I have made a very complex solution to something I feel should have a much simpler solution.
In short what I want:
I want to compute a new column containing the minimum value across 3 columns
I want to ignore zeros and NAs
If I only have zeros and NAs I want a zero
If I have only NAs I want a NA
Here is my solution, it works, but it is very complex and produces a warning.
> library(dplyr)
> df <- data.frame(
+ id = c(1, 2, 3, 4),
+ test1 = c( NA, NA, 2 , 3),
+ test2 = c( NA, 0, 1 , 1),
+ test3 = c(NA, NA, 0 , 2)
+ )
> df2 <- df %>%
+ mutate(nieuw = apply(across(test1:test3), 1, function(x) min(x[x>0]))) %>%
+ rowwise() %>%
+ mutate(nieuw = if_else(is.na(nieuw), max(across(test1:test3), na.rm = TRUE), nieuw)) %>%
+ mutate(nieuw = ifelse(is.infinite(nieuw), NA, nieuw))
> df
id test1 test2 test3
1 1 NA NA NA
2 2 NA 0 NA
3 3 2 1 0
4 4 3 1 2
> df2
# A tibble: 4 x 5
# Rowwise:
id test1 test2 test3 nieuw
<dbl> <dbl> <dbl> <dbl> <dbl>
1 1 NA NA NA NA
2 2 NA 0 NA 0
3 3 2 1 0 1
4 4 3 1 2 1
Warning message:
Problem while computing `nieuw = if_else(...)`.
i no non-missing arguments to max; returning -Inf
i The warning occurred in row 1.
You can create a helper function and then apply it rowwise:
library(dplyr)
safe <- function(x, f, ...) ifelse(all(is.na(x)), NA,
ifelse(all(is.na(x) | x == 0),
0, f(x[x > 0], na.rm = TRUE, ...)))
df %>%
rowwise() %>%
mutate(a = safe(c_across(test1:test3), min))
# A tibble: 4 × 5
# Rowwise:
id test1 test2 test3 a
<dbl> <dbl> <dbl> <dbl> <dbl>
1 1 NA NA NA NA
2 2 NA 0 NA 0
3 3 2 1 0 1
4 4 3 1 2 1
Here is another option. It leverages making zeros and NA's very large and then recodes them at the end:
library(tidyverse)
get_min <- function(data, cols){
data[is.na(data)] <- 1e6
data[data == 0] <- 1e5
nums <- do.call(pmin, select(data, all_of(cols)))
recode(nums, `1e+06` = NA_real_, `1e+05` = 0.)
}
df %>%
mutate(nieuw = get_min(., c("test1", "test2", "test3")))
#> id test1 test2 test3 nieuw
#> 1 1 NA NA NA NA
#> 2 2 NA 0 NA 0
#> 3 3 2 1 0 1
#> 4 4 3 1 2 1
Related
library(dplyr)
dat <-
data.frame(id = rep(c(1,2,3,4), each = 3),
value = c(NA, NA, NA, 0, 1, 2, 0, 1, NA, 1, 2,3))
dat %>%
dplyr::group_by(id) %>%
dplyr::summarise(value_sum = sum(value, na.rm = T))
# A tibble: 4 x 2
id value_sum
1 0
2 3
3 1
4 6
Is there any way I can return NA if all the entries in a group are NA. For e.g. id 1 has all the entries as NA so I want the value_sum to be NA as well.
# A tibble: 4 x 2
id value_sum
1 NA
2 3
3 1
4 6
One way is to use an if/else statement: If all is Na return NA else return sum():
dat %>%
dplyr::group_by(id) %>%
#dplyr::summarise(value_sum = sum(value, na.rm = F)) %>%
summarise(number = if(all(is.na(value))) NA_real_ else sum(value, na.rm = TRUE))
id number
<dbl> <dbl>
1 1 NA
2 2 3
3 3 1
4 4 6
We could use fsum
library(collapse)
fsum(dat$value, g = dat$id)
1 2 3 4
NA 3 1 6
Or with dplyr
library(dplyr)
dat %>%
group_by(id) %>%
summarise(number = fsum(value))
# A tibble: 4 × 2
id number
<dbl> <dbl>
1 1 NA
2 2 3
3 3 1
4 4 6
I want to replace value(s) with NA by group.
