data=data.frame(person=c(1,1,1,2,2,2,2,3,3,3,3),
t=c(3,NA,9,4,7,NA,13,3,NA,NA,12),
WANT=c(3,6,9,4,7,10,13,3,6,9,12))
So basically I am wanting to create a new variable 'WANT' which takes the PREVIOUS value in t and ADDS 3 to it, and if there are many NA in a row then it keeps doing this. My attempt is:
library(dplyr)
data %>%
group_by(person) %>%
mutate(WANT_TRY = fill(t) + 3)
Here's one way -
data %>%
group_by(person) %>%
mutate(
# cs = cumsum(!is.na(t)), # creates index for reference value; uncomment if interested
w = case_when(
# rle() gives the running length of NA
is.na(t) ~ t[cumsum(!is.na(t))] + 3*sequence(rle(is.na(t))$lengths),
TRUE ~ t
)
) %>%
ungroup()
# A tibble: 11 x 4
person t WANT w
<dbl> <dbl> <dbl> <dbl>
1 1 3 3 3
2 1 NA 6 6
3 1 9 9 9
4 2 4 4 4
5 2 7 7 7
6 2 NA 10 10
7 2 13 13 13
8 3 3 3 3
9 3 NA 6 6
10 3 NA 9 9
11 3 12 12 12
Here is another way. We can do linear interpolation with the imputeTS package.
library(dplyr)
library(imputeTS)
data2 <- data %>%
group_by(person) %>%
mutate(WANT2 = na.interpolation(WANT)) %>%
ungroup()
data2
# # A tibble: 11 x 4
# person t WANT WANT2
# <dbl> <dbl> <dbl> <dbl>
# 1 1 3 3 3
# 2 1 NA 6 6
# 3 1 9 9 9
# 4 2 4 4 4
# 5 2 7 7 7
# 6 2 NA 10 10
# 7 2 13 13 13
# 8 3 3 3 3
# 9 3 NA 6 6
# 10 3 NA 9 9
# 11 3 12 12 12
This is harder than it seems because of the double NA at the end. If it weren't for that, then the following:
ifelse(is.na(data$t), c(0, data$t[-nrow(data)])+3, data$t)
...would give you want you want. The simplest way, that uses the same logic but doesn't look very clever (sorry!) would be:
.impute <- function(x) ifelse(is.na(x), c(0, x[-length(x)])+3, x)
.impute(.impute(data$t))
...which just cheats by doing it twice. Does that help?
You can use functional programming from purrr and "NA-safe" addition from hablar:
library(hablar)
library(dplyr)
library(purrr)
data %>%
group_by(person) %>%
mutate(WANT2 = accumulate(t, ~.x %plus_% 3))
Result
# A tibble: 11 x 4
# Groups: person [3]
person t WANT WANT2
<dbl> <dbl> <dbl> <dbl>
1 1 3 3 3
2 1 NA 6 6
3 1 9 9 9
4 2 4 4 4
5 2 7 7 7
6 2 NA 10 10
7 2 13 13 13
8 3 3 3 3
9 3 NA 6 6
10 3 NA 9 9
11 3 12 12 12
Related
I have a data frame that looks like this :
a
b
c
1
2
10
2
2
10
3
2
10
4
2
10
5
2
10
I want to create a column with mutate function of something else under the dplyr framework of functions (or base) that will be sequence from b to c (i.e from 2 to 10 with length the number of rows of this tibble or data frame)
Ideally my new data frame I want to like like this :
a
b
c
c
1
2
10
2
2
2
10
4
3
2
10
6
4
2
10
8
5
2
10
10
How can I do this with R using dplyr ?
library(tidyverse)
n=5
a = seq(1,n,length.out=n)
b = rep(2,n)
c = rep(10,n)
data = tibble(a,b,c)
We may do
library(dplyr)
data %>%
rowwise %>%
mutate(new = seq(b, c, length.out = n)[a]) %>%
ungroup
-output
# A tibble: 5 × 4
a b c new
<dbl> <dbl> <dbl> <dbl>
1 1 2 10 2
2 2 2 10 4
3 3 2 10 6
4 4 2 10 8
5 5 2 10 10
If you want this done "by group" for each a value (creating many new rows), we can create the sequence as a list column and then unnest it:
data %>%
mutate(result = map2(b, c, seq, length.out = n)) %>%
unnest(result)
# # A tibble: 25 × 4
# a b c result
# <dbl> <dbl> <dbl> <dbl>
# 1 1 2 10 2
# 2 1 2 10 4
# 3 1 2 10 6
# 4 1 2 10 8
# 5 1 2 10 10
# 6 2 2 10 2
# 7 2 2 10 4
# 8 2 2 10 6
# 9 2 2 10 8
# 10 2 2 10 10
# # … with 15 more rows
# # ℹ Use `print(n = ...)` to see more rows
If you want to keep the same number of rows and go from the first b value to the last c value, we can use seq directly in mutate:
data %>%
mutate(result = seq(from = first(b), to = last(c), length.out = n()))
# # A tibble: 5 × 4
# a b c result
# <dbl> <dbl> <dbl> <dbl>
# 1 1 2 10 2
# 2 2 2 10 4
# 3 3 2 10 6
# 4 4 2 10 8
# 5 5 2 10 10
This one?
library(dplyr)
df %>%
mutate(c1 = a*b)
a b c c1
1 1 2 10 2
2 2 2 10 4
3 3 2 10 6
4 4 2 10 8
5 5 2 10 10
I am having some trouble getting mutate, across, and case_when to function properly, I've recreated a simple version of my problem here:
a <- c(1:10)
b <- c(2:11)
c <- c(3:12)
test <- tibble(a, b, c)
# A tibble: 10 x 3
a b c
<int> <int> <int>
1 1 2 3
2 2 3 4
3 3 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
My goal is to replace all of the 3's with 4's, and keep everything else the same. I have the following code:
test_1 <-
test %>%
mutate(across(a:c, ~ case_when(. == 3 ~ 4)))
# A tibble: 10 x 3
a b c
<dbl> <dbl> <dbl>
1 NA NA 4
2 NA 4 NA
3 4 NA NA
4 NA NA NA
5 NA NA NA
6 NA NA NA
7 NA NA NA
8 NA NA NA
9 NA NA NA
10 NA NA NA
It's close but I get NA values where I want to maintain the value in the original tibble. How do I maintain the original values using the mutate across structure?
Thank you in advance!
What about this?
> test %>%
+ mutate(across(a:c, ~ case_when(. == 3 ~ 4, TRUE ~ 1 * (.))))
# A tibble: 10 x 3
a b c
<dbl> <dbl> <dbl>
1 1 2 4
2 2 4 4
3 4 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
or
> test %>%
+ replace(. == 3, 4)
# A tibble: 10 x 3
a b c
<int> <int> <int>
1 1 2 4
2 2 4 4
3 4 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
In base R, we can do
test[test ==3] <- 4
This also works:
a <- c(1:10)
b <- c(2:11)
c <- c(3:12)
tibble(a, b, c) %>%
modify(~ ifelse(. == 3, 4, .))
# A tibble: 10 x 3
a b c
<dbl> <dbl> <dbl>
1 1 2 4
2 2 4 4
3 4 4 5
4 4 5 6
5 5 6 7
6 6 7 8
7 7 8 9
8 8 9 10
9 9 10 11
10 10 11 12
I have a dataset like:
df <- tibble(
id = 1:18,
class = rep(c(rep(1,3),rep(2,2),3),3),
var_a = rep(c("a","b"),9)
)
# A tibble: 18 x 3
id cluster var_a
<int> <dbl> <chr>
1 1 1 a
2 2 1 b
3 3 1 a
4 4 2 b
5 5 2 a
6 6 3 b
7 7 1 a
8 8 1 b
9 9 1 a
10 10 2 b
11 11 2 a
12 12 3 b
13 13 1 a
14 14 1 b
15 15 1 a
16 16 2 b
17 17 2 a
18 18 3 b
That dataset contains a number of observations in several classes. The classes are not balanced. In the sample above we can see, that only 3 observations are of class 3, while there are 6 observations of class 2 and 9 observations of class 1.
Now I want to automatically balance that dataset so that all classes are of the same size. So I want a dataset of 9 rows, 3 rows in each class. I can use the sample_n function from dplyr to do such a sampling.
I achieved to do so by first calculating the smallest class size..
min_length <- as.numeric(df %>%
group_by(class) %>%
summarise(n = n()) %>%
ungroup() %>%
summarise(min = min(n)))
..and then apply the sample_n function:
set.seed(1)
df %>% group_by(cluster) %>% sample_n(min_length)
# A tibble: 9 x 3
# Groups: cluster [3]
id cluster var_a
<int> <dbl> <chr>
1 15 1 a
2 7 1 a
3 13 1 a
4 4 2 b
5 5 2 a
6 17 2 a
7 18 3 b
8 6 3 b
9 12 3 b
I wondered If it's possible to do that (calculating the smallest class size and then sampling) in one go?
