I have the following dataframe:
data <- data.frame(week = c(rep("2014-01-06", 3), rep("2014-01-13", 3), rep("2014-01-20", 3)), values = c(1, 2, 3))
week values
1 2014-01-06 1
2 2014-01-06 2
3 2014-01-06 3
4 2014-01-13 1
5 2014-01-13 2
6 2014-01-13 3
7 2014-01-20 1
8 2014-01-20 2
9 2014-01-20 3
I'm wanting to create a column in data that counts the unique week and assigns it a sequential value, such that the df appears like this:
week values seq_value
1 2014-01-06 1 1
2 2014-01-06 2 1
3 2014-01-06 3 1
4 2014-01-13 1 2
5 2014-01-13 2 2
6 2014-01-13 3 2
7 2014-01-20 1 3
8 2014-01-20 2 3
9 2014-01-20 3 3
I guess the idiomatic way would be just to calculate the actual week of the year out of the date provided (in case your weeks are not starting from the first week of the year).
as.integer(format(as.Date(data$week), "%W"))
## [1] 1 1 1 2 2 2 3 3 3
Another base R solution would be using as.POSIXlt class and utilizing its yday attribute
as.POSIXlt(data$week)$yday %/% 7 + 1
## [1] 1 1 1 2 2 2 3 3 3
If you want a shorter syntax, data.table package (among many others - See #Kshashaas comment) offers a quick wrapper
library(data.table)
week(data$week)
## [1] 1 1 1 2 2 2 3 3 3
The nicest thing about this package is that you can create columns by reference (similar to #akruns last solution, but probably more efficient because doesn't require the by argument)
setDT(data)[, seq_value := week(week)]
You could use base R by converting the "week" column to factor and specifying the levels as the unique values of "week". Convert factor to numeric and get the numeric index of the levels.
data$seq_value <- with(data, as.numeric(factor(week,levels=unique(week) )))
data$seq_value
#[1] 1 1 1 2 2 2 3 3 3
Or match the "week" column to unique values of that column to get the numeric index.
with(data, match(week, unique(week)))
#[1] 1 1 1 2 2 2 3 3 3
Or using data.table, by first converting data.frame to data.table (setDT) and then get the index values (.GRP) of grouping variable 'week' and assign it to new column seq_value
library(data.table)
setDT(data)[,seq_value:=.GRP, week][]
A dplyr solution:
library(dplyr)
data %>%
mutate(seq_value = dense_rank(week))
Related
There is a wide data set, a simple example is
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(7,8,8,9,10,10),
"cx"=c(11,12,12,13,14,14))
I'm looking for a way to assign the values in "ax" to column "bx" and "cx". Here, imagine we have thousands of columns we intend to replace with "ax", so I want this to be done in an automated approach using R. The expected output look like
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(1,2,2,3,4,4),
"cx"=c(1,2,2,3,4,4))
I've thought of, and tried using mutate_at and ends_with, but this has not work for me. For example, I tried
df %>%
mutate_at(vars(ends_with("x")), labels = "ax")
and this prints an error. Not sure what's wrong or what's to be added to get this working, so I would like to request your help on this. Thank you very much!
A simple way using base R would be :
change_cols <- grep('x$', names(df))
df[change_cols] <- df$ax
df
# id ax bx cx
#1 1 1 1 1
#2 2 2 2 2
#3 3 2 2 2
#4 4 3 3 3
#5 5 4 4 4
#6 6 4 4 4
I would suggest this tidyverse approach using across() to select the range of variables you want:
library(tidyverse)
#Data
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(7,8,8,9,10,10),
"cx"=c(11,12,12,13,14,14))
#Mutate
df %>% mutate(across(c(bx:cx), ~ ax))
Output:
id ax bx cx
1 1 1 1 1
2 2 2 2 2
3 3 2 2 2
4 4 3 3 3
5 5 4 4 4
6 6 4 4 4
Another option with mutate_at()
df %>%
mutate_at(vars(matches("x$")), ~ax)
# id ax bx cx
# 1 1 1 1 1
# 2 2 2 2 2
# 3 3 2 2 2
# 4 4 3 3 3
# 5 5 4 4 4
# 6 6 4 4 4
My dataframe looks like this
data = data.frame(ID=c(1,2,3,4,5,6,7,8,9,10),
Gender=c('Male','Female','Female','Female','Male','Female','Male','Male','Female','Female'))
And I have a reference list that looks like this -
ref=list(Male=1,Female=2)
I'd like to replace values in the Gender column using this reference list, without adding a new column to my dataframe.
