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I have a dataframe with observations from three years time, with column df$week that indicates the week of the observation. (The week count of the second year continues from the count of the first, so the data contains 207 weeks).
I would like to divide the data to longer time periods, to df$period that would include all observations from several weeks' time.
If a period would be the length of three weeks, and I the data would include 13 observations in six weeks time, the I idea would be to divide
weeks <- c(1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 6)
into
periods <- c(1, 1, 1, 2, 2, 3, 3), c(4, 5, 5, 6, 6, 6)
periods
[1]
1 1 1 2 2 3 3
[2]
4 5 5 6 6 6
To look something like
> df
week period
1 1 1
2 1 1
3 1 1
4 2 1
5 2 1
6 3 1
7 3 1
8 4 2
9 5 2
10 5 2
11 6 2
12 6 2
13 6 2
>
The data contains +13k rows so would need to do some sort of map in style of
mapPeriod <- function(df, fun) {
out <- vector("vector_of_weeks", length(df))
for (i in seq_along(df)) {
out[i] <- fun(df[[i]])
}
out
}
I just don't know what to include in the fun to divide the weeks to the decided sequences of periods. Can function rep be of assistance here? How?
I would be very grateful for all input and suggestions.
split(weeks, f = (weeks - 1) %/% 3)
$`0`
[1] 1 1 1 2 2 3 3
$`1`
[1] 4 5 5 6 6 6
from comments below
weeks <- c(1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 6)
df <- data.frame(weeks)
library(data.table)
df$period <- data.table::rleid((weeks - 1) %/% 3)
# weeks period
# 1 1 1
# 2 1 1
# 3 1 1
# 4 2 1
# 5 2 1
# 6 3 1
# 7 3 1
# 8 4 2
# 9 5 2
# 10 5 2
# 11 6 2
# 12 6 2
# 13 6 2
df is a test dataframe and I need to sort the last three columns in ascending order (without hardcoding the order).
df <- data.frame(X = c(1, 2, 3, 4, 5),
Z = c(1, 2, 3, 4, 5),
Y = c(1, 2, 3, 4, 5),
A = c(1, 2, 3, 4, 5),
C = c(1, 2, 3, 4, 5),
B = c(1, 2, 3, 4, 5))
Desired output:
> df
X Z Y A B C
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
I'm aware of the order() function but I can't seem to find the right way to implement it to get the desired output.
Update:
Base R:
cbind(df[1:3],df[4:6][,order(colnames(df[4:6]))])
First answer:
We could use relocate from dplyr:
https://dplyr.tidyverse.org/reference/relocate.html
It is configured to arrange columns:
Here we relocate by the index.
We take last (index = 6) and put it before (position 5, which is C)
library(dplyr)
df %>%
relocate(6, .before = 5)
An alternative:
library(dplyr)
df %>%
select(order(colnames(df))) %>%
relocate(4:6, .before = 1)
X Z Y A B C
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
In base R, a selection on the first columns then sort the last 3 names :
df[, c(names(df)[1:(ncol(df)-3)], sort(names(df)[ncol(df)-2:0]))]
We want to reorder the columns based on the column names, so if we use names(df) as the argument to order, we can reorder the data frame as follows.
The complicating factor is that order() returns a vector of numbers, so if we want to reorder only a subset of the column names, we'll need an approach that retains the original sort order for the first three columns.
We accomplish this by creating a vector of the first 3 column names, the sorted remaining column names using a function that returns the values rather than locations in the vector, and then use this with the [ form of the extract operator.
df <- data.frame(X = c(1, 2, 3, 4, 5),
Z = c(1, 2, 3, 4, 5),
Y = c(1, 2, 3, 4, 5),
A = c(1, 2, 3, 4, 5),
C = c(1, 2, 3, 4, 5),
B = c(1, 2, 3, 4, 5))
df[,c(names(df[1:3]),sort(names(df[4:6])))]
...and the output:
> df[,c(names(df[1:3]),sort(names(df[4:6])))]
X Z Y A B C
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
to_order <- seq(ncol(df)) > ncol(df) - 3
df[order(to_order*order(names(df)))]
#> X Z Y A B C
#> 1 1 1 1 1 1 1
#> 2 2 2 2 2 2 2
#> 3 3 3 3 3 3 3
#> 4 4 4 4 4 4 4
#> 5 5 5 5 5 5 5
Created on 2021-12-24 by the reprex package (v2.0.1)
I have the following vector V1 <- c(1, 2, 3, 1, 4, 5, 5, 2, 1).
I would like to have a vector V2 of the same size that for each element indicates how often the corresponding element in V1 occurs, i.e. V2 should be c(3, 2, 1, 3, 1, 2, 2, 2, 3).
I know a multitude of ways to obtain this result, however all of them include using a loop, e.g.
V2 <- c()
for(k in 1:length(V1)){
V2[k] <- length(which(V1 == V1[k]))
}
My question is: Is there a more elegant way of doing it, i.e. without using a loop?
