如何在不手动输入变量名称的情况下使用 plotly 制作绘图？答案

【问题标题】：How to make a plot using plotly without typing manually the name of the variables?如何在不手动输入变量名称的情况下使用 plotly 制作绘图？
【发布时间】：2021-02-12 21:12:59
【问题描述】：

是否可以通过仅指定数据框来使用“plotly”包创建平行坐标图，以便使用所有变量。我可以找到的所有示例都通过使用“维度”选项指定要在绘图中使用的实际变量，这似乎是创建绘图的非常低效的方式。如果数据集有 20 个或更多变量并且您不想每次都键入它们怎么办。或者它是一个 Shiny 应用程序，它必须处理具有不同变量的不同用户数据集。

EXAMPLE3 使用“parcoords”包，它只需要数据框来创建绘图。这可能与情节有关吗？

谢谢！

library(tidyverse)
library(plotly)

df <- read.csv("https://raw.githubusercontent.com/bcdunbar/datasets/master/iris.csv")


# EXAMPLE1 - This example works
fig1 <- df %>% plot_ly(type = 'parcoords',
                      line = list(color = ~species_id,
                                  colorscale = list(c(0,'red'),c(0.5,'green'),c(1,'blue'))),
                      dimensions = list(
                        list(range = c(2,4.5),
                             label = 'Sepal Width', values = ~sepal_width),
                        list(range = c(4,8),
                             constraintrange = c(5,6),
                             label = 'Sepal Length', values = ~sepal_length),
                        list(range = c(0,2.5),
                             label = 'Petal Width', values = ~petal_width),
                        list(range = c(1,7),
                             label = 'Petal Length', values = ~petal_length)
                      )
)
fig1

# EXAMPLE2 - This example doesn't work
fig2 <- plot_ly(data = df,type = 'parcoords',
                      line = list(color = ~species_id,
                                  colorscale = list(c(0,'red'),c(0.5,'green'),c(1,'blue')))
)
fig2


# EXAMPLE3 - Example using parcoords library
library(parcoords)
parcoords(df, brushMode = "1d-axes", reorderable = TRUE)

【问题讨论】：

标签： r plotly r-plotly

【解决方案1】：

您不能省略维度参数。

但是，您可以通过lapply 以编程方式创建维度列表，从而大大减少您的打字工作量。请检查以下内容：

library(plotly)

# df <- read.csv("https://raw.githubusercontent.com/bcdunbar/datasets/master/iris.csv")
df <- structure(list(sepal_length = c(5.1, 4.9, 4.7, 4.6, 5, 7, 6.4, 
6.9, 5.5, 6.5, 6.3, 5.8, 7.1, 6.3, 6.5), sepal_width = c(3.5, 
3, 3.2, 3.1, 3.6, 3.2, 3.2, 3.1, 2.3, 2.8, 3.3, 2.7, 3, 2.9, 
3), petal_length = c(1.4, 1.4, 1.3, 1.5, 1.4, 4.7, 4.5, 4.9, 
4, 4.6, 6, 5.1, 5.9, 5.6, 5.8), petal_width = c(0.2, 0.2, 0.2, 
0.2, 0.2, 1.4, 1.5, 1.5, 1.3, 1.5, 2.5, 1.9, 2.1, 1.8, 2.2), 
    species = c("setosa", "setosa", "setosa", "setosa", "setosa", 
    "versicolor", "versicolor", "versicolor", "versicolor", "versicolor", 
    "virginica", "virginica", "virginica", "virginica", "virginica"
    ), species_id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 
    3L, 3L, 3L, 3L, 3L)), row.names = c(1L, 2L, 3L, 4L, 5L, 51L, 
52L, 53L, 54L, 55L, 101L, 102L, 103L, 104L, 105L), class = "data.frame")

dimensionColumns <- names(which(sapply(df, class) == "numeric")) # sapply(df, mode)

fig1 <- df %>% plot_ly(
  type = 'parcoords',
  line = list(color = ~ species_id,
              colorscale = list(c(0, 'red'), c(0.5, 'green'), c(1, 'blue'))),
  dimensions = lapply(dimensionColumns, function(x) {
    list(range = range(df[[x]]),
         label = x,
         values = as.formula(paste0("~", x)))
  })
)
fig1

【讨论】：