I wanted to create a vector of counts if possible.
For example: I have a vector
x <- c(3, 0, 2, 0, 0)
How can I create a frequency vector for all integers between 0 and 3? Ideally I wanted to get a vector like this:
> 3 0 1 1
which gives me the counts of 0, 1, 2, and 3 respectively.
Much appreciated!
You can do
table(factor(x, levels=0:3))
Simply using table(x) is not enough.
Or with tabulate which is faster
tabulate(factor(x, levels = min(x):max(x)))
You can do this using rle (I made this in minutes, so sorry if it's not optimized enough).
x = c(3, 0, 2, 0, 0)
r = rle(x)
f = function(x) sum(r$lengths[r$values == x])
s = sapply(FUN = f, X = as.list(0:3))
data.frame(x = 0:3, freq = s)
#> data.frame(x = 0:3, freq = s)
# x freq
#1 0 3
#2 1 0
#3 2 1
#4 3 1
You can just use table():
a <- table(x)
a
x
#0 2 3
#3 1 1
Then you can subset it:
a[names(a)==0]
#0
#3
Or convert it into a data.frame if you're more comfortable working with that:
u<-as.data.frame(table(x))
u
# x Freq
#1 0 3
#2 2 1
#3 3 1
Edit 1:
For levels:
a<- as.data.frame(table(factor(x, levels=0:3)))
Related
I have got a data set that looks like this:
COMPANY DATABREACH CYBERBACKGROUND
A 1 2
B 0 2
C 0 1
D 0 2
E 1 1
F 1 2
G 0 2
H 0 2
I 0 2
J 0 2
No I want to create the following: 40% of the cases that the column DATABREACH has the value of 1, I want the value CYBERBACKGROUND to take the value of 2. I figure there must be some function to do this, but I cannot find it.
ind <- which(df$DATABREACH == 1)
ind <- ind[rbinom(length(ind), 1, prob = 0.4) > 0]
df$CYBERBACKGROUND[ind] <- 2
The above is a bit more efficient in that it only pulls randomness for as many as strictly required. If you aren't concerned (11000 doesn't seem too high), you can reduce that to
df$CYBERBACKGROUND <-
ifelse(df$DATABREACH == 1 & rbinom(nrow(df), 1, prob = 0.4) > 0,
2, df$CYBERBACKGROUND)
We may use
library(dplyr)
df1 <- df1 %>%
mutate(CYBERBACKGROUND = replace(CYBERBACKGROUND,
sample(which(DATABREACH == 0), sum(ceiling(sum(DATABREACH) * 0.4))), 2))
I'm trying to create a function where I can pass a function as a variable to perform on a variable number of columns, after removing zeros. I'm not too comfortable with ellipses yet, and I'm guessing this is where the problem is arising. The function is using all the values in the specified rows, summarizing them based on the selected function, and then mutating that one value. I'd like to maintain the function across the row (e.g. rowMeans)
Example:
# Setup dataframe
a <- 1:5
b <- c(0, 4, 3, 0, 1)
c <- c(5:1)
d <- c(2, 0, 1, 0, 4)
df <- data.frame(a, b, c, d)
FUNexcludeZero <- function(function_name, ...){
# Match function name
FUN <- match.fun(function_name)
# get all the values - I'm sure this is the problem, need to somehow turn it back into a df?
vals <- unlist(list(...))
# Remove 0's and perform function
valsNo0 <- vals[vals != 0]
compiledVals <- FUN(valsNo0)
return(compiledVals)
}
df %>%
mutate(foo = FUNexcludeZero(function_name = 'sd', a, b))
a b c d foo
1 1 0 5 2 1.457738
2 2 4 4 0 1.457738
3 3 3 3 1 1.457738
4 4 0 2 0 1.457738
5 5 1 1 4 1.457738
df %>%
mutate(foo = FUNexcludeZero(function_name = 'min', a, b))
a b c d foo
1 1 0 5 2 1
2 2 4 4 0 1
3 3 3 3 1 1
4 4 0 2 0 1
5 5 1 1 4 1
# Try row-function (same error occurs with rowMeans)
df %>%
mutate(foo = FUNexcludeZero(function_name = 'pmin', a, b))
Error in mutate_impl(.data, dots) :
Column `foo` must be length 5 (the number of rows) or one, not 8
For function_name = 'sd' the column should be c(NA, 1.41, 0, NA, 2.828) and the min and pmin should be c(1, 2, 3, 4, 1). I'm 100% sure the error has something to do with the list/unlist, but any other way I try it I end up with an error.
