cut2 splits into unequal buckets - r

I am currently doing some data manipulation and have been searching for a way to create deciles with equal number of observations in each group. I ran into the Hmisc package and the cut2 function and was under the impression it should split the data into 10 buckets with equal numbers of observations in each by specifying g=10. However the output from this function has been quite a bit off. Am I using cut2 incorrectly?
The code I am using:
library(Hmisc)
testdata <- data.frame(rating= c(8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 8, 8, 6, 8, 8, 8, 8, 6, 8, 6, 8, 4, 8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 8, 8, 8, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 6, 8, 8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 8, 8, 6, 8, 8, 6, 4, 8, 8, 8, 8, 8, 6, 8, 8, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 2, 8, 6, 8, 8, 8, 6, 8, 8, 6, 6, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 8, 8, 6, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 6, 8, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 6, 8, 8, 8, 8, 8, 6, 8, 8, 8, 6)
,age=c(0, 0, 0, 0, 3, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29, 30, 30, 30, 31, 31, 32, 32, 32, 32, 32, 32, 32, 32, 33, 33, 34, 34, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 36, 36, 36, 36, 37, 37, 37, 37, 37, 38, 38, 38, 38, 38, 39, 39, 39, 40, 40, 41, 41, 41, 41, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 42, 43, 43, 43, 44, 44, 44, 44, 44, 44, 45, 45, 45, 45, 45, 46, 46, 46, 46, 47, 47, 47, 48, 48, 48, 54, 54, 54, 56, 56, 58, 59, 59, 59, 59, 60, 60, 60, 61, 66, 66, 70, 72))
cutcutcut <- cut2(testdata$age,g=10)
testtable <- table(cutcutcut)
and the output of unequal observations in each bucket
testtable
[ 0,13) [13,15) [15,20) [20,24) [24,26) [26,28) [28,33) [33,40) [40,46) [46,72]
46 16 35 28 33 35 26 31 31 28


The answer to your question lies in looking at the distribution of your data:
table(testdata$age)
# 0 3 4 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
# 4 1 4 6 4 3 4 2 2 16 9 7 5 10 6 7 7 13 4 2 9
# 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
# 23 10 18 17 8 5 3 2 8 2 2 5 9 5 5 3 2 8 7 3 6
# 45 46 47 48 54 56 58 59 60 61 66 70 72
# 5 4 3 3 3 2 1 4 3 1 2 1 1
We see that some ages have a large number of individuals at that age (e.g. there are 16 individuals with age 12 and 23 individuals with age 24). Since the cutting algorithm needs to put all individuals with the exact same age into the same bucket, this may lead to some imbalances in the buckets.
Since there are 309 total observations in your data and you seek 10 buckets, you would ideally want 31 observations in 9 of the buckets and 30 in the last. Right now the last bucket is defined as [46, 72], which contains 28 elements (too low). If you expanded this to [45, 72], it would contain 33 elements (too many). There is no way to split up the data to get exactly 30 or 31 observations in this last bucket because there are 5 elements with value 45.

Related

tidyverse and dplyr: Conditional replacement of values in a column based on other column [duplicate]

