I have a df like this one:
id <- c(1,1,2,2,3,3,4,4,5,5)
v1 <- c(3,1,2,3,4,5,6,1,5,4)
pos <- c(1,2,1,2,1,2,1,2,1,2)
df <- data.frame(id,v1,pos)
How can I "randomize" the values of v1 WHILE keeping the inherent order from the "Id" var and also the values of "pos" such as I get df with randomized values like this:
id v1 pos
1 1 1
1 3 2
2 2 1
2 3 2
3 5 1
3 4 2
4 6 1
4 1 2
5 5 1
5 4 2
Above and example of resulting df with id and pos staying as originally created and v1 randomized.
Thx!
Is sample what you're looking for?
df %>%
group_by(id) %>%
mutate(v1 = sample(v1, size = length(v1)))
# A tibble: 10 x 3
# Groups: id [5]
id v1 pos
<dbl> <dbl> <dbl>
1 1 3 1
2 1 1 2
3 2 3 1
4 2 2 2
5 3 4 1
6 3 5 2
7 4 1 1
8 4 6 2
9 5 5 1
10 5 4 2
Related
I have some results cluster labels from kmeans done on different ids (reprex example below). the problem is the kmeans clusters codes are not ordered consistently across ids although all ids have 3 clusters.
reprex = data.frame(id = rep(1:2, each = 41,
v1 = rep(seq(1:4), 2),
cluster = c(2,2,1,3,3,1,2,2))
reprex
id v1 cluster
1 1 1 2
2 1 2 2
3 1 3 1
4 1 4 3
5 2 1 3
6 2 2 1
7 2 3 2
8 2 4 2
what I want is that the variable cluster should always start with 1 within each ID. Note I don't want to reorder that dataframe by cluster, the order needs to remain the same. so the desired result would be:
reprex_desired<- data.frame(id = rep(1:2, each = 4),
v1 = rep(seq(1:4), 2),
cluster = c(2,2,1,3,3,1,2,2),
what_iWant = c(1,1,2,3,1,2,3,3))
reprex_desired
id v1 cluster what_iWant
1 1 1 2 1
2 1 2 2 1
3 1 3 1 2
4 1 4 3 3
5 2 1 3 1
6 2 2 1 2
7 2 3 2 3
8 2 4 2 3
We can use match after grouping by 'id'
library(dplyr)
reprex <- reprex %>%
group_by(id) %>%
mutate(what_IWant = match(cluster, unique(cluster))) %>%
ungroup
-output
reprex
# A tibble: 8 × 4
id v1 cluster what_IWant
<int> <int> <dbl> <int>
1 1 1 2 1
2 1 2 2 1
3 1 3 1 2
4 1 4 3 3
5 2 1 3 1
6 2 2 1 2
7 2 3 2 3
8 2 4 2 3
Here is a version with cumsum combined with lag:
library(dplyr)
df %>%
group_by(id) %>%
mutate(what_i_want = cumsum(cluster != lag(cluster, def = first(cluster)))+1)
id v1 cluster what_i_want
<int> <int> <dbl> <dbl>
1 1 1 2 1
2 1 2 2 1
3 1 3 1 2
4 1 4 3 3
5 2 1 3 1
6 2 2 1 2
7 2 3 2 3
8 2 4 2 3
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
I am trying to create a new variable in df that takes the mode of a variable with the same ID number. I want the output to look like this:
ID X MODE
1 3 2
1 2 2
1 2 2
2 1 3
2 2 3
2 1 3
2 3 3
2 3 3
3 5 5
3 5 5
Any suggestions?
Maybe try this:
library(dplyr)
#Function
getmode <- function(v) {
uniqv <- sort(unique(v),decreasing = T)
uniqv[which.max(tabulate(match(v, uniqv)))]
}
#Code
newdf <- df %>% group_by(ID) %>% mutate(Mode2=getmode(X))
Output:
# A tibble: 10 x 4
# Groups: ID [3]
ID X MODE Mode2
<int> <int> <int> <int>
1 1 3 2 2
2 1 2 2 2
3 1 2 2 2
4 2 1 3 3
5 2 2 3 3
6 2 1 3 3
7 2 3 3 3
8 2 3 3 3
9 3 5 5 5
10 3 5 5 5
I have a list let's say
k<-c(1,2,3,4)
I want to create a dataframe with let's say 5 rows using the same list in each row as shown below.
X1 X2 X3 X4
1 1 2 3 4
2 1 2 3 4
3 1 2 3 4
4 1 2 3 4
5 1 2 3 4
I tried doing:-
> rep(k, each = 5)
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4
However I am not able to get intended result. Any suggestions?