I am not sure if this is exactly what you what. You needed to perform a row wise operation on the two vectors, thus I used the apply function. This should work for any number of equal length vectors.
# Setup dataframe
a <- 1:5
b <- c(0, 4, 3, 0, 1)
c <- c(5:1)
d <- c(2, 0, 1, 0, 4)
#df <- data.frame(a, b, c, d) #not used
FUNexcludeZero <- function(function_name, ...){
# Match function name
FUN <- match.fun(function_name)
#combine the vectors into a matrix
df<-cbind(...)
#remove 0 from rows and apply function to the rows
compiledVals <- apply(df, 1, function(x) { x<-x[x!=0]
FUN(x)})
return(compiledVals)
}
FUNexcludeZero(function_name = 'sd', a, b)
#[1] NA 1.414214 0.000000 NA 2.828427
FUNexcludeZero(function_name = 'min', a, b)
#[1] 1 2 3 4 1
I am trying to create a column in a very large data frame (~ 2.2 million rows) that calculates the cumulative sum of 1's for each factor level, and resets when a new factor level is reached. Below is some basic data that resembles my own.
itemcode <- c('a1', 'a1', 'a1', 'a1', 'a1', 'a2', 'a2', 'a3', 'a4', 'a4', 'a5', 'a6', 'a6', 'a6', 'a6')
goodp <- c(0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1)
df <- data.frame(itemcode, goodp)
I would like the output variable, cum.goodp, to look like this:
cum.goodp <- c(0, 1, 2, 0, 1, 1, 2, 0, 0, 1, 1, 1, 2, 0, 1)
I get that there is a lot out there using the canonical split-apply-combine approach, which, conceptually is intuitive, but I tried using the following:
k <- transform(df, cum.goodp = goodp*ave(goodp, c(0L, cumsum(diff(goodp != 0)), FUN = seq_along, by = itemcode)))
When I try to run this code it's very very slow. I get that transform is part of the reason why (the 'by' doesn't help either). There are over 70K different values for the itemcode variable, so it should probably be vectorized. Is there a way to vectorize this, using cumsum? If not, any help whatsoever would be truly appreciated. Thanks so much.
A base R approach is to calculate cumsum over the whole vector, and capture the geometry of the sub-lists using run-length encoding. Figure out the start of each group, and create new groups
start <- c(TRUE, itemcode[-1] != itemcode[-length(itemcode)]) | !goodp
f <- cumsum(start)
Summarize these as a run-length encoding, and calculate the overall sum
r <- rle(f)
x <- cumsum(x)
Then use the geometry to get the offset that each embedded sum needs to be corrected by
offset <- c(0, x[cumsum(r$lengths)])
and calculate the updated value
x - rep(offset[-length(offset)], r$lengths)
Here's a function
cumsumByGroup <- function(x, f) {
start <- c(TRUE, f[-1] != f[-length(f)]) | !x
r <- rle(cumsum(start))
x <- cumsum(x)
offset <- c(0, x[cumsum(r$lengths)])
x - rep(offset[-length(offset)], r$lengths)
}
Here's the result applied to the sample data
> cumsumByGroup(goodp, itemcode)
[1] 0 1 2 0 1 1 2 0 0 1 1 1 2 0 1
and it's performance
> n <- 1 + rpois(1000000, 1)
> goodp <- sample(c(0, 1), sum(n), TRUE)
> itemcode <- rep(seq_along(n), n)
> system.time(cumsumByGroup(goodp, itemcode))
user system elapsed
0.55 0.00 0.55
The dplyr solution takes about 70s.
#alexis_laz solution is both elegant and 2 times faster than mine
cumsumByGroup1 <- function(x, f) {
start <- c(TRUE, f[-1] != f[-length(f)]) | !x
cs = cumsum(x)
cs - cummax((cs - x) * start)
}
With the modified example input/output you could use the following base R approach (among others):
transform(df, cum.goodpX = ave(goodp, itemcode, cumsum(goodp == 0), FUN = cumsum))
# itemcode goodp cum.goodp cum.goodpX
#1 a1 0 0 0
#2 a1 1 1 1
#3 a1 1 2 2
#4 a1 0 0 0
#5 a1 1 1 1
#6 a2 1 1 1
#7 a2 1 2 2
#8 a3 0 0 0
#9 a4 0 0 0
#10 a4 1 1 1
#11 a5 1 1 1
#12 a6 1 1 1
#13 a6 1 2 2
#14 a6 0 0 0
#15 a6 1 1 1
Note: I added column cum.goodp to the input df and created a new column cum.goodpX so you can easily compare the two.
But of course you can use many other approaches with packages, either what #MartinMorgan suggested or for example using dplyr or data.table, to name just two options. Those may be a lot faster than base R approaches for large data sets.
Here's how it would be done in dplyr:
library(dplyr)
df %>%
group_by(itemcode, grp = cumsum(goodp == 0)) %>%
mutate(cum.goodpX = cumsum(goodp))
A data.table option was already provided in the comments to your question.
I have a data.frame and would like to take a certain value from a cell if another is in a dataframe.
I tried the apply function.
n <- c(2, 3, 0 ,1)
s <- c(0, 1, 1, 2)
b <- c("THIS", "FALSE", "NOT", "THIS")
df <- data.frame(n, s, b)
df <- sapply(df$Vals, FUN=function(x){ if(b[x]=="THIS") ? n[x] : s[x] } )
My logic is:
if(b at position x is equal to "This") {
add n[x] to the column df$Vals
} else {
add s[x] to the column df$Vals
}
Whereas x is a single row.
Any recommendation what I am doing wrong?
I appreciate your reply!
Like this:
df$Vals = with(df, ifelse(b=="THIS", n, s))
Or giving direct the resulting data.frame:
transform(df, Vals=with(df, ifelse(b=="THIS", n, s)))
# n s b Vals
#1 2 0 THIS 2
#2 3 1 FALSE 1
#3 0 1 NOT 1
#4 1 2 THIS 1
With your additional conditions:
func=Vectorize(function(b, s, n){if(b=='THIS') return(n);if(b==F) return(n+s);s})
df$Vals = with(df, func(b,s,n))
Or you could use the row/column indexing
df$Vals <- df[1:2][cbind(1:nrow(df),(df$b!='THIS')+1)]
df
# n s b Vals
#1 2 0 THIS 2
#2 3 1 FALSE 1
#3 0 1 NOT 1
#4 1 2 THIS 1
I have a data frame containing a list vector with jagged entries:
df = data.frame(x = rep(c(1,2), 2), y = rep(c("a", "b"), each = 2))
L = list()
for (each in round(runif(4, 1,5))) L = c(L, list(1:each))
df$L = L
For example,
x y L
1 a 1
2 a 1, 2, 3, 4
1 b 1, 2, 3
2 b 1, 2, 3
How could I create a table which counts the values of L for each x, across the values of y? So, in this example it would output something like,
1 2 3 4
X
1 2 1 1 0
2 2 2 2 1
I had some luck using
tablist = function(L) table(unlist(L))
tapply(df$L, df$x, tablist)
which produces,
$`1`
1 2 3
2 1 1
$`2`
1 2 3 4
2 2 2 1
However, I'm not sure how to go from here to a single table. Also, I'm beggining to suspect that this approach might start taking an unruly amount of time for large data frames. Any thoughts / suggestions would be greatly appreciated!
Using pylr
library(plyr)
df = data.frame(x = rep(c(1,2), 2), y = rep(c("a", "b"), each = 2))
L = list()
set.seed(2)
for (each in round(runif(4, 1,5))) L = c(L, list(1:each))
df$L = L
> df
x y L
1 1 a 1, 2
2 2 a 1, 2, 3, 4
3 1 b 1, 2, 3
4 2 b 1, 2
table(ddply(df,.(x),summarize,unlist(L)))
> table(ddply(df,.(x),summarize,unlist(L)))
..1
x 1 2 3 4
1 2 2 1 0
2 2 2 1 1
If you're not into plyr...
vals <- unique(unlist(df$L))
names(vals) <- vals
do.call("rbind",
lapply(split(df,df$x),function(byx){
sapply(vals, function(i){
sum(unlist(sapply(byx$L,"==",i)))
})
})
)