data1=data.frame("Group1" = sample(1:2,100,r=T),
"Group2" = sample(c('a','b'),100,r=T),
"V1" = sample(1:3, 100, r=T),
"V2" = sample(0:1, 100, r=T),
"V3" = sample(1:5, 100, r=T),
"V4" = sample(1:2, 100, r=T))
data2=data.frame("Group1"=c(1,1,2,2),
"Group2"=c('a','b','a','b'),
"Size"=c(9,7,6,10),
"V1"=c(NA),
"V2"=c(NA),
"V3"=c(NA),
"V4"=c(NA))
I have 'data1' that contains my data. Then I have 'data2' which has 'Group1' and 'Group2' and 'Size'.
What I wish for is to group my data by ('Group1' and 'Group2') and take a random sample of size 'Size' from 'data1' to fill in V1-V4 in data2.
The hopeful output would look like this but with the NA values filled in based on 'data1'
library(dplyr);library(tidyr)
data3= data2 %>%
uncount(Size)
library(data.table)
setDT(data1)
setDT(data2)
# sample indices from each group
i <-
data2[data1, on = .(Group1, Group2)
][, .(i_samp = sample(.I, Size)), by = .(Group1, Group2, Size)
][, i_samp]
# subset to sampled indices
merge(data1[i], data2[, .(Group1, Group2, Size)])
# Group1 Group2 V1 V2 V3 V4 Size
# 1: 1 a 3 1 2 2 9
# 2: 1 a 3 1 5 1 9
# 3: 1 a 2 1 4 2 9
# 4: 1 a 3 1 1 1 9
# 5: 1 a 3 1 4 1 9
# 6: 1 a 1 0 3 1 9
# 7: 1 a 3 1 1 1 9
# 8: 1 a 1 1 1 2 9
# 9: 1 a 2 0 2 1 9
# 10: 1 b 2 0 5 2 7
# 11: 1 b 3 0 5 2 7
# 12: 1 b 3 1 4 2 7
# 13: 1 b 1 1 1 1 7
# 14: 1 b 1 1 4 1 7
# 15: 1 b 1 0 1 1 7
# 16: 1 b 1 0 3 1 7
# 17: 2 a 2 0 5 1 6
# 18: 2 a 1 0 5 1 6
# 19: 2 a 3 1 1 2 6
# 20: 2 a 1 0 2 1 6
# 21: 2 a 3 1 1 2 6
# 22: 2 a 1 1 3 2 6
# 23: 2 b 3 0 2 1 10
# 24: 2 b 2 1 5 1 10
# 25: 2 b 3 0 1 1 10
# 26: 2 b 3 1 2 1 10
# 27: 2 b 2 0 5 1 10
# 28: 2 b 2 0 2 1 10
# 29: 2 b 2 0 2 2 10
# 30: 2 b 1 0 1 1 10
# 31: 2 b 3 0 5 1 10
# 32: 2 b 3 0 5 1 10
# Group1 Group2 V1 V2 V3 V4 Size
Input data used:
data1=data.frame("Group1" = sample(1:2,100,r=T),
"Group2" = sample(c('a','b'),100,r=T),
"V1" = sample(1:3, 100, r=T),
"V2" = sample(0:1, 100, r=T),
"V3" = sample(1:5, 100, r=T),
"V4" = sample(1:2, 100, r=T))
data2=data.frame("Group1"=c(1,1,2,2),
"Group2"=c('a','b','a','b'),
"Size"=c(9,7,6,10),
"V1"=c(NA),
"V2"=c(NA),
"V3"=c(NA),
"V4"=c(NA))
Here's a more parameterized version where you explicitly set the columns you want to fill and the keys connecting the two tables
fill_key <- c('Group1', 'Group2')
columns_to_fill <- paste0('V', 1:4)
# sample indices from each group
i <-
data2[data1, on = (fill_key)
][, .(i_samp = sample(.I, Size)), by = c(fill_key, 'Size')
][, i_samp]
# subset to sampled indices
merge(data1[i, c(fill_key, columns_to_fill), with = FALSE],
data2[, c(fill_key, 'Size'), with = FALSE])
