In this type of dataframe:
df <- data.frame(
x = c(3,3,1,12,2,2,10,10,10,1,5,5,2,2,17,17)
)
how can I create a new column recording the run-length ID of only a subset of x values, say, 3-20?
My own attempt only succeeds at inserting NA where the run-length count should be interrupted; but internally it seems the count is uninterrupted:
library(data.table)
df %>%
mutate(rle = ifelse(x %in% 3:20, rleid(x), NA))
x rle
1 3 1
2 3 1
3 1 NA
4 12 3
5 2 NA
6 2 NA
7 10 5
8 10 5
9 10 5
10 1 NA
11 5 7
12 5 7
13 2 NA
14 2 NA
15 17 9
16 17 9
The expected result:
x rle
1 3 1
2 3 1
3 1 NA
4 12 2
5 2 NA
6 2 NA
7 10 3
8 10 3
9 10 3
10 1 NA
11 5 4
12 5 4
13 2 NA
14 2 NA
15 17 5
16 17 5
In base R:
df[df$x %in% 3:20, "rle"] <- data.table::rleid(df[df$x %in% 3:20, ])
x rle
1 3 1
2 3 1
3 1 NA
4 12 2
5 2 NA
6 2 NA
7 10 3
8 10 3
9 10 3
10 1 NA
11 5 4
12 5 4
13 2 NA
14 2 NA
15 17 5
16 17 5
With left_join:
left_join(df, df %>%
filter(x %in% 3:20) %>%
distinct() %>%
mutate(rle = row_number()))
Joining, by = "x"
x rle
1 3 1
2 3 1
3 1 NA
4 12 2
5 2 NA
6 2 NA
7 10 3
8 10 3
9 10 3
10 1 NA
11 5 4
12 5 4
13 2 NA
14 2 NA
15 17 5
16 17 5
With data.table:
library(data.table)
setDT(df)
df[x %between% c(3,20),rle:=rleid(x)][]
x rle
<num> <int>
1: 3 1
2: 3 1
3: 1 NA
4: 12 2
5: 2 NA
6: 2 NA
7: 10 3
8: 10 3
9: 10 3
10: 1 NA
11: 5 4
12: 5 4
13: 2 NA
14: 2 NA
15: 17 5
16: 17 5
Related
I have a data frame grouped by 'id' and a variable 'age' which contains missing values, NA.
Within each 'id', I want to replace missing values of 'age', but only "fill up" before the first non-NA value.
data <- data.frame(id=c(1,1,1,1,1,1,2,2,2,2,2,3,3,3,3,3),
age=c(NA,6,NA,8,NA,NA,NA,NA,3,8,NA,NA,NA,7,NA,9))
id age
1 1 NA
2 1 6 # first non-NA in id = 1. Fill up from here
3 1 NA
4 1 8
5 1 NA
6 1 NA
7 2 NA
8 2 NA
9 2 3 # first non-NA in id = 2. Fill up from here
10 2 8
11 2 NA
12 3 NA
13 3 NA
14 3 7 # first non-NA in id = 3. Fill up from here
15 3 NA
16 3 9
Expected output:
1 1 6
2 1 6
3 1 NA
4 1 8
5 1 NA
6 1 NA
7 2 3
8 2 3
9 2 3
10 2 8
11 2 NA
12 3 7
13 3 7
14 3 7
15 3 NA
16 3 9
I tried using fill with .direction = "up" like this:
library(dplyr)
library(tidyr)
data1 <- data %>% group_by(id) %>%
fill(!is.na(age[1]), .direction = "up")
You could use cumall(is.na(age)) to find the positions before the first non-NA value.