This question already has answers here:
Can dplyr package be used for conditional mutating?
(5 answers)
Closed 2 years ago.
I want to mutate a column A4 by A3 but reducing value of A3 by 1 if Total == 63. What am I doing wrong here?
tb1 %>%
mutate(A4 = replace(A3, Total == 63, A3-1))
The complete code with data is here
library(tidyverse)
tb1 <-
structure(
list(
A1 = c(16, 11, 16, 18, 20, 19, 16, 18, 20, 15,
17, 19, 19, 19, 16, 19, 16, 15, 19, 19, 16, 18, 18, 19, 19, 18,
20, 18, 19, 19, 19, 19, 17, 19, 17, 16, 18, 19, 16, 18, 17, 19,
19, 20, 17, 16, 18, 16, 15, 19, 19, 17, 20, 18, 16, 19, 19, 15,
17, 17, 19, 19, 16, 17, 18, 19, 17, 19, 17, 15, 19, 16, 17
)
, A2 = c(8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
)
, A3 = c(33, 34, 38, 36, 36, 34, 41, 36, 40, 38, 38, 41, 38, 34, 33, 36,
41, 40, 41, 38, 41, 33, 40, 38, 40, 38, 41, 41, 40, 41, 40,
38, 34, 40, 36, 41, 40, 40, 33, 38, 36, 41, 40, 40, 28, 41,
40, 41, 33, 41, 36, 36, 40, 34, 41, 41, 38, 38, 41, 38, 41,
41, 36, 40, 38, 38, 40, 41, 38, 22, 36, 34, 38
)
, Total = c(57, 53, 62, 62, 64, 61, 65, 62, 68, 61, 63, 68, 65, 61, 57, 63,
65, 63, 68, 65, 65, 59, 66, 65, 67, 64, 69, 67, 67, 68, 67,
65, 59, 67, 61, 65, 66, 67, 57, 64, 61, 68, 67, 68, 53, 65,
66, 65, 56, 68, 63, 61, 68, 60, 65, 68, 65, 61, 66, 63, 68,
68, 60, 65, 64, 65, 65, 68, 63, 45, 63, 58, 63
)
)
, class = "data.frame"
, row.names = c(NA, -73L)
)
tb1 %>%
filter(Total == 63)
#> A1 A2 A3 Total
#> 1 17 8 38 63
#> 2 19 8 36 63
#> 3 15 8 40 63
#> 4 19 8 36 63
#> 5 17 8 38 63
#> 6 17 8 38 63
#> 7 19 8 36 63
#> 8 17 8 38 63
tb2 <-
tb1 %>%
mutate(A4 = replace(A3, Total == 63, A3-1)) %>%
mutate(Total = A1 + A2 + A3)
#> Warning: Problem with `mutate()` input `A4`.
#> x number of items to replace is not a multiple of replacement length
#> ℹ Input `A4` is `replace(A3, Total == 63, A3 - 1)`.
tb2 %>%
filter(Total == 62)
#> A1 A2 A3 Total
#> 1 16 8 38 62
#> 2 18 8 36 62
#> 3 18 8 36 62

You are better using ifelse here :
library(dplyr)
tb1 %>% mutate(A4 = ifelse(Total == 63, A3 -1, A3))
As far as why replace does not work if you check the source code of replace :
replace
function (x, list, values)
{
x[list] <- values
x
}
It assigns values to x after subsetting for list.
When you use :
tb1 %>% mutate(A4 = replace(A3, Total == 63, A3-1))
your values is of length length(tb1$A3) but list is of length sum(tb1$Total == 63) which do not match hence you get the warning of number of items to replace is not a multiple of replacement length, since it tries recycling those values but still the length is unequal.
If you want to make replace work you can try :
tb1 %>% mutate(A4 = replace(A3, Total == 63, A3[Total == 63] -1))
but again as I mentioned it is easier to just use ifelse here.

Maintain order of time series with group_by

Suppose my time series consists only of two columns: signal and day
The signal variable is supposed to repeat itself in a cycle of 1 to 6. So I need to insert empty rows for each implicit missing Signal but with signal counting from 1 to 6. (Suppose I have more columns that should also be empty (NA)).
In other words, for each unique day, there should be 6 rows with signal counting from 1 to 6.
My dataframe:
df = structure(list(data.Signal = c(2, 3, 4, 5, 6, 1, 2, 3, 4, 6,
1, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4, 5, 6, 2, 3, 4,
5, 6, 1, 3, 4, 5, 6, 2, 3, 4, 5, 6, 3, 4, 6, 1, 3, 4, 5, 6, 1,
2, 3, 4, 5, 6, 1, 2, 3, 4, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5,
6, 1, 2, 3, 4, 5, 1, 2, 3, 4, 6, 2, 3, 6, 3, 4, 5, 6, 1, 3, 4,
5, 6, 1, 1, 2, 3, 4, 5, 3, 4, 1, 2, 3, 4, 5, 5, 1, 2, 3, 4),
data.day = c(18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 20,
20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 22, 23, 23, 23, 23,
23, 23, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 26, 26, 26,
26, 26, 27, 27, 27, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29,
29, 30, 30, 30, 30, 30, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2,
2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7,
7, 7, 7, 7, 8, 9, 9, 9, 9, 9, 10, 10, 11, 11, 11, 11, 11,
11, 12, 12, 12, 12)), class = "data.frame", row.names = c(NA,
-114L))
My approach:
data <- df %>%
group_by(gr=data.day) %>%
complete(data.Signal = 1:6) %>%
ungroup() %>%
select(-gr)
This however, sorts the days ascending. The order of the days is obviously meaningful in the time series data. How can I "re-sort" to the original order or is there another way of solving my problem? Thank you!