Here's my attempt
do.call(dplyr::recode, c(list(data), ref))
Which gives me the following error -
no applicable method for 'recode' applied to an object of class
"data.frame"
Any inputs would be greatly appreciated
An option would be do a left_join after stacking the 'ref' list to a two column data.frame
library(dplyr)
left_join(data, stack(ref), by = c('Gender' = 'ind')) %>%
select(ID, Gender = values)
A base R approach would be
unname(unlist(ref)[as.character(data$Gender)])
#[1] 1 2 2 2 1 2 1 1 2 2
In base R:
data$Gender = sapply(data$Gender, function(x) ref[[x]])
You can use factor, i.e.
factor(data$Gender, levels = names(ref), labels = ref)
#[1] 1 2 2 2 1 2 1 1 2 2
You can unlist ref to give you a named vector of codes, and then index this with your data:
transform(data,Gender=unlist(ref)[as.character(Gender)])
ID Gender
1 1 1
2 2 2
3 3 2
4 4 2
5 5 1
6 6 2
7 7 1
8 8 1
9 9 2
10 10 2
Surprisingly, that one works as well:
data$Gender <- ref[as.character(data$Gender)]
#> data
# ID Gender
# 1 1 1
# 2 2 2
# 3 3 2
# 4 4 2
# 5 5 1
# 6 6 2
# 7 7 1
# 8 8 1
# 9 9 2
# 10 10 2
I need to create (with R) a rolling index of pairs from a data set that includes groups. Consider the following data set:
times <- c(4,3,2)
V1 <- unlist(lapply(times, function(x) seq(1, x)))
df <- data.frame(group = rep(1:length(times), times = times),
V1 = V1,
rolling_index = c(1,1,2,2,3,3,4,5,5))
df
group V1 rolling_index
1 1 1 1
2 1 2 1
3 1 3 2
4 1 4 2
5 2 1 3
6 2 2 3
7 2 3 4
8 3 1 5
9 3 2 5
The data frame I have includes the variables group and V1. Within each group V1 designates a running index (that may or may not start at 1).
I want to create a new indexing variable that looks like rolling_index. This variable groups rows within the same group and consecutive V1 value, thus creating a new rolling index. This new index must be consecutive over groups. If there is an uneven amount of rows within a group (e.g. group 2), then the last, single row gets its own rolling index value.
You can try
library(data.table)
setDT(df)[, gr:=as.numeric(gl(.N, 2, .N)), group][,
rollindex:=cumsum(c(TRUE,abs(diff(gr))>0))][,gr:= NULL]
# group V1 rolling_index rollindex
#1: 1 1 1 1
#2: 1 2 1 1
#3: 1 3 2 2
#4: 1 4 2 2
#5: 2 1 3 3
#6: 2 2 3 3
#7: 2 3 4 4
#8: 3 1 5 5
#9: 3 2 5 5
Or using base R
indx1 <- !duplicated(df$group)
indx2 <- with(df, ave(group, group, FUN=function(x)
gl(length(x), 2, length(x))))
cumsum(c(TRUE,diff(indx2)>0)|indx1)
#[1] 1 1 2 2 3 3 4 5 5
Update
The above methods are based on the 'group' column. Suppose you already have a sequence column ('V1') by group as showed in the example, creation of rolling index is easier
cumsum(!!df$V1 %%2)
#[1] 1 1 2 2 3 3 4 5 5
As mentioned in the post, if the 'V1' column do not start at '1' for some groups, we can get the sequence from the 'group' and then do the cumsum as above
cumsum(!!with(df, ave(seq_along(group), group, FUN=seq_along))%%2)
#[1] 1 1 2 2 3 3 4 5 5
There is probably a simpler way but you can do:
rep_each <- unlist(mapply(function(q,r) {c(rep(2, q),rep(1, r))},
q=table(df$group)%/%2,
r=table(df$group)%%2))
df$rolling_index <- inverse.rle(x=list(lengths=rep_each, values=seq(rep_each)))
df$rolling_index
#[1] 1 1 2 2 3 3 4 5 5
The data comes from another question I was playing around with:
dt <- data.table(user=c(rep(3, 5), rep(4, 5)),
country=c(rep(1,4),rep(2,6)),
event=1:10, key="user")
# user country event
#1: 3 1 1
#2: 3 1 2
#3: 3 1 3
#4: 3 1 4
#5: 3 2 5
#6: 4 2 6
#7: 4 2 7
#8: 4 2 8
#9: 4 2 9
#10: 4 2 10
And here's the surprising behavior:
dt[user == 3, as.data.frame(table(country))]
# country Freq
#1 1 4
#2 2 1
dt[user == 4, as.data.frame(table(country))]
# country Freq
#1 2 5
dt[, as.data.frame(table(country)), by = user]
