R: Aggregate data in sliding window into new columns - r

let's say I have a dataframe like this:
df <- tibble(ID = c(1, 1, 1, 1, 1), v1 = c(3, 5, 1, 0, 1), v2 = c(10, 6, 1, 20, 23), Time = c(as.POSIXct("1900-01-01 10:00:00"), as.POSIXct("1900-01-01 11:00:00"), as.POSIXct("1900-01-01 13:00:00"), as.POSIXct("1900-01-01 16:00:00"), as.POSIXct("1900-01-01 20:00:00"))) %>% group_by(ID)
# A tibble: 5 x 4
# Groups: ID [1]
ID v1 v2 Time
<dbl> <dbl> <dbl> <dttm>
1 1 3 10 1900-01-01 10:00:00
2 1 5 6 1900-01-01 11:00:00
3 1 1 1 1900-01-01 13:00:00
4 1 0 20 1900-01-01 16:00:00
5 1 1 23 1900-01-01 20:00:00
In words, this is a simple timeseries of a specific ID with two values v1 and v2 per time.
As quite common in machine learning, I want to aggregate the last n timesteps into one feature vector. For all previous timesteps there should be a time reference in hours when this data point occured. For the first row, where no previous timestep is available, the data should be filled with zeros.
Let's make an example. In this case n=2, that is I want to aggregate the current time step (t2) and the prevopus (t1) together:
# A tibble: 5 x 6
ID v1_t1 v2_t1 time_t1 v1_t2 v2_t2
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 0 0 NA 3 10
2 1 3 10 1 5 6
3 1 5 6 2 1 1
4 1 1 1 3 0 20
5 1 0 20 4 1 23
I want to keep that as generic as possible, so that n can change and the number of data columns. Any idea how to do this?
Thanks :)

Using dplyr::lag and dplyr::across you could do:
library(dplyr, warn=FALSE)
library(lubridate, warn=FALSE)
df %>%
group_by(ID) %>%
mutate(time_t1 = lubridate::hour(Time) - lag(lubridate::hour(Time))) %>%
mutate(across(c(v1, v2), .fns = list(t2 = ~.x, t1 = ~lag(.x, default = 0)))) %>%
select(-v1, -v2, -Time)
#> # A tibble: 5 × 6
#> # Groups: ID [1]
#> ID time_t1 v1_t2 v1_t1 v2_t2 v2_t1
#> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 NA 3 0 10 0
#> 2 1 1 5 3 6 10
#> 3 1 2 1 5 1 6
#> 4 1 3 0 1 20 1
#> 5 1 4 1 0 23 20
UPDATE Here is a more generic approach which makes use of some function factories to create list of functions which could then be passed to the .fns argument of across. Haven't tested for the more general case but should work for any n or number of lags to include and also for any number of data columns.
library(dplyr, warn=FALSE)
library(lubridate, warn=FALSE)
fun_factory1 <- function(n) {
function(x) {
lubridate::hour(x) - lag(lubridate::hour(x), n = n)
}
}
fun_factory2 <- function(n) {
function(x) {
lag(x, n = n, default = 0)
}
}
n <- 2
fns1 <- lapply(seq(n - 1), fun_factory1)
names(fns1) <- paste0("t", seq(n - 1))
fns2 <- lapply(seq(n) - 1, fun_factory2)
names(fns2) <- paste0("t", seq(n))
df %>%
group_by(ID) %>%
mutate(across(Time, .fns = fns1)) %>%
mutate(across(c(v1, v2), .fns = fns2)) %>%
select(-v1, -v2, -Time)
#> # A tibble: 5 × 6
#> # Groups: ID [1]
#> ID Time_t1 v1_t1 v1_t2 v2_t1 v2_t2
#> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 NA 3 0 10 0
#> 2 1 1 5 3 6 10
#> 3 1 2 1 5 1 6
#> 4 1 3 0 1 20 1
#> 5 1 4 1 0 23 20

Related

How can a table be rearranged one step at a time so that two or more observations are listed in a row in successive columns?

