I have a dataframe including a column of factors that I would like to subset to select every nth row, after grouping by factor level. For example,
my_df <- data.frame(col1 = c(1:12), col2 = rep(c("A","B", "C"), 4))
my_df
col1 col2
1 1 A
2 2 B
3 3 C
4 4 A
5 5 B
6 6 C
7 7 A
8 8 B
9 9 C
10 10 A
11 11 B
12 12 C
Subsetting to select every 2nd row should yield my_new_df as,
col1 col2
1 4 A
2 10 A
3 5 B
4 11 B
5 6 C
6 12 C
I tried in dplyr:
my_df %>% group_by(col2) %>%
my_df[seq(2, nrow(my_df), 2), ] -> my_new_df
I get an error:
Error: Can't subset columns that don't exist.
x Locations 4, 6, 8, 10, and 12 don't exist.
ℹ There are only 2 columns.
To see if the nrow function was a problem, I tried using the number directly. So,
my_df %>% group_by(col2) %>%
my_df[seq(2, 4, 2), ] -> my_new_df
Also gave an error,
Error: Can't subset columns that don't exist.
x Location 4 doesn't exist.
ℹ There are only 2 columns.
Run `rlang::last_error()` to see where the error occurred.
My expectation was that it would run the subsetting on each group of data and then combine them into 'my_new_df'. My understanding of how group_by works is clearly wrong but I am stuck on how to move past this error. Any help would much appreciated.
Try:
my_df %>%
group_by(col2)%>%
slice(seq(from = 2, to = n(), by = 2))
# A tibble: 6 x 2
# Groups: col2 [3]
col1 col2
<int> <chr>
1 4 A
2 10 A
3 5 B
4 11 B
5 6 C
6 12 C
You might want to ungroup after slicing if you want to do other operations not based on col2.
Here is a data.table option:
library(data.table)
data <- as.data.table(my_df)
data[(rowid(col2) %% 2) == 0]
col1 col2
1: 4 A
2: 5 B
3: 6 C
4: 10 A
5: 11 B
6: 12 C
Or base R:
my_df[as.logical(with(my_df, ave(col1, col2, FUN = function(x)
seq_along(x) %% 2 == 0))), ]
col1 col2
4 4 A
5 5 B
6 6 C
10 10 A
11 11 B
12 12 C
Related
I have a dataframe in the following format with ID's and A/B's. The dataframe is very long, over 3000 ID's.
id
type
1
A
2
B
3
A
4
A
5
B
6
A
7
B
8
A
9
B
10
A
11
A
12
A
13
B
...
...
I need to remove all rows (A+B), where more than one A is behind another one or more. So I dont want to remove the duplicates. If there are a duplicate (2 or more A's), i want to remove all A's and the B until the next A.
id
type
1
A
2
B
6
A
7
B
8
A
9
B
...
...
Do I need a loop for this problem? I hope for any help,thank you!
This might be what you want:
First, define a function that notes the indices of what you want to remove:
row_sequence <- function(value) {
inds <- which(value == lead(value))
sort(unique(c(inds, inds + 1, inds +2)))
}
Apply the function to your dataframe by first extracting the rows that you want to remove into df1 and second anti_joining df1 with df to obtain the final dataframe:
library(dplyr)
df1 <- df %>% slice(row_sequence(type))
df2 <- df %>%
anti_join(., df1)
Result:
df2
id type
1 1 A
2 2 B
3 6 A
4 7 B
5 8 A
6 9 B
Data:
df <- data.frame(
id = 1:13,
type = c("A","B","A","A","B","A","B","A","B","A","A","A","B")
)
I imagined there is only one B after a series of duplicated A values, however if that is not the case just let me know to modify my codes:
library(dplyr)
library(tidyr)
library(data.table)
df %>%
mutate(rles = data.table::rleid(type)) %>%
group_by(rles) %>%
mutate(rles = ifelse(length(rles) > 1, NA, rles)) %>%
ungroup() %>%
mutate(rles = ifelse(!is.na(rles) & is.na(lag(rles)) & type == "B", NA, rles)) %>%
drop_na() %>%
select(-rles)
# A tibble: 6 x 2
id type
<int> <chr>
1 1 A
2 2 B
3 6 A
4 7 B
5 8 A
6 9 B
Data
df <- read.table(header = TRUE, text = "
id type
1 A
2 B
3 A
4 A
5 B
6 A
7 B
8 A
9 B
10 A
11 A
12 A
13 B")
Following a question I came across today, I would like to know how I can use bind_rows function in a pipe while avoiding duplication and NA values. Consider I have the following simple tibble:
df <- tibble(
col1 = c(3, 4, 5),
col2 = c(5, 3, 1),
col3 = c(6, 4, 9),
col4 = c(9, 6, 5)
)
I would like to bind col1 & col2 row-wise with col3 & col4 so that I have a tibble with 2 columns and 6 observations. In the end changing the names of the columns to colnew1 and colnew2.
