convert a character column into numeric in R - r

I wonder why my conversion of the "t5" column was not successful--
The "t5" column is all characters, I want to convert it into a numeric column, leave non-numeric value as NA, named as "t5.num" in the tibble.
My code below:
first of all I assigned the name, then trying to mutate the column, but it did not work--
d <- tibble(id = c(3, 7, 1, 10,100), t5 = c("10", "<1", "NA", "8","78"))
convert_column <- function(data, col_name) {
new_col_name <- paste0(rlang::enquo(col_name),".num")
data %>%
mutate(new_col_name = as.numeric(!!col_name))
}
d %>% convert_column("t5")
Can someone point out what is wrong with my code? thanks for your help!

To get new_col_name you don't need enquo. To assign new_col_name as name of the column use !! + :=. As you are passing col_name as a string we need to convert it to symbol (sym) and then evaluate (!!).
library(dplyr)
library(rlang)
convert_column <- function(data, col_name) {
new_col_name <- paste0(col_name,".num")
data %>% mutate(!!new_col_name := as.numeric(!!sym(col_name)))
}
d %>% convert_column("t5")
# A tibble: 5 x 3
# id t5 t5.num
# <dbl> <chr> <dbl>
#1 3 10 10
#2 7 <1 NA
#3 1 NA NA
#4 10 8 8
#5 100 78 78
Returns a warning while converting "<1" to numeric before turning it to NA.

Related

Extract first Non NA value over multiple columns

I'm still learning R and was wondering if I there was an elegant way of manipulating the below df to achieve df2.
I'm not sure if it's a loop that is supposed to be used for this, but basically I want to extract the first Non NA "X_No" Value if the "X_No" value is NA in the first row. This would perhaps be best described through an example from df to the desired df2.
A_ID <- c('A','B','I','N')
A_No <- c(11,NA,15,NA)
B_ID <- c('B','C','D','J')
B_No <- c(NA,NA,12,NA)
C_ID <- c('E','F','G','P')
C_No <- c(NA,13,14,20)
D_ID <- c('J','K','L','M')
D_No <- c(NA,NA,NA,40)
E_ID <- c('W','X','Y','Z')
E_No <- c(50,32,48,40)
df <- data.frame(A_ID,A_No,B_ID,B_No,C_ID,C_No,D_ID,D_No,E_ID,E_No)
ID <- c('A','D','F','M','W')
No <- c(11,12,13,40,50)
df2 <- data.frame(ID,No)
I'm hoping for an elegant solution to this as there are over a 1000 columns similar to the example provided.
I've looked all over the web for a similar example however to no avail that would reproduce the expected result.
Your help is very much appreciated.
Thankyou
I don't know if I'd call it "elegant", but here is a potential solution:
library(tidyverse)
A_ID <- c('A','B','I','N')
A_No <- c(11,NA,15,NA)
B_ID <- c('B','C','D','J')
B_No <- c(NA,NA,12,NA)
C_ID <- c('E','F','G','P')
C_No <- c(NA,13,14,20)
D_ID <- c('J','K','L','M')
D_No <- c(NA,NA,NA,40)
E_ID <- c('W','X','Y','Z')
E_No <- c(50,32,48,40)
df <- data.frame(A_ID,A_No,B_ID,B_No,C_ID,C_No,D_ID,D_No,E_ID,E_No)
ID <- c('A','D','F','M','W')
No <- c(11,12,13,40,50)
df2 <- data.frame(ID,No)
output <- df %>%
pivot_longer(everything(),
names_sep = "_",
names_to = c("Col", ".value")) %>%
drop_na() %>%
group_by(Col) %>%
slice_head(n = 1) %>%
ungroup() %>%
select(-Col)
df2
#> ID No
#> 1 A 11
#> 2 D 12
#> 3 F 13
#> 4 M 40
#> 5 W 50
output
#> # A tibble: 5 × 2
#> ID No
#> <chr> <dbl>
#> 1 A 11
#> 2 D 12
#> 3 F 13
#> 4 M 40
#> 5 W 50
all_equal(df2, output)
#> [1] TRUE
Created on 2023-02-08 with reprex v2.0.2
Using base R with max.col (assuming the columns are alternating with ID, No)
ind <- max.col(!is.na(t(df[c(FALSE, TRUE)])), "first")
m1 <- cbind(seq_along(ind), ind)
data.frame(ID = t(df[c(TRUE, FALSE)])[m1], No = t(df[c(FALSE, TRUE)])[m1])
ID No
1 A 11
2 D 12
3 F 13
4 M 40
5 W 50
Here is a data.table solution that should scale well to a (very) large dataset.
functionally
split the data.frame to a list of chunks of columns, based on their
names. So all columns startting with A_ go to
the first element, all colums startting with B_ to the second
Then, put these list elements on top of each other, using
data.table::rbindlist. Ignure the column-namaes (this only works if
A_ has the same number of columns as B_ has the same number of cols
as n_)
Now get the first non-NA value of each value in the first column
code
library(data.table)
# split based on what comes after the underscore
L <- split.default(df, f = gsub("(.*)_.*", "\\1", names(df)))
# bind together again
DT <- rbindlist(L, use.names = FALSE)
# extract the first value of the non-NA
DT[!is.na(A_No), .(No = A_No[1]), keyby = .(ID = A_ID)]
# ID No
# 1: A 11
# 2: D 12
# 3: F 13
# 4: G 14
# 5: I 15
# 6: M 40
# 7: P 20
# 8: W 50
# 9: X 32
#10: Y 48
#11: Z 40