One dplyr option could be:
data1 %>%
left_join(data2 %>%
select(-starts_with("V"))) %>%
group_by(Group1, Group2) %>%
sample_n(Size)
Group1 Group2 V1 V2 V3 V4 Size
<dbl> <fct> <int> <int> <int> <int> <dbl>
1 1 a 1 1 1 2 9
2 1 a 3 0 3 2 9
3 1 a 2 0 3 2 9
4 1 a 1 1 2 1 9
5 1 a 2 1 2 2 9
6 1 a 2 0 5 2 9
7 1 a 1 0 1 2 9
8 1 a 3 0 5 2 9
9 1 a 1 0 5 1 9
10 1 b 2 0 1 1 7
11 1 b 2 1 3 1 7
12 1 b 3 1 4 2 7
13 1 b 1 1 1 1 7
14 1 b 2 1 2 2 7
15 1 b 1 1 1 2 7
16 1 b 1 1 2 1 7
17 2 a 3 1 5 1 6
18 2 a 1 0 5 1 6
19 2 a 1 0 1 1 6
20 2 a 2 0 5 1 6
21 2 a 3 0 1 1 6
22 2 a 2 1 4 1 6
23 2 b 3 0 2 1 10
24 2 b 1 1 5 1 10
25 2 b 3 1 1 1 10
26 2 b 3 1 4 1 10
27 2 b 1 0 4 2 10
28 2 b 3 1 1 2 10
29 2 b 2 1 4 1 10
30 2 b 1 0 1 1 10
31 2 b 2 1 4 2 10
32 2 b 2 1 5 2 10
Related
I have a data frame like so:
ID <- c('A','A','A','A','A','A','A','A','A','A','A','B','B','B','B','B')
val1 <- c(0,1,2,3,4,5,6,7,8,9,10,11,0,1,2,3)
val2 <- c(0,1,2,3,4,5,0,1,0,1,2,0,1,0,1,2)
df <- data.frame(ID, val1, val2)
Output:
ID val1 val2
1 A 0 0
2 A 1 1
3 A 2 2
4 A 3 3
5 A 4 4
6 A 5 5
7 A 6 0
8 A 7 1
9 A 8 0
10 A 9 1
11 A 10 2
12 B 11 0
13 B 0 1
14 B 1 0
15 B 2 1
16 B 3 2
I am trying to create a third column (val 3) which is like an index. When val1 = 0 and val 2 = 0 it should be 1 (this is also grouped by ID). It should stay as one and then increment by 1 until val2 = 0 again, like so showing desired output:
ID val1 val2 val3
1 A 0 0 1
2 A 1 1 1
3 A 2 2 1
4 A 3 3 1
5 A 4 4 1
6 A 5 5 1
7 A 6 0 2
8 A 7 1 2
9 A 8 0 3
10 A 9 1 3
11 A 10 2 3
12 B 11 0 1
13 B 0 1 1
14 B 1 0 2
15 B 2 1 2
16 B 3 2 2
How can this be achieved? I tried:
df <- df %>%
group_by(ID, val2) %>%
mutate(val3 = row_number())
And:
df$val3 <- cumsum(c(1,diff(df$val2)==0))
But neither provide the desired outcome.
Inside cumsum use the logical comparison val2==0
df %>%
group_by(ID) %>%
mutate(val3 = cumsum(val2==0))
# A tibble: 16 × 4
# Groups: ID [2]
ID val1 val2 val3
<chr> <dbl> <dbl> <int>
1 A 0 0 1
2 A 1 1 1
3 A 2 2 1
4 A 3 3 1
5 A 4 4 1
6 A 5 5 1
7 A 6 0 2
8 A 7 1 2
9 A 8 0 3
10 A 9 1 3
11 A 10 2 3
12 B 11 0 1
13 B 0 1 1
14 B 1 0 2
15 B 2 1 2
16 B 3 2 2
Hi I have a data frame like this
df <-data.frame(x=rep(rep(seq(0,3),each=2),2 ),gr=gl(2,8))
x gr
1 0 1
2 0 1
3 1 1
4 1 1
5 2 1
6 2 1
7 3 1
8 3 1
9 0 2
10 0 2
11 1 2
12 1 2
13 2 2
14 2 2
15 3 2
16 3 2
I want to add a new column numbering sequence of numbers when the x value ==0
I tried
library(dplyr)
df%>%
group_by(gr)%>%
mutate(numbering=seq(2,8,2))
Error in mutate_impl(.data, dots) :
Column `numbering` must be length 8 (the group size) or one, not 4
?
