Question

以下图表接近我正在寻找的内容，但我想知道以下是否可行：

节点的左对齐而不是沿着x轴对齐，例如，仅具有2个节点的流将完成x轴的中途，而不是x-max（在我的非玩具sankey图中，这是左对齐的，但是，我无法解决差异）
仅删除节点上的hovertext（不在链接上）。我尝试了各种组合＆＃34;标签＆＃34;，＆＃34;文字＆＃34;，＆＃34;价值＆＃34;，＆＃34;百分比＆＃34;，＆＃34;名称＆＃34;加入＆＃34; +＆＃34;或者＆＃34;所有＆＃34;或＆＃34;无＆＃34;或者＆＃34;跳过＆＃34;，但这似乎没有任何区别。
使用NA来处理丢弃，例如，我不想看到从SA到Drop（蓝色节点）的链接，但是想要看到x = -1处的绿色条显示一个有人在他们的第一个假期去了南澳大利亚，并没有再度假。（如果我离开source = SA并且target = NA，则图表为空白）。我建议的解决方法是将DROP Node和SA-DROP链接设置为白色......

用所需的蓝色变化注释图像。

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：我最初没有如何用对应节点的断点标记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

您无法在Plotly中更改节点的位置，但如果您更改“已修复”的排列方式。自由形式＆＃39;您可以在渲染图表后手动将节点移动到任何位置。但是，每次渲染图表时，都必须由用户手动完成。目前无法在Plotly脚本中订购节点。

Answer 2

实际上，您可以手动覆盖节点的位置（所有节点或仅覆盖所需节点）。

您可以在节点列表中进行此操作，并添加x轴矢量和y轴矢量以及要更改的节点位置。如果要将节点保持在相同位置，只需将NA添加到该矢量位置即可。

node = list(
    label = sank_nodes$label_pad,
    color = sank_nodes$color,
    pad = 15,
    thickness = 15,
    line = list(
      color = "black",
      width = 0.5
    ), 
    x = c(NA, 0.35, 0.65, NA, NA, NA, NA, NA),
    y = c(NA, 0.10, 0.42, NA, NA, NA, NA, NA)
  )

Answer 3

实际上，这是完全可能的。

import plotly.graph_objects as go

fig = go.Figure(go.Sankey(
    arrangement = "snap",
    node = {
        "label": ["A", "B", "C", "D", "E", "F"],
        "x": [0.2, 0.1, 0.5, 0.7, 0.3, 0.5],
        "y": [0.7, 0.5, 0.2, 0.4, 0.2, 0.3],
        'pad':10},  # 10 Pixels
    link = {
        "source": [0, 0, 1, 2, 5, 4, 3, 5],
        "target": [5, 3, 4, 3, 0, 2, 2, 3],
        "value": [1, 2, 1, 1, 1, 1, 1, 2]}))

fig.show()

来自plotly.com的代码。

Plotly Sankey finetuning;节点沿x轴对齐，下降

3 个答案: