我有疾病和日子的数据。我必须为每种疾病及其感染天数绘制一个地势图。我使用ggplot2进行了尝试。但是,它结合了我不想要的相同疾病的天数。我有兴趣每天绘制每列而不管疾病类型如何。 l使用了以下代码。
original_datafile <-
structure(list(disease = structure(c(1L, 2L,
3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 6L, 6L, 6L),
.Label = c("AA", "BB", "CC", "DD", "EE", "FF"),
class = "factor"), days = c(5L, 5L, 9L, 2L,
3L, 4L, 4L, 5L, 7L, 15L, 3L, 7L, 7L, 15L)),
class = "data.frame", row.names = c(NA, -14L))
library(ggplot2)
ggplot(data = original_datafile, aes(x = disease, y = days)) +
geom_bar(stat = "identity") +
theme(axis.text.x = element_text(angle = 40, hjust = 1))
任何建议将不胜感激。
答案 0 :(得分:0)
这是我制定的几个解决方案。不能100%确定这是否是您要追求的目标,但希望这可以使您与您保持联系。为了为每个单独的行创建一个条,而不是将它们组合在一起,我创建了一个名为id
的新列,该列仅用作每种疾病的每一行的计数器。然后,我加入了两种可能的ggplot组合,我相信它们会接近您的需求。
original_datafile <-
structure(list(disease = structure(c(1L, 2L,
3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 6L, 6L, 6L),
.Label = c("AA", "BB", "CC", "DD", "EE", "FF"),
class = "factor"), days = c(5L, 5L, 9L, 2L,
3L, 4L, 4L, 5L, 7L, 15L, 3L, 7L, 7L, 15L)),
class = "data.frame", row.names = c(NA, -14L))
library(ggplot2)
# Modified data file adds an 'id' column to split each row individually.
modified_datafile <- original_datafile %>%
group_by(disease) %>%
mutate(id = row_number())
# Facetted ggplot - each disease has its own block
ggplot(data = modified_datafile, aes(x = id, y = days)) +
geom_bar(stat = 'identity', position = 'dodge') +
theme(axis.text.x = element_text(angle = 40, hjust = 1)) +
facet_wrap(. ~ disease, nrow = 2) +
theme(axis.text.x = element_blank()) +
labs(x = '', y = 'Days')
# Non facetted ggplot - closer to original, but each row has a bar.
ggplot(data = modified_datafile, aes(x = disease, y = days, group = id)) +
geom_bar(stat = 'identity', position = position_dodge2(preserve = 'single')) +
theme(axis.text.x = element_text(angle = 40, hjust = 1))