In R: subset or dplyr::filter with variable from vector

问题

df <- 
  data.frame(a=LETTERS[1:4],
             b=rnorm(4)
             )

vals <- c("B","D")

I can filter/subset df with values in val with:

dplyr::filter(df, a %in% vals)
subset(df, a %in% vals)

Both gives:

  a         b
2 B 0.4481627
4 D 0.2916513

What if I have a variable name in a vector, e.g.:

> names(df)[1]
[1] "a"

Then it doesnt work - I guess because its quoted

dplyr::filter(df, names(df)[1] %in% vals)
[1] a b
<0 rows> (or 0-length row.names)

How do you do this ?

UPDATE ( what if its dplyr::tbl_df(df) )

Answers below work fine for data.frames, but not for dplyr::tbl_df wrapped data:

df<-dplyr::tbl_df(df)
dplyr::filter(df, df[,names(df)[1]] %in% vals)

Does not work (I thought tbl_df was a simple wrap on top of df ? )

This does work again:

dplyr::filter(df, as.data.frame(df)[,names(df)[1]] %in% vals)

FINAL UPDATE: It works with tbl_df() using lazyeval::interp

See AndreyAkinshin's solution below.

回答1:

You can use df[,"a"] or df[,1]:

df <- data.frame(a = LETTERS[1:4], b = rnorm(4))
vals <- c("B","D")

dplyr::filter(df, df[,1] %in% vals)
#  a         b
# 2 B 0.4481627
# 4 D 0.2916513

subset(df, df[,1] %in% vals)
#  a         b
# 2 B 0.4481627
# 4 D 0.2916513

dplyr::filter(df, df[,"a"] %in% vals)
#  a         b
# 2 B 0.4481627
# 4 D 0.2916513

subset(df, df[,"a"] %in% vals)
#  a         b
# 2 B 0.4481627
# 4 D 0.2916513

Working with dplyr::tbl_df(df)

Some magic with lazyeval::interp helps us!

df <- dplyr::tbl_df(df)
expr <- lazyeval::interp(quote(x %in% y), x = as.name(names(df)[1]), y = vals)

df %>% filter_(expr)
# Source: local data frame [2 x 2]
#
#   a        b
# 1 B 0.4481627
# 2 D 0.2916513

回答2:

A simple way to solve this problem in the tidyverse:

library(tidyverse)
df <- data.frame(a = LETTERS[1:4], b = rnorm(4))
vals <- c("B","D")
df %>% filter(!!sym(names(.)[1]) %in% vals)

来源：https://stackoverflow.com/questions/31358953/in-r-subset-or-dplyrfilter-with-variable-from-vector

标签

subset

dplyr