Summing up the solutions provided
# #bouncyball
ave(V1, V1, FUN = length)
# [1] 3 2 1 3 1 2 2 2 3
# #AntoniosK
table(V1)[V1]
# V1
# 1 2 3 1 4 5 5 2 1
# 3 2 1 3 1 2 2 2 3
# #Roman
library(tidyverse);
as.tibble(V1) %>% group_by(value) %>% mutate(V2=n())
# A tibble: 9 x 2
# Groups: value [5]
# value V2
# <dbl> <int>
# 1 1 3
# 2 2 2
# 3 3 1
# 4 1 3
# 5 4 1
# 6 5 2
# 7 5 2
# 8 2 2
# 9 1 3
Say I have a longitudinal data set as below
ID <- c(1, 1, 2, 2, 3, 3, 4, 4)
time <- c(1, 2, 1, 2, 1, 2, 1, 2)
value <- c(7, 5, 9, 2, NA, 3, 7, NA)
mydata <- data.frame(ID, time, value)
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
5 3 1 NA
6 3 2 3
7 4 1 7
8 4 2 NA
In this data-set, we have 4 cases with data at two time-points (let's say pre and post treatment)
Something I want to do is set criteria to delete any case that are not complete for both time-points. In this example, I would want to delete ID3 (who is missing timepoint 1), and ID4 (who is missing timepoint 2). Like below:
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
I am not having much luck. I've tried variants of complete.cases() or which() to no avail
I'm still new to R, and would be hugely appreciative if anyone could help me out
Edit: Thank you Ronak for answering my question. Upon reflection of my real data, I have encountered a second problem. My actual data is more reflected by the below:
ID <- c(1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 8)
time <- c(1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1)
value <- c(7, 5, 9, 2, NA, 3, 7, NA, 8, 9, 7, 6)
mydata <- data.frame(ID, time, value)
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
5 3 1 NA
6 3 2 3
7 4 1 7
8 4 2 NA
9 5 1 8
10 6 1 9
11 7 1 7
12 8 1 6
Where I would also want to remove cases 5, 6, 7 and 8. These IDs have an entry for Time 1, but not Time 2. Hopefully this makes sense
Thanks a heap
If you switch your data to wide format (where each time point is represented as its own column), then you can use na.omit. Using dplyr and tidyr functions:
library(dplyr)
mydata <- mydata %>%
tidyr::spread(key=time, value=value) %>% # reformat to wide
na.omit() %>% # delete cases with missingness on any variable (i.e. any time point)
tidyr::gather(key="time", value="value", -ID) # put it back in long format
> mydata
ID time value
1 1 1 7
2 2 1 9
3 1 2 5
4 2 2 2
Note that this will work (it will keep only cases with complete data for both time 1 and time 2) even when you have a time point missing without an explicit NA present in the data, like this:
> mydata
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
5 3 1 NA
6 3 2 3
7 4 1 7
8 4 2 NA
9 5 1 8
10 6 1 9
11 7 1 7
12 8 1 6
You can do this easily with sqldf.
library(sqldf)
sqldf(' select * from (select ID, count(*) as cnt from mydata where value is not null group by id having cnt >1 ) t1 inner join mydata t2 on t1.ID=t2.ID')
You would select those id having a count greater than 1 and who doesn't have NA in their values and then join back with the original data.
#Ronak already provided
mydata[!mydata$ID %in% mydata$ID[is.na(mydata$value)], ]
For the second part, you can just group over each id and filter on their frequency
k2 <- data.frame(table(mydata$ID))
k2$Var1[k2$Freq > 1]
and then do something like
mydata[mydata$ID %in% k2$Var1[k2$Freq > 1],]
See the updated answer
# Eliminates ID cases with NA
mydata = mydata[!mydata$ID %in% mydata[!complete.cases(mydata) ,]$ID, ]
library(plyr)
# counts all the IDs
cnt = count(mydata, "ID")
# Eliminates any ID that doesn't have 2 observations
mydata[mydata$ID %in% cnt[cnt$freq == 2, ]$ID, ]
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
Extending this former question, how can I shuffle (randomize) the following vector
a1 = c(1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 5, 5)
in order to get something like this:
a2 = c(5, 5, 3, 3, 3, 3, 1, 1, 2, 4, 4, 4)
or even better like this:
a3 = c(4, 4, 4, 2, 3, 3, 3, 3, 1, 1, 5, 5)?
such that each element could randomly change to another but with keeping the number of each element constant?
You can try something like this: create a factor from a1 with randomly shuffled levels and then convert it to integers:
as.integer(factor(a1, levels = sample(unique(a1), length(unique(a1)))))
# [1] 5 5 4 4 4 4 3 3 2 1 1 1
The data:
a1 <- c(1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 5, 5)
First steps:
# extract values and their frequencies
val <- unique(a1)
tab <- table(a1)
freq <- tab[as.character(val)]
Keep original order of frequencies but sample values
rep(sample(val), freq)
# [1] 4 4 1 1 1 1 3 3 5 2 2 2
Keep original frequencies but sample order of values
rep(sa <- sample(val), freq[as.character(sa)])
# [1] 4 2 2 2 2 3 3 1 1 5 5 5
Seems like a perfect application for rle and its inverse rep:
rand_inverse_rle <- function(x) { x=sort(x)
ord=sample (length(rle(x)$values) )
unlist( mapply( rep, rle(x)$values[ord], rle(x)$lengths[ord]))}
rand_inverse_rle(a1)
#----------
[1] 3 3 4 5 5 5 2 2 2 2 1 1
This was my reading of a function needed to satisfy the natural language requirements:
> a1 = sample( c(1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 5, 5) )
> a1
[1] 5 2 5 2 5 1 3 4 2 2 3 1
> rand_inverse_rle(a1)
[1] 5 5 5 4 2 2 2 2 3 3 1 1
> rand_inverse_rle(a1)
[1] 1 1 3 3 5 5 5 2 2 2 2 4
> rand_inverse_rle(a1)
[1] 1 1 3 3 4 5 5 5 2 2 2 2