library(dplyr)
data %>%
group_by(id) %>%
mutate(age2 = replace(age, cumall(is.na(age)), age[!is.na(age)][1])) %>%
ungroup()
# A tibble: 16 × 3
id age age2
<dbl> <dbl> <dbl>
1 1 NA 6
2 1 6 6
3 1 NA NA
4 1 8 8
5 1 NA NA
6 1 NA NA
7 2 NA 3
8 2 NA 3
9 2 3 3
10 2 8 8
11 2 NA NA
12 3 NA 7
13 3 NA 7
14 3 7 7
15 3 NA NA
16 3 9 9
Another option (agnostic about where the missing and non-missing values start) could be:
data %>%
group_by(id) %>%
mutate(rleid = with(rle(is.na(age)), rep(seq_along(lengths), lengths)),
age2 = ifelse(rleid == min(rleid[is.na(age)]),
age[rleid == (min(rleid[is.na(age)]) + 1)][1],
age))
id age rleid age2
<dbl> <dbl> <int> <dbl>
1 1 NA 1 6
2 1 6 2 6
3 1 NA 3 NA
4 1 8 4 8
5 1 NA 5 NA
6 1 NA 5 NA
7 2 NA 1 3
8 2 NA 1 3
9 2 3 2 3
10 2 8 2 8
11 2 NA 3 NA
12 3 NA 1 7
13 3 NA 1 7
14 3 7 2 7
15 3 NA 3 NA
16 3 9 4 9
I have a list of dataframes with either 2 or 4 columns.
a <- data.frame(a=1:10,
b=1:10,
c=1:10,
d=1:10)
b <- data.frame(a=1:10,
b=1:10)
list_of_df <- list(a,b)
I want to add 2 empty columns to each dataframe with only 2 columns.
I've tried this lapply approach:
lapply(list_of_df, function(x) ifelse(ncol(x) < 4,x%>%add_column(empty=NA),x <- x))
Which does not work unfortunately. How can I fix this?
I came up with something similar:
add_col <- function(x){
col_to_add <- 4 - ncol(x)
if(col_to_add == 0) return(x)
z <- rep(NA, nrow(x))
for (i in 1:col_to_add){
x <- cbind(x, z)
}
x
}
lapply(list_of_df, add_col)
I would use a for loop to avoid copying the whole list:
for (i in seq_along(list_of_df)) {
n_columns = ncol(list_of_df[[i]])
if (n_columns == 2L) {
list_of_df[[i]][c('empty1', 'empty2')] <- NA
}
}
Result:
> list_of_df
[[1]]
a b c d
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
10 10 10 10 10
[[2]]
a b empty1 empty2
1 1 1 NA NA
2 2 2 NA NA
3 3 3 NA NA
4 4 4 NA NA
5 5 5 NA NA
6 6 6 NA NA
7 7 7 NA NA
8 8 8 NA NA
9 9 9 NA NA
10 10 10 NA NA
We could use bind_rows and then group_split and map from purrr to remove the id_Group column:
library(dplyr)
library(purrr)
bind_rows(list_of_df) %>%
group_split(id_Group =cumsum(a==1)) %>%
map(., ~ (.x %>% ungroup() %>%
select(-id_Group)))
[[1]]
# A tibble: 10 x 4
a b c d
<int> <int> <int> <int>
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
10 10 10 10 10
[[2]]
# A tibble: 10 x 4
a b c d
<int> <int> <int> <int>
1 1 1 NA NA
2 2 2 NA NA
3 3 3 NA NA
4 4 4 NA NA
5 5 5 NA NA
6 6 6 NA NA
7 7 7 NA NA
8 8 8 NA NA
9 9 9 NA NA
10 10 10 NA NA
i got this df:
df <- data.frame(month = c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4),
day = c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5),
flow = c(2,5,7,8,5,4,6,7,9,2,NA,1,6,10,2,NA,NA,NA,NA,NA))
and i want to reach this result:
month day flow dayofminflow
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
11 3 1 NA 2
12 3 2 1 2
13 3 3 6 2
14 3 4 10 2
15 3 5 2 2
16 4 1 NA NA
17 4 2 NA NA
18 4 3 NA NA
19 4 4 NA NA
20 4 5 NA NA
I was using this solution, but it returns NA when the first value is NA:
newdf <- df %>% group_by(month) %>% mutate(Val=day[flow==min(flow)][1])
And this solution returns an error when all data is NA:
library(dplyr)
df <- df %>%
group_by(month) %>%
mutate(dayminflowofthemonth = day[which.min(flow)]) %>%
ungroup
You would just change the default na.rm = TRUE in min() from the first solution to ignore NAs?
df %>%
group_by(month) %>%
mutate(dayofminflow = day[which(min(flow, na.rm = TRUE) == flow)][1])
# A tibble: 20 x 4
# Groups: month [4]
month day flow dayofminflow
<dbl> <dbl> <dbl> <dbl>
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
11 3 1 NA 2
12 3 2 1 2
13 3 3 6 2
14 3 4 10 2
15 3 5 2 2
16 4 1 NA NA
17 4 2 NA NA
18 4 3 NA NA
19 4 4 NA NA
20 4 5 NA NA
Though you get a warning no non-missing arguments to min; returning Inf from month 4 since all flow values are NA.