You can do it in one step, but it is cheating a little:
set.seed(42)
df %>%
group_by(class) %>%
sample_n(min(table(df$class))) %>%
ungroup()
# # A tibble: 9 x 3
# id class var_a
# <int> <dbl> <chr>
# 1 1 1 a
# 2 8 1 b
# 3 15 1 a
# 4 4 2 b
# 5 5 2 a
# 6 11 2 a
# 7 12 3 b
# 8 18 3 b
# 9 6 3 b
I say "cheating" because normally you would not want to reference df$ from within the pipe. However, because they property we're looking for is of the whole frame but the table function only sees one group at a time, we need to side-step that a little.
One could do
df %>%
mutate(mn = min(table(class))) %>%
group_by(class) %>%
sample_n(mn[1]) %>%
ungroup()
# # A tibble: 9 x 4
# id class var_a mn
# <int> <dbl> <chr> <int>
# 1 14 1 b 3
# 2 13 1 a 3
# 3 7 1 a 3
# 4 4 2 b 3
# 5 16 2 b 3
# 6 5 2 a 3
# 7 12 3 b 3
# 8 18 3 b 3
# 9 6 3 b 3
Though I don't think that that is any more elegant/readable.
I have two data frames of the same respondents, one from Time 1 and the next from Time 2. In each wave they nominated their friends, and I want to know:
1) how many friends are nominated in Time 2 but not in Time 1 (new friends)
2) how many friends are nominated in Time 1 but not in Time 2 (lost friends)
Sample data:
Time 1 DF
ID friend_1 friend_2 friend_3
1 4 12 7
2 8 6 7
3 9 NA NA
4 15 7 2
5 2 20 7
6 19 13 9
7 12 20 8
8 3 17 10
9 1 15 19
10 2 16 11
Time 2 DF
ID friend_1 friend_2 friend_3
1 4 12 3
2 8 6 14
3 9 NA NA
4 15 7 2
5 1 17 9
6 9 19 NA
7 NA NA NA
8 7 1 16
9 NA 10 12
10 7 11 9
So the desired DF would include these columns (EDIT filled in columns):
ID num_newfriends num_lostfriends
1 1 1
2 1 1
3 0 0
4 0 0
5 3 3
6 0 1
7 0 3
8 3 3
9 2 3
10 2 1
EDIT2:
I've tried doing an anti join
df3 <- anti_join(df1, df2)
But this method doesn't take into account friend id numbers that might appear in a different column in time 2 (For example respondent #6 friend 9 and 19 are in T1 and T2 but in different columns in each time)
Another option:
library(tidyverse)
left_join(
gather(df1, key, x, -ID),
gather(df2, key, y, -ID),
by = c("ID", "key")
) %>%
group_by(ID) %>%
summarise(
num_newfriends = sum(!y[!is.na(y)] %in% x[!is.na(x)]),
num_lostfriends = sum(!x[!is.na(x)] %in% y[!is.na(y)])
)
Output:
# A tibble: 10 x 3
ID num_newfriends num_lostfriends
<int> <int> <int>
1 1 1 1
2 2 1 1
3 3 0 0
4 4 0 0
5 5 3 3
6 6 0 1
7 7 0 3
8 8 3 3
9 9 2 3
10 10 2 2
Simple comparisons would be an option
library(tidyverse)
na_sums_old <- rowSums(is.na(time1))
na_sums_new <- rowSums(is.na(time2))
kept_friends <- map_dbl(seq(nrow(time1)), ~ sum(time1[.x, -1] %in% time2[.x, -1]))
kept_friends <- kept_friends - na_sums_old * (na_sums_new >= 1)
new_friends <- 3 - na_sums_new - kept_friends
lost_friends <- 3 - na_sums_old - kept_friends
tibble(ID = time1$ID, new_friends = new_friends, lost_friends = lost_friends)
# A tibble: 10 x 3
ID new_friends lost_friends
<int> <dbl> <dbl>
1 1 1 1
2 2 1 1
3 3 0 0
4 4 0 0
5 5 3 3
6 6 0 1
7 7 0 3
8 8 3 3
9 9 2 3
10 10 2 2
You can make anti_join work by first pivoting to a "long" data frame.
