How does one count the characters based on the order they appear in a single length string. Below is an minimal example:
x <- "abbccdddaab"
First thought was this but it only counts them irrespective of order:
table(unlist(strsplit(x, "\\b")))
a b c d
3 3 2 3
But the desired output is:
a b c d a b
1 2 2 3 2 1
I would imagine the solution would require a for loop?
We can use rle instead of table as rle returns the output as a list of values and lengths based on checking whether the adjacent elements are same or not
out <- rle(strsplit(x, "\\b")[[1]])
setNames(out$lengths, out$values)
# a b c d a b
# 1 2 2 3 2 1
Using data.table::rleid :
x <- "abbccdddaab"
tmp <- strsplit(x, "\\b")[[1]]
table(data.table::rleid(tmp))
#1 2 3 4 5 6
#1 2 2 3 2 1
Related
Given the following df:
a=c('a','b','c')
b=c(1,2,5)
c=c(2,3,4)
d=c(2,1,6)
df=data.frame(a,b,c,d)
a b c d
1 a 1 2 2
2 b 2 3 1
3 c 5 4 6
I'd like to apply a function that normally takes a vector (and returns a vector) like cummax row by row to the columns in position b to d.
Then, I'd like to have the output back in the df, either as a vector in a new column of the df, or replacing the original data.
I'd like to avoid writing it as a for loop that would iterate every row, pull out the content of the cells into a vector, do its thing and put it back.
Is there a more efficient way? I've given the apply family functions a go, but I'm struggling to first get a good way to vectorise content of columns by row and get the right output.
the final output could look something like that (imagining I've applied a cummax() function).
a b c d
1 a 1 2 2
2 b 2 3 3
3 c 5 5 6
or
a b c d output
1 a 1 2 2 (1,2,2)
2 b 2 3 1 (2,3,3)
3 c 5 4 6 (5,5,6)
where output is a vector.
Seems this would just be a simple apply problem that you want to cbind to df:
> cbind(df, apply(df[ , 4:2] # work with columns in reverse order
, 1, # do it row-by-row
cummax) )
a b c d 1 2 3
d a 1 2 2 2 1 6
c b 2 3 1 2 3 6
b c 5 4 6 2 3 6
Ouch. Bitten by failing to notice that this would be returned in a column oriented matrix and need to transpose that result; Such a newbie mistake. But it does show the value of having a question with a reproducible dataset I suppose.
> cbind(df, t(apply(df[ , 4:2] , 1, cummax) ) )
a b c d d c b
1 a 1 2 2 2 2 2
2 b 2 3 1 1 3 3
3 c 5 4 6 6 6 6
To destructively assign the result to df you would just use:
df <- # .... that code.
This does the concatenation with commas (and as a result no longer needs to be transposed:
> cbind(df, output=apply(df[ , 4:2] , 1, function(x) paste( cummax(x), collapse=",") ) )
a b c d output
1 a 1 2 2 2,2,2
2 b 2 3 1 1,3,3
3 c 5 4 6 6,6,6
We have a data frame as below :
raw<-data.frame(v1=c("A","B","C","D"),v2=c(NA,"B","C","A"),v3=c(NA,"A",NA,"D"),v4=c(NA,"D",NA,NA))
I need a result data frame in the following format :
result<-data.frame(v1=c("A","B","C","D"), v2=c(3,2,2,3))
Used the following code to get the count across one particular column :
count_raw<-sqldf("SELECT DISTINCT(v1) AS V1, COUNT(v1) AS count FROM raw GROUP BY v1")
This would return count of unique values across an individual column.
Any help would be highly appreciated.
Use this
table(unlist(raw))
Output
A B C D
3 2 2 3
For data frame type output wrap this with as.data.frame.table
as.data.frame.table(table(unlist(raw)))
Output
Var1 Freq
1 A 3
2 B 2
3 C 2
4 D 3
If you want a total count,
sapply(unique(raw[!is.na(raw)]), function(i) length(which(raw == i)))
#A B C D
#3 2 2 3
We can use apply with MARGIN = 1
cbind(raw[1], v2=apply(raw, 1, function(x) length(unique(x[!is.na(x)]))))
If it is for each column
sapply(raw, function(x) length(unique(x[!is.na(x)])))
Or if we need the count based on all the columns, convert to matrix and use the table
table(as.matrix(raw))
# A B C D
# 3 2 2 3
If you have only character values in your dataframe as you've provided, you can unlist it and use unique or to count the freq, use count
> library(plyr)
> raw<-data.frame(v1=c("A","B","C","D"),v2=c(NA,"B","C","A"),v3=c(NA,"A",NA,"D"),v4=c(NA,"D",NA,NA))
> unique(unlist(raw))
[1] A B C D <NA>
Levels: A B C D
> count(unlist(raw))
x freq
1 A 3
2 B 2
3 C 2
4 D 3
5 <NA> 6
This is what my dataframe looks like:
a <- c(1,1,4,4,5)
b <- c(1,2,3,3,5)
c <- c(1,4,4,4,5)
d <- c(2,2,4,4,5)
e <- c(1,5,3,3,5)
df <- data.frame(a,b,c,d,e)
I'd like to write something that returns all unique instances of vectors a,b,c,d that have a repeated value in vector e.