Dplyr: add multiple columns with mutate/across from character vector

I want to add several columns (filled with NA) to a data.frame using dplyr. I've defined the names of the columns in a character vector. Usually, with only one new column, you can use the following pattern:
test %>%
mutate(!!new_column := NA)
However, I don't get it to work with across:
library(dplyr)
test <- data.frame(a = 1:3)
add_cols <- c("col_1", "col_2")
test %>%
mutate(across(!!add_cols, ~ NA))
#> Error: Problem with `mutate()` input `..1`.
#> x Can't subset columns that don't exist.
#> x Columns `col_1` and `col_2` don't exist.
#> ℹ Input `..1` is `across(c("col_1", "col_2"), ~NA)`.
test %>%
mutate(!!add_cols := NA)
#> Error: The LHS of `:=` must be a string or a symbol
expected_output <- data.frame(
a = 1:3,
col_1 = rep(NA, 3),
col_2 = rep(NA, 3)
)
expected_output
#> a col_1 col_2
#> 1 1 NA NA
#> 2 2 NA NA
#> 3 3 NA NA
Created on 2021-10-05 by the reprex package (v1.0.0)
With the first approach, the column names are correctly created, but then it directly tries to find it in the existing column names. In the second approach, I can't use anything other than a single string.
Is there a tidyverse solution or do I need to resort to the good old for loop?
The !! works for a single element
for(nm in add_cols) test <- test %>%
mutate(!! nm := NA)
-output
> test
a col_1 col_2
1 1 NA NA
2 2 NA NA
3 3 NA NA
Or another option is
test %>%
bind_cols(setNames(rep(list(NA), length(add_cols)), add_cols))
a col_1 col_2
1 1 NA NA
2 2 NA NA
3 3 NA NA
In base R, this is easier
test[add_cols] <- NA
Which can be used in a pipe
test %>%
`[<-`(., add_cols, value = NA)
a col_1 col_2
1 1 NA NA
2 2 NA NA
3 3 NA NA
across works only if the columns are already present i.e. it is suggesting to loop across the columns present in the data and do some modification/create new columns with .names modification
We could make use add_column from tibble
library(tibble)
library(janitor)
add_column(test, !!! add_cols) %>%
clean_names %>%
mutate(across(all_of(add_cols), ~ NA))
a col_1 col_2
1 1 NA NA
2 2 NA NA
3 3 NA NA
Another approach:
library(tidyverse)
f <- function(x) df$x = NA
mutate(test, map_dfc(add_cols,~ f(.x)))

Find last of several columns that is not NA (tidyverse)

Not sure what I'm doing wrong but I'm struggling getting the index per row of the last column (among several columns) that is not NA.
Using tidyverse and across, I'm getting as many output columns as input columns where I'd expect one single output column with the index of the respective column.
dat <- data.frame(id = c(1, 2, 3),
x = c(1, NA, NA),
y = c(NA, NA, NA),
z = c(3, 1, NA))
I tried the following (among others, inspired by this one: Return last data frame column which is not NA):
dat %>%
mutate(last = across(-id, ~max.col(!is.na(.x), ties.method="last")))
Expected outcome would be:
id x y z last
1 1 1 NA 3 3
2 2 NA NA 1 3
3 3 NA NA NA NA
The problems with your current flow:
across is going to pass one column at a time to the function/expression; your code needs a row or a matrix/frame. For this, across is not appropriate.
Your desired output of NA for the last row is inconsistent with the logic: !is.na(.x) should return c(F,F,F), which still has a max. Your logic then requires a custom function, since you need to handle it differently.
Try this adaptation of max.col into a custom function:
max.col.notna <- function (m, ties.method = c("random", "first", "last")) {
ties.method <- match.arg(ties.method)
tieM <- which(ties.method == eval(formals()[["ties.method"]]))
out <- .Internal(max.col(as.matrix(m), tieM))
m[] <- !m %in% c(0,NA) # 'm[] <-' is required to maintain the matrix shape
replace(out, rowSums(m) == 0, NA_integer_)
}
dat %>%
mutate(last = max.col.notna(!is.na(select(., -id)), ties.method = "last"))
# id x y z last
# 1 1 1 NA 3 3
# 2 2 NA NA 1 3
# 3 3 NA NA NA NA
Note: I've edited/changed the function several times, trying to ensure a consistent API to the intent of this custom function. As it stands now, the notna in the function name to me reflects a sense of "emptiness" (either 0 or NA). With this logic, the function is usable with logical (as here) and numeric data. Perhaps it's overkill, but I prefer APIs that operate consistently/predictably across input classes.
tidyverse isn't really suitable for row-wise operation. Most of the times reshaping the data into long format (as shown in #Rui Barradas answer) is a good approach.
Here is one way using rowwise keeping the data wide.
library(dplyr)
dat %>%
rowwise() %>%
mutate(last = {ind = which(!is.na(c_across(x:z)));
if(length(ind)) tail(ind, 1) else NA})
# id x y z last
# <dbl> <dbl> <lgl> <dbl> <int>
#1 1 1 NA 3 3
#2 2 NA NA 1 3
#3 3 NA NA NA NA
An R base solution:
dat$last = apply(dat[,2:4], 1,
FUN = function(x) ifelse(max(which(is.na(x))) == length(x), NA, max(which(is.na(x)))+1 ))
dat
# id x y z last
# 1 1 1 NA 3 3
# 2 2 NA NA 1 3
# 3 3 NA NA NA NA
You want to use c_across() and rowwise() to do this. rowwise() works similar to group_by_all(), except it is more explicit. c_across() creates flat vectors out of columns (whereas across() creates tibbles).
If we first define a function seperately to pull out the last non-NA value, or return NA if there are none:
get_last <- function(x){
y <- c(NA,which(!is.na(x)))
y[length(y)]
}
We can then apply that function c_across() the variables we need, but only after converting into a rowwise_df using rowwise()
dat %>%
rowwise() %>%
mutate(last = get_last(c_across(x:z)))
base R
df <- data.frame(id = c(1, 2, 3),
x = c(1, NA, NA),
y = c(NA, NA, NA),
z = c(3, 1, NA))
df$last <- apply(df[-1], 1, function(x) max(as.vector(!is.na(x)) * seq_len(length(x))))
df$last[df$last == 0] <- NA
df
#> id x y z last
#> 1 1 1 NA 3 3
#> 2 2 NA NA 1 3
#> 3 3 NA NA NA NA
Created on 2020-12-29 by the reprex package (v0.3.0)
Starting with a vector of NAs, you could step through each col and if the given element passes your check_fun returning TRUE, assign the index of that col to that element. The difference from the other answers here is that this does not check the condition row-wise or create a matrix from the data. Not sure whether creating two new temp vectors for each column is better/worse than just converting the entire data to a matrix first though.
library(tidyverse) # purrr and dplyr
last_matching_ind <- function(dat, check_fun){
check_fun <- as_mapper(check_fun)
reduce2(dat, seq_along(dat), .init = NA_integer_,
function(prev, dat, ind) if_else(check_fun(dat), ind, prev) )
}
dat %>%
mutate(last = last_matching_ind(dat[-1], ~ !is.na(.x)))
# id x y z last
# 1 1 1 NA 3 3
# 2 2 NA NA 1 3
# 3 3 NA NA NA NA

