Question

这里是一些示例数据

sample = data.frame("col1" = c("val1", "val1", "val1", "val1", "val2", "val2", "val2", "val3", "val3", "val3", "val3"),
                    "col2" = c("this", "that", "some", "thing", "thing", "that", "some", "diff", "some", "this", "that"))

我想确定列col2的每个条目，它出现在列col1的每个唯一值中。这有点可能吗？这将是样本数据的结果：

result = c("that", "some")

提前致谢。

Answer 1

base R中的（快速且肮脏）解决方案：

sample_list <- split(sample, sample$col1)
for (i in 1:length(sample_list)) sample_list[[i]] <- sample_list[[i]]$col2
Reduce(intersect, sample_list)
[1] "that" "some"

编辑：

data.table解决方案受到马特dplyr回答的启发：

library(data.table)
setDT(sample)
n <- uniqueN(sample$col1)
sample[, .N, by = .(col1, col2)][, .N, by = col2][N == n, col2]
[1] that some

此解决方案将在大型数据集上快速完成。

编辑2：

使用dcast中提供的data.table：

present_in <- colSums(!is.na(dcast(sample, col1 ~ col2, value.var = "col2")))
names(present_in)[present_in == 3][-1]
[1] "some" "that"

Answer 2

这里有一点关于使用dplyr的方式。

require(dplyr)

sets <- length(unique(sample$col1))

s <- sample %>%
    group_by(col2) %>%
    summarise(n = n_distinct()) %>%
    filter(n == sets)

result <- s$col2
[1] some that

Answer 3

这是使用dplyr执行此操作的一种方法：

split(sample,sample$col1)%>%
Reduce(function(dtf1,dtf2) inner_join(dtf1,dtf2,by="col2"), .)%>%select(col2)%>%print()

  col2
1 that
2 some

Answer 4

您需要的是intersect。这是一种快速而又肮脏的方式：

<强> CODE

library(data.table)
dt <- as.data.table(sample) 

# Split data.table into different chunks based on unique values in col1
# output is a list where each entry is a data.table 
l <- split(dt, by = "col1")

# Find the intersection of all values in col2 
Reduce(intersect, lapply(1:length(l), function(z) as.character(l[[z]]$col2)))

<强>输出

> Reduce(intersect, lapply(1:length(l), function(z) as.character(l[[z]]$col2)))
[1] "that" "some"

Answer 5

另一个脏基R解决方案：

names(which(table(unlist(aggregate(sample$col2, list(sample$col1), unique)[, 2])) == length(unique(sample$col1))))

[1] "some" "that"

R在列中找到另一列的唯一值的相同值

5 个答案: