I have a grouped df and I would like to add additional rows to the top of the groups that match with a variable (item_code) from the df.
The additional rows do not have an id column. The additional rows should not be duplicated within the groups of df.
Example data:
df <- as.tibble(data.frame(id=rep(1:3,each=2),
item_code=c("A","A","B","B","B","Z"),
score=rep(1,6)))
additional_rows <- as.tibble(data.frame(item_code=c("A","Z"),
score=c(6,6)))
What I tried
I found this post and tried to apply it:
Add row in each group using dplyr and add_row()
df %>% group_by(id) %>% do(add_row(additional_rows %>%
filter(item_code %in% .$item_code)))
What I get:
# A tibble: 9 x 3
# Groups: id [3]
id item_code score
<int> <fct> <dbl>
1 1 A 6
2 1 Z 6
3 1 NA NA
4 2 A 6
5 2 Z 6
6 2 NA NA
7 3 A 6
8 3 Z 6
9 3 NA NA
What I am looking for:
# A tibble: 6 x 3
id item_code score
<int> <fct> <dbl>
1 1 A 6
2 1 A 1
3 1 A 1
4 2 B 1
5 2 B 1
6 3 B 1
7 3 Z 6
8 3 Z 1
This should do the trick:
library(plyr)
df %>%
join(subset(df, item_code %in% additional_rows$item_code, select = c(id, item_code)) %>%
join(additional_rows) %>%
subset(!duplicated(.)), type = "full") %>%
arrange(id, item_code, -score)
Not sure if its the best way, but it works
Edit: to get the score in the same order added the other arrange terms
Edit 2: alright, there should now be no duplicated rows added from the additional rows as per your comment
Related
I have a data frame like this:
Team
GF
A
3
B
5
A
2
A
3
B
1
B
6
Looking for output like this (just an additional column):
Team
x
avg(X)
A
3
0
B
5
0
A
2
3
A
3
2.5
B
1
5
B
6
3
avg(x) is the average of all previous instances of x where Team is the same. I have the following R code which gets the overall average, however I'm looking for the "step-wise" average.
new_df <- df %>% group_by(Team) %>% summarise(avg_x = mean(x))
Is there a way to vectorize this while only evaluating the previous rows on each "iteration"?
You want the cummean() function from dplyr, combined with lag():
df %>% group_by(Team) %>% mutate(avg_x = replace_na(lag(cummean(x)), 0))
Producing the following:
# A tibble: 6 × 3
# Groups: Team [2]
Team x avg_x
<chr> <dbl> <dbl>
1 A 3 0
2 B 5 0
3 A 2 3
4 A 3 2.5
5 B 1 5
6 B 6 3
As required.
Edit 1:
As #Ritchie Sacramento pointed out, the following is cleaner and clearer:
df %>% group_by(Team) %>% mutate(avg_x = lag(cummean(x), default = 0))
This question already has answers here:
Count number of rows per group and add result to original data frame
(11 answers)
Closed last year.
I have the following data frame:
df <- data.frame(catergory=c("a","b","b","b","b","a","c"), value=c(1,5,3,6,7,4,6))
and I want to record the number of occurrences of each category so the output would be:
df <- data.frame(catergory=c("a","b","b","b","b","a","c"), value=c(1,5,3,6,7,4,6),
category_count=c(2,4,4,4,4,2,1))
Is there a simple way to do this?
# load package
library(data.table)
# set as data.table
setDT(df)
# count by category
df[, category_count := .N, category]
With dplyr:
library(dplyr)
df %>%
group_by(category) %>%
mutate(category_count = n()) %>%
ungroup
# A tibble: 7 × 3
category value category_count
<chr> <dbl> <int>
1 a 1 2
2 b 5 4
3 b 3 4
4 b 6 4
5 b 7 4
6 a 4 2
7 c 6 1
base
df <- data.frame(catergory=c("a","b","b","b","b","a","c"), value=c(1,5,3,6,7,4,6),
category_count=c(2,4,4,4,4,2,1))
df$res <- with(df, ave(x = seq(nrow(df)), list(catergory), FUN = length))
df
#> catergory value category_count res
#> 1 a 1 2 2
#> 2 b 5 4 4
#> 3 b 3 4 4
#> 4 b 6 4 4
#> 5 b 7 4 4
#> 6 a 4 2 2
#> 7 c 6 1 1
Created on 2022-02-08 by the reprex package (v2.0.1)
Given the following data:
test = data.frame(x = c(NA,1,1,2,3,4),
y = c(NA,1,2,3,4,4))
I want to perform some calculations and store these as new columns. The calculations, however, might result in a variable amount of columns. E.g. suppose I want store for each row the column index of the column(s) that contain the minimum per row. E.g. in row 1, both columns contain the minimum, hence I need to create two columns.