Conditional replacement of NA based on Data type

I have a database of over 80 different variables in which most have NAs. Some of the variables are integers and some are factors.
What I am trying to do is develop a function that:
1. Looks through my column list;
2. Identifies column type;
3. If datatype contained in the column is factor, function replaces NA with "Others";
4. However, if datatype contained in the column is an integer, replace with the number 0.
Any ideas?
Thanks, guys.
FOO <- function(x){
if(is.numeric(x)){
x[is.na(x)] <- 0
}
if(is.factor(x)){
x[is.na(x)] <- "Others"
}
return(x)
}
Now just use lapply to loop over multiple columns of your data, e.g. df[1:10] <- lapply(df[1:10], FOO).
Note: This requires that the factor level "Others" is already present in every factor variable you want to change. If this is not the case, use
FOO <- function(x){
if(is.numeric(x)){
x[is.na(x)] <- 0
}
if(is.factor(x)){
x <- as.character(x)
x[is.na(x)] <- "Others"
x <- as.factor(x)
}
return(x)
}
This might rearrange the order of the factor levels, though.
Using the dplyr and forcats packages:
library(dplyr)
library(forcats)
# sample data frame
df <- data_frame(fac1 = as.factor(c('NY', NA, 'PA', 'MN', 'OH', 'TX', NA)),
int1 = as.integer(c(1,2,3,NA,NA,6,7)),
fac2 = as.factor(c('red', 'blue', NA, 'green', 'green', NA, 'yellow')),
int2 = as.integer(c(1,NA,3,4,5,NA,7)))
df %>%
mutate_if(is.integer, funs(replace(., is.na(.), 0))) %>%
mutate_if(is.factor, funs(fct_explicit_na(., na_level = 'Other')))
# A tibble: 7 x 4
fac1 int1 fac2 int2
<fctr> <dbl> <fctr> <dbl>
1 NY 1 red 1
2 Other 2 blue 0
3 PA 3 Other 3
4 MN 0 green 4
5 OH 0 green 5
6 TX 6 Other 0
7 Other 7 yellow 7

R - Replace specific value contents with NA [duplicate]

This question already has answers here:
Replacing character values with NA in a data frame
(7 answers)
Closed 3 years ago.
I have a fairly large data frame that has multiple "-" which represent missing data. The data frame consisted of multiple Excel files, which could not use the "na.strings =" or alternative function, so I had to import them with the "-" representation.
How can I replace all "-" in the data frame with NA / missing values? The data frame consists of 200 columns of characters, factors, and integers.
So far I have tried:
sum(df %in c("-"))
returns: [1] 0
df[df=="-"] <-NA #does not do anything
library(plyr)
df <- revalue(df, c("-",NA))
returns: Error in revalue(tmp, c("-", NA)) :
x is not a factor or a character vector.
library(anchors)
df <- replace.value(df,colnames(df),"-",as.character(NA))
Error in charToDate(x) :
character string is not in a standard unambiguous format
The data frame consists of 200 columns of characters, factors, and integers, so I can see why the last two do not work correctly. Any help would be appreciated.
Since you're already using tidyverse functions, you can easily use na_if from dplyr within your pipes.
For example, I have a dataset where 999 is used to fill in a non-answer:
df <- tibble(
alpha = c("a", "b", "c", "d", "e"),
val1 = c(1, 999, 3, 8, 999),
val2 = c(2, 8, 999, 1, 2))
If I wanted to change val1 so 999 is NA, I could do:
df %>%
mutate(val1 = na_if(val1, 999))
In your case, it sounds like you want to replace a value across multiple variables, so using across for multiple columns would be more appropriate:
df %>%
mutate(across(c(val1, val2), na_if, 999)) # or val1:val2
replaces all instances of 999 in both val1 and val2 with NA and now looks like this:
# A tibble: 5 x 3
alpha val1 val2
<chr> <dbl> <dbl>
1 a 1. 2.
2 b NA 8.
3 c 3. NA
4 d 8. 1.
5 e NA 2.
I believe the simplest solution is with base R function is.na<-. It's meant to solve precisely that issue.
First, make up some data. Then set the required values to NA.
set.seed(247) # make the results reproducible
df <- data.frame(X = 1:10, Y = sample(c("-", letters[1:2]), 10, TRUE))
is.na(df) <- df == "-"
df
# X Y
#1 1 a
#2 2 b
#3 3 b
#4 4 a
#5 5 <NA>
#6 6 b
#7 7 a
#8 8 <NA>
#9 9 b
#10 10 a
Here's a solution that will do it:
> library(dplyr)
> test <- tibble(x = c('100', '20.56', '0.003', '-', ' -'), y = 5:1)
> makeNA <- function(x) str_replace(x,'-',NA_character_)
> mutate_all(test, funs(makeNA))
# A tibble: 5 x 2
x y
<chr> <chr>
1 100 5
2 20.56 4
3 0.003 3
4 NA 2
5 NA 1

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