# user country Freq
#1: 3 1 4
#2: 3 2 1
#3: 4 1 5
# ^^^ - why is this 1 instead of 2?!
Thanks mnel and Victor K. The natural follow-up is - shouldn't it be 2, i.e. is this a bug? I expected
dt[, blah, by = user]
to return identical result to
rbind(dt[user == 3, blah], dt[user == 4, blah])
Is that expectation incorrect?
The idiomatic data.table approach is to use .N
dt[ , .N, by = list(user, country)]
This will be far quicker and it will also retain country as the same class as in the original.
As mnel noted in comments, as.data.frame(table(...)) produces a data frame where the first variable is a factor. For user == 4, there is only one level in the factor, which is stored internally as 1.
What you want is factor levels, but what you get is how factors are stored internally (as integers, starting from 1). The following provides the expected result:
> dt[, lapply(as.data.frame(table(country)), as.character), by = user]
user country Freq
1: 3 1 4
2: 3 2 1
3: 4 2 5
Update. Regarding your second question: no, I think data.table behaviour is correct. Same thing happens in plain R when you join two factors with different levels:
> a <- factor(3:5)
> b <- factor(6:8)
> a
[1] 3 4 5
Levels: 3 4 5
> b
[1] 6 7 8
Levels: 6 7 8
> c(a,b)
[1] 1 2 3 1 2 3
I need a sequence of repeated numbers, i.e. 1 1 ... 1 2 2 ... 2 3 3 ... 3 etc. The way I implemented this was:
nyear <- 20
names <- c(rep(1,nyear),rep(2,nyear),rep(3,nyear),rep(4,nyear),
rep(5,nyear),rep(6,nyear),rep(7,nyear),rep(8,nyear))
which works, but is clumsy, and obviously doesn't scale well.
How do I repeat the N integers M times each in sequence?
I tried nesting seq() and rep() but that didn't quite do what I wanted.
I can obviously write a for-loop to do this, but there should be an intrinsic way to do this!
You missed the each= argument to rep():
R> n <- 3
R> rep(1:5, each=n)
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
R>
so your example can be done with a simple
R> rep(1:8, each=20)
Another base R option could be gl():
gl(5, 3)
Where the output is a factor:
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
Levels: 1 2 3 4 5
If integers are needed, you can convert it:
as.numeric(gl(5, 3))
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
For your example, Dirk's answer is perfect. If you instead had a data frame and wanted to add that sort of sequence as a column, you could also use group from groupdata2 (disclaimer: my package) to greedily divide the datapoints into groups.
# Attach groupdata2
library(groupdata2)
# Create a random data frame
df <- data.frame("x" = rnorm(27))
# Create groups with 5 members each (except last group)
group(df, n = 5, method = "greedy")
x .groups
<dbl> <fct>
1 0.891 1
2 -1.13 1
3 -0.500 1
4 -1.12 1
5 -0.0187 1
6 0.420 2
7 -0.449 2
8 0.365 2
9 0.526 2
10 0.466 2
# … with 17 more rows
There's a whole range of methods for creating this kind of grouping factor. E.g. by number of groups, a list of group sizes, or by having groups start when the value in some column differs from the value in the previous row (e.g. if a column is c("x","x","y","z","z") the grouping factor would be c(1,1,2,3,3).