Question

我想这会有一个非常简单的答案。但是这里有。

长格式的数据。像这样

d <- data.frame(cbind(numbers = rnorm(10), 
                         year = rep(c(2008, 2009), 5), 
                         name = c("john", "David", "Tom", "Kristin", "Lisa","Eve","David","Tom","Kristin","Lisa")))

如何仅使用2008年和2009年出现的名称行获取新数据框？（即只有大卫，克里斯汀，丽莎和汤姆）。

提前致谢

Answer 1

简单方法：

subset(
    d,
    name %in% intersect(name[year==2008], name[year==2009])
)

Answer 2

一种方法是使用reshape包创建一个data.frame，其中包含列数和行数的年份：

library(reshape)
cast(d, name ~ year, value = "numbers")

然后，您可以使用complete.cases来提取感兴趣的行。

Answer 3

如果每年只有一条记录，只需计算每个人在数据集中出现的次数：

counts <- as.data.frame(table(name = d$name))

然后寻找两次出现的人：

subset(counts, Freq == 2)

Answer 4

这是另一个仅使用基数R的解决方案，并且不对每个人每年记录的数量做出任何假设：

d <- data.frame(cbind(numbers = rnorm(10), 
                      year = rep(c(2008, 2009), 5),
                      name = c("john", "David", "Tom", "Kristin",
                               "Lisa","Eve","David","Tom","Kristin",
                               "Lisa")))
# split data into 2 data.frames (1 for each year)
by.year <- split(d, d$year, drop=T)

# find the names that appear in both years
keep <- intersect(by.year[['2008']]$name, by.year[['2009']]$name)
# Or, if you had several years, use Reduce as a more general solution:
keep <- Reduce(intersect, lapply(by.year, '[[', 'name'))

# show the rows of the original dataset only if their $name field
# is in our 'keep' vector
d[d$name %in% keep,]

data.frame子集长格式

4 个答案: