剧情版本4.x

Question

我有点困难，想弄清楚如何使用plotly重建蜘蛛（或雷达）图表的下图。实际上，我甚至无法在最新版本的ggplot2中重新创建它，因为自1.0.1以来一直有重大变化。

这是一个示例图形：

这是构建它的原始函数：

http://pcwww.liv.ac.uk/~william/Geodemographic%20Classifiability/func%20CreateRadialPlot.r

以下是原始函数如何工作的示例：

http://rstudio-pubs-static.s3.amazonaws.com/5795_e6e6411731bb4f1b9cc7eb49499c2082.html

这里有一些不那么虚假的数据：

d <- structure(list(Year = rep(c("2015","2016"),each=24),
                    Response = structure(rep(1L:24L,2), 
                                         .Label = c("Trustworthy", "Supportive", "Leading",
                                                    "Strong", "Dependable", "Consultative",
                                                    "Knowledgeable", "Sensible", 
                                                    "Intelligent", "Consistent", "Stable", 
                                                    "Innovative", "Aggressive", 
                                                    "Conservative", "Visionary", 
                                                    "Arrogant", "Professional", 
                                                    "Responsive", "Confident", "Accessible", 
                                                    "Timely", "Focused", "Niche", "None"),
                                         class = "factor"), 
                    Proportion = c(0.54, 0.48, 0.33, 0.35, 0.47, 0.3, 0.43, 0.29, 0.36,
                                   0.38, 0.45, 0.32, 0.27, 0.22, 0.26,0.95, 0.57, 0.42, 
                                   0.38, 0.5, 0.31, 0.31, 0.12, 0.88, 0.55, 0.55, 0.31,
                                   0.4, 0.5, 0.34, 0.53, 0.3, 0.41, 0.41, 0.46, 0.34, 
                                   0.22, 0.17, 0.28, 0.94, 0.62, 0.46, 0.41, 0.53, 0.34, 
                                   0.36, 0.1, 0.84), n = rep(c(240L,258L),each=24)),
               .Names = c("Year", "Response", "Proportion", "n"), 
               row.names = c(NA, -48L), class = c("tbl_df", "tbl", "data.frame"))

这是我的尝试（不太好）

plot_ly(d, r = Proportion, t = Response, x = Response, 
        color = factor(Year), mode = "markers") %>%
layout(margin = list(l=50,r=0,b=0,t=0,pad = 4), showlegend = TRUE)

关于如何使用plotly重新创建这个的任何想法？

Answer 1

极坐标图可用的选项仍然有限。据我所知，没有任何方法可以在圆周上为类别标签添加文本到极坐标图。文本散点，注释和刻度标签（四个四分之一点除外）都与目前的极坐标兼容。

所以，我们需要有点创意。

一种可以很好地工作的极坐标系是使用方位角投影的精确地球的投影图。以下是您如何适应此问题的演示。

首先，将值转换为以南极为中心的纬度和经度：

scale <- 10   # multiply latitudes by a factor of 10 to scale plot to good size in initial view
d$lat <- scale*d$Proportion - 90
d$long <- (as.numeric(d$Response)-1) * 360/24

使用方位角等距投影绘制

p <- plot_ly(d[c(1:24,1,25:48,25),], lat=lat, lon=long, color = factor(Year), colors=c('#F8756B','#00BDC2'),
             type = 'scattergeo', mode = 'lines+markers', showlegend=T) %>%
layout(geo = list(scope='world', showland=F, showcoastlines=F, showframe=F,
             projection = list(type = 'azimuthal equidistant', rotation=list(lat=-90), scale=5)), 
             legend=list(x=0.7,y=0.85))

在

上添加一些标签

p %<>% add_trace(type="scattergeo",  mode = "text", lat=rep(scale*1.1-90,24), lon=long, 
                 text=Response, showlegend=F, textfont=list(size=10)) %>%
       add_trace(type="scattergeo",  mode = "text", showlegend=F, textfont=list(size=12),
                 lat=seq(-90, -90+scale,length.out = 5), lon=rep(0,5), 
                 text=c("","25%","50%","75%","100%"))

最后，添加网格线

l1 <- list(width = 0.5, color = rgb(.5,.5,.5), dash = "3px")
l2 <- list(width = 0.5, color = rgb(.5,.5,.5))
for (i in c(0.1, 0.25, 0.5, 0.75, 1)) 
    p <- add_trace(lat=rep(-90, 100)-scale*i, lon=seq(0,360, length.out=100), type='scattergeo', mode='lines', line=l1, showlegend=F, evaluate=T)
for (i in 1:24) 
    p <- add_trace(p,lat=c(-90+scale*0.1,-90+scale), lon=rep(i*360/24,2), type='scattergeo', mode='lines', line=l2, showlegend=F, evaluate=T)

剧情版本4.x

的更新

更改中的更改以图示意味着原始版本不再起作用而不进行一些修改以使其更新。这是一个更新版本：

library(data.table)
gridlines1 = data.table(lat = -90 + scale*(c(0.1, 0.25, 0.5, 0.75, 1)))
gridlines1 = gridlines1[, .(long = c(seq(0,360, length.out=100), NA)), by = lat]
gridlines1[is.na(long), lat := NA]

gridlines2 = data.table(long = seq(0,360, length.out=25)[-1])
gridlines2 = gridlines2[, .(lat = c(NA, -90, -90+scale, NA)), by = long]
gridlines2[is.na(lat), long := NA]

text.labels = data.table(
  lat=seq(-90, -90+scale,length.out = 5),
  long = 0,
  text=c("","25%","50%","75%","100%"))

p = plot_ly() %>%
add_trace(type="scattergeo", data = d[c(1:24, 1, 25:48, 25),], 
      lat=~lat, lon=~long, 
      color = factor(d[c(1:24, 1, 25:48, 25),]$Year), 
      mode = 'lines+markers')%>%
layout(geo = list(scope='world', showland=F, showcoastlines=F, showframe=F,
    projection = list(type = 'azimuthal equidistant', rotation=list(lat=-90), scale=5)), 
    legend = list(x=0.7, y=0.85)) %>%
add_trace(data = gridlines1, lat=~lat, lon=~long, 
    type='scattergeo', mode='lines', line=l1, 
    showlegend=F, inherit = F)  %>%
add_trace(data = gridlines2, lat=~lat, lon=~long,
    type='scattergeo', mode='lines', line=l2, showlegend=F) %>%
add_trace(data = text.labels, lat=~lat, lon=~long, 
  type="scattergeo", mode = "text", text=~text, textfont = list(size = 12),
    showlegend=F, inherit=F) %>%
add_trace(data = d, lat=-90+scale*1.2, lon=~long, 
    type="scattergeo", mode = "text", text=~Response, textfont = list(size = 10),
    showlegend=F, inherit=F) 

p

Answer 2

通过假装，我在这方面取得了一些进展。极地坐标似乎只是恨我：

数据：

df <- d <- structure(list(Year = c("2015", "2015", "2015", "2015", "2015", 
"2015", "2015", "2015", "2015", "2015", "2015", "2015", "2015", 
"2015", "2015", "2015", "2015", "2015", "2015", "2015", "2015", 
"2015", "2015", "2015", "2016", "2016", "2016", "2016", "2016", 
"2016", "2016", "2016", "2016", "2016", "2016", "2016", "2016", 
"2016", "2016", "2016", "2016", "2016", "2016", "2016", "2016", 
"2016", "2016", "2016"), Response = structure(c(1L, 2L, 3L, 4L, 
5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 
19L, 20L, 21L, 22L, 23L, 24L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 
9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 
22L, 23L, 24L), .Label = c("Trustworthy", "Supportive", "Leading", 
"Strong", "Dependable", "Consultative", "Knowledgeable", "Sensible", 
"Intelligent", "Consistent", "Stable", "Innovative", "Aggressive", 
"Conservative", "Visionary", "Arrogant", "Professional", "Responsive", 
"Confident", "Accessible", "Timely", "Focused", "Niche", "None"
), class = "factor"), Proportion = c(0.54, 0.48, 0.33, 0.35, 
0.47, 0.3, 0.43, 0.29, 0.36, 0.38, 0.45, 0.32, 0.27, 0.22, 0.26, 
0.95, 0.57, 0.42, 0.38, 0.5, 0.31, 0.31, 0.12, 0.88, 0.55, 0.55, 
0.31, 0.4, 0.5, 0.34, 0.53, 0.3, 0.41, 0.41, 0.46, 0.34, 0.22, 
0.17, 0.28, 0.94, 0.62, 0.46, 0.41, 0.53, 0.34, 0.36, 0.1, 0.84
), n = c(240L, 240L, 240L, 240L, 240L, 240L, 240L, 240L, 240L, 
240L, 240L, 240L, 240L, 240L, 240L, 240L, 240L, 240L, 240L, 240L, 
240L, 240L, 240L, 240L, 258L, 258L, 258L, 258L, 258L, 258L, 258L, 
258L, 258L, 258L, 258L, 258L, 258L, 258L, 258L, 258L, 258L, 258L, 
258L, 258L, 258L, 258L, 258L, 258L)), .Names = c("Year", "Response", 
"Proportion", "n"), row.names = c(NA, -48L), class = c("tbl_df", 
"tbl", "data.frame"))