But when I use bind_rows I got the following output with a lot of duplications and NA values.
df %>%
bind_rows(
select(., 1:2),
select(., 3:4)
)
# A tibble: 9 x 4
col1 col2 col3 col4
<dbl> <dbl> <dbl> <dbl>
1 3 5 6 9
2 4 3 4 6
3 5 1 9 5
4 3 5 NA NA
5 4 3 NA NA
6 5 1 NA NA
7 NA NA 6 9
8 NA NA 4 6
9 NA NA 9 5
# My desired output would be something like this:
f1 <- function(x) {
df <- x %>%
set_names(nm = rep(c("newcol1", "newcol2"), 2))
bind_rows(df[, c(1, 2)], df[, c(3, 4)])
}
f1(df)
# A tibble: 6 x 2
newcol1 newcol2
<dbl> <dbl>
1 3 5
2 4 3
3 5 1
4 6 9
5 4 6
6 9 5
I can get the desired output without a pipe but first I would like to know how I could use bind_rows in a pipe without getting NA values and duplications and second whether I could use select function in bind_rows as I remember once Hadley Wickham used filter function wrapped by bind_rows.
I would appreciate any explanation to this problem and thank you in advance.
Select the first two columns and bind_rows col3 col4 to col1 and col2 then use transmute
df1 <- df %>%
select(col1, col2) %>%
bind_rows(
df %>%
transmute(col1 = col3, col2 = col4)
)
Results:
# A tibble: 6 x 2
col1 col2
<dbl> <dbl>
1 3 5
2 4 3
3 5 1
4 6 9
5 4 6
6 9 5
I am struggling with one maybe easy question. I have a dataframe of 1 column with n rows (n is a multiple of 3). I would like to add a second column with integers like: 1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,.. How can I achieve this with dplyr as a general solution for different length of rows (all multiple of 3).
I tried this:
df <- tibble(Col1 = c(1:12)) %>%
mutate(Col2 = rep(1:4, each=3))
This works. But I would like to have a solution for n rows, each = 3 . Many thanks!
You can specify each and length.out parameter in rep.
library(dplyr)
tibble(Col1 = c(1:12)) %>%
mutate(Col2 = rep(row_number(), each=3, length.out = n()))
# Col1 Col2
# <int> <int>
# 1 1 1
# 2 2 1
# 3 3 1
# 4 4 2
# 5 5 2
# 6 6 2
# 7 7 3
# 8 8 3
# 9 9 3
#10 10 4
#11 11 4
#12 12 4
We can use gl
library(dplyr)
df %>%
mutate(col2 = as.integer(gl(n(), 3, n())))
As integer division i.e. %/% 3 over a sequence say 0:n will result in 0, 0, 0, 1, 1, 1, ... adding 1 will generate the desired sequence automatically, so simply this will also do
df %>% mutate(col2 = 1+ (row_number()-1) %/% 3)
# A tibble: 12 x 2
Col1 col2
<int> <dbl>
1 1 1
2 2 1
3 3 1
4 4 2
5 5 2
6 6 2
7 7 3
8 8 3
9 9 3
10 10 4
11 11 4
12 12 4
I have data table
Name Score
A 5
A 6
B 9
B 1
B 0
...
I want to calculate and add a column 'FScore'=max score to this table
My expected result
Name Score Fscore
A 5 6
A 6 6
B 9 9
B 1 9
B 0 9
Thank.
We can use the base R option ave
df$Fscore <- ave(df$Score, df$Name, FUN = max)
df
# Name Score Fscore
#1 A 5 6
#2 A 6 6
#3 B 9 9
#4 B 1 9
#5 B 0 9
If you are trying to find the maximum score for each Name value, you can use data.table as below.
# example data
d <- data.table(Name = c("A", "A", "B", "B", "B"),
Score = c(5, 6, 9, 1, 0))
# find max for each Name and save the value in a new column, Fscore
d[ , Fscore := max(Score), by=Name]
Result:
> print(d)
Name Score Fscore
1: A 5 6
2: A 6 6
3: B 9 9
4: B 1 9
5: B 0 9
Another option using dplyr could be:
df = data.frame(Name = c('a', 'a', 'b','b','b'), Score = c(5,6,9,1,0))
df %>% group_by(Name) %>% mutate(Fscore = max(Score))
Source: local data frame [5 x 3]
Groups: Name [2]
Name Score FScore
<fctr> <dbl> <dbl>
1 a 5 6
2 a 6 6
3 b 9 9
4 b 1 9
5 b 0 9
There are many answers for how to split a dataframe, for example How to split a data frame?
However, I'd like to split a dataframe so that the smaller dataframes contain the last row of the previous dataframe and the first row of the following dataframe.