have <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,4,5,6,7))
want1 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,NA,5,6,NA))
want2 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,NA,NA,5,NA,NA))
want1 corresponds to replacing the last obs of value with NA and want2 corresponds to replacing last obs of value & last 2nd value with NA. I'm currently trying to do with with dplyr package but can't seem to get any traction. Any help would be much appreciated. Thanks!
We can use row_number() to test the current row against n() the total rows in the group.
have |>
group_by(id) |>
mutate(
last1 = ifelse(row_number() == n(), NA, value),
last2 = ifelse(row_number() >= n() - 1, NA, value)
)
# # A tibble: 7 × 4
# # Groups: id [2]
# id value last1 last2
# <dbl> <dbl> <dbl> <dbl>
# 1 1 1 1 1
# 2 1 2 2 2
# 3 1 3 3 NA
# 4 1 4 NA NA
# 5 2 5 5 5
# 6 2 6 6 NA
# 7 2 7 NA NA
And a general way to provide variants as different data frames.
lapply(
1:2,
function(k) {
have %>%
group_by(id) %>%
mutate(value=ifelse(row_number() <= (n() - k), value, NA))
}
)
[[1]]
# A tibble: 7 × 2
# Groups: id [2]
id value
<dbl> <dbl>
1 1 1
2 1 2
3 1 3
4 1 NA
5 2 5
6 2 6
7 2 NA
[[2]]
# A tibble: 7 × 2
# Groups: id [2]
id value
<dbl> <dbl>
1 1 1
2 1 2
3 1 NA
4 1 NA
5 2 5
6 2 NA
7 2 NA
Here is a base R way.
have <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,4,5,6,7))
want1 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,NA,5,6,NA))
want2 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,NA,NA,5,NA,NA))
with(have, ave(value, id, FUN = \(x){
x[length(x)] <- NA
x
}))
#> [1] 1 2 3 NA 5 6 NA
with(have, ave(value, id, FUN = \(x){
x[length(x)] <- NA
if(length(x) > 1)
x[length(x) - 1L] <- NA
x
}))
#> [1] 1 2 NA NA 5 NA NA
Created on 2022-06-09 by the reprex package (v2.0.1)
Then reassign these results to column value.
I have a database like this:
id <- c(rep(1,3), rep(2, 3), rep(3, 3))
condition <- c(0, 0, 1, 0, 0, 1, 1, 1, 0)
time_point1 <- c(1, 1, NA)
time_point2 <- c(NA, 1, NA)
time_point3 <- c(NA, NA, NA)
time_point4 <- c(1, NA, NA, 1, NA, NA, NA, NA, 1)
data <- data.frame(id, condition, time_point1, time_point2, time_point3, time_point4)
data
id condition time_point1 time_point2 time_point3 time_point4
1 1 0 1 NA NA 1
2 1 0 1 1 NA NA
3 1 1 NA NA NA NA
4 2 0 1 NA NA 1
5 2 0 1 1 NA NA
6 2 1 NA NA NA NA
7 3 1 1 NA NA NA
8 3 1 1 1 NA NA
9 3 0 NA NA NA 1
I want to make a table with how many have the condition == 1 (n_x) and also how many are in each time point (n_t). In case there is none also I want a 0. I tried this:
data %>%
pivot_longer(cols = contains("time_point")) %>%
filter (!is.na(value)) %>%
group_by(name) %>%
mutate(n_t = n_distinct(id)) %>%
ungroup() %>%
filter(condition == 1) %>%
group_by(name) %>%
summarise(n_x = n_distinct(id), n_t = first(n_t))
Obtaining this:
name n_x n_t
<chr> <int> <int>
1 time_point1 1 3
2 time_point2 1 3
Desired Outcome: I want this type of table that considers the cases with condition and without it:
name n_x n_t
1 time_point1 2 6
2 time_point2 1 3
3 time_point3 0 0
4 time_point4 0 3
Thank you!