使用基础知识在散点图上创建循环映射：

df$degree <- seq(0,345,15) # 24 responses, equals 15 degrees per response
df$o <- df$Proportion * sin(df$degree * pi / 180) # SOH
df$a <- df$Proportion * cos(df$degree * pi / 180) # CAH
df$o100 <- 1 * sin(df$degree * pi / 180) # Outer ring x
df$a100 <- 1 * cos(df$degree * pi / 180) # Outer ring y 
df$a75 <- 0.75 * cos(df$degree * pi / 180) # 75% ring y
df$o75 <- 0.75 * sin(df$degree * pi / 180) # 75% ring x
df$o50 <- 0.5 * sin(df$degree * pi / 180) # 50% ring x
df$a50 <- 0.5 * cos(df$degree * pi / 180) # 50% ring y

并绘制情节。我在这里作弊让他们通过双重绘制第1行和第25行来连接到最后一个位置：

p = plot_ly()

for(i in 1:24) {
  p <- add_trace(
    p, 
    x = c(d$o100[i],0), 
    y = c(d$a100[i],0), 
    evaluate = TRUE,
    line = list(color = "#d3d3d3", dash = "3px"),
    showlegend = FALSE
    )
}

p %>% 
  add_trace(data = d[c(1:48,1,25),], x = o, y = a, color = Year, 
            mode = "lines+markers",
            hoverinfo = "text", 
            text = paste(Year, Response,round(Proportion * 100), "%")) %>% 
  add_trace(data = d, x = o100, y = a100, 
            text = Response,
            hoverinfo = "none",
            textposition = "top middle", mode = "lines+text", 
            line = list(color = "#d3d3d3", dash = "3px", shape = "spline"),
            showlegend = FALSE) %>% 
  add_trace(data = d, x = o50, y = a50, mode = "lines", 
            line = list(color = "#d3d3d3", dash = "3px", shape = "spline"), 
            hoverinfo = "none",
            showlegend = FALSE) %>% 
  add_trace(data = d, x = o75, y = a75, mode = "lines", 
            line = list(color = "#d3d3d3", dash = "3px", shape = "spline"), 
            hoverinfo = "none",
            showlegend = FALSE) %>%
  layout(
    autosize = FALSE,
    hovermode = "closest",     
    autoscale = TRUE,
    width = 800,
    height = 800,
    xaxis = list(range = c(-1.25,1.25), showticklabels = FALSE, zeroline = FALSE, showgrid = FALSE),
    yaxis = list(range = c(-1.25,1.25), showticklabels = FALSE, zeroline = FALSE, showgrid = FALSE))

~~正如你所看到的，除了最后连接线的和从原点传递到响应文本的行之外，我得到了它。~~

Answer 3

据我所知，你已经用ggplot2（示例图片）获得了你的情节。如果这是真的，那么最简单的想法是你应该在你的ggplot对象上运行ggplotly()，如下例所示：

install.packages(c("ggplot2","plotly"))
library(ggplot2)
library(plotly)

plot <- ggplot(data =mtcars, aes(x =  mpg, y = cyl))+
 geom_point()

ggplotly (plot)

将产生以下交互式情节：

我可以在剧情中重新创建这个极坐标蜘蛛图吗？

3 个答案:

剧情版本4.x