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Imagine I have a tidy dataset with 1 variable and 10 observations. The values of the variable are e.g. 3, 5, 7, 9, 13, 17, 29, 33, 34, 67. How do I recode it so that the 3 will be 1, the 5 will be 2 (...) and the 67 will be 10?
One possibility is to use rank: in a ´dplyr` setting it could look like this:
library(dplyr)
tibble(x = c(3, 5, 7, 9, 13, 17, 29, 33, 34, 67)) %>%
mutate(y = rank(x))
Here is one way -
x <- c(3, 5, 7, 9, 13, 17, 29, 33, 67, 34)
x1 <- sort(x)
y <- match(x1, unique(x1))
y
#[1] 1 2 3 4 5 6 7 8 9 10
Changed the order of last 2 values so that it also works when the data is not in order.
Another way:
x <- c(3, 5, 7, 9, 13, 17, 29, 33, 67, 34)
x <- sort(x)
seq_along(x)
# 1 2 3 4 5 6 7 8 9 10
For example I have a vector like this :
a <- c(4, 8, 9, 15, 21, 21, 24, 25, 26, 28, 29, 34)
and I want to do this:
Step 1:
Partition into equal-frequency (equi-depth)
Bins:
Bin 1: 4, 8, 9, 15
Bin 2: 21, 21, 24, 25
Bin 3: 26, 28, 29, 34
Step2:
Smoothing by bin means:
Bin 1: 9, 9, 9, 9
Bin 2: 23, 23, 23, 23
Bin 3: 29, 29, 29, 29
Output :
9,9,9,9,23,23,23,23,29,29,29,29
We can create groups by dividing length of a in equal number of bins and use ave to calculate rounded mean in each group.
no_of_bins <- 4
round(ave(a, rep(1:length(a), each = no_of_bins, length.out = length(a))))
#[1] 9 9 9 9 23 23 23 23 29 29 29 29
PS -
ave has default function as mean so it has not been explicitly applied.
Try this (take Orange$age predefined R variable as your input, 10 is the bin size)
v=split(Orange$age, ceiling(seq_along(Orange$age)/10))
lapply(v, function(item){rep(mean(item), length(item))})
I am trying to write code that subtracts a given value from a variable until each row has a predicted probability at or above .05.
train <- data.frame('cost'= c(120, 3, 2, 4, 10, 110, 200, 43, 1, 51, 22, 14),
'price' = c(120, 20, 10, 4, 3, 4, 30, 43, 56, 88, 75, 44),
'dich' = c(0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0))
train$dich <- as.factor(train$dich)
test <- data.frame('cost'= c(13, 5, 32, 22, 14, 145, 54, 134, 11, 14, 33, 21),
'price' = c(32, 11, 210, 6, 3, 7, 22, 423, 19, 99, 192, 32)
)
model <- glm(dich ~ cost + price,
data = train,
family = "binomial")
pred <- predict(model, test, type = "response")
1 2 3 4
3.001821e-01 4.442316e-01 4.507495e-04 6.310900e-01
5 6 7 8
5.995459e-01 9.888085e-01 7.114101e-01 1.606681e-06
9 10 11 12
4.096450e-01 2.590474e-02 9.908167e-04 3.572890e-01
So in the above output the cases 4, 5, 6, and 7 would remain the same because they are already above .05 but for the rest of the cases I would like to subtract 1 from the price column and then run the prediction again and repeat until all cases have a probability at or above .05.
