I have a list of dataframes that contain a subset of columns. Excluding the identifier columns (id and n), all possible column names are included in a character vector called colors. I want to make so that if a dataframe, in this list of dataframes, is missing a column from colors, I want to put those missing columns into that dataframe and make those originally-missing columns NA.
This SO post demonstrates how to do this one column at a time, but considering the size of my real dataframe, I ideally would like to do it for all columns at once. I have included a reprex below with a working function that only does one color at a time and a non-working function that contains a loop.
set.seed(123)
library(tidyverse)
n <- c(1:10)
red <- sample(1:10, 10, replace = TRUE)
blue <- sample(1:10, 10, replace = TRUE)
yellow <- sample(1:10, 10, replace = TRUE)
green <- sample(1:10, 10, replace = TRUE)
pink <- sample(1:10, 10, replace = TRUE)
df1 <- data.frame(id = "1", n, red, blue, green)
df2 <- data.frame(id = "2", n, red, blue, green, pink)
df3 <- data.frame(id = "3", n, blue, yellow, green, pink)
df4 <- data.frame(id = "4", n, blue, yellow, pink)
df5 <- data.frame(id = "5", n, green)
lst <- list(df1, df2, df3, df4, df5)
# this works but only does red
red_func <- function(x) {
if (!'red' %in% names(x)) x <- x %>% add_column(red = NA)
else x
}
# failed attempted at loop
color_func <- function(z) {
colors <- c("red", "blue", "yellow", "green", "pink")
for (x in colors) {
if (!x %in% names(z)) z <- z %>% add_column(x = NA)
else z
}
}
test <- lapply(lst, color_func)
print(test[[1]])
#> NULL
CodePudding user response：
We could use := with !!
color_func <- function(z) {
colors <- c("red", "blue", "yellow", "green", "pink")
for (x in colors) {
if (!x %in% names(z)) z <- z %>% add_column(!!x := NA)
else z
}
return(z)
}
-testing
> lapply(lst, color_func)
[[1]]
id n red blue green yellow pink
1 1 1 3 5 10 NA NA
2 1 2 3 3 7 NA NA
3 1 3 10 9 5 NA NA
4 1 4 2 9 7 NA NA
5 1 5 6 9 5 NA NA
6 1 6 5 3 6 NA NA
7 1 7 4 8 9 NA NA
8 1 8 6 10 2 NA NA
9 1 9 9 7 5 NA NA
10 1 10 10 10 8 NA NA
[[2]]
id n red blue green pink yellow
1 2 1 3 5 10 2 NA
2 2 2 3 3 7 1 NA
3 2 3 10 9 5 9 NA
4 2 4 2 9 7 9 NA
5 2 5 6 9 5 6 NA
6 2 6 5 3 6 5 NA
7 2 7 4 8 9 9 NA
8 2 8 6 10 2 10 NA
9 2 9 9 7 5 4 NA
10 2 10 10 10 8 6 NA
[[3]]
id n blue yellow green pink red
1 3 1 5 9 10 2 NA
2 3 2 3 3 7 1 NA
3 3 3 9 4 5 9 NA
4 3 4 9 1 7 9 NA
5 3 5 9 7 5 6 NA
6 3 6 3 5 6 5 NA
7 3 7 8 10 9 9 NA
8 3 8 10 7 2 10 NA
9 3 9 7 9 5 4 NA
10 3 10 10 9 8 6 NA
[[4]]
id n blue yellow pink red green
1 4 1 5 9 2 NA NA
2 4 2 3 3 1 NA NA
3 4 3 9 4 9 NA NA
4 4 4 9 1 9 NA NA
5 4 5 9 7 6 NA NA
6 4 6 3 5 5 NA NA
7 4 7 8 10 9 NA NA
8 4 8 10 7 10 NA NA
9 4 9 7 9 4 NA NA
10 4 10 10 9 6 NA NA
[[5]]
id n green red blue yellow pink
1 4 1 10 NA NA NA NA
2 4 2 7 NA NA NA NA
3 4 3 5 NA NA NA NA
4 4 4 7 NA NA NA NA
5 4 5 5 NA NA NA NA
6 4 6 6 NA NA NA NA
7 4 7 9 NA NA NA NA
8 4 8 2 NA NA NA NA
9 4 9 5 NA NA NA NA
10 4 10 8 NA NA NA NA
It can be done more easily, i.e. without a loop
colors <- c("red", "blue", "yellow", "green", "pink")