Here is the code for my example dataset.
df = data.frame("group" =c(rep(1,5),rep(1,6),rep(2,4),rep(2,3)), "time" = c(rep(NA,5),seq(1,6),rep(NA,4),seq(1,3)), "p" = seq(1,18) )
group time p
1 1 NA 1
2 1 NA 2
3 1 NA 3
4 1 NA 4
5 1 NA 5
6 1 1 6
7 1 2 7
8 1 3 8
9 1 4 9
10 1 5 10
11 1 6 11
12 2 NA 12
13 2 NA 13
14 2 NA 14
15 2 NA 15
16 2 1 16
17 2 2 17
18 2 3 18
I would like to figure out how to apply a function by group to only the values that have time then append the result as a new column in the data frame. Here is my example function I would like to apply.
pfunc <- function(p){
p+5
}
The output I am hoping to obtain would look as follows.
group time p new_p
1 1 NA 1 NA
2 1 NA 2 NA
3 1 NA 3 NA
4 1 NA 4 NA
5 1 NA 5 NA
6 1 1 6 11
7 1 2 7 12
8 1 3 8 13
9 1 4 9 14
10 1 5 10 15
11 1 6 11 16
12 2 NA 12 NA
13 2 NA 13 NA
14 2 NA 14 NA
15 2 NA 15 NA
16 2 1 16 21
17 2 2 17 22
18 2 3 18 23
You can try this:
library(dplyr)
df %>% group_by(group) %>%
mutate(pnew=ifelse(is.na(time),time,time+5))
# A tibble: 18 x 4
# Groups: group [2]
group time p pnew
<dbl> <int> <int> <dbl>
1 1 NA 1 NA
2 1 NA 2 NA
3 1 NA 3 NA
4 1 NA 4 NA
5 1 NA 5 NA
6 1 1 6 6
7 1 2 7 7
8 1 3 8 8
9 1 4 9 9
10 1 5 10 10
11 1 6 11 11
12 2 NA 12 NA
13 2 NA 13 NA
14 2 NA 14 NA
15 2 NA 15 NA
16 2 1 16 6
17 2 2 17 7
18 2 3 18 8
Update
You can use this function:
increase <- function(data,n)
{
data %>% group_by(group) %>%
mutate(pnew=ifelse(is.na(time),time,time+n)) -> result
return(result)
}
increase(df,n = 10)
# A tibble: 18 x 4
# Groups: group [2]
group time p pnew
<dbl> <int> <int> <dbl>
1 1 NA 1 NA
2 1 NA 2 NA
3 1 NA 3 NA
4 1 NA 4 NA
5 1 NA 5 NA
6 1 1 6 11
7 1 2 7 12
8 1 3 8 13
9 1 4 9 14
10 1 5 10 15
11 1 6 11 16
12 2 NA 12 NA
13 2 NA 13 NA
14 2 NA 14 NA
15 2 NA 15 NA
16 2 1 16 11
17 2 2 17 12
18 2 3 18 13
Update 2
I hope this helps:
df %>% group_by(group) %>% rowwise() %>% mutate(pnew=ifelse(is.na(time),NA,pfunc(time)))
# A tibble: 18 x 4
# Rowwise: group
group time p pnew
<dbl> <int> <int> <dbl>
1 1 NA 1 NA
2 1 NA 2 NA
3 1 NA 3 NA
4 1 NA 4 NA
5 1 NA 5 NA
6 1 1 6 6
7 1 2 7 7
8 1 3 8 8
9 1 4 9 9
10 1 5 10 10
11 1 6 11 11
12 2 NA 12 NA
13 2 NA 13 NA
14 2 NA 14 NA
15 2 NA 15 NA
16 2 1 16 6
17 2 2 17 7
18 2 3 18 8
I'm searching the web for a few a days now and I can't find a solution to my (probably easy to solve) problem.
I have huge data frames with 4 variables and over a million observations each. Now I want to select 100 rows before, all rows while and 1000 rows after a specific condition is met and fill the rest with NA's. I tried it with a for loop and if/ifelse but it doesn't work so far. I think it shouldn't be a big thing, but in the moment I just don't get the hang of it.