Using the tidyverse approach, I know I can use the set_names argument when passing my function as a list. But this doesn't work when I don't know the number of columns my calculation will create. See also here: https://community.rstudio.com/t/how-to-handle-lack-of-names-with-unnest-wider/40496
My approach for the calculations:
library(tidyverse)
test %>%
rowwise() %>%
mutate(dist = min(c_across(everything())),
code = list(which(c_across(cols = c(everything(), -dist)) == dist))) %>%
ungroup() %>%
unnest_wider(code)
which automatically names the unnested columns with "...1" and "...2":
# A tibble: 6 x 5
x y dist ...1 ...2
<dbl> <dbl> <dbl> <int> <int>
1 NA NA NA NA NA
2 1 1 1 1 2
3 1 2 1 1 NA
4 2 3 2 1 NA
5 3 4 3 1 NA
6 4 4 4 1 2
But that's not what I want. I also tried to use the named_repair argument within the unnest_wider, i.e. unnest_wider(code, names_repair = ~paste0("code", .x)) but this renames all columns.
Any ideas (preferably in the tidyverse approach)? Expected outcome:
# A tibble: 6 x 5
x y dist code_1 code_2
<dbl> <dbl> <dbl> <int> <int>
1 NA NA NA NA NA
2 1 1 1 1 2
3 1 2 1 1 NA
4 2 3 2 1 NA
5 3 4 3 1 NA
6 4 4 4 1 2
EDITED to add an example where one row contains only missings.
Edit 2: this is my current solution. But it is really ugly and requires to stop half way through. Problem here is that the rename_with function doesn't recognize the on-the-fly generated "length_code" column when I put everything into one pipe.
test2 <- test %>%
rowwise() %>%
mutate(dist = min(c_across(everything())),
code = list(which(c_across(cols = c(everything(), -dist)) == dist)),
length_code = length(code)) %>%
ungroup() %>%
unnest_wider(code) %>%
test3 <- test2 %>%
rename_with(.cols = starts_with("..."), .fn = ~paste0("code_", 1:max(test2$length_code)))
which gives:
# A tibble: 6 x 6
x y dist code_1 code_2 length_code
<dbl> <dbl> <dbl> <int> <int> <int>
1 NA NA NA NA NA 0
2 1 1 1 1 2 2
3 1 2 1 1 NA 1
4 2 3 2 1 NA 1
5 3 4 3 1 NA 1
6 4 4 4 1 2 2
I have a tibble containing time series of various blood parameters like CRP over the course of several days. The tibble is tidy, with each time series in one column, as well as a column for the day of measurement. The tibble contains another column with a day of infection. I want to replace each blood parameter with NA if the Day variable is greater-equal than the InfectionDay. Since I have a lot of variables, I'd like to have a function which accepts the column name dynamically and creates a new column name by appending "_censored" to the old one. I've tried the following:
censor.infection <- function(df, colname){
newcolname <- paste0(colname, "_censored")
return(df %>% mutate(!!newcolname := ifelse( Day < InfectionDay, !!colname, NA)))
}
data = tibble(Day=1:5, InfectionDay=3, CRP=c(3,2,5,4,1))
data = censor.infection(data, "CRP")
Running this, I expected
# A tibble: 5 x 4
Day InfectionDay CRP CRP_censored
<int> <dbl> <dbl> <chr>
1 1 3 3 3
2 2 3 2 2
3 3 3 5 NA
4 4 3 4 NA
5 5 3 1 NA
but I get
# A tibble: 5 x 4
Day InfectionDay CRP CRP_censored
<int> <dbl> <dbl> <chr>
1 1 3 3 CRP
2 2 3 2 CRP
3 3 3 5 NA
4 4 3 4 NA
5 5 3 1 NA
You can add sym() to the column name in mutate to convert to symbol before evaluating
censor.infection <- function(df, colname){
newcolname <- paste0(colname, "_censored")
return(df %>% mutate(!!newcolname := ifelse( Day < InfectionDay, !! sym(colname), NA)))
}
data = tibble(Day=1:5, InfectionDay=3, CRP=c(3,2,5,4,1))
data = censor.infection(data, "CRP")
We can select columns on which we want to apply the function (cols) and use mutate_at which will also automatically rename the columns. Added an extra column in the data to show renaming.
library(dplyr)
cols <- c("CRP", "CRP1")
data %>%
mutate_at(cols, list(censored = ~replace(., Day >= InfectionDay, NA)))
# A tibble: 5 x 6
# Day InfectionDay CRP CRP1 CRP_censored CRP1_censored
# <int> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 3 3 3 3 3
#2 2 3 2 2 2 2
#3 3 3 5 5 NA NA
#4 4 3 4 4 NA NA
#5 5 3 1 1 NA NA
data
data <- tibble(Day=1:5, InfectionDay=3, CRP=c(3,2,5,4,1), CRP1 = c(3,2,5,4,1))
The data frame x has a column in which the values are periodic. For each unique value in that column, I want to calculate summation of the second column. If x is something like this:
x <- data.frame(a=c(1:2,1:2,1:2),b=c(1,4,5,2,3,4))
a b
1 1 1
2 2 4
3 1 5
4 2 2
5 1 3
6 2 4
The output I want is the following data frame:
a b
1 9
2 10
Using aggregate as follows will get you your desired result
aggregate(b ~ a, x, sum)
Here is the option with dplyr
library(dplyr)
x %>%
group_by(a) %>%
summarise(b = sum(b))
# A tibble: 2 x 2
# a b
# <int> <dbl>
#1 1 9.00
#2 2 10.0