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如何在循环filter()函数中使用colnames()后删除双引号

dplyr filter dataframe r

0

Sung Joo Lee · 技术社区 · 1 周前

我想用 filter 函数从符合条件的数据帧中提取行。

我的数据帧如下所示:

head(EIA)
  length location  stype  rock  dist  ftype  depth aspect  degree  drain  TWI
1   len1     loc3 stype2 rock3 dist3 ftype1 depth2   asp2 degree2 drain4 TWI1
2   len4     loc2 stype2 rock3 dist1 ftype4 depth3   asp4 degree3 drain4 TWI3
3   len2     loc2 stype2 rock2 dist1 ftype4 depth3   asp1 degree2 drain4 TWI2
4   len4     loc2 stype2 rock3 dist2 ftype4 depth3   asp4 degree2 drain4 TWI2
5   len4     loc2 stype1 rock3 dist1 ftype2 depth3   asp1 degree3 drain4 TWI2
6   len4     loc3 stype2 rock2 dist1 ftype2 depth3   asp4 degree3 drain1 TWI2

等等。总行数:10560

但是,由于我试图在for循环中完成这项工作,所以filter函数中的每个元素都必须与变量(colcolcol、a a、bb、cc)一起传递:

 colcol
    [1] "length"   "location" "stype"    
 aa;bb;cc
    [1] "len1"
    [1] "loc2"
    [1] "stype1"

我试着这样做过滤功能,但没有成功:

filter(EIA, colcol[1] == aa & colcol[2] == bb & colcol[3] == cc)

结果必须提取两行。我想这是因为colcol[1]元素周围有双引号。我试图用 as.symbol , as.name ,和 noquote 但功能不起作用。

这个问题有什么解决办法吗?

提前谢谢你。

1 回复 | 直到 1 周前

1

0

Ronak Shah 1 周前

将其转换为符号,然后计算

library(dplyr)
filter(EIA,!!sym(colcol[1]) == aa & !!sym(colcol[2]) == bb & !!sym(colcol[3]) == cc)

例如,当我们评估第一个条件时,我们得到

filter(EIA, !!sym(colcol[1]) == aa)
#  length location  stype  rock  dist  ftype  depth aspect  degree  drain  TWI
#1   len1     loc3 stype2 rock3 dist3 ftype1 depth2   asp2 degree2 drain4 TWI1

数据

EIA <- structure(list(length = structure(c(1L, 3L, 2L, 3L, 3L, 3L), 
.Label = c("len1", "len2", "len4"), class = "factor"), location = structure(c(2L, 
1L, 1L, 1L, 1L, 2L), .Label = c("loc2", "loc3"), class = "factor"), 
stype = structure(c(2L, 2L, 2L, 2L, 1L, 2L), .Label = c("stype1", 
"stype2"), class = "factor"), rock = structure(c(2L, 2L, 
1L, 2L, 2L, 1L), .Label = c("rock2", "rock3"), class = "factor"), 
dist = structure(c(3L, 1L, 1L, 2L, 1L, 1L), .Label = c("dist1", 
"dist2", "dist3"), class = "factor"), ftype = structure(c(1L, 
3L, 3L, 3L, 2L, 2L), .Label = c("ftype1", "ftype2", "ftype4"
), class = "factor"), depth = structure(c(1L, 2L, 2L, 2L, 
2L, 2L), .Label = c("depth2", "depth3"), class = "factor"), 
aspect = structure(c(2L, 3L, 1L, 3L, 1L, 3L), .Label = c("asp1", 
"asp2", "asp4"), class = "factor"), degree = structure(c(1L, 
2L, 1L, 1L, 2L, 2L), .Label = c("degree2", "degree3"), class = "factor"), 
drain = structure(c(2L, 2L, 2L, 2L, 2L, 1L), .Label = c("drain1", 
"drain4"), class = "factor"), TWI = structure(c(1L, 3L, 2L, 
2L, 2L, 2L), .Label = c("TWI1", "TWI2", "TWI3"), class = "factor")), 
class = "data.frame", row.names = c("1", "2", "3", "4", "5", "6"))

colcol <- c("length",   "location", "dist")
aa <- "len1"
bb <- "loc2"
cc <- "stype1"

2

1

Tushar Lad 1 周前

子集是此类问题的最佳解

 aa <- subset(EIA,EIA$lenght=='len1')

上面的例子你可能会喜欢,简单而聪明的解决方案。也可以使用and(&)条件提取行。

 aa <- subset(EIA,EIA$lenght=='len1' & EIA$location=='loc2')