I have a file like this in R.
**0 1**
0 2
**0 3**
0 4
0 5
0 6
0 7
0 8
0 9
0 10
**1 0**
1 11
1 12
1 13
1 14
1 15
1 16
1 17
1 18
1 19
**3 0**
As we can see, there are similar unordered pairs in this ( marked pairs ), like,
1 0
and
0 1
I wish to remove these pairs. And I want to count the number of such pairs that I have and append the count in front of the tow that is repeated. If not repeated, then 1 should be written in the third column.
For example ( A sample of the output file )
0 1 2
0 2 1
0 3 2
0 4 1
0 5 1
0 6 1
0 7 1
0 8 1
0 9 1
0 10 1
1 11 1
1 12 1
1 13 1
1 14 1
1 15 1
1 16 1
1 17 1
1 18 1
1 19 1
How can I achieve it in R?
Here is a way using transform, pmin and pmax to reorder the data by row, and then aggregate to provide a count:
# data
x <- data.frame(a=c(rep(0,10),rep(1,10),3),b=c(1:10,0,11:19,0))
#logic
aggregate(count~a+b,transform(x,a=pmin(a,b), b=pmax(a,b), count=1),sum)
a b count
1 0 1 2
2 0 2 1
3 0 3 2
4 0 4 1
5 0 5 1
6 0 6 1
7 0 7 1
8 0 8 1
9 0 9 1
10 0 10 1
11 1 11 1
12 1 12 1
13 1 13 1
14 1 14 1
15 1 15 1
16 1 16 1
17 1 17 1
18 1 18 1
19 1 19 1
Here's one approach:
First, create a vector of the columns sorted and then pasted together.
x <- apply(mydf, 1, function(x) paste(sort(x), collapse = " "))
Then, use ave to create the counts you are looking for.
mydf$count <- ave(x, x, FUN = length)
Finally, you can use the "x" vector again, this time to detect and remove duplicated values.
mydf[!duplicated(x), ]
# V1 V2 count
# 1 0 1 2
# 2 0 2 1
# 3 0 3 2
# 4 0 4 1
# 5 0 5 1
# 6 0 6 1
# 7 0 7 1
# 8 0 8 1
# 9 0 9 1
# 10 0 10 1
# 12 1 11 1
# 13 1 12 1
# 14 1 13 1
# 15 1 14 1
# 16 1 15 1
# 17 1 16 1
# 18 1 17 1
# 19 1 18 1
# 20 1 19 1
Related
I need a recoding help. Here how my dataset looks like.
df <- data.frame(id = c(1,1,1,1,1, 2,2,2,2,2, 3,3,3,3,3, 4,4,4,4,4),
score = c(0,1,0,1,0, 0,2,0,2,2, 0,3,3,0,0, 0,1,3,1,3))
> df
id score
1 1 0
2 1 1
3 1 0
4 1 1
5 1 0
6 2 0
7 2 2
8 2 0
9 2 2
10 2 2
11 3 0
12 3 3
13 3 3
14 3 0
15 3 0
16 4 0
17 4 1
18 4 3
19 4 1
20 4 3
Some ids have missing score categories. So if this is the case per id, I would like to recode score category. So:
a) if the score options are `0,1,2` and `1` score is missing, then `2` need to be recoded as `1`,
b) if the score options are `0,1,2,3` and `1,2` score is missing, then `3` need to be recoded as `1`,
c) if the score options are `0,1,2,3` and `2` score is missing, then `2,3` need to be recoded as `1,2`,
the idea is there should not be any missing score categories in between.
