I have a data frame that looks like this:
df <- data.frame(
id = c(1, 2, 3, 4, 5),
generation = as.factor(c(3, 2, 4, 3, 4)),
income = as.factor(c(4, 3, 3, 7, 3)),
fem = as.factor(c(0, 0, 1, 0, 1))
)
where id is an identifier for individuals in the data set and generation, income and fem are categorical characteristics of the individuals. Now, I want to put the individuals into cohorts ("groups") based on the individual characteristics, where individuals with the exact same values for the individual characteristics should get the same cohort_id. Hence, I want the following result:
data.frame(
id = c(1, 2, 3, 4, 5),
generation = as.factor(c(3, 2, 4, 3, 4)),
income = as.factor(c(4, 3, 3, 7, 3)),
fem = as.factor(c(0, 0, 1, 0, 1)),
cohort_id = as.factor(c(1, 2, 3, 4, 3))
)
Note that id = 3 and id = 5 get the same cohort_id as they have the same characteristcs.
My question is whether there is a fast way to create the cohort_ids without using multiple case_when or ifelse over and over again? This can get quite tedious if you want to build many cohorts. A solution using dplyr would be nice but is not necessary.
There are multiple ways to do this - one option is to paste the columns and match with the unique values
library(dplyr)
library(stringr)
df %>%
mutate(cohort_id = str_c(generation, income, fem),
cohort_id = match(cohort_id, unique(cohort_id)))
-output
id generation income fem cohort_id
1 1 3 4 0 1
2 2 2 3 0 2
3 3 4 3 1 3
4 4 3 7 0 4
5 5 4 3 1 3
The following code will create an index 'cohort_id' with values a little different from the provided expected, but compliant with the grouping rules:
library(dplyr)
df %>% group_by(generation, income, fem) %>%
mutate(cohort_id = cur_group_id())%>%
ungroup()
# A tibble: 5 × 5
id generation income fem cohort_id
<dbl> <fct> <fct> <fct> <int>
1 1 3 4 0 2
2 2 2 3 0 1
3 3 4 3 1 4
4 4 3 7 0 3
5 5 4 3 1 4
Related
My dataframe contains data about political careers, such as a unique identifier (called: ui) column for each politician and the electoral term(called: electoral_term) in which they were elected. Since a politician can be elected in multiple electoral terms, there are multiple rows that contain the same ui.
Now I would like to add another column to my dataframe, that counts how many times the politician got re-elected.
So e.g. the politician with ui=1 was re-elected 2 times, since he occured in 3 electoral_terms.
I already tried
df %>% count(ui)
But that only gives out a table which can't be added into my dataframe.
Thanks in advance!
We may use base R
df$reelected <- with(df, ave(ui, ui, FUN = length)-1)
-output
> df
ui electoral reelected
1 1 1 2
2 1 2 2
3 1 3 2
4 2 2 0
5 3 7 1
6 3 9 1
data
df <- structure(list(ui = c(1, 1, 1, 2, 3, 3), electoral = c(1, 2,
3, 2, 7, 9)), class = "data.frame", row.names = c(NA, -6L))
mydf <- tibble::tribble(~ui, ~electoral, 1, 1, 1, 2, 1, 3, 2, 2, 3, 7, 3, 9)
library(dplyr)
df |>
add_count(ui, name = "re_elected") |>
mutate(re_elected = re_elected - 1)
# A tibble: 6 × 3
ui electoral re_elected
<dbl> <dbl> <dbl>
1 1 1 2
2 1 2 2
3 1 3 2
4 2 2 0
5 3 7 1
6 3 9 1
library(tidyverse)
df %>%
group_by(ui) %>%
mutate(re_elected = n() - 1)
# A tibble: 6 × 3
# Groups: ui [3]
ui electoral re_elected
<dbl> <dbl> <dbl>
1 1 1 2
2 1 2 2
3 1 3 2
4 2 2 0
5 3 7 1
6 3 9 1
I have 5 data frames. I want to recode all variables ending with "_comfort", "_agree", and "effective" using the same rules for each data frame. As is, the values in each column are 1:5 and I want is to recode 5's to 1, 4's to 2, 2's to 4, and 5's to 1 (3 will stay the same).
I do not want the end result to one merged dataset, but instead to apply the same recoding rules across all 5 independent data frames. For simplicity sake, let's just assume I have 2 data frames:
df1 <- data.frame(a_comfort = c(1, 2, 3, 4, 5),
b_comfort = c(1, 2, 3, 4, 5),
c_effective = c(1, 2, 3, 4, 5))
df2 <- data.frame(a_comfort = c(1, 2, 3, 4, 5),
b_comfort = c(1, 2, 3, 4, 5),
c_effective = c(1, 2, 3, 4, 5))
What I want is:
df1 <- data.frame(a_comfort = c(5, 4, 3, 2, 1),
b_comfort = c(5, 4, 3, 2, 1),
c_effective = c(5, 4, 3, 2, 1))
df2 <- data.frame(a_comfort = c(5, 4, 3, 2, 1),
b_comfort = c(5, 4, 3, 2, 1),
c_effective = c(5, 4, 3, 2, 1))
Conventionally, I would use dplyr's mutate_atand ends_withto achieve my goal, but have not been successful with this method across multiple data frames. I am thinking a combination of the purr and dplyr packages will work, but haven't nailed down the exact technique.