I create the data using:
foo<-data.frame(t = 1:15, a = sample(1:15), b = c(1,1,1,1,1,4,4,4,4,1,1,1,1,1,1), c = sample(1:15))
My Data looks like this:
ID t a b c
1 1 4 1 7
2 2 7 1 10
3 3 10 1 6
4 4 2 1 4
5 5 13 1 9
6 6 15 4 3
7 7 8 4 15
8 8 3 4 1
9 9 9 4 2
10 10 14 1 8
11 11 5 1 11
12 12 11 1 13
13 13 12 1 5
14 14 6 1 14
15 15 1 1 12
What I want is to pick the value of a (in this example) 2 rows before, all rows while and 3 rows after the value of b is >1 and fill the rest with NA's. [Because this is just an example I guess you can imagine that after these 15 rows there are more rows with the value for b changing from 1 to 4 several times (I did not post it, so I won't spam the question with unnecessary data).]
So I want to get something like:
ID t a b c d
1 1 4 1 7 NA
2 2 7 1 10 NA
3 3 10 1 6 NA
4 4 2 1 4 2
5 5 13 1 9 13
6 6 15 4 3 15
7 7 8 4 15 8
8 8 3 4 1 3
9 9 9 4 2 9
10 10 14 1 8 14
11 11 5 1 11 5
12 12 11 1 13 11
13 13 12 1 5 NA
14 14 6 1 14 NA
15 15 1 1 12 NA
I'm thankful for any help.
Thank you.
Best regards,
Chris
here is the same attempt as missuse, but with data.table:
library(data.table)
foo<-data.frame(t = 1:11, a = sample(1:11), b = c(1,1,1,4,4,4,4,1,1,1,1), c = sample(1:11))
DT <- setDT(foo)
DT[ unique(c(DT[,.I[b>1] ],DT[,.I[b>1]+3 ],DT[,.I[b>1]-2 ])), d := a]
t a b c d
1: 1 10 1 2 NA
2: 2 6 1 10 6
3: 3 5 1 7 5
4: 4 11 4 4 11
5: 5 4 4 9 4
6: 6 8 4 5 8
7: 7 2 4 8 2
8: 8 3 1 3 3
9: 9 7 1 6 7
10: 10 9 1 1 9
11: 11 1 1 11 NA
Here
unique(c(DT[,.I[b>1] ],DT[,.I[b>1]+3 ],DT[,.I[b>1]-2 ]))
gives you your desired indixes : the unique indices of the line for your condition, the same indices+3 and -2.
Here is an attempt.
Get indexes that satisfy the condition b > 1
z <- which(foo$b > 1)
get indexes for (z - 2) : (z + 3)
ind <- unique(unlist(lapply(z, function(x){
g <- pmax(x - 2, 1) #if x - 2 is negative
g : (x + 3)
})))
create d column filled with NA
foo$d <- NA
replace elements with appropriate indexes with foo$a
foo$d[ind] <- foo$a[ind]
library(dplyr)
library(purrr)
# example dataset
foo<-data.frame(t = 1:15,
a = sample(1:15),
b = c(1,1,1,1,1,4,4,4,4,1,1,1,1,1,1),
c = sample(1:15))
# function to get indices of interest
# for a given index x go 2 positions back and 3 forward
# keep only positive indices
GetIDsBeforeAfter = function(x) {
v = (x-2) : (x+3)
v[v > 0]
}
foo %>% # from your dataset
filter(b > 1) %>% # keep rows where b > 1
pull(t) %>% # get the positions
map(GetIDsBeforeAfter) %>% # for each position apply the function
unlist() %>% # unlist all sets indices
unique() -> ids_to_remain # keep unique ones and save them in a vector
foo$d = foo$c # copy column c as d
foo$d[-ids_to_remain] = NA # put NA to all positions not in our vector
foo
# t a b c d
# 1 1 5 1 8 NA
# 2 2 6 1 14 NA
# 3 3 4 1 10 NA
# 4 4 1 1 7 7
# 5 5 10 1 5 5
# 6 6 8 4 9 9
# 7 7 9 4 15 15
# 8 8 3 4 6 6
# 9 9 7 4 2 2
# 10 10 12 1 3 3
# 11 11 11 1 1 1
# 12 12 15 1 4 4
# 13 13 14 1 11 NA
# 14 14 13 1 13 NA
# 15 15 2 1 12 NA