The desired output would be:
> df.1
id score score.recoded
1 1 0 0
2 1 1 1
3 1 0 0
4 1 1 1
5 1 0 0
6 2 0 0
7 2 2 1
8 2 0 0
9 2 2 1
10 2 2 1
11 3 0 0
12 3 3 1
13 3 3 1
14 3 0 0
15 3 0 0
16 4 0 0
17 4 1 1
18 4 3 2
19 4 1 1
20 4 3 2
df %>%
group_by(id)%>%
mutate(score = as.numeric(factor(score)) - 1)
# A tibble: 20 x 2
# Groups: id [4]
id score
<dbl> <dbl>
1 1 0
2 1 1
3 1 0
4 1 1
5 1 0
6 2 0
7 2 1
8 2 0
9 2 1
10 2 1
11 3 0
12 3 1
13 3 1
14 3 0
15 3 0
16 4 0
17 4 1
18 4 2
19 4 1
20 4 2
Using data.table
library(data.table)
setDT(df)[, score.recoded := 0][
score >0, score.recoded := match(score, score), id]
-output
> df
id score score.recoded
<num> <num> <int>
1: 1 0 0
2: 1 1 1
3: 1 0 0
4: 1 1 1
5: 1 0 0
6: 2 0 0
7: 2 2 1
8: 2 0 0
9: 2 2 1
10: 2 2 1
11: 3 0 0
12: 3 3 1
13: 3 3 1
14: 3 0 0
15: 3 0 0
16: 4 0 0
17: 4 1 1
18: 4 3 2
19: 4 1 1
20: 4 3 2
I have some time series with corresponding number for each date as 0 or 1. For example:
date value
1 0
2 0
3 1
4 1
5 1
6 0
7 1
8 1
So I want to count the consecutive 1´s like for date 3-5 the sum should be 3 and then start at date 7 again to count. And if this sum is below 6 the 1´s should be transformed to 0´s.
library(dplyr)
data.frame(
date = 1:8,
value = c(0,0,1,1,1,0,1,1)
) %>%
mutate(
count = rle(value) %>%
{list(.$lengths * .$values, .$lengths)} %>%
{rep(x = .[[1]], times = .[[2]])},
count_1 = ifelse(count < 6, 0, count)
)
gives:
date value count count_1
1 1 0 0 0
2 2 0 0 0
3 3 1 3 0
4 4 1 3 0
5 5 1 3 0
6 6 0 0 0
7 7 1 2 0
8 8 1 2 0
I would first create a grouping variable and then use this to aggregate the dataset.
d = data.frame("date"=1:12,
"value"=c(1,1,0,0,1,1,1,1,0,0,1,0))
d$group = 1
for(i in 2:dim(d)[1]){
if(d$value[i]==d$value[i-1]){
d$group[i]=d$group[i-1]
} else {
d$group[i]=d$group[i-1]+1
}
}
nd = data.frame("Group"=unique(d$group),
"Start"=aggregate(d$date~d$group,FUN=min)[,2],
"End"=aggregate(d$date~d$group,FUN=max)[,2],
"Count"=aggregate(d$value~d$group,FUN=sum)[,2])
The output for this data would be:
> d ## Input data
date value
1 1 1
2 2 1
3 3 0
4 4 0
5 5 1
6 6 1
7 7 1
8 8 1
9 9 0
10 10 0
11 11 1
12 12 0
> nd ## All groups
Group Start End Count
1 1 1 2 2
2 2 3 4 0
3 3 5 8 4
4 4 9 10 0
5 5 11 11 1
6 6 12 12 0
> nd[nd$Count>0,] ## Just the groups with 1 in them:
Group Start End Count
1 1 1 2 2
3 3 5 8 4
5 5 11 11 1
Another solution which looks like what you expected :
d = data.frame("date"=1:20,"value"=c(1,1,0,0,1,1,1,1,0,0,1,0,1,1,1,1,1,1,1,0))
repl <- rle(d$value)
rep_lengths <- rep(repl$lengths, repl$lengths)
rep_lengths[rep_lengths < 6] <- 0
d$value <- rep_lengths
returns
> d
date value
1 1 0
2 2 0
3 3 0
4 4 0
5 5 0
6 6 0
7 7 0
8 8 0
9 9 0
10 10 0
11 11 0
12 12 0
13 13 7
14 14 7
15 15 7
16 16 7
17 17 7
18 18 7
19 19 7
20 20 0
You can use rle to count the consecutive and use ifelse to set those lower 6 to 0:
y <- rle(x$value)
y[[2]] <- y[[1]] * y[[2]]
y[[2]] <- ifelse(y[[2]] < 6, 0, y[[2]])
inverse.rle(y)
#[1] 0 0 0 0 0 0 0 0
Data:
x <- data.frame(date = 1:8, value = c(0,0,1,1,1,0,1,1))
I have a dataframe with repeated IDs, I want to get a colunm that contains the very first occurance of each ID.
For instance:
ID <- as.character( "ae61_10", "ae61_10", "ae61_10", "ae61_10", "ae61_10", "ae61_1", "ae61_1", "ae61_1", "ae61_11", "ae61_2", "ae61_2", "ae61_3", "ae61_4", "ae61_4", "ae61_5", "ae61_6", "ae61_7", "ae61_8", "ae61_8", "ae61_8", "ae61_9")
df <- data.frame(ID)
df
ID
1 ae61_10
2 ae61_10
3 ae61_10
4 ae61_10
5 ae61_10
6 ae61_1
7 ae61_1
8 ae61_1
9 ae61_11
10 ae61_2
11 ae61_2
12 ae61_3
13 ae61_4
14 ae61_4
15 ae61_5
16 ae61_6
17 ae61_7
18 ae61_8
19 ae61_8
20 ae61_8
21 ae61_9
I want to get the colunm FIRST_OC, which returns 1 if that is the first ID occurance, and 0 if it is not. Like this:
ID FIRST_OC
1 ae61_10 1
2 ae61_10 0
3 ae61_10 0
4 ae61_10 0
5 ae61_10 0
6 ae61_1 1
7 ae61_1 0
8 ae61_1 0
9 ae61_11 1
10 ae61_2 1
11 ae61_2 0
12 ae61_3 1
13 ae61_4 1
14 ae61_4 0
15 ae61_5 1
16 ae61_6 1
17 ae61_7 1
18 ae61_8 1
19 ae61_8 0
20 ae61_8 0
21 ae61_9 1
I have tried this command, but it did not work:
df$FIRST_OC <- 0
FIRST_OC <- df[match(unique(df$ID), df$ID),]
df$FIRST_OC[which(df$ID %in% FIRST_OC)] <- 1
Could somebody help me to figure out what is going wrong?
Many thanks.
duplicated is the function to use!
df$FIRST_OC <- ifelse(!duplicated(df$ID), 1, 0)
# OR
df$FIRST_OC <- as.numeric(!duplicated(df$ID))
We can use duplicated
+(!duplicated(df$ID))
#[1] 1 0 0 0 0 1 0 0 1 1 0 1 1 0 1 1 1 1 0 0 1
I want to accumulate the values of a column till the end of the group, though starting the addition when a specific value occurs in another column. I am only interested in the first instance of the specific value within a group. So if that value occurs again within the group, the addition column should continue to add the values. I know this sounds like a rather strange problem, so hopefully the example table makes sense.
The following data frame is what I have now:
> df = data.frame(group = c(1,1,1,1,2,2,2,2,2,3,3,3,4,4,4),numToAdd = c(1,1,3,2,4,2,1,3,2,1,2,1,2,3,2),occurs = c(0,0,1,0,0,1,0,0,0,0,1,1,0,0,0))
> df
group numToAdd occurs
1 1 1 0
2 1 1 0
3 1 3 1
4 1 2 0
5 2 4 0
6 2 2 1
7 2 1 0
8 2 3 0
9 2 2 0
10 3 1 0
11 3 2 1
12 3 1 1
13 4 2 0
14 4 3 0
15 4 2 0
Thus, whenever a 1 occurs within a group, I want a cumulative sum of the values from the column numToAdd, until a new group starts. This would look like the following:
> finalDF = data.frame(group = c(1,1,1,1,2,2,2,2,2,3,3,3,4,4,4),numToAdd = c(1,1,3,2,4,2,1,3,2,1,2,1,2,3,2),occurs = c(0,0,1,0,0,1,0,0,0,0,1,1,0,0,0),added = c(0,0,3,5,0,2,3,6,8,0,2,3,0,0,0))
> finalDF
group numToAdd occurs added
1 1 1 0 0
2 1 1 0 0
3 1 3 1 3
4 1 2 0 5
5 2 4 0 0
6 2 2 1 2
7 2 1 0 3
8 2 3 0 6
9 2 2 0 8
10 3 1 0 0
11 3 2 1 2
12 3 1 1 3
13 4 2 0 0
14 4 3 0 0
15 4 2 0 0
Thus, the added column is 0 until a 1 occurs within the group, then accumulates the values from numToAdd until it moves to a new group, turning the added column back to 0. In group three, a value of 1 is found a second time, yet the cumulated sum continues. Additionally, in group 4, a value of 1 is never found, thus the value within the added column remains 0.