So far I have done this to achieve the desired result:
# A tibble: 4 x 2
frag treat
<dbl> <dbl>
1 1 1
2 2 1
3 1 2
4 2 2
treat_1 <- tab_example %>% filter(treat == "1")
treat_2 <- tab_example %>% filter(treat == "2")
new_tab_example <- full_join(treat_1, treat_2, by = "frag")
> new_tab_example
# A tibble: 2 x 3
frag treat.x treat.y
<dbl> <dbl> <dbl>
1 1 1 2
2 2 1 2
Is there a way to do it in one step?
You can use pivot_wider :
tidyr::pivot_wider(tab_example, names_from = treat,
names_prefix = 'treat', values_from = treat)
# frag treat1 treat2
# <dbl> <dbl> <dbl>
#1 1 1 2
#2 2 1 2
There is a way using spread() function:
library(dplyr)
library(tidyr)
# Yours data
df = tibble(frag = c(1, 2, 1, 2), treat = c(1,1,2,2) )
dfnew = df %>%
mutate(treat_name = case_when(treat==1 ~ 'treat.x', # Build names of columns
treat==2 ~ 'treat.y')
) %>%
spread(treat_name, treat) # Use spread function
If you print the result:
print(dfnew)
# A tibble: 2 x 3
frag treat.x treat.y
<dbl> <dbl> <dbl>
1 1 1 2
2 2 1 2

How to Create Iterative Forumla to calculate Z Score in R?

