我很难用ggplotly设置手动颜色。
library(ggplot2)
library(plotly)
set.seed(1)
data.frame(x = 1:10, y = rnorm(10)) %>%
ggplot(aes(x, y, fill = factor(x > 5), alpha = y)) +
geom_bar(stat = "identity") +
scale_fill_manual(values = c("red", "blue"))
ggplotly()我知道错误:
Error in setNames(as.numeric(x), c("red", "green", "blue", "alpha")) :
'names' attribute [4] must be the same length as the vector [1]有追溯力:
14: setNames(as.numeric(x), c("red", "green", "blue", "alpha"))
13: FUN(X[[i]], ...)
12: lapply(X = X, FUN = FUN, ...)
11: sapply(valz, function(x) {
x <- setNames(as.numeric(x), c("red", "green", "blue", "alpha"))
x[["alpha"]] <- x[["alpha"]] * 255
do.call(grDevices::rgb, c(x, list(maxColorValue = 255)))
})
10: rgb2hex(x[idx])
9: toRGB(aes2plotly(data, params, "fill"), aes2plotly(data, params,
"alpha"))
8: geom2trace.GeomBar(dots[[1L]][[1L]], dots[[2L]][[1L]], dots[[3L]][[1L]])
7: (function (data, params, p)
{
UseMethod("geom2trace")
})(dots[[1L]][[1L]], dots[[2L]][[1L]], dots[[3L]][[1L]])
6: mapply(FUN = f, ..., SIMPLIFY = FALSE)
5: Map(geom2trace, dl, paramz[i], list(p))
4: layers2traces(data, prestats_data, panel$layout, p)
3: gg2list(p, width = width, height = height, tooltip = tooltip,
source = source)
2: ggplotly.ggplot()
1: ggplotly()当使用ggplotly时,设置alpha和填充/颜色的正确方法是什么?如果不使用alpha或不手动设置颜色,则不存在此错误。
发布于 2016-05-28 11:10:01
我相信alpha = y才是问题所在。
library(plotly)
# Generate sample data
df <- data.frame(x = 1:10,
y = sample(1:5, size = 10, replace = T),
col = sample(LETTERS[1:4], size = 10, replace = T))第一个geom_point (工作良好)
# ggplot2 syntax
ggplot(df, aes(x, y, color = col, alpha = y)) +
geom_point()
ggplotly()

下一次geom_bar (中断)
ggplot(df, aes(x, y, fill = col, alpha = y)) +
geom_bar(stat = "identity")
ggplotly()

用alpha作为因子修改geom_bar (有效)
ggplot(df, aes(x, y, fill = col, alpha = factor(y))) +
geom_bar(stat = "identity")
ggplotly()

我相信这与plotly.js有关。当alpha被指定为一组离散值时,每个条都被绘制为一个单独的跟踪,为整个跟踪指定一个alpha值,而不是特定的标记。

我认为条形图目前不支持连续的alpha值集,因为它确实适用于散点图。
希望这能帮上忙。
https://stackoverflow.com/questions/37494785
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