【问题标题】:scale_fill_manual based on another factor in ggplot2scale_fill_manual 基于 ggplot2 中的另一个因素
【发布时间】:2013-09-28 14:33:31
【问题描述】:

我正在尝试根据用于在 ggplot2 中“填充”我的 geom_bar 的因素的更广泛分类来对我的图例进行颜色编码。我的情节看起来像这样: 我使用这个 R 代码得到的:

ggplot(df, aes(year, TOTALshark, fill=fishery)) + geom_bar(width=.5,stat="identity", position="dodge")+ facet_wrap(~div)

这是我的数据集的一个输入样本:

> dput(smpl)
df <- structure(list(X1 = structure(c(6L, 11L, 22L, 27L, 10L, 10L, 
6L, 11L, 6L, 10L, 8L, 6L, 6L, 4L, 22L, 18L, 10L, 10L, 11L, 6L
), .Label = c("AMERICAN PLAICE", "BIGEYE TUNA", "BIVALVE", "BLUEFIN TUNA", 
"CAPELIN", "COD(ATL)", "CRAB(SNOW,QUEEN)", "HADDOCK", "HAGFISH(ATL)", 
"HALIBUT(ATL)", "HALIBUT(GREENLAND)", "HERRING(ATL)", "JONAH CRAB (CANC.BOR.)", 
"LOBSTER", "LONGHORN SCULPIN", "LUMPFISH", "MACKEREL(ATL)", "MONKFISH", 
"PAND.BOR.", "PAND.MON.", "POLLOCK", "REDFISH", "SCALLOP", "SEA URCHINS", 
"SEACU", "SILVER HAKE", "SWORDFISH", "WHELK", "WHITE HAKE", "WINTER FLOUNDER", 
"WITCH FLOUNDER", "YELLOWFIN TUNA", "YELLOWTAIL FLOUNDER"), class = "factor"), 
    X2 = structure(c(2L, 2L, 8L, 5L, 5L, 5L, 5L, 8L, 5L, 5L, 
    5L, 2L, 5L, 5L, 8L, 2L, 5L, 5L, 2L, 2L), .Label = c("Dredge", 
    "Gillnet", "Hook", "Jigger", "Line", "Seine", "Trap", "Trawlb", 
    "Trawlm"), class = "factor"), fishery = structure(c(12L, 
    25L, 43L, 50L, 24L, 24L, 15L, 27L, 15L, 24L, 21L, 12L, 15L, 
    9L, 43L, 36L, 24L, 24L, 25L, 12L), .Label = c("AMERICAN PLAICE-Gillnet", 
    "AMERICAN PLAICE-Line", "AMERICAN PLAICE-Trawlb", "BIGEYE TUNA-Jigger", 
    "BIGEYE TUNA-Line", "BIVALVE-Dredge", "BLUEFIN TUNA-Hook", 
    "BLUEFIN TUNA-Jigger", "BLUEFIN TUNA-Line", "CAPELIN-Seine", 
    "CAPELIN-Trap", "COD(ATL)-Gillnet", "COD(ATL)-Hook", "COD(ATL)-Jigger", 
    "COD(ATL)-Line", "COD(ATL)-Trap", "COD(ATL)-Trawlb", "CRAB(SNOW,QUEEN)-Trap", 
    "CUSK-Line", "HADDOCK-Gillnet", "HADDOCK-Line", "HADDOCK-Trawlb", 
    "HAGFISH(ATL)-Trap", "HALIBUT(ATL)-Line", "HALIBUT(GREENLAND)-Gillnet", 
    "HALIBUT(GREENLAND)-Line", "HALIBUT(GREENLAND)-Trawlb", "HERRING(ATL)-Seine", 
    "HERRING(ATL)-Trawlm", "JONAH CRAB (CANC.BOR.)-Trap", "LOBSTER-Trap", 
    "LONGHORN SCULPIN-Trawlb", "LUMPFISH-Gillnet", "MACKEREL(ATL)-Seine", 
    "MACKEREL(ATL)-Trawlm", "MONKFISH-Gillnet", "MONKFISH-Trawlb", 
    "PAND.BOR.-Trawlb", "PAND.MON.-Trawlb", "POLLOCK-Gillnet", 
    "POLLOCK-Trawlb", "REDFISH-Gillnet", "REDFISH-Trawlb", "REDFISH-Trawlm", 
    "SCALLOP-Dredge", "SEA URCHINS-Dredge", "SEACU-Dredge", "SILVER HAKE-Trawlb", 
    "SWORDFISH-Jigger", "SWORDFISH-Line", "SWORDFISH-unk", "WHELK-Trap", 
    "WHITE HAKE-Gillnet", "WHITE HAKE-Line", "WINTER FLOUNDER-Gillnet", 
    "WINTER FLOUNDER-Trawlb", "WITCH FLOUNDER-Trawlb", "YELLOWFIN TUNA-Line", 
    "YELLOWTAIL FLOUNDER-Trawlb"), class = "factor"), year = c(2008L, 
    2008L, 2009L, 2009L, 2008L, 2009L, 2009L, 2008L, 2006L, 2007L, 
    2007L, 2007L, 2007L, 2007L, 2008L, 2008L, 2009L, 2009L, 2009L, 
    2009L), div = structure(c(6L, 19L, 2L, 4L, 5L, 10L, 3L, 19L, 
    9L, 10L, 3L, 9L, 6L, 4L, 3L, 9L, 6L, 11L, 7L, 9L), .Label = c("5Z", 
    "5Y", "4X", "4W", "4V", "4T", "4S", "4R", "3P", "3O", "3N", 
    "3M", "3L", "3K", "2J", "2H", "2G", "1F", "0B", "1B", "0A"
    ), class = "factor"), TOTALshark = c(3369.72, 12243.2, 6080.06, 
    316646.05, 18786.8, 6565.91, 1339771.2, 45841.03, 41329.64, 
    6411.86, 204980.36, 67608.78, 2617.05, 61547.64, 447349.44, 
    13226.4, 1362.55, 6012.23, 13152.51, 1067.92), cat = structure(c(1L, 
    1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 
    1L, 1L, 1L, 1L), .Label = c("groundfish", "largepelagic", 
    "bivalve", "smallpelagic", "crabs/lobsters", "shrimps", "others"
    ), class = "factor")), .Names = c("X1", "X2", "fishery", 
"year", "div", "TOTALshark", "cat"), class = "data.frame", row.names = c(70L, 
278L, 500L, 554L, 242L, 245L, 131L, 315L, 106L, 224L, 194L, 60L, 
115L, 37L, 489L, 385L, 249L, 244L, 284L, 75L))

我希望有相同的图例,但根据渔业属于哪个类别“猫”变量(即远洋、底层鱼)使用一些颜色。

【问题讨论】:

  • dput样本数据
  • “我希望有相同的图例,但根据渔业属于哪个类别的“猫”变量(即远洋、底层鱼)使用一些颜色。你的意思是你想要同样的鱼,但颜色因渔业而异?如果是这样,将您的 ggplot 调用更改为 ggplot(df, aes(year, TOTALshark, fill=cat)) + geom_bar(width=.5,stat="identity", position="dodge")+ facet_wrap(~div) 不会完成这项工作吗?
  • 没有。我想要的是上面的图例(带有渔业),但根据猫有 n 种颜色(这是针对主要鱼类的更广泛的类别)。例如,所有针对“大型中上层”(如金枪鱼)的渔业(BIGEYE TUNA-Line、BLUEFIN TUNA-Line 等)将是一种颜色,所有针对“底层鱼”的渔业(COD(ATL)-刺网、COD(ATL)-线等)将是另一种颜色,依此类推。我想根据猫对上面的图例进行颜色编码。希望这能澄清我想要做什么!谢谢。

标签: r ggplot2 legend


【解决方案1】:

这是你想要的吗?

library(ggplot2)
library(plyr)
library(gridExtra)