You can pivot_longer() to be able to group_by() time_points and then summarise just adding up the values. For conditions only sum values where the column values != NA.
data %>%
pivot_longer(cols=c(3:6),names_to = 'point', values_to='values') %>%
group_by(point) %>%
summarise(n_x = sum(condition[!is.na(values)]), n_t = sum(values, na.rm = TRUE))
Output:
# A tibble: 4 x 3
point n_x n_t
<chr> <dbl> <dbl>
1 time_point1 2 6
2 time_point2 1 3
3 time_point3 0 0
4 time_point4 0 3
I have the following data
df <- tibble(Type=c(1,2,2,1,1,2),ID=c(6,4,3,2,1,5))
Type ID
1 6
2 4
2 3
1 2
1 1
2 5
For each of the type 2 rows, I want to find the IDs of the type 1 rows just below and above them. For the above dataset, the output will be:
Type ID IDabove IDbelow
1 6 NA NA
2 4 6 2
2 3 6 2
1 2 NA NA
1 1 NA NA
2 5 1 NA
Naively, I can write a for loop to achieve this, but that would be too time consuming for the dataset I am dealing with.
One approach using dplyr lead,lag to get next and previous value respectively and data.table's rleid to create groups of consecutive Type values.
library(dplyr)
library(data.table)
df %>%
mutate(IDabove = ifelse(Type == 2, lag(ID), NA),
IDbelow = ifelse(Type == 2, lead(ID), NA),
grp = rleid(Type)) %>%
group_by(grp) %>%
mutate(IDabove = first(IDabove),
IDbelow = last(IDbelow)) %>%
ungroup() %>%
select(-grp)
# Type ID IDabove IDbelow
# <dbl> <dbl> <dbl> <dbl>
#1 1 6 NA NA
#2 2 4 6 2
#3 2 3 6 2
#4 1 2 NA NA
#5 1 1 NA NA
#6 2 5 1 NA
A dplyr only solution:
You could create your own rleid function then apply the logic provided by Ronak(Many thanks. Upvoted).
library(dplyr)
my_func <- function(x) {
x <- rle(x)$lengths
rep(seq_along(x), times=x)
}
# this part is the same as provided by Ronak.
df %>%
mutate(IDabove = ifelse(Type == 2, lag(ID), NA),
IDbelow = ifelse(Type == 2, lead(ID), NA),
grp = my_func(Type)) %>%
group_by(grp) %>%
mutate(IDabove = first(IDabove),
IDbelow = last(IDbelow)) %>%
ungroup() %>%
select(-grp)
Output:
Type ID IDabove IDbelow
<dbl> <dbl> <dbl> <dbl>
1 1 6 NA NA
2 2 4 6 2
3 2 3 6 2
4 1 2 NA NA
5 1 1 NA NA
6 2 5 1 NA
I have a dataframe with a lot of variables seen in multiple conditions. I'd like to merge each variable by condition.
The example data frame is a simplified version of what I have (3 variables over 2 conditions).
VAR.B_1 <- c(1, 2, 3, 4, 5, 'NA', 'NA', 'NA', 'NA', 'NA')
VAR.B_2 <- c(2, 2, 3, 4, 5,'NA', 'NA', 'NA', 'NA', 'NA')
VAR.B_3 <- c(1, 1, 1, 1, 1,'NA', 'NA', 'NA', 'NA', 'NA')
VAR.E_1 <- c(NA, NA, NA, NA, NA, 1, 1, 1, 1, 1)
VAR.E_2 <- c(NA, NA, NA, NA, NA, 1, 2, 3, 4, 5)
VAR.E_3 <- c(NA, NA, NA, NA, NA, 1, 1, 1, 1, 1)
Condition <- c("B", "B","B","B","B","E","E","E","E","E")
#Example dataset
data<-as.data.frame(cbind(VAR.B_1,VAR.B_2,VAR.B_3, VAR.E_1,VAR.E_2, VAR.E_3, Condition))
I want to end up with this, appended to the original data frame:
VAR_1 VAR_2 VAR_3
1 2 1
2 2 1
3 3 1
4 4 1
5 5 1
1 1 1
1 2 1
1 3 1
1 4 1
1 5 1
I understand that R won't work with i inside the variable name, but I have an example of the kind of for loop I was trying to do. I would rather not call variables by column location, since there will be a lot of variables.
