I have a problem figuering this out:
suppose this is how my data looks like:
Num condition y
1 a 1
2 a 2
3 a 3
4 b 4
5 b 5
6 b 6
7 c 7
8 c 8
9 c 9
10 b 10
11 b 11
12 b 12
I now want to make calculation (e.g., mean) on b, depending on whether value was in the row before b, in this example a or c?
Thanks for any help!!!
Angelika
Is this what you want?
# in order to separate between different runs of condition 'b',
# get length and value of runs of equal values of 'condition'
rl <- rle(x = df$condition)
df$run <- rep(x = seq_len(length(rl$lengths)), times = rl$lengths)
# calculate sum of y, on data grouped by condition and run, and where condition is 'b'
aggregate(y ~ condition + run, data = df, subset = condition == "b", sum)
You can add a "lagged" condition column to your dataframe (assuming DF) using
> DF <- within(DF, lag_cond <- c(NA, head(as.character(condition), -1)))
Result:
Num condition y lag_cond
1 a 1 <NA>
2 a 2 a
3 a 3 a
4 b 4 a
5 b 5 b
6 b 6 b
7 c 7 b
8 c 8 c
9 c 9 c
10 b 10 c
11 b 11 b
12 b 12 b
Now you can identify rows you want like this:
> DF[with(DF, condition=="b" & lag_cond %in% c("a","c")),]
Num condition y lag_cond
4 b 4 a
10 b 10 c
Related
Say we have a data frame with a factor (Group) that is a grouping variable for a list of IDs:
set.seed(123)
data <- data.frame(Group = factor(sample(5,10, replace = T)),
ID = c(1:10))
In this example, the ID's belong to one of 5 Groups, labeled 1:5. We simply want to replace 1:5 with A:E. In other words, if Group == 1, we want to change it to A, if Group == 2, we want to change it to B, and so on. What is the simplest way to achieve this?
You may assign new labels= in a names list using factor once again.
data$Group1 <- factor(data$Group, labels=list("1"="A", "2"="B", "3"="C", "4"="D", "5"="E"))
## more succinct:
data$Group2 <- factor(data$Group, labels=setNames(list("A", "B", "C", "D", "E"), 1:5))
data
# Group ID Group1 Group2 Group3
# 1 3 1 C C C
# 2 3 2 C C C
# 3 2 3 B B B
# 4 2 4 B B B
# 5 3 5 C C C
# 6 5 6 E E E
# 7 4 7 D D D
# 8 1 8 A A A
# 9 2 9 B B B
# 10 3 10 C C C
This for general, if indeed capital letters are wanted see #RonakShah's solution.
You can use the built-in constant in R LETTERS :
data$new_group <- LETTERS[data$Group]
data
# Group ID new_group
#1 3 1 C
#2 3 2 C
#3 2 3 B
#4 2 4 B
#5 3 5 C
#6 5 6 E
#7 4 7 D
#8 1 8 A
#9 2 9 B
#10 3 10 C
Created a new column (new_group) here for comparison purposes. You can overwrite the same column if you wish to.
I have a dataframe in the following format
> x <- data.frame("a" = c(1,1),"b" = c(2,2),"c" = c(3,4))
> x
a b c
1 1 2 3
2 1 2 4
I'd like to add 3 new columns which is a cumulative product of the columns a b c, however I need a reverse cumulative product i.e. the output should be
row 1:
result_d = 1*2*3 = 6 , result_e = 2*3 = 6, result_f = 3
and similarly for row 2
The end result will be
a b c result_d result_e result_f
1 1 2 3 6 6 3
2 1 2 4 8 8 4
the column names do not matter this is just an example. Does anyone have any idea how to do this?
as per my comment, is it possible to do this on a subset of columns? e.g. only for columns b and c to return:
a b c results_e results_f
1 1 2 3 6 3
2 1 2 4 8 4
so that column "a" is effectively ignored?
One option is to loop through the rows and apply cumprod over the reverse of elements and then do the reverse
nm1 <- paste0("result_", c("d", "e", "f"))
x[nm1] <- t(apply(x, 1,
function(x) rev(cumprod(rev(x)))))
x
# a b c result_d result_e result_f
#1 1 2 3 6 6 3
#2 1 2 4 8 8 4
Or a vectorized option is rowCumprods
library(matrixStats)
x[nm1] <- rowCumprods(as.matrix(x[ncol(x):1]))[,ncol(x):1]
temp = data.frame(Reduce("*", x[NCOL(x):1], accumulate = TRUE))
setNames(cbind(x, temp[NCOL(temp):1]),
c(names(x), c("res_d", "res_e", "res_f")))
# a b c res_d res_e res_f
#1 1 2 3 6 6 3
#2 1 2 4 8 8 4
I have a dataset with several variables, but I want to keep the rows that are the same based on two columns. Here is an example of what I want to do:
a <- c(rep('A',3), rep('B', 3), rep('C',3))
b <- c(1,1,2,4,4,4,5,5,5)
df <- data.frame(a,b)
a b
1 A 1
2 A 1
3 A 2
4 B 4
5 B 4
6 B 4
7 C 5
8 C 5
9 C 5
I know that if I use the duplicated function I can get:
df[!duplicated(df),]
a b
1 A 1
3 A 2
4 B 4
7 C 5
But since the level 'A' on column a does not have a unique value in b, I want to drop both observations to get a new data.frame as this:
a b
4 B 4
7 C 5
I don't mind to have repeated values across b, as long as for every same level on a there is the same value in b.
Is there a way to do this? Thanks!
This one maybe?
ag <- aggregate(b~a, df, unique)
ag[lengths(ag$b)==1,]
# a b
#2 B 4
#3 C 5
Maybe something like this:
> ind <- apply(sapply(with(df, split(b,a)), diff), 2, function(x) all(x==0) )
> out <- df[!duplicated(df),]
> out[out$a %in% names(ind)[ind], ]
a b
4 B 4
7 C 5
Here is another option with data.table
library(data.table)
setDT(df)[, if(uniqueN(b)==1) .SD[1L], by = a]
# a b
#1: B 4
#2: C 5
I've been playing with some data in order to obtain the ratios between two levels within one variable and taking into account two other variables. I've been using the function aggregate(), which is very useful to calculate means and sums. However, I'm stuck when I want to calculate some ratios (divisions).
Here you find a dataframe very similar to my data:
w<-c("A","B","C","D","E","F","A","B","C","D","E","F")
x<-c(1,1,1,1,1,1,2,2,2,2,2,2)
y<-c(3,4,5,6,8,10,3,4,5,7,9,10)
z<-runif(12)
df<-data.frame(w,x,y,z)
df
w x y z
1 A 1 3 0.93767621
2 B 1 4 0.09169992
3 C 1 5 0.49012926
4 D 1 6 0.90886690
5 E 1 8 0.37058120
6 F 1 10 0.83558267
7 A 2 3 0.42670001
8 B 2 4 0.05656252
9 C 2 5 0.70694423
10 D 2 7 0.13634309
11 E 2 9 0.92065671
12 F 2 10 0.56276176
What I want is to obtain the ratios of z from the two levels of x and taking into account the variables w and y. So the level "A" from the variable "w" in the level "3" from the variable "y" should be:
df$z[1]/df$z[7]
With aggregate function should be something like this:
final<-aggregate(z~y:w, data=df)
However, I know that I miss something because in the variable y there are some classes that not appear in the two categories of w (e.g. 7, 8 and 9).
Any help will be welcomed!
We can use data.table. We convert the 'data.frame' to 'data.table' (setDT(df)), grouped by 'w', 'y', if the nrow (.N) is 2, we divide the first value by the second or else return the 'z'. Assign (:=) the output to a new column 'z1'.
library(data.table)
setDT(df)[,z1 :=if(.N==2) z[1]/z[2] else z , by = .(w,y)]
df
# w x y z z1
# 1: A 1 3 0.93767621 2.1975069
# 2: B 1 4 0.09169992 1.6212135
# 3: C 1 5 0.49012926 0.6933068
# 4: D 1 6 0.90886690 0.9088669
# 5: E 1 8 0.37058120 0.3705812
# 6: F 1 10 0.83558267 1.4847894
# 7: A 2 3 0.42670001 2.1975069
# 8: B 2 4 0.05656252 1.6212135
# 9: C 2 5 0.70694423 0.6933068
#10: D 2 7 0.13634309 0.1363431
#11: E 2 9 0.92065671 0.9206567
#12: F 2 10 0.56276176 1.4847894
If we just want the summary output we don't need to use :=
setDT(df)[, list(z=if(.N==2) z[1]/z[2] else z) , by = .(w,y)]
Or using aggregate
aggregate(z~w+y, df, FUN=function(x)
if(length(x)==2) x[1]/x[2] else x)
I apologize for the title of this question. I can't figure out how a good way to briefly describe what I want to do.
I have something like this, with >8000 rows:
x y value_xy
A B 7
A C 2
B A 3
B C 6
C A 2
C B 1
I want to create a new column, value_yx, that looks like this:
x y value_xy value_yx
A B 7 3
A C 2 2
B A 3 7
B C 1 1
C A 2 2
C B 1 1
For each value of x and y, I want to have a new column that finds the value of y to x (as y appears later in the x column). Sometimes these values are equal, other times they aren't.
I have explored using for loops, ave(), and several other functions, but I haven't been able to make it work.
Try merge. The by.x and by.y arguments specify columns to be matched, and here the order of matching columns is reversed in by.y:
merge(x = df, y = df, by.x = c("x", "y"), by.y = c("y", "x"))
# x y value_xy.x value_xy.y
# 1 A B 7 3
# 2 A C 2 2
# 3 B A 3 7
# 4 B C 6 1
# 5 C A 2 2
# 6 C B 1 6
Looks like I was beat to it but an alternative solution with mapply
df$value_yx = mapply(function(x_flip, y_flip) df[df$x == y_flip & df$y == x_flip,]$value_xy, df$x, df$y)
# x y value_xy value_yx
#1 A B 7 3
#2 A C 2 2
#3 B A 3 7
#4 B C 6 1
#5 C A 2 2
#6 C B 1 6
xtabs will return a value-matrix that can be indexed by a two-column, character-valued matrix formed from the first two columns and are probably factors (hence the need for the as.character()-conversion:
> dfrm$value_yx <- xtabs(value_xy~x+y, dfrm)[
sapply(dfrm[2:1],as.character) ]
> dfrm
x y value_xy value_yx
1 A B 7 3
2 A C 2 2
3 B A 3 7
4 B C 6 1
5 C A 2 2
6 C B 1 6
--- See what is being indexed
> xtabs(value_xy~x+y, dfrm)
y
x A B C
A 0 7 2
B 3 0 6
C 2 1 0