绿色条以显示一个人去了 SA他们的第一个假期，还没有过另一个假期。 (如果我保留 source=SA 和 target=NA，图表是空白的)。我建议的解决方法是将 DROP 节点和 SA-DROP 链接着色为白色...

用蓝色的所需变化对图像进行了注释。

require(dplyr); require(plotly); require(RColorBrewer); require(stringr)

# Summarise flow data
dat <- data.frame(customer = c(1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5),
              holiday_loc = c("SA", "SA", "AB", "SA", "SA", "SA", "SA", "AB", "AB", "SA", "SA", "SA")) %>%
  group_by(customer) %>%
          mutate(holiday_num = seq_along(customer), 
                 source=paste0(holiday_loc, '_', holiday_num), 
                 target = lead(source),
                 last_hol = ifelse(holiday_num == n(), 'Y', 'N')) %>%
  filter(last_hol== 'N'| holiday_num == 1) %>%
  select(-last_hol)

 sank_links <-  dat %>%
   group_by(source, target) %>%
   summarise(n=n()) %>%
   mutate(target=ifelse(is.na(target), "DROP", target)) # is there another option here?

# obtain colours for nodes
f <- function(pal) brewer.pal(brewer.pal.info[pal, "maxcolors"], pal)
cols <- f("Set1")

# set up nodes
sank_nodes <- data.frame(
                      name = factor(sort(unique(c(as.character(sank_links$source), 
                                   as.character(sank_links$target)))))
                      ) %>%    
                        mutate(label=sub("_[0-9]$", "", name), 
                              # for some unknown reason, plotly allows only three labels to be the same
                              label_pad=sub("_[1-3]$", "", name),
                              label_pad=sub("_[4-6]$", " ", label_pad)) %>%
                        arrange(label) %>%
                        mutate(color = cols[cumsum(1-duplicated(label))])

# update links to get index of node and name (without holiday_num)
sank_links <- sank_links %>%
          mutate(source_num = match(source, sank_nodes$name) -1 , 
                 source_name = str_replace(source, "_[0-9]$", ""),
                 target_num = match(target, sank_nodes$name) - 1,
                 target_name = str_replace(target, "_[0-9]$", ""))


# diagram
p <- plot_ly(
  type = "sankey",
  domain = c(
    x =  c(0,1),
    y =  c(0,1)
  ),
  orientation = "h",
  valueformat = ".0f",
  valuesuffix = "Customers",
  arrangement="fixed",


  node = list(
    label = sank_nodes$label_pad,
    color = sank_nodes$color,
    pad = 15,
    thickness = 15,
    line = list(
      color = "black",
      width = 0.5
    )
  ),

  link = list(
    source = sank_links$source_num,
    target = sank_links$target_num,
    value =  sank_links$n
  )
) %>% 
  layout(
    title = "",
    font = list(
      size = 10
    ),
    xaxis = list(showgrid = F, zeroline = F),
    yaxis = list(showgrid = F, zeroline = F)
  )

p

编辑 :我最初不知道如何使用与节点对应的中断来标记 x 轴并为 x 轴提供标题；代码如下:

    %>% 
  layout(
    title = "",
    font = list(
      size = 10
    ),
    xaxis = list(showgrid = F, zeroline = F, title="Holiday Number", tickvals=-1:4, ticktext=1:6),
    yaxis = list(showgrid = F, zeroline = F, showticklabels=FALSE)
  )

来源: https://plot.ly/r/reference/#layout-xaxis-tickformat

Answer 1

您无法更改 Plotly 中节点的位置，但如果您将排列从“固定”更改为“自由形式”，您可以在渲染图表后手动将节点移动到您想要的任何位置。但是，每次呈现图表时，这都必须由用户手动完成。目前无法在 Plotly 脚本中对节点进行排序。