Just for side note mutate(numbering=rep(seq(2,8,2),each=2)) would work for this minimal example but for the general case its better to look x value change from 0!
the expected output
x gr numbering
1 0 1 2
2 0 1 2
3 1 1 4
4 1 1 4
5 2 1 6
6 2 1 6
7 3 1 8
8 3 1 8
9 0 2 2
10 0 2 2
11 1 2 4
12 1 2 4
13 2 2 6
14 2 2 6
15 3 2 8
16 3 2 8
Do you mean something like this?
library(tidyverse);
df %>%
group_by(gr) %>%
mutate(numbering = cumsum(c(1, diff(x) != 0)))
## A tibble: 16 x 3
## Groups: gr [2]
# x gr numbering
# <int> <fct> <dbl>
# 1 0 1 1.
# 2 0 1 1.
# 3 1 1 2.
# 4 1 1 2.
# 5 2 1 3.
# 6 2 1 3.
# 7 3 1 4.
# 8 3 1 4.
# 9 0 2 1.
#10 0 2 1.
#11 1 2 2.
#12 1 2 2.
#13 2 2 3.
#14 2 2 3.
#15 3 2 4.
#16 3 2 4.
Or if you must have a numbering sequence 2,4,6,... instead of 1,2,3,... you can do
df %>%
group_by(gr) %>%
mutate(numering = 2 * cumsum(c(1, diff(x) != 0)));
## A tibble: 16 x 3
## Groups: gr [2]
# x gr numering
# <int> <fct> <dbl>
# 1 0 1 2.
# 2 0 1 2.
# 3 1 1 4.
# 4 1 1 4.
# 5 2 1 6.
# 6 2 1 6.
# 7 3 1 8.
# 8 3 1 8.
# 9 0 2 2.
#10 0 2 2.
#11 1 2 4.
#12 1 2 4.
#13 2 2 6.
#14 2 2 6.
#15 3 2 8.
#16 3 2 8.
Here is an option using match to get the index and then pass on the seq values to fill
df %>%
group_by(gr) %>%
mutate(numbering = seq(2, length.out = n()/2, by = 2)[match(x, unique(x))])
# A tibble: 16 x 3
# Groups: gr [2]
# x gr numbering
# <int> <fct> <dbl>
# 1 0 1 2
# 2 0 1 2
# 3 1 1 4
# 4 1 1 4
# 5 2 1 6
# 6 2 1 6
# 7 3 1 8
# 8 3 1 8
# 9 0 2 2
#10 0 2 2
#11 1 2 4
#12 1 2 4
#13 2 2 6
#14 2 2 6
#15 3 2 8
#16 3 2 8
dfin <-
ID SEQ GRP C1 C2 C3 T1 T2 T3
1 1 1 0 5 8 0 1 2
1 2 1 5 10 15 5 6 7
2 1 2 20 25 30 0 1 2
C1 is the concentration (CONC) at T1 (TIME) and so on. This is what I want as an output:
dfout <-
ID SEQ GRP CONC TIME
1 1 1 0 0
1 1 1 5 1
1 1 1 8 2
1 2 1 5 5
1 2 1 10 6
1 2 1 15 7
2 1 2 20 0
2 1 2 25 1
2 1 2 30 2
The dfin has much more columns for Cx and Tx where x is the number of concentration readings.