I have a number of large data frames that have the following basic format, where the final two rows are a mean (d) and standard deviation (e) - although these are calculated elsewhere.
a b c
a 4 3 4
b 3 2 6
c 2 1 8
d 3 2 6
e 1 1 2
I would like to create an iterative function that converts each raw data point into a z-score via the mean and sd value in d and e per column. The formula I would like to apply is ((x-mean)/SD).
The result would be the following:
a b c
a 1 1 1
b 0 0 0
c -1 -1 -1
I don't mind if this is added to the end, created as a new dataframe or the data is converted.
Thanks!
Here is one approach, note that I do not use the mean/sd provided in the data but re-calculate it on the fly.
Also note that usually the data should be in a tidy data representation, which in your case would mean that a, b, c would be in columns and then mean/sd would be either calculated on the fly or be in a separate column (note that this would reshaping the data, not shown here).
# your input data
raw_data <- data.frame(
a = c(4, 3, 2, 3, 1),
b = c(3, 2, 1, 2, 1),
c = c(4, 6, 8, 6, 2),
row.names = c("a", "b", "c", "d", "e")
)
raw_data
#> a b c
#> a 4 3 4
#> b 3 2 6
#> c 2 1 8
#> d 3 2 6
#> e 1 1 2
# remove the mean/sd values
data <- raw_data[!rownames(raw_data) %in% c("d", "e"), ]
data
#> a b c
#> a 4 3 4
#> b 3 2 6
#> c 2 1 8
# quick way to recalculate the values
means <- apply(data, 2, mean)
means
#> a b c
#> 3 2 6
sds <- apply(data, 2, sd)
sds
#> a b c
#> 1 1 2
z_scores <- apply(data, 2, function(x) (x - mean(x)) / sd(x))
z_scores
#> a b c
#> a 1 1 -1
#> b 0 0 0
#> c -1 -1 1
Created on 2021-01-07 by the reprex package (v0.3.0)
Edit / Full Code
The following code is a bit longer but most of it is spent on getting the data into the right (long/tidy) format.
If you have any questions, feel free to use the comments.
Note that the tidyverse is really helpful, but might need some time to get used to. The code used here is mostly dplyr (included in the tidyverse).
If you understand the functions: %>% (pipe), group_by(), mutate(), summarise(), and pivot_longer/wider() you got everything.
library(tidyverse)
# use your original dataset again
raw_data <- data.frame(
a = c(4, 3, 2, 3, 1),
b = c(3, 2, 1, 2, 1),
c = c(4, 6, 8, 6, 2),
row.names = c("a", "b", "c", "d", "e")
)
### 1) Turn the data into a nicer format
# match-table how to rename the variables
var_match <- c(d = "mean", e = "sd")
# convert the raw data into a nicer format, first we do some minor changes
# (variable names, etc)
data_mixed <- raw_data %>%
# have the rownames as explicit variable
rownames_to_column("metric") %>%
# nicer printing etc
as_tibble() %>%
# replace variable names with mean/sd
mutate(metric = ifelse(metric %in% c("d", "e"),
var_match[metric], metric))
data_mixed
#> # A tibble: 5 x 4
#> metric a b c
#> <chr> <dbl> <dbl> <dbl>
#> 1 a 4 3 4
#> 2 b 3 2 6
#> 3 c 2 1 8
#> 4 mean 3 2 6
#> 5 sd 1 1 2
# separate the dataset into two:
# data holds the values
# data_vars holds the metrics mean and sd
data <- data_mixed %>% filter(!metric %in% var_match) %>% select(-metric)
data_vars <- data_mixed %>% filter(metric %in% var_match)
data
#> # A tibble: 3 x 3
#> a b c
#> <dbl> <dbl> <dbl>
#> 1 4 3 4
#> 2 3 2 6
#> 3 2 1 8
data_vars
#> # A tibble: 2 x 4
#> metric a b c
#> <chr> <dbl> <dbl> <dbl>
#> 1 mean 3 2 6
#> 2 sd 1 1 2
# turn the value dataset into its longer form, makes it easier to work with it later
data_long <- data %>%
pivot_longer(everything(), names_to = "var", values_to = "val")
data_long
#> # A tibble: 9 x 2
#> var val
#> <chr> <dbl>
#> 1 a 4
#> 2 b 3
#> 3 c 4
#> 4 a 3
#> 5 b 2
#> 6 c 6
#> 7 a 2
#> 8 b 1
#> 9 c 8
# turn the metric dataset into another long form, allowing easy combination in the next step
data_vars2 <- data_vars %>%
pivot_longer(-metric, names_to = "var", values_to = "val") %>%
pivot_wider(var, names_from = metric, values_from = val)
data_vars2
#> # A tibble: 3 x 3
#> var mean sd
#> <chr> <dbl> <dbl>
#> 1 a 3 1
#> 2 b 2 1
#> 3 c 6 2
# combine the datasets
data_all <- left_join(data_long, data_vars2, by = "var")
data_all
#> # A tibble: 9 x 4
#> var val mean sd
#> <chr> <dbl> <dbl> <dbl>
#> 1 a 4 3 1
#> 2 b 3 2 1
#> 3 c 4 6 2
#> 4 a 3 3 1
#> 5 b 2 2 1
#> 6 c 6 6 2
#> 7 a 2 3 1
#> 8 b 1 2 1
#> 9 c 8 6 2
## 2) calculate the z-score
# now comes the actual number crunchin!
# per variable var (a, b, c) compute the variable val_z as the z-score
data_res <- data_all %>%
group_by(var) %>%
mutate(val_z = (val - mean) / sd)
data_res
#> # A tibble: 9 x 5
#> # Groups: var [3]
#> var val mean sd val_z
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 a 4 3 1 1
#> 2 b 3 2 1 1
#> 3 c 4 6 2 -1
#> 4 a 3 3 1 0
#> 5 b 2 2 1 0
#> 6 c 6 6 2 0
#> 7 a 2 3 1 -1
#> 8 b 1 2 1 -1
#> 9 c 8 6 2 1
## 3) make the results more readable
# lastly pivot the results to its original form
data_res_wide <- data_res %>%
select(var, val_z) %>%
group_by(var) %>%
mutate(id = 1:n()) %>% # needed for easier identification of values
pivot_wider(id, names_from = var, values_from = val_z)
data_res_wide
#> # A tibble: 3 x 4
#> id a b c
#> <int> <dbl> <dbl> <dbl>
#> 1 1 1 1 -1
#> 2 2 0 0 0
#> 3 3 -1 -1 1
Created on 2021-01-07 by the reprex package (v0.3.0)