Thanks in advance for any help!
You can use get() and assign() in a loop:
library(dplyr)
for (df_name in c("df1", "df2")) {
df <- mutate(
get(df_name),
across(
ends_with(c("_comfort", "_agree", "_effective")),
\(x) 6 - x
)
)
assign(df_name, df)
}
Result:
#> df1
a_comfort b_comfort c_effective
1 5 5 5
2 4 4 4
3 3 3 3
4 2 2 2
5 1 1 1
#> df2
a_comfort b_comfort c_effective
1 5 5 5
2 4 4 4
3 3 3 3
4 2 2 2
5 1 1 1
Note, however, it’s often better practice to keep multiple related dataframes in a list than loose in the global environment (see). In this case, you can use purrr::map() (or base::lapply()):
library(dplyr)
library(purrr)
dfs <- list(df1, df2)
dfs <- map(
dfs,
\(df) mutate(
df,
across(
ends_with(c("_comfort", "_agree", "_effective")),
\(x) 6 - x
)
)
)
Result:
#> dfs
[[1]]
a_comfort b_comfort c_effective
1 5 5 5
2 4 4 4
3 3 3 3
4 2 2 2
5 1 1 1
[[2]]
a_comfort b_comfort c_effective
1 5 5 5
2 4 4 4
3 3 3 3
4 2 2 2
5 1 1 1
You can use ls(pattern = 'df\\d+') to find all objects whose names match a certain pattern. Then store them into a list and pass to purrr::map or lapply for recoding.
library(dplyr)
df.lst <- purrr::map(
mget(ls(pattern = 'df\\d+')),
~ .x %>% mutate(6 - across(ends_with(c("_comfort", "_agree", "effective"))))
)
# $df1
# a_comfort b_comfort c_effective
# 1 5 5 5
# 2 4 4 4
# 3 3 3 3
# 4 2 2 2
# 5 1 1 1
#
# $df2
# a_comfort b_comfort c_effective
# 1 5 5 5
# 2 4 4 4
# 3 3 3 3
# 4 2 2 2
# 5 1 1 1
You can further overwrite those dataframes in your workspace from the list through list2env().
list2env(df.lst, .GlobalEnv)
Please try the below code where i convert the columns to factor and then recode them
data
a_comfort b_comfort c_effective
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
code
library(tidyverse)
df1 %>% mutate(across(c(ends_with('comfort'),ends_with('effective')), ~ factor(.x, levels=c('1','2','3','4','5'), labels=c('5','4','3','2','1'))))
output
a_comfort b_comfort c_effective
1 5 5 5
2 4 4 4
3 3 3 3
4 2 2 2
5 1 1 1
I have the following dataframe:
df1 <- data.frame(ID = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2),
var1 = c(0, 2, 3, 4, 2, 5, 6, 10, 11, 0, 1, 2, 1, 5, 7, 10))
I want to select only the rows containing values up to 5, once 5 is reached I want it to go to the next ID and select only values up to 5 for that group so that the final result would look like this:
ID var1
1 0
1 2
1 3
1 4
1 2
1 5
2 0
2 1
2 2
2 1
2 5
I would like to try something with dplyr as it is what I am most familiar with.
You could use which.max() to find the first occurrence of var1 >= 5, and then extract those rows whose row numbers are before it.
library(dplyr)
df1 %>%
group_by(ID) %>%
filter(row_number() <= which.max(var1 >= 5)) %>%
ungroup()
or
df1 %>%
group_by(ID) %>%
slice(1:which.max(var1 >= 5)) %>%
ungroup()
# # A tibble: 11 × 2
# ID var1
# <dbl> <dbl>
# 1 1 0
# 2 1 2
# 3 1 3
# 4 1 4
# 5 1 2
# 6 1 5
# 7 2 0
# 8 2 1
# 9 2 2
# 10 2 1
# 11 2 5
I have the following data.frame:
df <- data.frame(date = c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3),
id = c(4, 4, 2, 4, 1, 2, 3, 1, 2, 2, 1, 1))
And I want to add a new column grp which, for each date, ranks the IDs. Ties should have the same value, but there should be no omitted values. That is, if there are two values which are equally minimum, they should both get rank 1, and the next lowest values should get rank 2.
The expected result would therefore look like this. Note that, as mentioned, the groups are for each date, so the operation must be grouped by date.
data.frame(date = c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3),
id = c(4, 4, 2, 4, 1, 2, 3, 1, 2, 2, 1, 1),
grp = c(2, 2, 1, 2, 1, 2, 3, 1, 2, 2, 1, 1))
I'm sure there's a trivial way to do this but I haven't found it: none of the options for tie.method behave in this way (data.table::frank also doesn't help, since it only adds a dense rank).