data.frame(t(replicate(5, k)))
#OR
data.frame(matrix(rep(k, each = 5), 5))
#OR
data.frame(t(sapply(1:5, function(x) k)))
# X1 X2 X3 X4
#1 1 2 3 4
#2 1 2 3 4
#3 1 2 3 4
#4 1 2 3 4
#5 1 2 3 4
Here is one option by converting the vector to list with as.list, change it to data.frame (as.data.frame and replicate the rows
as.data.frame(as.list(k))[rep(1, 5),]
# X1 X2 X3 X4
#1 1 2 3 4
#1.1 1 2 3 4
#1.2 1 2 3 4
#1.3 1 2 3 4
#1.4 1 2 3 4
Or another option is to take the transpose of the vector to get a row matrix, replicate the rows and convert to data.frame
as.data.frame(t(k)[rep(1, 5),])
In tidyverse, one option is to convert to tibble and then uncount
library(dplyr)
library(tidyr)
library(stringr)
as.list(k) %>%
set_names(str_c("X", seq_along(k))) %>%
as_tibble %>%
uncount(5)
# A tibble: 5 x 4
# X1 X2 X3 X4
# <dbl> <dbl> <dbl> <dbl>
#1 1 2 3 4
#2 1 2 3 4
#3 1 2 3 4
#4 1 2 3 4
#5 1 2 3 4
purrr::map_dfc(k, rep, 5)
# # A tibble: 5 x 4
# V1 V2 V3 V4
# <dbl> <dbl> <dbl> <dbl>
# 1 1 2 3 4
# 2 1 2 3 4
# 3 1 2 3 4
# 4 1 2 3 4
# 5 1 2 3 4
Using data.table:
k = c(1,2,3,4)
n = 5 # Number of rows
df = data.table()
df = df[, lapply(1:length(k), function(x) rep(k[x], n))]
> df
V1 V2 V3 V4
1: 1 2 3 4
2: 1 2 3 4
3: 1 2 3 4
4: 1 2 3 4
5: 1 2 3 4
I have 2 dataframes. One has a ID column with alot of arranged IDs.
The other one has just specific rows of the first column. Those are my markers.
I need to get the sum of the of the values in a specific column based on the id values of the second column.
The first column may be
id goals cards group
1 2 2 1
2 3 2 1
3 4 2 1
4 5 1 1
5 1 2 1
1 2 2 2
2 3 2 2
3 4 2 2
4 5 1 3
5 1 2 3
the second one:
id goals cards group
2 3 2 1
5 1 2 1
2 3 2 2
3 4 2 2
5 1 2 3
what i need to get:
id goals cards group points
1 2 2 1 2-(2+2)
2 3 2 1 0 cause in second list
3 4 2 1 4-(2+1+2)
4 5 1 1 5-(1+2)
5 1 2 1 0 cause in second list
1 2 2 2 2-(2+2)
2 3 2 2 0
3 4 2 2 0
4 5 1 3 5-(1+2)
5 1 2 3 0
Something like: ??
df1<- df1%>%
rowwise() %>%
mutate(points=
goals
-(sum( df1$cards[df1$id <= df2$id & df1$id>df1$id])))
df1 = read.table(text = "
id goals cards
1 2 2
2 3 2
3 4 2
4 5 1
5 1 2
", header=T)
df2 = read.table(text = "
id goals cards
2 3 2
5 1 2
", header=T)
library(dplyr)
# function that gets an id and returns the sum of cards based on df2
GetSumOfCards = function(x) {
ids = min(df2$id[df2$id >= x]) # for a given id of df1 find the minimum id in df2 that is bigger than this id
ifelse(x %in% df2$id, # if the given id exists in df2
0, # sum of cards is zero
sum(df1$cards[df1$id >= x & df1$id <= ids])) # otherwise get sum of cards in df1 from this id until the id obtained before
}
# update function to be vectorised
GetSumOfCards = Vectorize(GetSumOfCards)
df1 %>%
mutate(sum_cards = GetSumOfCards(id), # get sum of cards for each id using the function
points = goals - sum_cards) # get the points
# id goals cards sum_cards points
# 1 1 2 2 4 -2
# 2 2 3 2 0 3
# 3 3 4 2 5 -1
# 4 4 5 1 3 2
# 5 5 1 2 0 1
Based on your updated question, applying a similar function to every row makes the process very slow. So, this solution groups data in a way that you can just count the cards on chunks of data/rows:
df1 = read.table(text = "
id goals cards group
1 2 2 1
2 3 2 1
3 4 2 1
4 5 1 1
5 1 2 1
1 2 2 2
2 3 2 2
3 4 2 2
4 5 1 3
5 1 2 3
", header=T)
df2 = read.table(text = "
id goals cards group
2 3 2 1
5 1 2 1
2 3 2 2
3 4 2 2
5 1 2 3
", header=T)
library(dplyr)
df1 %>%
arrange(group, desc(id)) %>% # order by group and id descending (this will help with counting the cards)
left_join(df2 %>% # join specific columns of df2 and add a flag to know that this row exists in df2
select(id, group) %>%
mutate(flag = 1), by=c("id","group")) %>%
mutate(flag = ifelse(is.na(flag), 0, flag), # replace NA with 0
flag2 = cumsum(flag)) %>% # this flag will create the groups we need to count cards
group_by(group, flag2) %>% # for each new group (we need both as the card counting will change when we have a row from df2, or if group changes)
mutate(sum_cards = ifelse(flag == 1, 0, cumsum(cards))) %>% # get cummulative sum of cards unless the flag = 1, where we need 0 cards
ungroup() %>% # forget the grouping
arrange(group, id) %>% # back to original order
mutate(points = goals - sum_cards) %>% # calculate points
select(-flag, -flag2) # remove flags
# # A tibble: 10 x 6
# id goals cards group sum_cards points
# <int> <int> <int> <int> <dbl> <dbl>
# 1 1 2 2 1 4 -2
# 2 2 3 2 1 0 3
# 3 3 4 2 1 5 -1
# 4 4 5 1 1 3 2
# 5 5 1 2 1 0 1
# 6 1 2 2 2 4 -2
# 7 2 3 2 2 0 3
# 8 3 4 2 2 0 4
# 9 4 5 1 3 3 2
# 10 5 1 2 3 0 1