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)
Related
I am trying to convert my data to a long format using one variable that indicates a day of the update.
I have the following variables:
baseline temperature variable "temp_b";
time-varying temperature variable "temp_v" and
the number of days "n_days" when the varying variable is updated.
I want to create a long format using the carried forward approach and a max follow-up time of 5 days.
Example of data
df <- structure(list(id=1:3, temp_b=c(20L, 7L, 7L), temp_v=c(30L, 10L, NA), n_days=c(2L, 4L, NA)), class="data.frame", row.names=c(NA, -3L))
# id temp_b temp_v n_days
# 1 1 20 30 2
# 2 2 7 10 4
# 3 3 7 NA NA
df_long <- structure(list(id=c(1,1,1,1,1, 2,2,2,2,2, 3,3,3,3,3),
days_cont=c(1,2,3,4,5, 1,2,3,4,5, 1,2,3,4,5),
long_format=c(20,30,30,30,30,7,7,7,10,10,7,7,7,7,7)),
class="data.frame", row.names=c(NA, -15L))
# id days_cont long_format
# 1 1 1 20
# 2 1 2 30
# 3 1 3 30
# 4 1 4 30
# 5 1 5 30
# 6 2 1 7
# 7 2 2 7
# 8 2 3 7
# 9 2 4 10
# 10 2 5 10
# 11 3 1 7
# 12 3 2 7
# 13 3 3 7
# 14 3 4 7
# 15 3 5 7
You could repeat each row 5 times with tidyr::uncount():
library(dplyr)
df %>%
tidyr::uncount(5) %>%
group_by(id) %>%
transmute(days_cont = 1:n(),
temp = ifelse(row_number() < n_days | is.na(n_days), temp_b, temp_v)) %>%
ungroup()
# # A tibble: 15 × 3
# id days_cont temp
# <int> <int> <int>
# 1 1 1 20
# 2 1 2 30
# 3 1 3 30
# 4 1 4 30
# 5 1 5 30
# 6 2 1 7
# 7 2 2 7
# 8 2 3 7
# 9 2 4 10
# 10 2 5 10
# 11 3 1 7
# 12 3 2 7
# 13 3 3 7
# 14 3 4 7
# 15 3 5 7
Here's a possibility using tidyverse functions. First, pivot_longer and get rid of unwanted values (that will not appear in the final df, i.e. values with temp_v == NA), then group_by id, and mutate the n_days variable to match the number of rows it will have in the final df. Finally, uncount the dataframe.
library(tidyverse)
df %>%
replace_na(list(n_days = 6)) %>%
pivot_longer(-c(id, n_days)) %>%
filter(!is.na(value)) %>%
group_by(id) %>%
mutate(n_days = case_when(name == "temp_b" ~ n_days - 1,
name == "temp_v" ~ 5 - (n_days - 1))) %>%
uncount(n_days) %>%
mutate(days_cont = row_number()) %>%
select(id, days_cont, long_format = value)
id days_cont long_format
<int> <int> <int>
1 1 1 20
2 1 2 30
3 1 3 30
4 1 4 30
5 1 5 30
6 2 1 7
7 2 2 7
8 2 3 7
9 2 4 10
10 2 5 10
11 3 1 7
12 3 2 7
13 3 3 7
14 3 4 7
15 3 5 7
I have a series of rows in a single dataframe. I'm trying to aggregate the first two rows for each ID- i.e. - I want to combine events 1 and 2 for ID 1 into a single row, events 1 and 2 for ID 2 into a singlw row etc, but leave event 3 completely untouched.
id <- c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5)
event <- c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3)
score <- c(3,NA,1,3,NA,2,6,NA,1,8,NA,2,4,NA,1)
score2 <- c(NA,4,1,NA,5,2,NA,0,3,NA,5,6,NA,8,7)
df <- tibble(id, event, score, score2)
# A tibble: 15 x 4
id event score score2
<dbl> <dbl> <dbl> <dbl>
1 1 1 3 NA
2 1 2 NA 4
3 1 3 1 1
4 2 1 3 NA
5 2 2 NA 5
6 2 3 2 2
7 3 1 6 NA
8 3 2 NA 0
9 3 3 1 3
10 4 1 8 NA
11 4 2 NA 5
12 4 3 2 6
13 5 1 4 NA
14 5 2 NA 8
15 5 3 1 7
I've tried :
df_merged<- df %>% group_by (id) %>% summarise_all(funs(min(as.character(.),na.rm=TRUE))),
which aggregates these nicely, but then I struggle to merge these back into the orignal dataframe/tibble (there are really about 300 different "score" columns in the full dataset, so a right_join is a headache with score.x, score.y, score2.x, score2.y all over the place...)
Ideally, the situation would need to be dplyr as the rest of my code runs on this!