If you want to subtract 1 for each row (or "customer") individually, rather than 1 across the board:
test$pred_prob <- NA
for (n in 1:nrow(test)) {
print("-----------------------------")
print(n)
while (TRUE) {
pred <- predict(model, test[n,], type = "response")
print(pred)
test$pred_prob[n] <- pred
if (sum(pred > 0.05) == length(pred)) {
print(test$price[n])
break
}
test$price[n] <- test$price[n] - 1
}
print(test)
}
# cost price pred_prob
# 1 13 32 0.30018209
# 2 5 11 0.44423163
# 3 32 96 0.05128337
# 4 22 6 0.63109001
# 5 14 3 0.59954586
# 6 145 7 0.98880854
# 7 54 22 0.71141007
# 8 134 175 0.05074762
# 9 11 19 0.40964501
# 10 14 82 0.05149897
# 11 33 97 0.05081947
# 12 21 32 0.35728897
I see what you are trying to do but the results are quite hilarious. This is if you want to subtract 1 from all elements of price each time:
x <- 1
while (TRUE) {
print("----------------------------------------")
print(x)
test$price <- test$price - 1
pred <- predict(model, test, type = "response")
print(pred)
x <- x + 1
if (sum(pred > 0.05) == length(pred)) {
print(test)
break
}
}
# ... loops 247 times
# [1] "----------------------------------------"
# [1] 248
# 1 2 3 4 5 6 7 8 9 10 11 12
# 0.99992994 0.99996240 0.93751936 0.99998243 0.99997993 0.99999966 0.99998781 0.05074762 0.99995669 0.99887117 0.97058913 0.99994594
# cost price
# 1 13 -216
# 2 5 -237
# 3 32 -38
# 4 22 -242
# 5 14 -245
# 6 145 -241
# 7 54 -226
# 8 134 175
# 9 11 -229
# 10 14 -149
# 11 33 -56
# 12 21 -216
In case anyone else wants to run the same thing with a xgboost model.
train <- data.frame('cost'= c(120, 3, 2, 4, 10, 110, 200, 43, 1, 51, 22, 14),
'price' = c(120, 20, 10, 4, 3, 4, 30, 43, 56, 88, 75, 44))
label <- data.frame('dich' = c(0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0))
train <- as.matrix(train)
label <- as.matrix(label)
model <- xgboost(data = train,
label = label,
max.depth = 3,
nround = 1,
objective = "binary:logistic")
test <- data.frame('cost'= c(13, 5, 32, 22, 14, 145, 54, 134, 11, 14, 33, 21),
'price' = c(32, 11, 210, 6, 3, 7, 22, 423, 19, 99, 192, 32)
)
test <- as.matrix(test)
#FOR A MATRIX
test <- cbind(test, rep(NA, nrow(test)))
colnames(test)[3] <- c("pred_prob")
for (n in 1:nrow(test)) {
print("-----------------------------")
print(n)
while (TRUE) {
pred <- predict(model, t(test[n,]), type = "response")
print(pred)
test[,"pred_prob"][n] <- pred
if (sum(pred > 0.5) == length(pred)) {
print(test[,"pred_prob"][n])
break
}
test[,"price"][n] <- test[,"price"][n] - .01
}
print(test)
}
It seems to take a while to run on 12 rows. I need to do some thinking about the thresholds of a tree model and how that'll effect a range of different changes in the price to obtain at or above .5 probability (which I meant in my first question but I wrote .05 haha).
Summary (tldr)
I need to perform a rolling regression on an irregular time series (i.e. the interval may not even be periodic and go from 0, 1, 2, 3... to ...7, 20, 24, 28...) that's simple numeric and does not necessarily require date/time, but the rolling window needs be by time. So if I have a timeseries that is irregularly sampled for 600 seconds and the window is 30, the regression is performed every 30 seconds, and not every 30 samples.
I've read examples, and while I could replicate doing rolling sums and medians by time, I can't seem to figure it out for regression.
The problem
First of all, I have read some of the other questions with regards to performing rolling functions on irregular time series data, such as this: optimized rolling functions on irregular time series with time-based window, and this: Rolling window over irregular time series.
The issue is that the examples provided, so far, are simple for equations like sum or median, but I have not yet figured out how to perform a simple rolling regression, i.e. using lm, that is still based on the same caveat that the window is based on an irregular time series. Also, my timeseries is much, much simpler; no date is necessary, it's simply time "elapsed".
Anyway, getting this right is important to me because with irregular time - for example, a skip in the time interval - may give an over- or underestimate of the coefficients in the rolling regression, as the sample window will include additional time.
So I was wondering if anyone can help me with creating a function that does this in the simplest way? The dataset is based on measuring a variable over time i.e. 2 variables: time, and response. Time is measured every x time elapsed units (seconds, minutes, so not date/time formatted), but once in a while it becomes irregular.
For every row in the function, it should perform a linear regression based on a width of n time units. The width should never exceed n units, but may be floored (i.e. reduced) to accomodate irregular time sampling. So for example, if the width is specified at 20 seconds, but time is sampled every 6 seconds, then the window will be rounded to 18, not 24 seconds.
I have looked at the question here: How to calculate the average slope within a moving window in R, and I tested that code on an irregular time series, but it looks like it's based on regular time series.
Sample data:
sample <-
structure(list(x = c(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48,
49), y = c(50, 49, 48, 47, 46, 47, 46, 45, 44, 43, 44, 43, 42,
41, 40, 41, 40, 39, 38, 37, 38, 37, 36, 35, 34, 35, 34, 33, 32,
31, 30, 29, 28, 29, 28, 27, 26, 25, 26, 25, 24, 23, 22, 21, 20,
19)), .Names = c("x", "y"), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -46L))
My current code (based on a previous question I referred to). I know it's not subsetting by time:
library(zoo)
clm <- function(z) coef(lm(y ~ x, as.data.frame(z)))
rollme <- rollapplyr(zoo(sample), 10, clm, by.column = F, fill = NA)
The expected output (manually calculated) is below. The output is different from a regular rolling regression -- the numbers are different as soon as the time interval skips at 29 (secs):
NA
NA
NA
NA
NA
NA
NA
NA
NA
-0.696969697
-0.6
-0.551515152
-0.551515152
-0.6
-0.696969697
-0.6
-0.551515152
-0.551515152
-0.6
-0.696969697
-0.6
-0.551515152
-0.551515152
-0.6
-0.696969697
-0.6
-0.551515152
-0.551515152
-0.6
-0.696969697
-0.605042017
-0.638888889
-0.716981132
-0.597560976
-0.528301887
-0.5
-0.521008403
-0.642857143
-0.566666667
-0.551515152
-0.551515152
-0.6
-0.696969697
-0.605042017
-0.638888889
-0.716981132
I hope I'm providing enough information, but let me know (or give me a guide to a good example somewhere) for me to try this?
Other things I have tried:
I've tried converting the time to POSIXct format but I don't know how to perform lm on that:
require(lubridate)
x <- as.POSIXct(strptime(sample$x, format = "%S"))
Update : Added tldr section.
Try this:
# time interval is 1
sz=10
pl2=list()
for ( i in 1:nrow(sample)){
if (i<sz) period=sz else
period=length(sample$x[sample$x>(sample$x[i]-sz) & sample$x<=sample$x[i]])-1
pl2[[i]]=seq(-period,0)
}
#update for time interval > 1
sz=10
tint=1
pl2=list()
for ( i in 1:nrow(sample)){
if (i<sz) period=sz else
period=length(sample$x[sample$x>(sample$x[i]-sz*tint) & sample$x<=sample$x[i]])-1
pl2[[i]]=seq(-period,0)
}
rollme3 <- rollapplyr(zoo(sample), pl2, clm, by.column = F, fill = NA)
> tail(rollme3)
(Intercept) x
41 47.38182 -0.5515152
42 49.20000 -0.6000000
43 53.03030 -0.6969697
44 49.26050 -0.6050420
45 50.72222 -0.6388889
46 54.22642 -0.7169811
For the sake of completeness, here is an answer which uses data.table to aggregate in a non-equi join.
Although there many similar questions, e.g., r calculating rolling average with window based on value (not number of rows or date/time variable), this question deserves an answer on its own as the OP is looking for the coefficients of a rolling regression.
library(data.table)
ws <- 10 # size of sliding window in time units
setDT(sample)[.(start = x - ws, end = x), on = .(x > start, x <= end),
as.list(coef(lm(y ~ x.x))), by = .EACHI]
x x (Intercept) x.x
1: -10 0 50.00000 NA
2: -9 1 50.00000 -1.0000000
3: -8 2 50.00000 -1.0000000
4: -7 3 50.00000 -1.0000000
5: -6 4 50.00000 -1.0000000
6: -5 5 49.61905 -0.7142857
7: -4 6 49.50000 -0.6428571
8: -3 7 49.50000 -0.6428571
9: -2 8 49.55556 -0.6666667
10: -1 9 49.63636 -0.6969697
11: 0 10 49.20000 -0.6000000
12: 1 11 48.88485 -0.5515152
13: 2 12 48.83636 -0.5515152
14: 3 13 49.20000 -0.6000000
15: 4 14 50.12121 -0.6969697
16: 5 15 49.20000 -0.6000000
17: 6 16 48.64242 -0.5515152
18: 7 17 48.59394 -0.5515152
19: 8 18 49.20000 -0.6000000
20: 9 19 50.60606 -0.6969697
21: 10 20 49.20000 -0.6000000
22: 11 21 48.40000 -0.5515152
23: 12 22 48.35152 -0.5515152
24: 13 23 49.20000 -0.6000000
25: 14 24 51.09091 -0.6969697
26: 15 25 49.20000 -0.6000000
27: 16 26 48.15758 -0.5515152
28: 17 27 48.10909 -0.5515152
29: 18 28 49.20000 -0.6000000
30: 19 29 51.57576 -0.6969697
31: 22 32 49.18487 -0.6050420
32: 23 33 50.13889 -0.6388889
33: 24 34 52.47170 -0.7169811
34: 25 35 48.97561 -0.5975610
35: 26 36 46.77358 -0.5283019
36: 27 37 45.75000 -0.5000000
37: 28 38 46.34454 -0.5210084
38: 29 39 50.57143 -0.6428571
39: 30 40 47.95556 -0.5666667
40: 31 41 47.43030 -0.5515152
41: 32 42 47.38182 -0.5515152
42: 33 43 49.20000 -0.6000000
43: 34 44 53.03030 -0.6969697
44: 37 47 49.26050 -0.6050420
45: 38 48 50.72222 -0.6388889
46: 39 49 54.22642 -0.7169811
x x (Intercept) x.x
Please note that rows 10 to 30 where the time series is regularly spaced are identical to OP's rollme.
The call to as.list() forces the result of coef(lm(...)) to appear in separate columns.
The code above uses a right aligned rolling window. However, the code can be easily adapted to support a left aligned window as well:
# left aligned window
setDT(sample)[.(start = x, end = x + ws), on = .(x >= start, x < end),
as.list(coef(lm(y ~ x.x))), by = .EACHI]
With runner one can apply any R function in irregular time series. User has to specify put data to x argument and vector of dates to idx argument (to make windows time dependent). Window width k can be a integer k = 30 or character like in seq.POSIXt k = "30 secs".
First example shows how to obtain both parameters from lm function - output will be a matrix
library(runner)
runner(
x = sample,
k = "30 secs",
idx = sample$datetime,
function(x) {
coefficients(lm(y ~ x, data = x))
}
)
Or one can execute runner separately for each parameter
library(runner)
sample$intercept <- runner(
sample,
k = "30 secs",
idx = sample$datetime,
function(x) {
coefficients(lm(y ~ x, data = x))[1]
}
)
sample$slope <- runner(
sample,
k = "30 secs",
idx = sample$datetime,
function(x) {
coefficients(lm(y ~ x, data = x))[2]
}
)
head(sample, 15)
# datetime x y intercept slope
# 1 2020-04-13 09:27:20 0 50 50.00000 NA
# 2 2020-04-13 09:27:21 1 49 50.00000 -1.0000000
# 3 2020-04-13 09:27:25 2 48 50.00000 -1.0000000
# 4 2020-04-13 09:27:29 3 47 50.00000 -1.0000000
# 5 2020-04-13 09:27:29 4 46 50.00000 -1.0000000
# 6 2020-04-13 09:27:32 5 47 49.61905 -0.7142857
# 7 2020-04-13 09:27:34 6 46 49.50000 -0.6428571
# 8 2020-04-13 09:27:38 7 45 49.50000 -0.6428571
# 9 2020-04-13 09:27:38 8 44 49.55556 -0.6666667
# 10 2020-04-13 09:27:41 9 43 49.63636 -0.6969697
# 11 2020-04-13 09:27:44 10 44 49.45455 -0.6363636
# 12 2020-04-13 09:27:47 11 43 49.38462 -0.6153846
# 13 2020-04-13 09:27:48 12 42 49.38462 -0.6153846
# 14 2020-04-13 09:27:49 13 41 49.42857 -0.6263736
# 15 2020-04-13 09:27:50 14 40 49.34066 -0.6263736
Data with datetime column
sample <- structure(
list(
datetime = c(3, 1, 4, 4, 0, 3, 2, 4, 0, 3, 3, 3, 1, 1, 1, 3, 0, 2, 4, 2, 2,
3, 0, 1, 2, 4, 0, 1, 4, 4, 1, 2, 1, 3, 0, 4, 4, 1, 3, 0, 0, 2,
1, 0, 2, 0) + Sys.time(),
x = c(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 33, 34, 35, 36, 37, 38,
39, 40, 41, 42, 43, 44, 47, 48, 49),
y = c(50, 49, 48, 47, 46, 47, 46, 45, 44, 43, 44, 43, 42, 41, 40, 41, 40, 39,
38, 37, 38, 37, 36, 35, 34, 35, 34, 33, 32, 31, 30, 29, 28, 29, 28, 27,
26, 25, 26, 25, 24, 23, 22, 21, 20,19)
),
.Names = c("x", "y"),
class = c("tbl_df", "tbl", "data.frame"),
row.names = c(NA, -46L)
)
I have the following data and nested for loop:
x <- c(12, 27, 21, 16, 12, 21, 18, 16, 20, 23, 21, 10, 15, 26, 21, 22, 22, 19, 26, 26)
y <- c(8, 10, 7, 7, 9, 5, 7, 7, 10, 4, 10, 3, 9, 6, 4, 2, 4, 2, 3, 6)
a <- c(20,25)
a.sub <- c()
df <- c()
for(j in 1:length(a)){
a.sub <- which(x >= a[j])
for(i in 1:length(a.sub)){
df[i] <- y[a.sub[i]]
}
print(df)
}
I'd like the loop to return values for df as:
[1] 10 6 3 6 4 10 6 4 2 4 3 6
[1] 10 6 3 6
As I have it, however, the loop returns the same values twice of df for a <- 20 but not a <- 25:
[1] 10 7 5 10 4 10 6 4 2 4 3 6
[1] 10 6 3 6 4 10 6 4 2 4 3 6
for(i in 1:length(a.sub)){
df[i] <- y[a.sub[i]]
}
can become
df <- y[a.sub]
neither a.sub nor df need to be predefined then and thus...
x <- c(12, 27, 21, 16, 12, 21, 18, 16, 20, 23, 21, 10, 15, 26, 21, 22, 22, 19, 26, 26)
y <- c(8, 10, 7, 7, 9, 5, 7, 7, 10, 4, 10, 3, 9, 6, 4, 2, 4, 2, 3, 6)
a <- c(20,25)
for(j in 1:length(a)){
a.sub <- which(x >= a[j])
df <- y[a.sub]
print(df)
}
It could be made shorter. df is unnecessary if you're just printing the subset of y anyway. Just print it directly. And the selector is so short it wouldn't make a single line confusing. Furthermore, why use length of a and index.. loop through a directly. So, it could be...
a <- c(20,25)
for(ax in a){
print( y[ which(x >= ax) ] )
}
Not sure if this is a simplified version of a more complex problem, but I'd probably solve this using some direct indexing and an apply function. Something like this:
z <- cbind(x,y)
sapply(c(20,25), function(x) z[z[, 1] >= x, 2])
[[1]]
[1] 10 7 5 10 4 10 6 4 2 4 3 6
[[2]]
[1] 10 6 3 6