我有一个数据集,我试图通过计算一个类别来仅选择前n个,然后使用数据集中的其他变量进行绘图-基本是前n个聚集的一个级别,但需要返回到全部数据绘制在ggplot
中。
因此在下面的问题中,我想要两个最常见的examName
,然后根据facetwrap
的数量绘制和year
。
ap <-
tribble(
~year, ~examName,
2014, "Statistics",
2015, "Statistics",
2016, "Statistics",
2016, "Statistics",
2016, "Statistics",
2016, "Statistics",
2017, "Statistics",
2017, "Statistics",
2017, "Statistics",
2017, "Statistics",
2017, "Statistics",
2013, "Macroeconomics",
2013, "Macroeconomics",
2014, "Macroeconomics",
2015, "Macroeconomics",
2016, "Macroeconomics",
2016, "Macroeconomics",
2016, "Macroeconomics",
2016, "Macroeconomics",
2016, "Macroeconomics",
2017, "Macroeconomics",
2017, "Macroeconomics",
2017, "Macroeconomics",
2017, "Macroeconomics",
2017, "Macroeconomics",
2017, "Macroeconomics",
2013, "Calculus",
2014, "Calculus",
2015, "Calculus",
2016, "Calculus",
2017, "Calculus",
2017, "Psychology",
2017, "Psychology",
2017, "Psychology",
2017, "Psychology",
2017, "Psychology",
2018, "Psychology",
2018, "Psychology")
ap_top <- ap %>%
count(examName, sort = TRUE) %>%
head(2) %>%
inner_join(ap, by = "examName") %>%
select(-n)
ap_top %>%
count(examName, year) %>%
ggplot(aes(x = year, y = n, group = examName)) +
geom_line() +
facet_wrap(~ examName)
我的想法是让我的前n名,然后inner_join
回到原始数据集中。然后使用它进行绘图;本质上是使用内部联接作为过滤器。
我知道有更好的方法可以做到这一点,我希望有一个更优雅的解决方案!我全是耳朵!给出的示例数据集(很长很抱歉)。
答案 0 :(得分:5)
您不需要inner_join()
,我只需要在单独的语句中确定前两项考试,然后对这些考试进行过滤即可。
top_exams <- count(ap, examName) %>%
top_n(2, n) %>% pull(examName)
ap %>%
filter(examName %in% top_exams) %>%
count(year, examName) %>%
ggplot(aes(x = year, y = n, group = examName)) +
geom_line() +
facet_wrap(~ examName)
答案 1 :(得分:2)
另一种可能性:
ap %>%
group_by(examName) %>%
mutate(temp = n()) %>%
ungroup() %>%
mutate(temp = dense_rank(desc(temp))) %>%
filter(temp %in% c(1,2)) %>%
select(-temp) %>%
count(year, examName) %>%
ggplot(aes(x = year, y = n, group = examName)) +
geom_line() +
facet_wrap(~ examName)
它根据“ examName”对个案进行计数,并对计数进行排名。然后,它过滤具有最大和第二大计数的案例。