I need to find a running maximum of a variable by group using R. The variable is sorted by time within group using df[order(df$group, df$time),].
My variable has some NA's but I can deal with it by replacing them with zeros for this computation.
this is how the data frame df looks like:
(df <- structure(list(var = c(5L, 2L, 3L, 4L, 0L, 3L, 6L, 4L, 8L, 4L),
group = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
.Label = c("a", "b"), class = "factor"),
time = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L)),
.Names = c("var", "group","time"),
class = "data.frame", row.names = c(NA, -10L)))
# var group time
# 1 5 a 1
# 2 2 a 2
# 3 3 a 3
# 4 4 a 4
# 5 0 a 5
# 6 3 b 1
# 7 6 b 2
# 8 4 b 3
# 9 8 b 4
# 10 4 b 5
And I want a variable curMax as:
var | group | time | curMax
5 a 1 5
2 a 2 5
3 a 3 5
4 a 4 5
0 a 5 5
3 b 1 3
6 b 2 6
4 b 3 6
8 b 4 8
4 b 5 8
Please let me know if you have any idea how to implement it in R.
We can try data.table. Convert the 'data.frame' to 'data.table' (setDT(df1)), grouped by 'group' , we get the cummax of 'var' and assign (:=) it to a new variable ('curMax')
library(data.table)
setDT(df1)[, curMax := cummax(var), by = group]
As commented by #Michael Chirico, if the data is not ordered by 'time', we can do that in the 'i'
setDT(df1)[order(time), curMax:=cummax(var), by = group]
Or with dplyr
library(dplyr)
df1 %>%
group_by(group) %>%
mutate(curMax = cummax(var))
If df1 is tbl_sql explicit ordering might be required, using arrange
df1 %>%
group_by(group) %>%
arrange(time, .by_group=TRUE) %>%
mutate(curMax = cummax(var))
or dbplyr::window_order
library(dbplyr)
df1 %>%
group_by(group) %>%
window_order(time) %>%
mutate(curMax = cummax(var))
you can do it so:
df$curMax <- ave(df$var, df$group, FUN=cummax)
Related
Despite using R and dplyr on a regular basis, I encountered the issue of not being able to calculate the sum of the absolute differences between all columns:
sum_diff=ABS(A-B)+ABS(B-C)+ABS(C-D)...
A
B
C
D
sum_diff
1
2
3
4
3
2
1
3
4
4
1
2
1
1
2
4
1
2
1
5
I know I could iterate using a for loop over all columns, but given the size of my data frame, I prefer a more elegant and fast solution.
Any help?
Thank you
We may remove the first and last columns, get the difference, and use rowSums on the absolute values in base R. This could be very efficient compared to a package solution
df1$sum_diff <- rowSums(abs(df1[-ncol(df1)] - df1[-1]))
-output
> df1
A B C D sum_diff
1 1 2 3 4 3
2 2 1 3 4 4
3 1 2 1 1 2
4 4 1 2 1 5
Or another option is rowDiffs from matrixStats
library(matrixStats)
rowSums(abs(rowDiffs(as.matrix(df1))))
[1] 3 4 2 5
data
df1 <- structure(list(A = c(1L, 2L, 1L, 4L), B = c(2L, 1L, 2L, 1L),
C = c(3L, 3L, 1L, 2L), D = c(4L, 4L, 1L, 1L)), row.names = c(NA,
-4L), class = "data.frame")
Daata from akrun (many thanks)!
This is complicated the idea is to generate a list of the combinations, I tried it with combn but then I get all possible combinations. So I created by hand.
With this combinations we then could use purrrs map_dfc and do some data wrangling after that:
library(tidyverse)
combinations <-list(c("A", "B"), c("B", "C"), c("C","D"))
purrr::map_dfc(combinations, ~{df <- tibble(a=data[[.[[1]]]]-data[[.[[2]]]])
names(df) <- paste0(.[[1]],"_v_",.[[2]])
df}) %>%
transmute(sum_diff = rowSums(abs(.))) %>%
bind_cols(data)
sum_diff A B C D
<dbl> <int> <int> <int> <int>
1 3 1 2 3 4
2 4 2 1 3 4
3 2 1 2 1 1
4 5 4 1 2 1
data:
data <- structure(list(A = c(1L, 2L, 1L, 4L), B = c(2L, 1L, 2L, 1L),
C = c(3L, 3L, 1L, 2L), D = c(4L, 4L, 1L, 1L)), row.names = c(NA,
-4L), class = "data.frame")
Here is a dplyrs version of #akrun's elegant aproach that calculates the diff of the dataframe with it's shifted variant:
df %>%
mutate(sum_diff = rowSums(abs(identity(.) %>% select(1:last_col(1))
- identity(.) %>% select(2:last_col()))))
And here we have the rowwise variant, which basicly follows the same idea but this time every row is used as a vector that get's substracted by it's shifted self.
df %>%
rowwise() %>%
mutate(sum_diff = map2_int(c_across(1:last_col(1)),
c_across(2:last_col()),
~ abs(.x - .y)) %>% sum())
I need to find a running maximum of a variable by group using R. The variable is sorted by time within group using df[order(df$group, df$time),].