How to transform a tibble from one column to two columns with repeated observations

I tried to transform df into df2. I have done it through a very patchy way using df3, Is there a simpler and more elegant way of doing it?
library(tidyverse)
# I want to transform df
df <- tibble(id = c(1, 2, 1, 2, 1, 2),
time = c('t1', 't1', 't2', 't2', 't3', 't3'),
value = c(2, 3, 6, 4, 5, 7))
df
#> # A tibble: 6 x 3
#> id time value
#> <dbl> <chr> <dbl>
#> 1 1 t1 2
#> 2 2 t1 3
#> 3 1 t2 6
#> 4 2 t2 4
#> 5 1 t3 5
#> 6 2 t3 7
# into df2
df2 <- tibble(id = c(1, 2, 1, 2),
t = c(2, 3, 6, 4),
r = c(6, 4, 5, 7))
df2
#> # A tibble: 4 x 3
#> id t r
#> <dbl> <dbl> <dbl>
#> 1 1 2 6
#> 2 2 3 4
#> 3 1 6 5
#> 4 2 4 7
# This is how I did it, but I think it should be a better way
df3 <- df %>% pivot_wider(names_from = time, values_from = value)
b <- tibble(id = numeric(), t = numeric(), r = numeric())
for (i in 2:3){
a <- df3[,c(1,i,i+1)]
colnames(a) <- c('id', 't', 'r')
b <- bind_rows(a, b)
}
b
#> # A tibble: 4 x 3
#> id t r
#> <dbl> <dbl> <dbl>
#> 1 1 6 5
#> 2 2 4 7
#> 3 1 2 6
#> 4 2 3 4
Created on 2020-11-25 by the reprex package (v0.3.0)
For each id you can use lead to select next value and create r column and drop NA rows.
library(dplyr)
df %>%
group_by(id) %>%
mutate(t = value,
r = lead(value)) %>%
na.omit() %>%
select(id, t, r)
# id t r
# <dbl> <dbl> <dbl>
#1 1 2 6
#2 2 3 4
#3 1 6 5
#4 2 4 7
We can use summarise from dplyr version >= 1.0. Previously, it had the constraint of returning only single observation per group. From version >= 1.0, it is no longer the case. Can return any number of rows i.e. it can be shorter or longer than the original number of rows
library(dplyr)
df %>%
group_by(id) %>%
summarise(t = value[-n()], r = value[-1], .groups = 'drop')
-output
# A tibble: 4 x 3
# id t r
# <dbl> <dbl> <dbl>
#1 1 2 6
#2 1 6 5
#3 2 3 4
#4 2 4 7