lapply(lst, function(x) {x[setdiff(colors, names(x))] <- NA; x})
CodePudding user response：
You can use data.table::rbindlist with split()
library(data.table)
split(rbindlist(lst,fill=TRUE), by="id"))
or dplyr::bind_rows() with group_split()
library(dplyr)
group_split(bind_rows(lst), id)
Output:
$`1`
id n red blue green pink yellow
1: 1 1 10 4 8 NA NA
2: 1 2 5 5 1 NA NA
3: 1 3 5 7 5 NA NA
4: 1 4 7 10 2 NA NA
5: 1 5 2 5 1 NA NA
6: 1 6 5 3 7 NA NA
7: 1 7 8 6 2 NA NA
8: 1 8 3 1 5 NA NA
9: 1 9 4 10 4 NA NA
10: 1 10 1 10 5 NA NA
$`2`
id n red blue green pink yellow
1: 2 1 10 4 8 6 NA
2: 2 2 5 5 1 10 NA
3: 2 3 5 7 5 4 NA
4: 2 4 7 10 2 2 NA
5: 2 5 2 5 1 7 NA
6: 2 6 5 3 7 8 NA
7: 2 7 8 6 2 7 NA
8: 2 8 3 1 5 7 NA
9: 2 9 4 10 4 9 NA
10: 2 10 1 10 5 8 NA
$`3`
id n red blue green pink yellow
1: 3 1 NA 4 8 6 4
2: 3 2 NA 5 1 10 6
3: 3 3 NA 7 5 4 10
4: 3 4 NA 10 2 2 1
5: 3 5 NA 5 1 7 1
6: 3 6 NA 3 7 8 7
7: 3 7 NA 6 2 7 6
8: 3 8 NA 1 5 7 10
9: 3 9 NA 10 4 9 3
10: 3 10 NA 10 5 8 2
$`4`
id n red blue green pink yellow
1: 4 1 NA 4 NA 6 4
2: 4 2 NA 5 NA 10 6
3: 4 3 NA 7 NA 4 10
4: 4 4 NA 10 NA 2 1
5: 4 5 NA 5 NA 7 1
6: 4 6 NA 3 NA 8 7
7: 4 7 NA 6 NA 7 6
8: 4 8 NA 1 NA 7 10
9: 4 9 NA 10 NA 9 3
10: 4 10 NA 10 NA 8 2
$`5`
id n red blue green pink yellow
1: 5 1 NA NA 8 NA NA
2: 5 2 NA NA 1 NA NA
3: 5 3 NA NA 5 NA NA
4: 5 4 NA NA 2 NA NA
5: 5 5 NA NA 1 NA NA
6: 5 6 NA NA 7 NA NA
7: 5 7 NA NA 2 NA NA
8: 5 8 NA NA 5 NA NA
9: 5 9 NA NA 4 NA NA
10: 5 10 NA NA 5 NA NA
CodePudding user response：
What about completely base R approach?:
First, let's create a function that does the job for a single dataframe:
colorize_columns = function(df) {
cols = c("red", "blue", "yellow", "green", "pink")
cols_df = names(df)[-c(1, 2)] # ignore the id and n columns
if (all(cols %in% cols_df)) { # do nothing if the color names match
return(df)
} else {
cols_unmatched = cols[!(cols %in% cols_df)] # colors that do not match
for (col in cols_unmatched) {
df[[col]] = NA
}
return(df)
}
}
Then, we can do:
lapply(lst, colorize_columns)
Output:
[[1]]
id n red blue green yellow pink
1 1 1 3 5 10 NA NA
2 1 2 3 3 7 NA NA
3 1 3 10 9 5 NA NA
4 1 4 2 9 7 NA NA
5 1 5 6 9 5 NA NA
6 1 6 5 3 6 NA NA
7 1 7 4 8 9 NA NA
8 1 8 6 10 2 NA NA
9 1 9 9 7 5 NA NA
10 1 10 10 10 8 NA NA
[[2]]
id n red blue green pink yellow
1 2 1 3 5 10 2 NA
2 2 2 3 3 7 1 NA
3 2 3 10 9 5 9 NA
4 2 4 2 9 7 9 NA
5 2 5 6 9 5 6 NA
6 2 6 5 3 6 5 NA
7 2 7 4 8 9 9 NA
8 2 8 6 10 2 10 NA
9 2 9 9 7 5 4 NA
10 2 10 10 10 8 6 NA
[[3]]
id n blue yellow green pink red
1 3 1 5 9 10 2 NA
2 3 2 3 3 7 1 NA
3 3 3 9 4 5 9 NA
4 3 4 9 1 7 9 NA
5 3 5 9 7 5 6 NA
6 3 6 3 5 6 5 NA
7 3 7 8 10 9 9 NA
8 3 8 10 7 2 10 NA
9 3 9 7 9 5 4 NA
10 3 10 10 9 8 6 NA
[[4]]
id n blue yellow pink red green
1 4 1 5 9 2 NA NA
2 4 2 3 3 1 NA NA
3 4 3 9 4 9 NA NA
4 4 4 9 1 9 NA NA
5 4 5 9 7 6 NA NA
6 4 6 3 5 5 NA NA
7 4 7 8 10 9 NA NA
8 4 8 10 7 10 NA NA
9 4 9 7 9 4 NA NA
10 4 10 10 9 6 NA NA
[[5]]
id n green red blue yellow pink
1 5 1 10 NA NA NA NA
2 5 2 7 NA NA NA NA
3 5 3 5 NA NA NA NA
4 5 4 7 NA NA NA NA
5 5 5 5 NA NA NA NA
6 5 6 6 NA NA NA NA
7 5 7 9 NA NA NA NA
8 5 8 2 NA NA NA NA
9 5 9 5 NA NA NA NA
10 5 10 8 NA NA NA NA