Here's an example
n <- 1:9
group <- rep(c("a","b","c"), each = 3)
data.frame(n = n, group)
n group
1 1 a
2 2 a
3 3 a
4 4 b
5 5 b
6 6 b
7 7 c
8 8 c
9 9 c
I'd like the output to look like:
d1 <- data.frame(n = 1:4, group = c(rep("a",3),"b"))
d2 <- data.frame(n = 3:7, group = c("a",rep("b",3),"c"))
d3 <- data.frame(n = 6:9, group = c("b",rep("c",3)))
d <- list(d1, d2, d3)
d
[[1]]
n group
1 1 a
2 2 a
3 3 a
4 4 b
[[2]]
n group
1 3 a
2 4 b
3 5 b
4 6 b
5 7 c
[[3]]
n group
1 6 b
2 7 c
3 8 c
4 9 c
What is an efficient way to accomplish this task?
Suppose DF is the original data.frame, the one with columns n and group. Let n be the number of rows in DF. Now define a function extract which given a sequence of indexes ix enlarges it to include the one prior to the first and after the last and then returns those rows of DF. Now that we have defined extract, split the vector 1, ..., n by group and apply extract to each component of the split.
n <- nrow(DF)
extract <- function(ix) DF[seq(max(1, min(ix) - 1), min(n, max(ix) + 1)), ]
lapply(split(seq_len(n), DF$group), extract)
$a
n group
1 1 a
2 2 a
3 3 a
4 4 b
$b
n group
3 3 a
4 4 b
5 5 b
6 6 b
7 7 c
$c
n group
6 6 b
7 7 c
8 8 c
9 9 c
Or why not try good'ol by, which "[a]ppl[ies] a Function to a Data Frame Split by Factors [INDICES]".
by(data = df, INDICES = df$group, function(x){
id <- c(min(x$n) - 1, x$n, max(x$n) + 1)
na.omit(df[id, ])
})
# df$group: a
# n group
# 1 1 a
# 2 2 a
# 3 3 a
# 4 4 b
# --------------------------------------------------------------------------------
# df$group: b
# n group
# 3 3 a
# 4 4 b
# 5 5 b
# 6 6 b
# 7 7 c
# --------------------------------------------------------------------------------
# df$group: c
# n group
# 6 6 b
# 7 7 c
# 8 8 c
# 9 9 c
Although the print method of by creates a 'fancy' output, the (default) result is a list, with elements named by the levels of the grouping variable (just try str and names on the resulting object).
I was going to comment under #cdetermans answer but its too late now.
You can generalize his approach using data.table::shift (or dyplr::lag) in order to find the group indices and then run a simple lapply on the ranges, something like
library(data.table) # v1.9.6+
indx <- setDT(df)[, which(group != shift(group, fill = TRUE))]
lapply(Map(`:`, c(1L, indx - 1L), c(indx, nrow(df))), function(x) df[x,])
# [[1]]
# n group
# 1: 1 a
# 2: 2 a
# 3: 3 a
# 4: 4 b
#
# [[2]]
# n group
# 1: 3 a
# 2: 4 b
# 3: 5 b
# 4: 6 b
# 5: 7 c
#
# [[3]]
# n group
# 1: 6 b
# 2: 7 c
# 3: 8 c
# 4: 9 c
Could be done with data.frame as well, but is there ever a reason not to use data.table? Also this has the option to be executed with parallelism.
library(data.table)
n <- 1:9
group <- rep(c("a","b","c"), each = 3)
df <- data.table(n = n, group)
df[, `:=` (group = factor(df$group))]
df[, `:=` (group_i = seq_len(.N), group_N = .N), by = "group"]
library(doParallel)
groups <- unique(df$group)
foreach(i = seq(groups)) %do% {
df[group == groups[i] | (as.integer(group) == i + 1 & group_i == 1) | (as.integer(group) == i - 1 & group_i == group_N), c("n", "group"), with = FALSE]
}
[[1]]
n group
1: 1 a
2: 2 a
3: 3 a
4: 4 b
[[2]]
n group
1: 3 a
2: 4 b
3: 5 b
4: 6 b
5: 7 c
[[3]]
n group
1: 6 b
2: 7 c
3: 8 c
4: 9 c
Here is another dplyr way:
library(dplyr)
data =
data_frame(n = n, group) %>%
group_by(group)
firsts =
data %>%
slice(1) %>%
ungroup %>%
mutate(new_group = lag(group)) %>%
slice(-1)
lasts =
data %>%
slice(n()) %>%
ungroup %>%
mutate(new_group = lead(group)) %>%
slice(-n())
bind_rows(firsts, data, lasts) %>%
mutate(final_group =
ifelse(is.na(new_group),
group,
new_group) ) %>%
arrange(final_group, n) %>%
group_by(final_group)