Reputation: 7041
Suppose I have a dataframe such that:
df<-data.frame(id=1:8,group=c(1,0,0,1,1,0,1,0),rep=c(rep("d1",4),rep("d2",4)),value=rbinom(8,1,0.6))
df
id group rep value
1 1 1 d1 0
2 2 0 d1 0
3 3 0 d1 0
4 4 1 d1 1
5 5 1 d2 1
6 6 0 d2 0
7 7 1 d2 1
8 8 0 d2 1
What's the best way to get the cumulative sum by group
and rep
such that:
cumsum
group d1 d1+d2 d1+d2+d3
0 0 1 ...
1 1 3 ...
Upvotes: 21
Views: 21188
Reputation: 103898
I'd recommend working with the tidy form of the data. Here's an approach with dplyr, but it would be trivial to translate to data.table or base R.
First I'll create the dataset, setting the random seed to make the example reproducible:
set.seed(1014)
df <- data.frame(
id = 1:8,
group = c(1, 0, 0, 1, 1, 0, 1, 0),
rep = c(rep("d1", 4), rep("d2", 4)),
value = rbinom(8, 1, 0.6)
)
df
%> id group rep value
%> 1 1 1 d1 1
%> 2 2 0 d1 0
%> 3 3 0 d1 0
%> 4 4 1 d1 1
%> 5 5 1 d2 1
%> 6 6 0 d2 1
%> 7 7 1 d2 1
%> 8 8 0 d2 1
Next, using dplyr, I'll first collapse to individual rows by group, and then compute the cumulative sum:
library(dplyr)
df <- df %>%
group_by(group, rep) %>%
summarise(value = sum(value)) %>%
mutate(csum = cumsum(value))
df
%> Source: local data frame [4 x 4]
%> Groups: group
%>
%> group rep value csum
%> 1 0 d1 0 0
%> 2 0 d2 2 2
%> 3 1 d1 2 2
%> 4 1 d2 2 4
For most cases, you're best of leaving the data in this form (it will be easier to work for), but you can reshape if you need to:
library(reshape2)
dcast(df, group ~ rep, value.var = "csum")
%> group d1 d2
%> 1 0 0 2
%> 2 1 2 4
Upvotes: 17
Reputation: 49448
library(data.table)
# convert to data.table in place
setDT(df)
# dcast and do individual sums
dt.cast = dcast.data.table(df, group ~ rep, value.var = 'value',
fun.aggregate = sum)
dt.cast
# group d1 d2
#1: 0 0 1
#2: 1 1 2
# cumsum
dt.cast[, as.list(cumsum(unlist(.SD))), by = group]
# group d1 d2
#1: 0 0 1
#2: 1 1 3
Upvotes: 13