df1 <- df1 %>%
pivot_longer(starts_with("friend_"), values_to = "friend") %>%
drop_na()
df2 <- df2 %>%
pivot_longer(starts_with("friend_"), values_to = "friend") %>%
drop_na()
head(df1)
#> # A tibble: 6 x 3
#> ID name friend
#> <int> <chr> <int>
#> 1 1 friend_1 4
#> 2 1 friend_2 12
#> 3 1 friend_3 7
#> 4 2 friend_1 8
#> 5 2 friend_2 6
#> 6 2 friend_3 7
lost_friends <- anti_join(df1, df2, by = c("ID", "friend"))
new_fiends <- anti_join(df2, df1, by = c("ID", "friend"))
respondents <- distinct(df1, ID)
respondents %>%
full_join(
count(lost_friends, ID, name = "num_lost_friends")
) %>%
full_join(
count(new_fiends, ID, name = "num_new_friends")
) %>%
mutate_at(vars(starts_with("num_")), replace_na, 0)
#> Joining, by = "ID"
#> Joining, by = "ID"
#> # A tibble: 10 x 3
#> ID num_lost_friends num_new_friends
#> <int> <dbl> <dbl>
#> 1 1 1 1
#> 2 2 1 1
#> 3 3 0 0
#> 4 4 0 0
#> 5 5 3 3
#> 6 6 1 0
#> 7 7 3 0
#> 8 8 3 3
#> 9 9 3 2
#> 10 10 2 2
Created on 2019-11-01 by the reprex package (v0.3.0)
Similar to this question but I want to use tidy evaluation instead.
df = data.frame(group = c(1,1,1,2,2,2,3,3,3),
date = c(1,2,3,4,5,6,7,8,9),
speed = c(3,4,3,4,5,6,6,4,9))
> df
group date speed
1 1 1 3
2 1 2 4
3 1 3 3
4 2 4 4
5 2 5 5
6 2 6 6
7 3 7 6
8 3 8 4
9 3 9 9
The task is to create a new column (newValue) whose values equals to the values of the date column (per group) with one condition: speed == 4. Example: group 1 has a newValue of 2 because date[speed==4] = 2.
group date speed newValue
1 1 1 3 2
2 1 2 4 2
3 1 3 3 2
4 2 4 4 4
5 2 5 5 4
6 2 6 6 4
7 3 7 6 8
8 3 8 4 8
9 3 9 9 8
It worked without tidy evaluation
df %>%
group_by(group) %>%
mutate(newValue=date[speed==4L])
#> # A tibble: 9 x 4
#> # Groups: group [3]
#> group date speed newValue
#> <dbl> <dbl> <dbl> <dbl>
#> 1 1 1 3 2
#> 2 1 2 4 2
#> 3 1 3 3 2
#> 4 2 4 4 4
#> 5 2 5 5 4
#> 6 2 6 6 4
#> 7 3 7 6 8
#> 8 3 8 4 8
#> 9 3 9 9 8
But had error with tidy evaluation
my_fu <- function(df, filter_var){
filter_var <- sym(filter_var)
df <- df %>%
group_by(group) %>%
mutate(newValue=!!filter_var[speed==4L])
}
my_fu(df, "date")
#> Error in quos(..., .named = TRUE): object 'speed' not found
Thanks in advance.
We can place the evaluation within brackets. Otherwise, it may try to evaluate the whole expression (filter_var[speed = 4L]) instead of filter_var alone
library(rlang)
library(dplyr)
my_fu <- function(df, filter_var){
filter_var <- sym(filter_var)
df %>%
group_by(group) %>%
mutate(newValue=(!!filter_var)[speed==4L])
}
my_fu(df, "date")
# A tibble: 9 x 4
# Groups: group [3]
# group date speed newValue
# <dbl> <dbl> <dbl> <dbl>
#1 1 1 3 2
#2 1 2 4 2
#3 1 3 3 2
#4 2 4 4 4
#5 2 5 5 4
#6 2 6 6 4
#7 3 7 6 8
#8 3 8 4 8
#9 3 9 9 8
Also, you can use from sqldf. Join df with a constraint on that:
library(sqldf)
df = data.frame(group = c(1,1,1,2,2,2,3,3,3),
date = c(1,2,3,4,5,6,7,8,9),
speed = c(3,4,3,4,5,6,6,4,9))
sqldf("SELECT df_origin.*, df4.`date` new_value FROM
df df_origin join (SELECT `group`, `date` FROM df WHERE speed = 4) df4
on (df_origin.`group` = df4.`group`)")