I've played around with dplyr, but can't get it to work. The following solution only outputs the total sum, and not the increasing cumulated number at each row.
library(dplyr)
df =
df %>%
mutate(added=ifelse(occurs == 1,cumsum(numToAdd),0)) %>%
group_by(group)
Try
df %>%
group_by(group) %>%
mutate(added= cumsum(numToAdd*cummax(occurs)))
# group numToAdd occurs added
# 1 1 1 0 0
# 2 1 1 0 0
# 3 1 3 1 3
# 4 1 2 0 5
# 5 2 4 0 0
# 6 2 2 1 2
# 7 2 1 0 3
# 8 2 3 0 6
# 9 2 2 0 8
# 10 3 1 0 0
# 11 3 2 1 2
# 12 3 1 1 3
# 13 4 2 0 0
# 14 4 3 0 0
# 15 4 2 0 0
Or using data.table
library(data.table)#v1.9.5+
i1 <-setDT(df)[, .I[(rleid(occurs) + (occurs>0))>1], group]$V1
df[, added:=0][i1, added:=cumsum(numToAdd), by = group]
Or a similar option as in dplyr
setDT(df)[,added := cumsum(numToAdd * cummax(occurs)) , by = group]
You can use split-apply-combine in base R with something like:
df$added <- unlist(lapply(split(df, df$group), function(x) {
y <- rep(0, nrow(x))
pos <- cumsum(x$occurs) > 0
y[pos] <- cumsum(x$numToAdd[pos])
y
}))
df
# group numToAdd occurs added
# 1 1 1 0 0
# 2 1 1 0 0
# 3 1 3 1 3
# 4 1 2 0 5
# 5 2 4 0 0
# 6 2 2 1 2
# 7 2 1 0 3
# 8 2 3 0 6
# 9 2 2 0 8
# 10 3 1 0 0
# 11 3 2 1 2
# 12 3 1 1 3
# 13 4 2 0 0
# 14 4 3 0 0
# 15 4 2 0 0
To add another base R approach:
df$added <- unlist(lapply(split(df, df$group), function(x) {
c(x[,'occurs'][cumsum(x[,'occurs']) == 0L],
cumsum(x[,'numToAdd'][cumsum(x[,'occurs']) != 0L]))
}))
# group numToAdd occurs added
# 1 1 1 0 0
# 2 1 1 0 0
# 3 1 3 1 3
# 4 1 2 0 5
# 5 2 4 0 0
# 6 2 2 1 2
# 7 2 1 0 3
# 8 2 3 0 6
# 9 2 2 0 8
# 10 3 1 0 0
# 11 3 2 1 2
# 12 3 1 1 3
# 13 4 2 0 0
# 14 4 3 0 0
# 15 4 2 0 0
Another base R:
df$added <- unlist(lapply(split(df,df$group),function(x){
cumsum((cumsum(x$occurs) > 0) * x$numToAdd)
}))
I have a dataframe with many rows, but the structure looks like this:
year factor
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 1
10 0
11 0
12 0
13 0
14 0
15 0
16 0
17 1
18 0
19 0
20 0
I would need to add a counter as a third column. It should count the cumulative cells that contains zero until it set again to zero once the value 1 is encountered. The result should look like this:
year factor count
1 0 0
2 0 1
3 0 2
4 0 3
5 0 4
6 0 5
7 0 6
8 0 7
9 1 0
10 0 1
11 0 2
12 0 3
13 0 4
14 0 5
15 0 6
16 0 7
17 1 0
18 0 1
19 0 2
20 0 3
I would be glad to do it in a quick way, avoiding loops, since I have to do the operations for hundreds of files.
You can copy my dataframe, pasting the dataframe in "..." here:
dt <- read.table( text="...", , header = TRUE )
Perhaps a solution like this with ave would work for you:
A <- cumsum(dt$factor)
ave(A, A, FUN = seq_along) - 1
# [1] 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3
Original answer:
(Missed that the first value was supposed to be "0". Oops.)
x <- rle(dt$factor == 1)
y <- sequence(x$lengths)
y[dt$factor == 1] <- 0
y
# [1] 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 0 1 2 3