Question

我有一个像这样的data.frame：

20021  K08975 K09735 0.929
20022  K08979 K09735 0.934
20023  K09140 K09735 0.901
20024  K09142 K09735 0.938
20025  K09152 K09735 0.947
20026  K09482 K09735 0.919
20027  K09716 K09735 0.944
20028  K09723 K09735 0.949
20029  K09726 K09735 0.915
20030  K06875 K09736 0.905
20031  K09149 K09736 0.901
20032  K09721 K09736 0.903
20033 OTU0001 K09738 0.908
20034 OTU0095 K09738 0.906
20035  K00952 K09738 0.904
20036  K01622 K09738 0.907
20037  K06875 K09738 0.912
20038  K06963 K09738 0.923
20039  K07060 K09738 0.934

共有三列：var1，var2和corr

var1和var2可以使用值“ KOXXXX”或“ OTUXXXX”。

我想保留var1和var2不同的行，我的意思是仅显示KOXXXX OTUXXXX或OTUXXXX KOXXXX的行

Answer 1

也许这很幼稚，但可以帮助您

# here you take only the rows where the first two character of var1 and var2
# are different
df[substr(df$var1,1,2) != substr(df$var2,1,2),]

         var1   var2  corr
20033 OTU0001 K09738 0.908
20034 OTU0095 K09738 0.906

Answer 2

大概是

subset(df, grepl("^K0", var1) & grepl("^OTU", var2) | 
           grepl("^OTU", var1) & grepl("^K0", var2))

#         var1   var2  corr
#20033 OTU0001 K09738 0.908
#20034 OTU0095 K09738 0.906

或使用startsWith

subset(df, startsWith(var1, "K0") & startsWith(var2, "OTU") | 
           startsWith(var1, "OTU") & startsWith(var2, "K0"))

或者使用dplyr，我们可以将grepl / str_detect与filter一起使用

library(dplyr)
library(stringr)

df %>%
  filter(str_detect(var1, "^K0") & str_detect(var2, "^OTU") | 
         str_detect(var1, "^OTU") & str_detect(var2, "^K0"))

数据

df <- structure(list(var1 = c("K08975", "K08979", "K09140", "K09142", 
"K09152", "K09482", "K09716", "K09723", "K09726", "K06875", "K09149", 
"K09721", "OTU0001", "OTU0095", "K00952", "K01622", "K06875", 
"K06963", "K07060"), var2 = c("K09735", "K09735", "K09735", "K09735", 
"K09735", "K09735", "K09735", "K09735", "K09735", "K09736", "K09736", 
"K09736", "K09738", "K09738", "K09738", "K09738", "K09738", "K09738", 
"K09738"), corr = c(0.929, 0.934, 0.901, 0.938, 0.947, 0.919, 
0.944, 0.949, 0.915, 0.905, 0.901, 0.903, 0.908, 0.906, 0.904, 
0.907, 0.912, 0.923, 0.934)), row.names = 20021:20039, class = 
"data.frame")

Answer 3

我们也可以在base R中这样做

df[Reduce(`!=`, lapply(df[1:2], substr, 1, 2)),]
#          var1   var2  corr
#20033 OTU0001 K09738 0.908
#20034 OTU0095 K09738 0.906

如何删除data.frame中的某些行？

3 个答案: