在时间轴上为二元因变量绘制十字。

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英文:

Plot crosses on time axis for a binary dependent variable

问题

我想使用ggplot2来绘制随时间变化的二元变量，以便每当我们观察到给定个体（id）的事件（y = 1）时，x轴上会有一个交叉点（或点）。

这是一些数据：

  1. access <- rbinom(n=500,size=1,prob=0.40)
  2. id <- rep(1:10, each = 50)   
  3. day_id <- rep(1:50, 10)
  4. da <- data.frame(access = access, id = id, day_id = day_id)

目前我正在使用ggplot2的以下方式：

  1. da %>% ggplot(aes(x = day_id, y = access)) + geom_point() + facet_wrap(~id)

这会导致这个结果：

在时间轴上为二元因变量绘制十字。

但我希望它看起来像这样（交叉点或点反映是否有访问权限；x轴反映每个个体的50天序列）：

在时间轴上为二元因变量绘制十字。

其中每条线代表一个id。

英文:

I want to use ggplot2 to plot a binary variable over time, such that whenever we observe the event (y = 1) for a given individual (id), there is a cross (or point) on the x-axis.

Here is some data:

  1. access &lt;- rbinom(n=500,size=1,prob=0.40)
  2. id &lt;- rep(1:10, each = 50)   
  3. day_id &lt;- rep(1:50, 10)
  4. da &lt;- data.frame(access = access, id = id, day_id = day_id)

Currently I'm using ggplot2 in the following way:

  1.   da %&gt;% ggplot(aes(x = day_id, y = access)) + geom_point() + facet_wrap(~id)

This leads to this result:
在时间轴上为二元因变量绘制十字。

But I want it to look like this (with crosses or points reflecting whether there was access; and the x_axis reflecting a sequence of 50 days per individual):
在时间轴上为二元因变量绘制十字。

Where each line represents one id.

答案1

得分: 2

Here is the translated code:

  1. library(ggplot2)
  2. library(dplyr)
  4. da %>% ggplot(aes(x = day_id, y = factor(id))) + 
  5.   geom_point(data=filter(da, access==1)) +
  6.   geom_line() + 
  7.   labs(y="id") + 
  8.   theme_classic()

Please note that I've translated the code part, and I won't provide answers to specific translation questions.

英文: 
  1. library(ggplot2)
  2. library(dplyr)
  4. da %&gt;% ggplot(aes(x = day_id, y = factor(id))) + 
  5.   geom_point(data=filter(da, access==1)) +
  6.   geom_line() + 
  7.   labs(y=&quot;id&quot;) + 
  8.   theme_classic()

在时间轴上为二元因变量绘制十字。

答案2

得分: 2

Points

  1. uid <- unique(da$id)
  3. plot.new()
  4. plot.window(c(0, nrow(da[da$id == uid[1], ])), c(1, length(uid)))
  5. for (i in seq_along(uid)) {
  6.   y <- with(da, access[id == uid[i]])
  7.   points(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), col=y, pch=20)
  8.   lines(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), lwd=1.5)
  9. } 
  10. mtext(LETTERS[rev(seq_along(uid))], 2, at=seq_along(uid), las=1)

Crosses

  1. plot.new()
  2. plot.window(c(0, nrow(da[da$id == uid[1], ])), c(1, length(uid)))
  3. for (i in seq_along(uid)) {
  4.   y <- with(da, access[id == uid[i]])
  5.   points(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), col=y, pch=3)
  6.   lines(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), lwd=1.5)
  7. } 
  8. mtext(LETTERS[rev(seq_along(uid))], 2, at=seq_along(uid), las=1)

Alternative visualization

  1. split(da$access, da$id) |> rev() |> do.call(what='cbind') |> image(col=0:1, axes=FALSE) +
  2.   mtext(LETTERS[rev(unique(da$id))], 2, 1, at=seq.int(0, 1, length.out=length(unique(da$id))), las=1) + box()

Data

  1. set.seed(42)
  2. access <- rbinom(n=500,size=1,prob=0.40)
  3. id <- rep(1:10, each=50)   
  4. day_id <- rep.int(1:50, 10)
  5. da <- data.frame(access=access, id=id, day_id=day_id)
  2. <details>
  3. <summary>英文:</summary>
  5.     Points
  7.     uid &lt;- unique(da$id)
  8.     
  9.     plot.new()
  10.     plot.window(c(0, nrow(da[da$id == uid[1], ])), c(1, length(uid)))
  11.     for (i in seq_along(uid)) {
  12.       y &lt;- with(da, access[id == uid[i]])
  13.       points(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), col=y, pch=20)
  14.       lines(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), lwd=1.5)
  15.     } 
  16.     mtext(LETTERS[rev(seq_along(uid))], 2, at=seq_along(uid), las=1)
  18. [![enter image description here][1]][1]
  20. *Crosses*
  22.     plot.new()
  23.     plot.window(c(0, nrow(da[da$id == uid[1], ])), c(1, length(uid)))
  24.     for (i in seq_along(uid)) {
  25.       y &lt;- with(da, access[id == uid[i]])
  26.       points(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), col=y, pch=3)
  27.       lines(seq_along(y), rep.int(length(uid) - i + 1L, length(y)), lwd=1.5)
  28.     } 
  29.     mtext(LETTERS[rev(seq_along(uid))], 2, at=seq_along(uid), las=1)
  31. [![enter image description here][2]][2]
  33. *Alternative visualization*
  35.     split(da$access, da$id) |&gt; rev() |&gt; do.call(what=&#39;cbind&#39;) |&gt; image(col=0:1, axes=FALSE) +
  36.       mtext(LETTERS[rev(unique(da$id))], 2, 1, at=seq.int(0, 1, length.out=length(unique(da$id))), las=1) + box()
  39. [![enter image description here][3]][3]
  41. ----
  43. *Data:*
  45.     set.seed(42)
  46.     access &lt;- rbinom(n=500,size=1,prob=0.40)
  47.     id &lt;- rep(1:10, each=50)   
  48.     day_id &lt;- rep.int(1:50, 10)
  49.     da &lt;- data.frame(access=access, id=id, day_id=day_id)
  52.   [1]: https://i.stack.imgur.com/mjiNa.png
  53.   [2]: https://i.stack.imgur.com/tYzfi.png
  54.   [3]: https://i.stack.imgur.com/VcoKN.png
  56. </details>

huangapple
  • 本文由 发表于 2023年6月29日 15:31:45
  • 转载请务必保留本文链接：https://go.coder-hub.com/76578906.html
  • binary
  • ggplot2
  • plot
  • r
  • time
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