##Example of how I want to merge - this code does not work
for(i in 1:3) {
data$VAR_[,i] <-ifelse(data$Condition == "B", VAR.B_[,i],
ifelse(data$Condition == "E", VAR.E_[,i], NA))
}
This might work for your situation:
library(tidyverse)
library(stringr)
data %>%
mutate_all(as.character) %>%
gather(key, value, -Condition) %>%
filter(!is.na(value), value != "NA") %>%
mutate(key = str_replace(key, paste0("\\.", Condition), "")) %>%
group_by(Condition, key) %>%
mutate(rowid = 1:n()) %>%
spread(key, value) %>%
bind_cols(data)
#> # A tibble: 10 x 12
#> # Groups: Condition [2]
#> Condition rowid VAR_1 VAR_2 VAR_3 VAR.B_1 VAR.B_2 VAR.B_3 VAR.E_1
#> <chr> <int> <chr> <chr> <chr> <fctr> <fctr> <fctr> <fctr>
#> 1 B 1 1 2 1 1 2 1 NA
#> 2 B 2 2 2 1 2 2 1 NA
#> 3 B 3 3 3 1 3 3 1 NA
#> 4 B 4 4 4 1 4 4 1 NA
#> 5 B 5 5 5 1 5 5 1 NA
#> 6 E 1 1 1 1 NA NA NA 1
#> 7 E 2 1 2 1 NA NA NA 1
#> 8 E 3 1 3 1 NA NA NA 1
#> 9 E 4 1 4 1 NA NA NA 1
#> 10 E 5 1 5 1 NA NA NA 1
#> # ... with 3 more variables: VAR.E_2 <fctr>, VAR.E_3 <fctr>,
#> # Condition1 <fctr>
data.frame(lapply(split.default(data[-NCOL(data)], gsub("\\D+", "", head(names(data), -1))),
function(a){
a = sapply(a, function(x) as.numeric(as.character(x)))
rowSums(a, na.rm = TRUE)
}))
# X1 X2 X3
#1 1 2 1
#2 2 2 1
#3 3 3 1
#4 4 4 1
#5 5 5 1
#6 1 1 1
#7 1 2 1
#8 1 3 1
#9 1 4 1
#10 1 5 1
#Warning messages:
#1: In FUN(X[[i]], ...) : NAs introduced by coercion
#2: In FUN(X[[i]], ...) : NAs introduced by coercion
#3: In FUN(X[[i]], ...) : NAs introduced by coercion
Your data appears to have two kinds of NA values in it. It has NA, or R's NA value, and it also has the string 'NA'. In my solution below, I replace both with zero, cast each column in the data frame to numeric, and then just sum together like-numbered VAR columns. Then, drop the original columns which you don't want anymore.
data <- as.data.frame(cbind(VAR.B_1,VAR.B_2,VAR.B_3, VAR.E_1,VAR.E_2, VAR.E_3),
stringsAsFactors=FALSE)
data[is.na(data)] <- 0
data[data == 'NA'] <- 0
data <- as.data.frame(lapply(data, as.numeric))
data$VAR_1 <- data$VAR.B_1 + data$VAR.E_1
data$VAR_2 <- data$VAR.B_2 + data$VAR.E_2
data$VAR_3 <- data$VAR.B_3 + data$VAR.E_3
data <- data[c("VAR_1", "VAR_2", "VAR_3")]
Demo