For example:
a b c d e
1 1 1 1 2 1
2 1 2 4 2 5
3 4 3 4 4 3
4 4 3 4 4 3
5 5 5 5 5 5
Rows 3 and 4 are exactly the same till vector d (having a combination of 4344) so only one instance of those should be returned, but they have 2 repeated values in vector e. I would want to get a count on those - so the combination of 4344 has 2 repeated values in vector e.
The expected output would me how many times a certain combination such as 4344 had repeated values in vector e. So in this case it would be something like:
a b c d e
4 3 4 4 2
Both R and SQL work, whatever does the job.
Again, see my comments above, but I believe the following gives you a start on your first question. First, create a "key" variable (in this case named key_abcd which uses tidyr::unite to unite columns a, b, c, and d). Then, count up e by this key_abcd variable. The group_by is implicit.
library(tidyr)
library(dplyr)
df <- data.frame(a,b,c,d,e,f,g)
df %>%
unite(key_abcd, a, b, c, d) %>%
count(key_abcd, e)
# key_abcd e n
# (chr) (dbl) (int)
# 1 1_1_1_2 1 1
# 2 1_2_4_2 5 1
# 3 4_3_4_4 3 2
# 4 5_5_5_5 5 1
It appears from how you've worded the question, you are only interested in "more than one" combinations, therefore, you could add %>% filter(n > 1) to the above code.
In a dataframe I'd like to replace values in a series where they exceed a given threshold.
For example, within a group ('ID') in a series designated by 'time', if 'value' ever exceeds 3, I'd like to make all following entries also equal 3.
ID <- as.factor(c(rep("A", 3), rep("B",3), rep("C",3)))
time <- rep(1:3, 3)
value <- c(c(1,1,2), c(2,3,2), c(3,3,2))
dat <- cbind.data.frame(ID, time, value)
dat
ID time value
A 1 1
A 2 1
A 3 2
B 1 2
B 2 3
B 3 2
C 1 3
C 2 3
C 3 2
I'd like it to be:
ID time value
A 1 1
A 2 1
A 3 2
B 1 2
B 2 3
B 3 3
C 1 3
C 2 3
C 3 3
This should be easy, but I can't figure it out. Thanks!
The ave function makes this very easy by allowing you to apply a function to each of the groupings. In this case, we will adapth the cummax (cumulative maximum) to see if we've seen a 3 yet.
dat$value2<-with(dat, ave(value, ID, FUN=
function(x) ifelse(cummax(x)>=3, 3, x)))
dat;
# ID time value value2
# 1 A 1 1 1
# 2 A 2 1 1
# 3 A 3 2 2
# 4 B 1 2 2
# 5 B 2 3 3
# 6 B 3 2 3
# 7 C 1 3 3
# 8 C 2 3 3
# 9 C 3 2 3
You could also just use FUN=cummax if you want never-decreasing values. I wasn't sure about the sequence c(1,2,1) if you wanted to keep that unchanged or not.
If you can assume your data are sorted by group, then this should be fast, essentially relying on findInterval() behind the scenes:
library(IRanges)
id <- Rle(ID)
three <- which(value>=3L)
ir <- reduce(IRanges(three, end(id)[findRun(three, id)])))
dat$value[as.integer(ir)] <- 3L
This avoids looping over the groups.
I am trying to identify duplicates based of a match of elements in two vectors. Using duplicate() provides a vector of all matches, however I would like to index which are matches with each other or not. Using the following code as an example:
x <- c(1,6,4,6,4,4)
y <- c(3,2,5,2,5,5)
frame <- data.frame(x,y)
matches <- duplicated(frame) | duplicated(frame, fromLast = TRUE)
matches
[1] FALSE TRUE TRUE TRUE TRUE TRUE
Ultimately, I would like to create a vector that identifies elements 2 and 4 are matches as well as 3,5,6. Any thoughts are greatly appreciated.
Another data.table answer, using the group counter .GRP to assign every distinct element a label:
d <- data.table(frame)
d[,z := .GRP, by = list(x,y)]
# x y z
# 1: 1 3 1
# 2: 6 2 2
# 3: 4 5 3
# 4: 6 2 2
# 5: 4 5 3
# 6: 4 5 3
How about this with plyr::ddply()
ddply(cbind(index=1:nrow(frame),frame),.(x,y),summarise,count=length(index),elems=paste0(index,collapse=","))
x y count elems
1 1 3 1 1
2 4 5 3 3,5,6
3 6 2 2 2,4
NB = the expression cbind(index=1:nrow(frame),frame) just adds an element index to each row
Using merge against the unique possibilities for each row, you can get a result:
labls <- data.frame(unique(frame),num=1:nrow(unique(frame)))
result <- merge(transform(frame,row = 1:nrow(frame)),labls,by=c("x","y"))
result[order(result$row),]
# x y row num
#1 1 3 1 1
#5 6 2 2 2
#2 4 5 3 3
#6 6 2 4 2
#3 4 5 5 3
#4 4 5 6 3
The result$num vector gives the groups.