I am making several graphs based on the clustering data from DAPC. I need the colors to be the same across all the graphs, and I'd like to use specific colors for the largest groups. The important thing for this question, is I get a data set from DAPC like this:
my_df <- data.frame(
ID = c(1:10),
Group = c("a", "b", "b", "c", "a", "b", "a", "b", "b", "c")
)
> my_df
ID Group
1 a
2 b
3 b
4 c
5 a
6 b
7 a
8 b
9 b
10 c
I know how to find the group with the most members like this:
freqs <- table(my_df$Group)
freqs <- freqs[order(freqs, decreasing = TRUE)]
>freqs
b a c
5 3 2
Is there a way to change the values based on their frequency? Each time I rerun DAPC, it changes the groups, so I'd like to write code that does this automatically instead of having to redo it manually. Here's how I'd like the dataframe to be changed:
> my_df > my_new_df
ID Group ID Group
1 a 1 '2nd'
2 b 2 '1st'
3 b 3 '1st'
4 c 4 '3rd'
5 a 5 '2nd'
6 b 6 '1st'
7 a 7 '2nd'
8 b 8 '1st'
9 b 9 '1st'
10 c 10 '3rd'
You may use ave and create a factor out of it with the corresponding labels=. To avoid hard-coding, define the labels in a vector lb beforehand.
lb <- c("1st", "2nd", "3rd", paste0(4:10, "th"))
with(my_df, factor(as.numeric(ave(as.character(Group), as.character(Group), FUN=table)),
labels=rev(lb[1:length(unique(table(Group)))])))
# [1] 2nd 1st 1st 3rd 2nd 1st 2nd 1st 1st 3rd
# Levels: 3rd 2nd 1st
To convert more columns like this, use sapply.
sapply(my_df[selected.columns], function(x) {
factor(as.numeric(ave(as.character(x), as.character(x), FUN=table)),
labels=rev(lb[1:length(unique(table(x)))]))
})
Do you mean something like this:
my_df %>% left_join(my_df %>% group_by(Group) %>% summarise(N=n())) %>%
arrange(desc(N)) %>% select(-N)
ID Group
1 2 B
2 3 B
3 6 B
4 8 B
5 9 B
6 1 A
7 5 A
8 7 A
9 4 C
10 10 C
Update
This can be useful. I hope this helps.
my_df %>% left_join(my_df %>% group_by(Group) %>% summarise(N=n()) %>% arrange(desc(N)) %>%
bind_cols(my_df %>% select(Group) %>% distinct() %>% rename(key=Group)) %>%
rename(NewGroup=Group,Group=key)) %>%
select(-c(Group,N)) %>% rename(Group=NewGroup)
ID Group
1 1 B
2 2 A
3 3 A
4 4 C
5 5 B
6 6 A
7 7 B
8 8 A
9 9 A
10 10 C
Related
I have the following dataframe called df (dput below):
> df
group value
1 A 5
2 A 1
3 A 1
4 A 5
5 B 8
6 B 2
7 B 2
8 B 3
9 C 10
10 C 1
11 C 1
12 C 8
I would like to filter groups based on the difference between their highest value (max) and second highest value. The difference should be smaller equal than 2 (<=2), this means that group B should be removed because the highest value is 8 and the second highest value is 3 which is a difference of 5. The desired output should look like this:
group value
1 A 5
2 A 1
3 A 1
4 A 5
5 C 10
6 C 1
7 C 1
8 C 8
So I was wondering if anyone knows how to filter groups based on the difference between their highest and second-highest value?
dput of df:
df<-structure(list(group = c("A", "A", "A", "A", "B", "B", "B", "B",
"C", "C", "C", "C"), value = c(5, 1, 1, 5, 8, 2, 2, 3, 10, 1,
1, 8)), class = "data.frame", row.names = c(NA, -12L))
Using dplyr
library(dplyr)
df %>%
group_by(group) %>%
filter(abs(diff(sort(value, decreasing=T)[1:2])) <= 2) %>%
ungroup()
# A tibble: 8 × 2
group value
<chr> <int>
1 A 5
2 A 1
3 A 1
4 A 5
5 C 10
6 C 1
7 C 1
8 C 8
A base R alternative
grp <- na.omit(aggregate(. ~ group, df, function(x)
abs(diff(sort(x, decreasing=T)[1:2])) <= 2))
do.call(rbind, c(mapply(function(g, v)
list(df[df$group == g & v,]), grp$group, grp$value), make.row.names=F))
group value
1 A 5
2 A 1
3 A 1
4 A 5
5 C 10
6 C 1
7 C 1
8 C 8
I possibility would be to first create a vector with the groups that achieve your condition and then filter in the original data.frame. Here how I thought:
library(dplyr)
group_to_keep <-
df %>%
group_by(group) %>%
slice_max(n = 2,value) %>%
filter(abs(diff(value)) <= 2) %>%
pull(group) %>%
unique()
df %>%
filter(group %in% group_to_keep)
You can use ave.
df[ave(df$value, df$group, FUN=\(x) diff(sort(c(-x, Inf)))[1]) <= 2,]
# group value
#1 A 5
#2 A 1
#3 A 1
#4 A 5
#9 C 10
#10 C 1
#11 C 1
#12 C 8
In case you can sure that you have all the time at least two values you can use.
df[ave(df$value, df$group, FUN=\(x) diff(tail(sort(x), 2))) <= 2,]
df[ave(df$value, df$group, FUN=\(x) diff(sort(-x)[1:2])) <= 2,]
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")
I have a list here, and I wish to mutate a new column with unique values for each list relative to the mutation. For example, I want to mutate a column named ID as n >= 1.