My variable has some NA's but I can deal with it by replacing them with zeros for this computation.
this is how the data frame df looks like:
(df <- structure(list(var = c(5L, 2L, 3L, 4L, 0L, 3L, 6L, 4L, 8L, 4L),
group = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
.Label = c("a", "b"), class = "factor"),
time = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L)),
.Names = c("var", "group","time"),
class = "data.frame", row.names = c(NA, -10L)))
# var group time
# 1 5 a 1
# 2 2 a 2
# 3 3 a 3
# 4 4 a 4
# 5 0 a 5
# 6 3 b 1
# 7 6 b 2
# 8 4 b 3
# 9 8 b 4
# 10 4 b 5
And I want a variable curMax as:
var | group | time | curMax
5 a 1 5
2 a 2 5
3 a 3 5
4 a 4 5
0 a 5 5
3 b 1 3
6 b 2 6
4 b 3 6
8 b 4 8
4 b 5 8
Please let me know if you have any idea how to implement it in R.
We can try data.table. Convert the 'data.frame' to 'data.table' (setDT(df1)), grouped by 'group' , we get the cummax of 'var' and assign (:=) it to a new variable ('curMax')
library(data.table)
setDT(df1)[, curMax := cummax(var), by = group]
As commented by #Michael Chirico, if the data is not ordered by 'time', we can do that in the 'i'
setDT(df1)[order(time), curMax:=cummax(var), by = group]
Or with dplyr
library(dplyr)
df1 %>%
group_by(group) %>%
mutate(curMax = cummax(var))
If df1 is tbl_sql explicit ordering might be required, using arrange
df1 %>%
group_by(group) %>%
arrange(time, .by_group=TRUE) %>%
mutate(curMax = cummax(var))
or dbplyr::window_order
library(dbplyr)
df1 %>%
group_by(group) %>%
window_order(time) %>%
mutate(curMax = cummax(var))
you can do it so:
df$curMax <- ave(df$var, df$group, FUN=cummax)
I am working with a gigantic person-period file and I thought that
a good way to deal with a large dataset is by using sampling and re-sampling technique.
My person-period file look like this
id code time
1 1 a 1
2 1 a 2
3 1 a 3
4 2 b 1
5 2 c 2
6 2 b 3
7 3 c 1
8 3 c 2
9 3 c 3
10 4 c 1
11 4 a 2
12 4 c 3
13 5 a 1
14 5 c 2
15 5 a 3
I have actually two distinct issues.
The first issue is that I am having trouble in simply sampling a person-period file.
For example, I would like to sample 2 id-sequences such as :
id code time
1 a 1
1 a 2
1 a 3
2 b 1
2 c 2
2 b 3
The following line of code is working for sampling a person-period file
dt[which(dt$id %in% sample(dt$id, 2)), ]
However, I would like to use a dplyr solution because I am interested in resampling and in particular I would like to use replicate.
I am interested in doing something like replicate(100, sample_n(dt, 2), simplify = FALSE)
I am struggling with the dplyr solution because I am not sure what should be the grouping variable.
library(dplyr)
dt %>% group_by(id) %>% sample_n(1)
gives me an incorrect result because it does not keep the full sequence of each id.
Any clue how I could both sample and re-sample person-period file ?
data
dt = structure(list(id = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L,
3L, 4L, 4L, 4L, 5L, 5L, 5L), .Label = c("1", "2", "3", "4", "5"
), class = "factor"), code = structure(c(1L, 1L, 1L, 2L, 3L,
2L, 3L, 3L, 3L, 3L, 1L, 3L, 1L, 3L, 1L), .Label = c("a", "b",
"c"), class = "factor"), time = structure(c(1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("1", "2",
"3"), class = "factor")), .Names = c("id", "code", "time"), row.names = c(NA,
-15L), class = "data.frame")
I think the idiomatic way would probably look like
set.seed(1)
samp = df %>% select(id) %>% distinct %>% sample_n(2)
left_join(samp, df)
id code time
1 2 b 1
2 2 c 2
3 2 b 3
4 5 a 1
5 5 c 2
6 5 a 3
This extends straightforwardly to more grouping variables and fancier sampling rules.