You can do this with data.table::melt, with its capability of melting the table into multiple columns based on the columns pattern:
library(data.table)
melt(
setDT(df),
id.vars=c("ID", "SEQ", "GRP"),
# columns starts with C and T should be melted into two separate columns
measure.vars=patterns("^C", "^T"),
value.name=c('CONC', 'TIME')
)[order(ID, SEQ)][, variable := NULL][]
# ID SEQ GRP CONC TIME
#1: 1 1 1 0 0
#2: 1 1 1 5 1
#3: 1 1 1 8 2
#4: 1 2 1 5 5
#5: 1 2 1 10 6
#6: 1 2 1 15 7
#7: 2 1 2 20 0
#8: 2 1 2 25 1
#9: 2 1 2 30 2
Or if the value column names follow the pattern [CT][0-9], you can use reshape from base R by specifying the sep="" which will split the value columns name by the letter/digit separation due to this default setting (from ?reshape):
split = if (sep == "") {
list(regexp = "[A-Za-z][0-9]", include = TRUE)
} else {
list(regexp = sep, include = FALSE, fixed = TRUE)}
reshape(df, varying=-(1:3), idvar=c("ID", "SEQ", "GRP"),
dir="long", sep="", v.names=c("CONC", "TIME"))
# ID SEQ GRP time CONC TIME
#1: 1 1 1 1 0 5
#2: 1 2 1 1 5 10
#3: 2 1 2 1 20 25
#4: 1 1 1 2 8 0
#5: 1 2 1 2 15 5
#6: 2 1 2 2 30 0
#7: 1 1 1 3 1 2
#8: 1 2 1 3 6 7
#9: 2 1 2 3 1 2
Somewhat new to R and I find myself needing to delete rows based on multiple criteria. The data frame has 3 columns and I need to delete rows where bid=99 and there are values less than 99 grouping by rid and qid. The desired output at an rid and qid level are bid has multiple values less than 99 or bid=99.
rid qid bid
1 1 5
1 1 6
1 1 99
1 2 6
2 1 7
2 1 99
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
3 2 99
4 1 2
4 1 6
4 2 1
4 2 2
4 2 99
5 1 99
5 2 99
The expected output...
rid qid bid
1 1 5
1 1 6
1 2 6
2 1 7
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
4 1 2
4 1 6
4 2 1
4 2 2
5 1 99
5 2 99
Any assistance would be appreciated.
You can use the base R function ave to generate a dropping variable like this:
df$dropper <- with(df, ave(bid, rid, qid, FUN= function(i) i == 99 & length(i) > 1))
ave calculates a function on bid, grouping by rid and qid. The function tests if each element of the grouped bid values i is 99 and if i has a length greater than 1. Also, with is used to reduce typing.
which returns
df
rid qid bid dropper
1 1 1 5 0
2 1 1 6 0
3 1 1 99 1
4 1 2 6 0
5 2 1 7 0
6 2 1 99 1
7 2 2 2 0
8 2 2 3 0
9 3 1 7 0
10 3 1 8 0
11 3 2 1 0
12 3 2 99 1
13 4 1 2 0
14 4 1 6 0
15 4 2 1 0
16 4 2 2 0
17 4 2 99 1
18 5 1 99 0
19 5 2 99 0
then drop the undesired observations with df[dropper == 0, 1:3] which will simultaneously drop the new variable.
If you want to just delete rows where bid = 99 then use dplyr.
library(dplyr)
df <- df %>%
filter(bid != 99)
Where df is your data frame. and != means not equal to
Updated solution using dplyr
df %>%
group_by(rid, qid) %>%
mutate(tempcount = n())%>%
ungroup() %>%
mutate(DropValue =ifelse(bid == 99 & tempcount > 1, 1,0) ) %>%
filter(DropValue == 0) %>%
select(rid,qid,bid)
Here is another option with all and if condition in data.table to subset the rows after grouping by 'rid' and 'qid'
library(data.table)
setDT(df1)[, if(all(bid==99)) .SD else .SD[bid!= 99], .(rid, qid)]
# rid qid bid
# 1: 1 1 5
# 2: 1 1 6
# 3: 1 2 6
# 4: 2 1 7
# 5: 2 2 2
# 6: 2 2 3
# 7: 3 1 7
# 8: 3 1 8
# 9: 3 2 1
#10: 4 1 2
#11: 4 1 6
#12: 4 2 1
#13: 4 2 2
#14: 5 1 99
#15: 5 2 99
Or without using the if
setDT(df1)[df1[, .I[all(bid==99) | bid != 99], .(rid, qid)]$V1]
Here is a solution using dplyr, which is a very expressive framework for this kind of problems.
