添加列(如果不存在) [英] Adding column if it does not exist
问题描述
我有一堆具有不同变量的数据框。我想将它们读入R并向缺少几个变量的列中添加列,以使它们都具有一组通用的标准变量,即使其中一些未被观察到也是如此。
I have a bunch of data frames with different variables. I want to read them into R and add columns to those that are short of a few variables so that they all have a common set of standard variables, even if some are unobserved.
换句话说...当不存在列时,是否可以在tidyverse中添加 NA
列?我当前的尝试是在不存在该列的地方添加新变量( top_speed
),但是在该列已经存在时失败( mpg
)(将所有观测值设置为第一个值,马自达RX4
)。
In other words... Is there a way to add columns of NA
in the tidyverse when a column does not exist? My current attempt works for adding new variables where the column doesn't exist (top_speed
) but fails when the column already exists (mpg
) (it sets all observations to the first value, Mazda RX4
).
library(tidyverse)
mtcars %>%
tbl_df() %>%
rownames_to_column("car") %>%
mutate(top_speed = ifelse("top_speed" %in% names(.), top_speed, NA),
mpg = ifelse("mpg" %in% names(.), mpg, NA)) %>%
select(car, top_speed, mpg, everything())
# # A tibble: 32 x 13
# car top_speed mpg cyl disp hp drat wt qsec vs am gear carb
# <chr> <lgl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 Mazda RX4 NA 21 6 160.0 110 3.90 2.620 16.46 0 1 4 4
# 2 Mazda RX4 Wag NA 21 6 160.0 110 3.90 2.875 17.02 0 1 4 4
# 3 Datsun 710 NA 21 4 108.0 93 3.85 2.320 18.61 1 1 4 1
# 4 Hornet 4 Drive NA 21 6 258.0 110 3.08 3.215 19.44 1 0 3 1
# 5 Hornet Sportabout NA 21 8 360.0 175 3.15 3.440 17.02 0 0 3 2
# 6 Valiant NA 21 6 225.0 105 2.76 3.460 20.22 1 0 3 1
# 7 Duster 360 NA 21 8 360.0 245 3.21 3.570 15.84 0 0 3 4
# 8 Merc 240D NA 21 4 146.7 62 3.69 3.190 20.00 1 0 4 2
# 9 Merc 230 NA 21 4 140.8 95 3.92 3.150 22.90 1 0 4 2
# 10 Merc 280 NA 21 6 167.6 123 3.92 3.440 18.30 1 0 4 4
推荐答案
不需要使用tibble的 add_column
创建辅助函数(或已经完整的data.frame):
Another option that does not require creating a helper function (or an already complete data.frame) using tibble's add_column
:
library(tibble)
cols <- c(top_speed = NA_real_, nhj = NA_real_, mpg = NA_real_)
add_column(mtcars, !!!cols[setdiff(names(cols), names(mtcars))])
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