# create data that links colour per 'cat' with 'fishery'
# the 'cat' colours will be used as manually set fill colours. 

# get 'cat' colours

# alt. 1: grab 'cat' colours from plot object
# create a plot with fill = fishery *and* colour = cat
g1 <- ggplot(df, aes(x = year, y = TOTALshark, fill = fishery, colour = cat)) +
  geom_bar(width = 0.5, stat = "identity", position = "dodge") +
  facet_wrap(~ div)

g1

# grab 'cat' colours for each 'fishery' from plot object
# to be used in manual fill
cat_cols <- unique(ggplot_build(g1)[["data"]][[1]]$colour)

# unique 'cat'
cat <- unique(df$cat)

# create data frame with one colour per 'cat'
df2 <- data.frame(cat = cat, cat_cols)
df2


# alt 2: create your own 'cat' colours
# number of unique 'cat'
n <- length(cats)

# select one colour per 'cat', from e.g. brewer_pal or other palette tools 
cat_cols <- brewer_pal(type = "qual")(n)
# cat_cols <- rainbow(n)

# create data frame with one colour per 'cat'
df2 <- data.frame(cat, cat_cols)
df2

# select unique 'fishery' and 'cat' combinations
# in the order they show up in the legend, i.e. ordered ('arranged') by fishery
df3 <- unique(arrange(df[, c("fishery", "cat")], fishery))
df3

# add 'cat' colours to 'fishery'
# use 'join' to keep order
df3 <- join(df3, df2)
df3

# plot with fill by 'fishery' creates a fill scale by fishery,
# but colours are set manually using scale_fill_manual and the 'cat' colours from above
g2 <- ggplot(df, aes(x = year, y = TOTALshark, fill = fishery)) +
  geom_bar(width = 0.5, stat = "identity", position = "dodge") +
  facet_wrap(~ div, nrow = 5) +
  scale_fill_manual(values = as.character(df3$cat_cols))

g2

# create plot with both 'fishery' and 'cat' legend.

# extract 'fisheries' legend
tmp <- ggplot_gtable(ggplot_build(g2))
leg <- which(sapply(tmp$grobs, function(x) x$name) ==  "guide-box")
legend_fish <- tmp$grobs[[leg]]

# create a non-sense plot just to get a 'fill = cat' legend
g3 <- ggplot(df, aes(x = year, y = TOTALshark, fill = cat)) +
  geom_bar(stat = "identity") +
  scale_fill_manual(values = as.character(df3$cat_cols))

# extract 'cat' legend
tmp <- ggplot_gtable(ggplot_build(g3))
leg <- which(sapply(tmp$grobs, function(x) x$name) ==  "guide-box")
legend_cat <- tmp$grobs[[leg]]


# arrange plot and legends

library(gridExtra)

# quick and dirty with grid.arrange
# in the first column, put the plot (g2) without legend (removed using the 'theme' code)
# put the two legends in the second column
grid.arrange(g2 + theme(legend.position = "none"),
             arrangeGrob(legend_fish, legend_cat), ncol = 2) 


# arrange with viewports
# define plotting regions (viewports)
grid.newpage()
vp_plot <- viewport(x = 0.25, y = 0.5,
                    width = 0.5, height = 1)

vp_legend <- viewport(x = 0.75, y = 0.7,
                      width = 0.5, height = 0.5)

vp_sublegend <- viewport(x = 0.7, y = 0.25,
                         width = 0.5, height = 0.3)


print(g2 + theme(legend.position = "none"), vp = vp_plot)
upViewport(0)

pushViewport(vp_legend)
grid.draw(legend_fish)

upViewport(0)
pushViewport(vp_sublegend)
grid.draw(legend_cat)  

另见@mnel 的答案here 以替换绘图对象中的值。在这里也可能值得尝试。您也可以查看gtable 安排grobs 的方法。

【讨论】:

  • 非常感谢@Henrik 的回答和链接。这就是我要找的东西!
  • 正如您在最后一个绘图的视口代码中所见,两个图例沿 x 轴的(几乎)对齐是硬编码的。我想我已经看到了关于 SO 的示例,其中不同 grobs 的位置是根据它们的宽度计算的。但那是另一回事了……祝你好运!
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