df <- read.table(text =
" rid qid bid
1 1 5
1 1 6
1 1 99
1 2 6
2 1 7
2 1 99
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
3 2 99
4 1 2
4 1 6
4 2 1
4 2 2
4 2 99
5 1 99
5 2 99",
header = TRUE, stringsAsFactors = FALSE)
Dplyr verbs allow to express the program in a way that is close to the very terms of your questions:
library(dplyr)
res <-
df %>%
group_by(rid, qid) %>%
filter(!(any(bid < 99) & bid == 99)) %>%
ungroup()
# # A tibble: 15 × 3
# rid qid bid
# <int> <int> <int>
# 1 1 1 5
# 2 1 1 6
# 3 1 2 6
# 4 2 1 7
# 5 2 2 2
# 6 2 2 3
# 7 3 1 7
# 8 3 1 8
# 9 3 2 1
# 10 4 1 2
# 11 4 1 6
# 12 4 2 1
# 13 4 2 2
# 14 5 1 99
# 15 5 2 99
Let's check we get the desired output:
desired_output <- read.table(text =
" rid qid bid
1 1 5
1 1 6
1 2 6
2 1 7
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
4 1 2
4 1 6
4 2 1
4 2 2
5 1 99
5 2 99",
header = TRUE, stringsAsFactors = FALSE)
identical(as.data.frame(res), desired_output)
# [1] TRUE
Given the following first two columns(id and time_diff), i want to generate the 'block' column
test
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5
The data is already sorted by id and time. The time_diff was computed based on the difference of the previous time and the time value for the row, given the same id. I want to create a block id which is an auto-increment value and increases when a new ID or a time_diff of >10 with the same id is encountered.
How can I achieve this in R?
Importing your data as a data frame with something like:
df = read.table(text='
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5')
You can do a one-liner like this to get occurrences satisfying your two conditions:
> new_col = as.vector(cumsum(
na.exclude(
c(F,diff(as.numeric(as.factor(df$id)))) | # change of id OR
df$time_diff > 10 # time_diff greater than 10
)
))
> new_col
[1] 0 0 0 0 0 1 2 2 2 2 3 3 4 4 4
And finally append this new column to your dataframe with cbind:
> cbind(df, block = c(0,new_col))
id time_diff block block
1 a NA 1 0
2 a 1 1 0
3 a 1 1 0
4 a 1 1 0
5 a 3 1 0
6 a 3 1 0
7 b NA 2 1
8 b 11 3 2
9 b 1 3 2
10 b 1 3 2
11 b 1 3 2
12 b 12 4 3
13 b 1 4 3
14 c NA 5 4
15 c 4 5 4
16 c 7 5 4
You will notice an offset between your wanted block variable and mine: correcting it is easy and can be done at several different step, I will leave it to you :)
Another variation of #Jealie's method would be:
with(test, cumsum(c(TRUE,id[-1]!=id[-nrow(test)])|time_diff>10))
#[1] 1 1 1 1 1 1 2 3 3 3 3 4 4 5 5 5
After learning from Jealie and akrun, I came up with this idea.
mydf %>%
mutate(group = cumsum(time_diff > 10 |!duplicated(id)))
# id time_diff block group
#1 a NA 1 1
#2 a 1 1 1
#3 a 1 1 1
#4 a 1 1 1
#5 a 3 1 1
#6 a 3 1 1
#7 b NA 2 2
#8 b 11 3 3
#9 b 1 3 3
#10 b 1 3 3
#11 b 1 3 3
#12 b 12 4 4
#13 b 1 4 4
#14 c NA 5 5
#15 c 4 5 5
#16 c 7 5 5
Here is an approach using dplyr:
require(dplyr)
set.seed(999)
test <- data.frame(
id = rep(letters[1:4], each = 3),
time_diff = sample(4:15)
)
test %>%
mutate(
b = as.integer(id) - lag(as.integer(id)),
more10 = time_diff > 10,
increment = pmax(b, more10, na.rm = TRUE),
increment = ifelse(row_number() == 1, 1, increment),
block = cumsum(increment)
) %>%
select(id, time_diff, block)
Try:
> df
id time_diff
1 a NA
2 a 1
3 a 1
4 a 1
5 a 3
6 a 3
7 b NA
8 b 11
9 b 1
10 b 1
11 b 1
12 b 12
13 b 1
14 c NA
15 c 4
16 c 7
block= c(1)
for(i in 2:nrow(df))
block[i] = ifelse(df$time_diff[i]>10 || df$id[i]!=df$id[i-1],
block[i-1]+1,
block[i-1])
df$block = block
df
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5