How to mutate a variable in a tibble based on columns named as numbers in R

I have a tibble with columns named as numbers (e.g. 1). I created a function to compute differences between columns, but I don't know how to do it with that type of columns:
<!-- language-all: lang-r -->
library(tidyverse)
df <- tibble(`1` = c(1,2,3), `2` = c(2,4,6))
# This works
df %>%
mutate(diff = `1` - `2`)
#> # A tibble: 3 x 3
#> `1` `2` diff
#> <dbl> <dbl> <dbl>
#> 1 1 2 -1
#> 2 2 4 -2
#> 3 3 6 -3
# But this doesn't
calc_diffs <- function(x, y){
df %>%
mutate(diff := !!x - !!y)
}
calc_diffs(1, 2)
#> # A tibble: 3 x 3
#> `1` `2` diff
#> <dbl> <dbl> <dbl>
#> 1 1 2 -1
#> 2 2 4 -1
#> 3 3 6 -1
<sup>Created on 2020-10-14 by the [reprex package](https://reprex.tidyverse.org) (v0.3.0)</sup>
We can convert to a symbol and evaluate
calc_diffs <- function(x, y){
df %>%
mutate(diff := !! rlang::sym(x) - !!rlang::sym(y))
}
Then, we just pass a string as argument
calc_diffs("1", "2")
# A tibble: 3 x 3
# `1` `2` diff
# <dbl> <dbl> <dbl>
#1 1 2 -1
#2 2 4 -2
#3 3 6 -3
Column names are strings. We could pass index to subset the column, but here the column name is an unusual name that starts with number. So, either we can wrap it with backreference using paste or just pass a string, convert to symbol and evaluate (!!)
Does this work:
> df <- tibble(`1` = c(1,2,3), `2` = c(2,4,6))
> df
# A tibble: 3 x 2
`1` `2`
<dbl> <dbl>
1 1 2
2 2 4
3 3 6
> calc_diffs <- function(x, y){
+ df %>%
+ mutate(diff = {{x}} - {{y}})
+ }
> calc_diffs(`1`,`2`)
# A tibble: 3 x 3
`1` `2` diff
<dbl> <dbl> <dbl>
1 1 2 -1
2 2 4 -2
3 3 6 -3
>

loop to multiply across columns

I have a data frame with columns labeled sales1, sales2, price1, price2 and I want to calculate revenues by multiplying sales1 * price1 and so-on across each number in an iterative fashion.
data <- data_frame(
"sales1" = c(1, 2, 3),
"sales2" = c(2, 3, 4),
"price1" = c(3, 2, 2),
"price2" = c(3, 3, 5))
data
# A tibble: 3 x 4
# sales1 sales2 price1 price2
# <dbl> <dbl> <dbl> <dbl>
#1 1 2 3 3
#2 2 3 2 3
#3 3 4 2 5
Why doesn't the following code work?
data %>%
mutate (
for (i in seq_along(1:2)) {
paste0("revenue",i) = paste0("sales",i) * paste0("price",i)
}
)
Assuming your columns are already ordered (sales1, sales2, price1, price2). We can split the dataframe in two parts and then multiply them
data[grep("sales", names(data))] * data[grep("price", names(data))]
# sales1 sales2
#1 3 6
#2 4 9
#3 6 20
If the columns are not already sorted according to their names, we can sort them by using order and then use above command.
data <- data[order(names(data))]
This answer is not brief. For that, #RonakShah's existing answer is the one to look at!
My response is intended to address a broader concern regarding the difficulty of trying to do this in the tidyverse. My understanding is this is difficult because the data is not currently in a "tidy" format. Instead, you can create a tidy data frame like so:
library(tidyverse)
tidy_df <- data %>%
rownames_to_column() %>%
gather(key, value, -rowname) %>%
extract(key, c("variable", "id"), "([a-z]+)([0-9]+)") %>%
spread(variable, value)
Which then makes the final calculation straightforward
tidy_df %>% mutate(revenue = sales * price)
#> # A tibble: 6 x 5
#> rowname id price sales revenue
#> <chr> <chr> <dbl> <dbl> <dbl>
#> 1 1 1 3 1 3
#> 2 1 2 3 2 6
#> 3 2 1 2 2 4
#> 4 2 2 3 3 9
#> 5 3 1 2 3 6
#> 6 3 2 5 4 20
If you need to get the data back into the original format you can although this feels clunky to me (I'm sure this can be improved in someway).
tidy_df %>% mutate(revenue = sales * price) %>%
gather(key, value, -c(rowname, id)) %>%
unite(key, key, id, sep = "") %>%
spread(key, value) %>%
select(starts_with("price"),
starts_with("sales"),
starts_with("revenue"))
#> # A tibble: 3 x 6
#> price1 price2 sales1 sales2 revenue1 revenue2
#> * <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 3 3 1 2 3 6
#> 2 2 3 2 3 4 9
#> 3 2 5 3 4 6 20

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