EDIT:
Ideally, my expected output would be:
# A tibble: 10 x 4
id event score score2
<dbl> <dbl> <dbl> <dbl>
1 1 1 3 4
3 1 3 1 1
4 2 1 3 5
6 2 3 2 2
7 3 1 6 0
9 3 3 1 3
10 4 1 8 5
12 4 3 2 6
13 5 1 4 8
15 5 3 1 7
We may change the order of NA elements with replace
library(dplyr)
df %>%
group_by(id) %>%
mutate(across(starts_with('score'),
~replace(., 1:2, .[1:2][order(is.na(.[1:2]))]))) %>%
ungroup %>%
filter(if_all(starts_with('score'), Negate(is.na)))
-output
# A tibble: 10 x 4
id event score score2
<dbl> <dbl> <dbl> <dbl>
1 1 1 3 4
2 1 3 1 1
3 2 1 3 5
4 2 3 2 2
5 3 1 6 0
6 3 3 1 3
7 4 1 8 5
8 4 3 2 6
9 5 1 4 8
10 5 3 1 7
Here is an alternative way to achieve your task with fill from tidyr package:
library(dplyr)
library(tidyr)
df %>%
group_by(id) %>%
fill(everything(), .direction = "down") %>%
fill(everything(), .direction = "up") %>%
slice(1,3)
id event score score2
<dbl> <dbl> <dbl> <dbl>
1 1 1 3 4
2 1 3 1 1
3 2 1 3 5
4 2 3 2 2
5 3 1 6 0
6 3 3 1 3
7 4 1 8 5
8 4 3 2 6
9 5 1 4 8
10 5 3 1 7
How about this?
library(dplyr)
df_e12 <- df %>%
filter(event %in% c(1, 2)) %>%
group_by(id) %>%
mutate(across(starts_with("score"), ~min(.x, na.rm = TRUE))) %>%
ungroup() %>%
distinct(id, .keep_all = TRUE)
df_e3 <- df %>%
filter(event == 3)
df <- bind_rows(df_e12, df_e3) %>%
arrange(id, event)
df
> df
# A tibble: 10 x 4
id event score score2
<dbl> <dbl> <dbl> <dbl>
1 1 1 3 4
2 1 3 1 1
3 2 1 3 5
4 2 3 2 2
5 3 1 6 0
6 3 3 1 3
7 4 1 8 5
8 4 3 2 6
9 5 1 4 8
10 5 3 1 7
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
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`)")
I want to create two variables giving me the total number of positive and negative values by id, hopefully using dplyr.
Example data:
library(dplyr)
set.seed(42)
df <- data.frame (id=rep(1:10,each=10),
ff=rnorm(100, 0,14 ))
> head(df,20)
id ff
1 1 19.1934183
2 1 -7.9057744
3 1 5.0837978
4 1 8.8600765
5 1 5.6597565
6 1 -1.4857432
7 1 21.1613080
8 1 -1.3252265
9 1 28.2579320
10 1 -0.8779974
11 2 18.2681752
12 2 32.0130355
13 2 -19.4440498
14 2 -3.9030427
15 2 -1.8664987
16 2 8.9033056
17 2 -3.9795409
18 2 -37.1903759
19 2 -34.1665370
20 2 18.4815868
the resulting dataset should look like:
> head(df,20)
id ff pos neg
1 1 19.1934183 6 4
2 1 -7.9057744 6 4
3 1 5.0837978 6 4
4 1 8.8600765 6 4
5 1 5.6597565 6 4
6 1 -1.4857432 6 4
7 1 21.1613080 6 4
8 1 -1.3252265 6 4
9 1 28.2579320 6 4
10 1 -0.8779974 6 4
11 2 18.2681752 4 6
12 2 32.0130355 4 6
13 2 -19.4440498 4 6
14 2 -3.9030427 4 6
15 2 -1.8664987 4 6
16 2 8.9033056 4 6
17 2 -3.9795409 4 6
18 2 -37.1903759 4 6
19 2 -34.1665370 4 6
20 2 18.4815868 4 6
I have thought something similar to this will work:
df<-df%>% group_by(id) %>% mutate(pos= nrow(ff>0)) %>% ungroup()
Any help would be great, thanks.
You need sum():
df %>% group_by(id) %>%
mutate(pos = sum(ff>0),
neg = sum(ff<0))
For a fun (and a fast) solution data.table can also be used:
library(data.table)
setDT(df)
df[, ":="(pos = sum(ff > 0), neg = sum(ff < 0)), by = id]
Here's an answer that add the ifelse part of your question:
df <- df %>% group_by(id) %>%
mutate(pos = sum(ff>0), neg = sum(ff<0)) %>%
group_by(id) %>%
mutate(any_neg=ifelse(any(ff < 0), 1, 0))
Output:
> head(df, 20)
Source: local data frame [20 x 5]
Groups: id [2]
id ff pos neg any_neg
<int> <dbl> <int> <int> <dbl>
1 1 19.1934183 6 4 1
2 1 -7.9057744 6 4 1
3 1 5.0837978 6 4 1
4 1 8.8600765 6 4 1
5 1 5.6597565 6 4 1
6 1 -1.4857432 6 4 1
7 1 21.1613080 6 4 1
8 1 -1.3252265 6 4 1
9 1 28.2579320 6 4 1
10 1 -0.8779974 6 4 1
11 2 18.2681752 4 6 1
12 2 32.0130355 4 6 1
13 2 -19.4440498 4 6 1
14 2 -3.9030427 4 6 1
15 2 -1.8664987 4 6 1
16 2 8.9033056 4 6 1
17 2 -3.9795409 4 6 1
18 2 -37.1903759 4 6 1
19 2 -34.1665370 4 6 1
20 2 18.4815868 4 6 1