Question

我有一个table(1)看起来像这样（它是所有距离matrix transformed到标签分隔列表中）：

sample1    sample2    405
sample3    sample4    400
sample5    sample6    1
sample7    sample8    20
sample1    sample3    40

我有另一个table(2)，其中包含符合特定条件的样本：

sample1
sample2
sample8

如何解析第一个table(1)以仅提取1中可以找到2和table(2)列中的变量的那些行？

即所需的比较仅为：

sample1    sample2    405
sample2    sample8    40
sample8    sample1    100

Answer 1

我尝试使用表格（1）的数据框和表格（2）的矢量进行类似的设置。

table_one <- data.frame(col_1 = c("a", "b", "c", "d"),
       col_2 = c("b", "d", "f", "g"),
       col_3 = c(1, 2, 3, 4))
table_two <- c("b", "d")

当你以这种方式设置时，这样的事情应该有效：

library(tidyverse)
table_one %>% filter(col_1 %in% table_two,
                     col_2 %in% table_two)

Answer 2

以下是基础R解决方案：

final = merged.pivot(index='PERMNO', columns='FROMDATE', values='MORET').reset_index()
# final: 
FROMDATE  PERMNO  20131010  20131231
0          79702       NaN  0.012283
1          85576       NaN  0.038766
2          85751     -0.01       NaN
3          93044     -0.02       NaN

...和输出：

rawData1 <- "first second distance
 sample1    sample2    405
 sample3    sample4    400
 sample5    sample6    1
 sample7    sample8    20
 sample1    sample3    40"

rawData2 <- "sample
 sample1
 sample2
 sample8"

a <- read.table(textConnection(rawData1),stringsAsFactors=FALSE,header=TRUE)
b <- read.table(textConnection(rawData2),stringsAsFactors=FALSE,header=TRUE)

a[a$first %in% b$sample & a$second %in% b$sample, ]

Answer 3

最佳选项可能是inner_join两次，包括第一列和第二列，然后执行两个结果集的intersect。

library(dplyr)

df1 <- read.table(text = "Samp1 Samp2  Val
sample1    sample2    405
sample3    sample4    400
sample5    sample6    1
sample7    sample8    20
sample1    sample3    40", header = TRUE, stringsAsFactors = FALSE)
> df1
    Samp1   Samp2 Val
1 sample1 sample2 405
2 sample3 sample4 400
3 sample5 sample6   1
4 sample7 sample8  20
5 sample1 sample3  40

df2 <- data.frame(Samp = c("sample1",
                           "sample2",
                           "sample8"), stringsAsFactors = FALSE)
> df2
     Samp
1 sample1
2 sample2
3 sample8

#use inner_join between Samp1 with Samp and then again Samp2 with Samp
intersect(inner_join(df1,df2, by = c("Samp1" = "Samp")),
      inner_join(df1,df2, by = c("Samp2" = "Samp")))

The result will be:
    Samp1   Samp2 Val
1 sample1 sample2 405

根据R中的另一个表过滤一个表

3 个答案: