在ggplot中对齐标签
Align labels in ggplot
我正在用 ggplot 绘图，但我在使用标签时遇到了一些问题。这是我的代码：
ggplot(data = prueba[prueba$Prueba=="lectura",], aes(x = año,
y = Promedio,
group = Tipo,
color = Tipo))+
geom_point(shape = 18,
size = 4, position = position_dodge(0.9)) +
geom_errorbar(aes(ymin = Promedio - Desviacion,
ymax = Promedio + Desviacion,
width = 0.4),
position = position_dodge(0.9),
size = 1.3,
show.legend = F) +
ylim(60,140) +
scale_color_manual(values = c("#A32986",
"#A4589E"),
breaks = c("Nacional",
"Sena")) +
theme(
panel.background = element_rect(fill = "white",
color = "#C6C6C6",
size = 1,
linetype = "solid"),
panel.grid.minor = element_line(size = 0.1,
linetype = "dashed"),
axis.title = element_text(color = "#575756",
size = 13),
axis.text = element_text(color = "#575756",
size = 11),
legend.title = element_blank(),
legend.position = "bottom",
legend.text = element_text(color = "#575756",
size = 13)
) + geom_text(aes(label = round(Promedio)),
position = position_dodge(0.9),
hjust = -0.5,
size = 5,
color = "#575756") +
geom_text(aes(label = paste("(", round(Desviacion), ")")),
color = "#575756",
position = position_dodge(0.9),
hjust = -0.1,
vjust = 2,
size = 5) + xlab(NULL)
结果是
enter image description here
如您所见，标签未正确对齐。我想组织 de 标签，这样上面的标签与下面的标签围绕同一轴居中。有谁知道我该怎么做？
我的数据可以被
重现
structure(list(Tipo = c("Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional"), año = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L), .Label = c("2016", "2017", "2018", "2019"
), class = "factor"), Promedio = c(96.57, 98.03, 98.51, 100.22,
98.17, 94.57, 98.07, 98.8, 87.38, 101.45, 99, 99.35, 96.11, 98.82,
97.1, 95.8, 97.49, 99.86, 90.59, 97.88, 90.6, 95.13, 97.56, 98.25,
94.03, 89.12, 91.94, 97.21, 94.77, 98.8, 99.04, 98.49, 99, 99,
100, 100, 100, 100, 99, 100, 98, 101, 100, 96, 100, 90, 102,
98, 100, 100, 100, 95, 132, 92, 100, 93, 102, 146, 148, 100,
147, 96, 95, 100, 90, 98, 93, 100, 134, 143), Desviacion = c(19.96,
20.12, 20.27, 19.78, 19.7, 18.21, 19.85, 20.17, 29.58, 22.13,
21.98, 19.35, 20.65, 19.78, 21.17, 25.08, 22.66, 21.62, 20.9,
20.52, 21.32, 18.83, 19.49, 21.09, 21.46, 20.54, 23.93, 21.3,
20.09, 26.99, 19.72, 21.3, 21, 21, 21, 20, 20, 20, 21, 20, 22,
23, 21, 22, 20, 28, 23, 25, 23, 22, 20, 22, 31, 21, 22, 22, 22,
27, 28, 20, 28, 21, 21, 22, 21, 22, 25, 28, 30, 28), Prueba = c("mantenimiento",
"lectura", "competencias", "promoción", "inglés", "ensamblaje",
"razonamiento", "comunicación", "mantenimiento", "promoción",
"comunicación", "lectura", "razonamiento", "inglés", "competencias",
"ensamblaje", "competencias", "comunicación", "mantenimiento",
"inglés", "razonamiento", "promoción", "ensamblaje", "lectura",
"mantenimiento", "razonamiento", "competencias", "lectura", "promoción",
"comunicación", "ensamblaje", "inglés", "competencias", "razonamiento",
"inglés", "promoción", "comunicación", "mantenimiento", "lectura",
"ensamblaje", "competencias", "comunicación", "inglés", "razonamiento",
"lectura", "mantenimiento", "promoción", "ensamblaje", "competencias",
"inglés", "ensamblaje", "promoción", "formar", "mantenimiento",
"lectura", "razonamiento", "comunicación", "enseñar", "evaluar",
"ensamblaje", "evaluar", "promoción", "mantenimiento", "inglés",
"razonamiento", "lectura", "competencias", "comunicación", "formar",
"enseñar")), row.names = c(56L, 57L, 58L, 59L, 60L, 61L, 62L,
63L, 451L, 452L, 453L, 454L, 455L, 456L, 457L, 458L, 934L, 935L,
936L, 937L, 938L, 939L, 940L, 941L, 2036L, 2037L, 2038L, 2039L,
2040L, 2041L, 2042L, 2043L, 24800L, 24801L, 24802L, 24803L, 24804L,
24805L, 24806L, 24807L, 68059L, 68060L, 68061L, 68062L, 68063L,
68064L, 68065L, 68066L, 171328L, 171329L, 171330L, 171331L, 171332L,
171333L, 171334L, 171335L, 171336L, 171337L, 171338L, 339673L,
339674L, 339675L, 339676L, 339677L, 339678L, 339679L, 339680L,
339681L, 339682L, 339683L), class = "data.frame")
提前致谢
我想到了以下内容（请注意，我没有使用你所有的审美，但应该足以让大家明白这一点，找到一种可能的解决方法）。
将上面的标签放在恒定高度
ggplot(data=subset(df, Prueba=='lectura'), aes(x=año, y=Promedio, group=Tipo, color=Tipo)) +
geom_point(shape=18, size=4, position=position_dodge(0.9)) +
geom_errorbar(aes(
ymin=Promedio-Desviacion, ymax=Promedio+Desviacion, width=0.4
), position=position_dodge(0.9), size=1.3, show.legend=F) +
ylim(60,140) +
geom_text(aes(
label = paste0(round(Promedio), "\n(", round(Desviacion), ")"), y=100),
vjust=-4.2, position=position_dodge(0.9)
)
给你这个：
我找不到将文本与每个栏的右侧或左侧对齐的好方法。由于您正在对数据集使用 position_dodge，因此您无法使用 nudge_x 来平均移动所有内容 "to the right"。 It's not allowed to use both in the geom。玩 hjust 种作品，但你可以期待有趣的事情发生在这里与你已经体验过的两条线。
另请注意，我已将您的两个 geom_text 调用合并为一个，您可以在其中使用 paste0() 或 paste() 进行组合并即时使用 "make your own label" .
vjust 全面使用允许您将它们全部放在相同的高度 - 我认为这就是您所指的，对吗？
同样 - 颜色与您的不同，因为我没有包括您的 scale_color_manual 和 theme 电话。
将标签放在一边，轻推 + Position_dodge
如果你想把标签放在一边，你不能用 nudge_x 做到这一点，因为你无法在 [=16= 中组合 nudge_x 和 position 参数].解决方法是基于 df$año 创建一个用于标签美学的新变量，然后在 geom_text:
中的 aes(... 调用中使用该变量
df$año_new <- as.numeric(df$año) + 0.1 # nudge of "0.1". Had to force numeric, since + does not work on factors
然后你保留原始 plot 调用中的所有内容，但只更改最后一个 geom_text 调用（注意 hjust=0 的重要添加，它将所有内容左对齐）：
geom_text(aes(
x=año_new, y=100,
label = paste0(round(Promedio), "\n(", round(Desviacion), ")")
),
hjust=0,
position=position_dodge(0.9)
)
我正在用 ggplot 绘图，但我在使用标签时遇到了一些问题。这是我的代码：
ggplot(data = prueba[prueba$Prueba=="lectura",], aes(x = año,
y = Promedio,
group = Tipo,
color = Tipo))+
geom_point(shape = 18,
size = 4, position = position_dodge(0.9)) +
geom_errorbar(aes(ymin = Promedio - Desviacion,
ymax = Promedio + Desviacion,
width = 0.4),
position = position_dodge(0.9),
size = 1.3,
show.legend = F) +
ylim(60,140) +
scale_color_manual(values = c("#A32986",
"#A4589E"),
breaks = c("Nacional",
"Sena")) +
theme(
panel.background = element_rect(fill = "white",
color = "#C6C6C6",
size = 1,
linetype = "solid"),
panel.grid.minor = element_line(size = 0.1,
linetype = "dashed"),
axis.title = element_text(color = "#575756",
size = 13),
axis.text = element_text(color = "#575756",
size = 11),
legend.title = element_blank(),
legend.position = "bottom",
legend.text = element_text(color = "#575756",
size = 13)
) + geom_text(aes(label = round(Promedio)),
position = position_dodge(0.9),
hjust = -0.5,
size = 5,
color = "#575756") +
geom_text(aes(label = paste("(", round(Desviacion), ")")),
color = "#575756",
position = position_dodge(0.9),
hjust = -0.1,
vjust = 2,
size = 5) + xlab(NULL)
结果是
enter image description here
如您所见，标签未正确对齐。我想组织 de 标签，这样上面的标签与下面的标签围绕同一轴居中。有谁知道我该怎么做？
我的数据可以被
重现structure(list(Tipo = c("Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena", "Sena",
"Sena", "Sena", "Sena", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional", "Nacional", "Nacional",
"Nacional", "Nacional", "Nacional", "Nacional"), año = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L), .Label = c("2016", "2017", "2018", "2019"
), class = "factor"), Promedio = c(96.57, 98.03, 98.51, 100.22,
98.17, 94.57, 98.07, 98.8, 87.38, 101.45, 99, 99.35, 96.11, 98.82,
97.1, 95.8, 97.49, 99.86, 90.59, 97.88, 90.6, 95.13, 97.56, 98.25,
94.03, 89.12, 91.94, 97.21, 94.77, 98.8, 99.04, 98.49, 99, 99,
100, 100, 100, 100, 99, 100, 98, 101, 100, 96, 100, 90, 102,
98, 100, 100, 100, 95, 132, 92, 100, 93, 102, 146, 148, 100,
147, 96, 95, 100, 90, 98, 93, 100, 134, 143), Desviacion = c(19.96,
20.12, 20.27, 19.78, 19.7, 18.21, 19.85, 20.17, 29.58, 22.13,
21.98, 19.35, 20.65, 19.78, 21.17, 25.08, 22.66, 21.62, 20.9,
20.52, 21.32, 18.83, 19.49, 21.09, 21.46, 20.54, 23.93, 21.3,
20.09, 26.99, 19.72, 21.3, 21, 21, 21, 20, 20, 20, 21, 20, 22,
23, 21, 22, 20, 28, 23, 25, 23, 22, 20, 22, 31, 21, 22, 22, 22,
27, 28, 20, 28, 21, 21, 22, 21, 22, 25, 28, 30, 28), Prueba = c("mantenimiento",
"lectura", "competencias", "promoción", "inglés", "ensamblaje",
"razonamiento", "comunicación", "mantenimiento", "promoción",
"comunicación", "lectura", "razonamiento", "inglés", "competencias",
"ensamblaje", "competencias", "comunicación", "mantenimiento",
"inglés", "razonamiento", "promoción", "ensamblaje", "lectura",
"mantenimiento", "razonamiento", "competencias", "lectura", "promoción",
"comunicación", "ensamblaje", "inglés", "competencias", "razonamiento",
"inglés", "promoción", "comunicación", "mantenimiento", "lectura",
"ensamblaje", "competencias", "comunicación", "inglés", "razonamiento",
"lectura", "mantenimiento", "promoción", "ensamblaje", "competencias",
"inglés", "ensamblaje", "promoción", "formar", "mantenimiento",
"lectura", "razonamiento", "comunicación", "enseñar", "evaluar",
"ensamblaje", "evaluar", "promoción", "mantenimiento", "inglés",
"razonamiento", "lectura", "competencias", "comunicación", "formar",
"enseñar")), row.names = c(56L, 57L, 58L, 59L, 60L, 61L, 62L,
63L, 451L, 452L, 453L, 454L, 455L, 456L, 457L, 458L, 934L, 935L,
936L, 937L, 938L, 939L, 940L, 941L, 2036L, 2037L, 2038L, 2039L,
2040L, 2041L, 2042L, 2043L, 24800L, 24801L, 24802L, 24803L, 24804L,
24805L, 24806L, 24807L, 68059L, 68060L, 68061L, 68062L, 68063L,
68064L, 68065L, 68066L, 171328L, 171329L, 171330L, 171331L, 171332L,
171333L, 171334L, 171335L, 171336L, 171337L, 171338L, 339673L,
339674L, 339675L, 339676L, 339677L, 339678L, 339679L, 339680L,
339681L, 339682L, 339683L), class = "data.frame")
提前致谢
我想到了以下内容（请注意，我没有使用你所有的审美，但应该足以让大家明白这一点，找到一种可能的解决方法）。
将上面的标签放在恒定高度
ggplot(data=subset(df, Prueba=='lectura'), aes(x=año, y=Promedio, group=Tipo, color=Tipo)) +
geom_point(shape=18, size=4, position=position_dodge(0.9)) +
geom_errorbar(aes(
ymin=Promedio-Desviacion, ymax=Promedio+Desviacion, width=0.4
), position=position_dodge(0.9), size=1.3, show.legend=F) +
ylim(60,140) +
geom_text(aes(
label = paste0(round(Promedio), "\n(", round(Desviacion), ")"), y=100),
vjust=-4.2, position=position_dodge(0.9)
)
给你这个：
我找不到将文本与每个栏的右侧或左侧对齐的好方法。由于您正在对数据集使用 position_dodge，因此您无法使用 nudge_x 来平均移动所有内容 "to the right"。 It's not allowed to use both in the geom。玩 hjust 种作品，但你可以期待有趣的事情发生在这里与你已经体验过的两条线。
另请注意，我已将您的两个 geom_text 调用合并为一个，您可以在其中使用 paste0() 或 paste() 进行组合并即时使用 "make your own label" .
vjust 全面使用允许您将它们全部放在相同的高度 - 我认为这就是您所指的，对吗？
同样 - 颜色与您的不同，因为我没有包括您的 scale_color_manual 和 theme 电话。
将标签放在一边，轻推 + Position_dodge
如果你想把标签放在一边，你不能用 nudge_x 做到这一点，因为你无法在 [=16= 中组合 nudge_x 和 position 参数].解决方法是基于 df$año 创建一个用于标签美学的新变量，然后在 geom_text:
aes(... 调用中使用该变量
df$año_new <- as.numeric(df$año) + 0.1 # nudge of "0.1". Had to force numeric, since + does not work on factors
然后你保留原始 plot 调用中的所有内容，但只更改最后一个 geom_text 调用（注意 hjust=0 的重要添加，它将所有内容左对齐）：
geom_text(aes(
x=año_new, y=100,
label = paste0(round(Promedio), "\n(", round(Desviacion), ")")
),
hjust=0,
position=position_dodge(0.9)
)