Naturally, on a dataframe I would do this:
dat %>% mutate(id = row_number())
For a list, I would do this:
dat%>% map(~ mutate(., ID = row_number()))
And I would get an output likeso:
dat <- list(data.frame(x=c("a", "b" ,"c", "d", "e" ,"f" ,"g") ), data.frame(y=c("p", "lk", "n", "m", "g", "f", "t")))
[[1]]
x id
1 a 1
2 b 2
3 c 3
4 d 4
5 e 5
6 f 6
7 g 7
[[2]]
y id
1 p 1
2 lk 2
3 n 3
4 m 4
5 g 5
6 f 6
7 t 7
Though, how would I mutate a new column ID such that the row number continues from the first list.
Expected output:
[[1]]
x id
1 a 1
2 b 2
3 c 3
4 d 4
5 e 5
6 f 6
7 g 7
[[2]]
y id
1 p 8
2 lk 9
3 n 10
4 m 11
5 g 12
6 f 13
7 t 14
An option is to bind them into a single dataset, create the 'id' with row_number(), split by 'grp', loop over the list and remove any columns that have all NA values
library(dplyr)
library(purrr)
dat %>%
bind_rows(.id = 'grp') %>%
mutate(id = row_number()) %>%
group_split(grp) %>%
map(~ .x %>%
select(where(~ any(!is.na(.))), -grp))
-output
#[[1]]
# A tibble: 7 x 2
# x id
# <chr> <int>
#1 a 1
#2 b 2
#3 c 3
#4 d 4
#5 e 5
#6 f 6
#7 g 7
#[[2]]
# A tibble: 7 x 2
# y id
# <chr> <int>
#1 p 8
#2 lk 9
#3 n 10
#4 m 11
#5 g 12
#6 f 13
#7 t 14
Or an easier approach is to unlist (assuming single column), get the sequence, add a new column with map2
map2(dat, relist(seq_along(unlist(dat)), skeleton = dat),
~ .x %>% mutate(id = .y))
Or using a for loop
dat[[1]]$id <- seq_len(nrow(dat[[1]]))
for(i in seq_along(dat)[-1]) dat[[i]]$id <-
seq(tail(dat[[i-1]]$id, 1) + 1, length.out = nrow(dat[[i]]), by = 1)
How can I keep my duplicates, but remove unique values based on one column(qol)?
ID qol Sat
A 7 6
A 7 5
B 3 3
B 3 4
B 1 7
C 2 7
c 1 2
But I need this:
ID qol Sat
A 7 6
A 7 5
B 3 3
B 3 4
What can I do?
dplyr solution:
library(dplyr)
ID <- c("A", "A", "B", "B", "B", "C", "c")
qol <- c(7,7,3,3,1,2,1)
Sat <- c(6,5,3,4,7,7,2)
test_df <- data.frame(cbind(ID, qol, Sat))
filtered_df <- test_df %>%
group_by(qol) %>%
filter(n()>1)
Please note that this will return
ID qol Sat
1 A 7 6
2 A 7 5
3 B 3 3
4 B 3 4
5 B 1 7
6 c 1 2
If you also want to remove the two rows where qol == 1 but the IDs are different, just do:
filtered_df <- test_df %>%
group_by(ID, qol) %>%
filter(n()>1)
This will return the sample output you supplied in the question.
I want to sum values of rows which belongs to group other than the row's group. For example using this sample data
> df <- data.frame(id=1:5, group=c("A", "A", "B", "B", "A"), val=seq(9, 1, -2))
> df
id group val
1 1 A 9
2 2 A 7
3 3 B 5
4 4 B 3
5 5 A 1
Summarizing with dplyr by group
> df %>% group_by(group) %>% summarize(sumval = sum(val))
Source: local data frame [2 x 2]
group sumval
(fctr) (dbl)
1 A 17
2 B 8
What I want is the value for rows belonging to group A to use sumval of not group A. i.e. the final result is
id group val notval
1 1 A 9 8
2 2 A 7 8
3 3 B 5 17
4 4 B 3 17
5 5 A 1 8
Is there a way to do this in dplyr? Preferrably in a single chain?
We can do this with base R
s1 <- sapply(unique(df$group), function(x) sum(df$val[df$group !=x]))
s1[with(df, match(group, unique(group)))]
#[1] 8 8 17 17 8
Or using data.table
library(data.table)
setDT(df)[,notval := sum(df$val[df$group!=group]) ,group]
#akrun answers are best. But if you want to do in dplyr, this is a round about way.
df <- data.frame(id=1:5, group=c("A", "A", "B", "B", "A"), val=seq(9, 1, -2))
df %>% mutate(TotalSum = sum(val)) %>% group_by(group) %>%
mutate(valsumval = TotalSum - sum(val))
Source: local data frame [5 x 5]
Groups: group [2]
id group val TotalSum valsumval
(int) (fctr) (dbl) (dbl) (dbl)
1 1 A 9 25 8
2 2 A 7 25 8
3 3 B 5 25 17
4 4 B 3 25 17
5 5 A 1 25 8
This also works even if there are more than two groups.
Also Just this works
df %>% group_by(group) %>% mutate(notval = sum(df$val)- sum(val))