If you need to do this many times...
nrep = 100
ng = 2
samps = df %>% select(id) %>% distinct %>%
slice(rep(1:n(), nrep)) %>% mutate(r = rep(1:nrep, each = n()/nrep)) %>%
group_by(r) %>% sample_n(ng)
repdat = left_join(samps, df)
# then do stuff with it:
repdat %>% group_by(r) %>% do_stuff
I imagine you are doing some simulations and may want to do the subsetting many times. You probably also want to try this data.table method and utilize the fast binary search feature on the key column:
library(data.table)
setDT(dt)
setkey(dt, id)
replicate(2, dt[list(sample(unique(id), 2))], simplify = F)
#[[1]]
# id code time
#1: 3 c 1
#2: 3 c 2
#3: 3 c 3
#4: 5 a 1
#5: 5 c 2
#6: 5 a 3
#[[2]]
# id code time
#1: 3 c 1
#2: 3 c 2
#3: 3 c 3
#4: 4 c 1
#5: 4 a 2
#6: 4 c 3
We can use filter with sample
dt %>%
filter(id %in% sample(unique(id),2, replace = FALSE))
NOTE: The OP specified using dplyr method and this solution does uses the dplyr.
If we need to do replicate one option would be using map from purrr
library(purrr)
dt %>%
distinct(id) %>%
replicate(2, .) %>%
map(~sample(., 2, replace=FALSE)) %>%
map(~filter(dt, id %in% .))
#$id
# id code time
#1 1 a 1
#2 1 a 2
#3 1 a 3
#4 4 c 1
#5 4 a 2
#6 4 c 3
#$id
# id code time
#1 4 c 1
#2 4 a 2
#3 4 c 3
#4 5 a 1
#5 5 c 2
#6 5 a 3
I need to find a running maximum of a variable by group using R. The variable is sorted by time within group using df[order(df$group, df$time),].
My variable has some NA's but I can deal with it by replacing them with zeros for this computation.
this is how the data frame df looks like:
(df <- structure(list(var = c(5L, 2L, 3L, 4L, 0L, 3L, 6L, 4L, 8L, 4L),
group = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L),
.Label = c("a", "b"), class = "factor"),
time = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L)),
.Names = c("var", "group","time"),
class = "data.frame", row.names = c(NA, -10L)))
# var group time
# 1 5 a 1
# 2 2 a 2
# 3 3 a 3
# 4 4 a 4
# 5 0 a 5
# 6 3 b 1
# 7 6 b 2
# 8 4 b 3
# 9 8 b 4
# 10 4 b 5
And I want a variable curMax as:
var | group | time | curMax
5 a 1 5
2 a 2 5
3 a 3 5
4 a 4 5
0 a 5 5
3 b 1 3
6 b 2 6
4 b 3 6
8 b 4 8
4 b 5 8
Please let me know if you have any idea how to implement it in R.
We can try data.table. Convert the 'data.frame' to 'data.table' (setDT(df1)), grouped by 'group' , we get the cummax of 'var' and assign (:=) it to a new variable ('curMax')
library(data.table)
setDT(df1)[, curMax := cummax(var), by = group]
As commented by #Michael Chirico, if the data is not ordered by 'time', we can do that in the 'i'
setDT(df1)[order(time), curMax:=cummax(var), by = group]
Or with dplyr
library(dplyr)
df1 %>%
group_by(group) %>%
mutate(curMax = cummax(var))
If df1 is tbl_sql explicit ordering might be required, using arrange
df1 %>%
group_by(group) %>%
arrange(time, .by_group=TRUE) %>%
mutate(curMax = cummax(var))
or dbplyr::window_order
library(dbplyr)
df1 %>%
group_by(group) %>%
window_order(time) %>%
mutate(curMax = cummax(var))
you can do it so:
df$curMax <- ave(df$var, df$group, FUN=cummax)
I have a dataframe which looks like -
Id Result
A 1
B 2
C 1
B 1
C 1
A 2
B 1
B 2
C 1
A 1
B 2
Now I need to calculate how many 1's and 2's are there for each Id and then select the number whose frequency of occurrence is the greatest.
Id Result
A 1
B 2
C 1
How can I do that? I have tried using the table function in some way but not able to use it effectively. Any help would be appreciated.
Here you can use aggregate in one step:
df <- structure(list(Id = structure(c(1L, 2L, 3L, 2L, 3L, 1L, 2L, 2L,
3L, 1L, 2L), .Label = c("A", "B", "C"), class = "factor"),
Result = c(1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 2L)),
.Names = c("Id", "Result"), class = "data.frame", row.names = c(NA, -11L)
)
res <- aggregate(Result ~ Id, df, FUN=function(x){which.max(c(sum(x==1), sum(x==2)))})
res
Result:
Id Result
1 A 1
2 B 2
3 C 1
With data.table you can try (df is your data.frame):
require(data.table)
dt<-as.data.table(df)
dt[,list(times=.N),by=list(Id,Result)][,list(Result=Result[which.max(times)]),by=Id]
# Id Result
#1: A 1
#2: B 2
#3: C 1
Using dplyr, you can try
library(dplyr)
df %>% group_by(Id, Result) %>% summarize(n = n()) %>% group_by(Id) %>%
filter(n == max(n)) %>% summarize(Result = Result)
Id Result
1 A 1
2 B 2
3 C 1
An option using table and ave
subset(as.data.frame(table(df1)),ave(Freq, Id, FUN=max)==Freq, select=-3)
# Id Result
# 1 A 1
# 3 C 1
# 5 B 2