I thought of doing a normal rank and then using data.table::rleid, but that doesn't work if there are duplicate values separated by other values during the same day.
I also thought of grouping by date and id and then using a group-ID, but the lowest values each day must start at rank 1, so that won't work either.
The only functional solution I've found is to create another table with the unique ids per day and then join that table to this one:
suppressPackageStartupMessages(library(dplyr))
df <- data.frame(date = c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3),
id = c(4, 4, 2, 4, 1, 2, 3, 1, 2, 2, 1, 1))
uniques <- df %>%
group_by(
date
) %>%
distinct(
id
) %>%
mutate(
grp = rank(id)
)
df <- df %>% left_join(
unique
) %>% print()
#> Joining, by = c("date", "id")
#> date id grp
#> 1 1 4 2
#> 2 1 4 2
#> 3 1 2 1
#> 4 1 4 2
#> 5 2 1 1
#> 6 2 2 2
#> 7 2 3 3
#> 8 2 1 1
#> 9 3 2 2
#> 10 3 2 2
#> 11 3 1 1
#> 12 3 1 1
Created on 2020-05-08 by the reprex package (v0.3.0)
However, this seems quite inelegant and convoluted for what seems like a simple operation, so I'd rather see if other solutions are available.
Curious to see data.table solutions if available, but unfortunately the solution must be in dplyr.
We can use dense_rank
library(dplyr)
df %>%
group_by(date) %>%
mutate(grp = dense_rank(id))
# A tibble: 12 x 3
# Groups: date [3]
# date id grp
# <dbl> <dbl> <int>
# 1 1 4 2
# 2 1 4 2
# 3 1 2 1
# 4 1 4 2
# 5 2 1 1
# 6 2 2 2
# 7 2 3 3
# 8 2 1 1
# 9 3 2 2
#10 3 2 2
#11 3 1 1
#12 3 1 1
Or with frank
library(data.table)
setDT(df)[, grp := frank(id, ties.method = 'dense'), date]
I'm trying to generate random number by group with multiple times.
For example,
> set.seed(1002)
> df<-data.frame(ID=LETTERS[seq(1:5)],num=sample(c(2,3,4), size=5, replace=TRUE))
> df
ID num
1 A 3
2 B 4
3 C 3
4 D 2
5 E 3
In ID, I want to generate sequential random number without replacement with (for example) 4 times.
If ID is A, it will randomly select numbers among 1:3 4 times. So, this will be
sample(c(1,2,3,1,2,3,1,2,3),replace=FALSE)
or
ep(sample(c(1:4), replace=FALSE),times=4)
If the results is 3 2 1 2 1 3 2 3 3 1 1 2, then the data will be
ID num
1 A 3
2 A 2
3 A 2
4 A 1
5 A 1
6 A 3
7 A 2
8 A 1
9 A 3
I tried several things, like
df%>%group_by(ID)%>%mutate(random=sample(rep(1:num,times=4),replace=FALSE))
It failed. The warning appeared with In 1:num
I also tried this.
ddply(df,.(ID),function(x) sample(rep(1:num,times=4),replace=FALSE))
The error appeared again, with NA/NaN.
I would really appreciate if you let me know how to solve this problem.
We can create a list-column and then unnest it to have separate rows.
n <- 4
library(dplyr)
df %>%
group_by(ID) %>%
mutate(num = list(sample(rep(seq_len(num), n)))) %>%
tidyr::unnest(num)
# ID num
# <fct> <int>
# 1 A 2
# 2 A 2
# 3 A 2
# 4 A 3
# 5 A 3
# 6 A 1
# 7 A 3
# 8 A 1
# 9 A 1
#10 A 3
# … with 50 more rows
I'm not quite clear on your expected output.
The following samples num elements from 1:num with replacement, and stores samples in a list column sample.
library(tidyverse)
set.seed(2018)
df %>% mutate(sample = map(num, ~sample(1:.x, replace = T)))
# ID num sample
#1 A 2 1, 1
#2 B 4 3, 4, 1, 2
#3 C 2 1, 1
#4 D 4 3, 3, 4, 4
#5 E 2 2, 2
Or if you want to repeat sampling num elements (with replacement) 4 times, you can do
set.seed(2018)
df %>%
mutate(sample = map(num, ~as.numeric(replicate(4, sample(1:.x, replace = T)))))
#ID num sample
#1 A 2 1, 1, 1, 2, 1, 2, 1, 1
#2 B 4 3, 3, 4, 4, 4, 4, 4, 2, 3, 4, 3, 3, 2, 1, 1, 2
#3 C 2 1, 1, 1, 1, 1, 1, 1, 2
#4 D 4 2, 3, 2, 1, 3, 4, 1, 2, 1, 2, 2, 1, 1, 1, 2, 1
#5 E 2 2, 1, 2, 2, 1, 1, 1, 2