Convert data.day to factor before using complete
library(dplyr)
df %>%
group_by(gr = factor(data.day, levels = unique(data.day))) %>%
tidyr::complete(data.Signal = 1:6) %>%
ungroup() %>%
select(-gr)
# data.Signal data.day
# <dbl> <dbl>
# 1 1 NA
# 2 2 18
# 3 3 18
# 4 4 18
# 5 5 18
# 6 6 18
# 7 1 19
# 8 2 19
# 9 3 19
#10 4 19
# … with 141 more rows
If you want those NA's filled you can use this version.
df %>%
mutate(grp = factor(data.day, levels = unique(data.day))) %>%
complete(grp, data.Signal = 1:6) %>%
ungroup() %>%
select(-data.day)

separate data in a column into multiple columns

I have some data which looks like:
> head(d)
V1 V2 V3 V4 V5 V6 V7 V8 V9
50 28 79 4 6 48 2 17 4 20
51 28 79 4 6 48 2 17 4 21
52 28 79 4 6 48 2 17 4 22
53 28 79 4 6 48 2 17 4 23
54 28 79 4 6 48 2 17 4 24
55 28 79 4 6 48 2 17 4 25
V10
50 000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V
51 000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.5V000.8V000.8V000.5V000.4V000.4V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.6V
52 000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.4V000.6V000.4V000.4V000.5V000.5V000.5V000.4V000.4V000.4V000.5V000.5V000.5V000.7V000.7V000.8V
53 000.7V000.5V000.4V000.5V000.5V000.4V000.4V000.4V000.4V000.5V000.5V000.4V000.4V000.3V000.4V000.4V000.4V000.3V000.4V000.3V000.4V000.9V001.0V001.0V
54 000.8V000.5V000.3V000.4V000.4V000.4V000.7V001.2V001.2V001.0N000.5V000.4V000.4V000.4V000.4V000.3V000.3V000.3V000.4V000.4V000.5V000.5V000.6V000.6V
55 000.5V000.5V000.3V000.3V000.3V000.3V000.3V000.4V000.5V000.6V000.5V000.5V000.4V000.3V000.4V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.3V000.4V
Where column V10 is a large column separated by 24 V's and N's. The V's are "valid" observations and the N's are non-valid observations.
I want to separate the V10 column. I tried using the following but it does not solve the N's problem.
sep_data <- df %>%
separate(V10, into = paste("x",1:24, sep = "_"), sep = "V")
Looking at the tail() of the sep_data:
V1 V2 V3 V4 V5 x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8 x_9 x_10 x_11 x_12
174 28079008 6 48 2 17042 4002.0 001.1 000.6 000.4 000.4 000.4 000.4 000.5 000.7 000.9 000.7N000.5 000.5
175 28079008 6 48 2 17042 5000.7 000.5 000.4 000.3 000.3 000.3 000.3 000.4 000.5 000.6 000.5 000.5
176 28079008 6 48 2 17042 6000.4 000.3 000.3 000.3 000.3 000.3 000.3 000.4 000.5 000.6 000.5 000.4
177 28079008 6 48 2 17042 7000.3 000.3 000.2 000.2 000.2 000.2 000.2 000.3 000.4 000.5 000.4 000.4
178 28079008 6 48 2 17042 8000.3 000.3 000.3 000.3 000.3 000.3 000.3 000.4 000.5 000.5 000.5 000.5
179 28079008 6 48 2 17042 9000.3 000.3 000.3 000.3 000.3 000.3 000.3 000.3 000.3 000.3 000.3 000.3
x_13 x_14 x_15 x_16 x_17 x_18 x_19 x_20 x_21 x_22 x_23 x_24 nchar
174 000.4 000.5 000.5 000.5 000.5 000.5 000.6 000.6 000.6 000.6 000.7 145
175 000.5 000.5 000.5 000.5 000.5 000.5 000.5 000.6 000.6 000.6 000.6 000.5 145
176 000.4 000.4 000.5 000.5 000.5 000.5 000.5 000.5 000.5 000.4 000.4 000.3 145
177 000.4 000.4 000.4 000.4 000.4 000.5 000.5 000.6 000.5 000.5 000.5 000.4 145
178 000.4 000.4 000.4 000.5 000.5 000.5 000.5 000.4 000.4 000.4 000.4 000.4 145
179 000.4 000.3 000.3 000.3 000.3 000.3 000.4 000.5 000.5 000.5 000.6 000.5 145
I have 000.7N000.5.
How can I use separate() to separate based on V OR N.
Data:
structure(list(V1 = c(28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28,
28, 28, 28, 28, 28, 28, 28, 28), V2 = c(79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79,
79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79, 79), V3 = c(4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8), V4 = c(6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6, 6, 6, 6), V5 = c(48, 48, 48, 48, 48, 48, 48,
48, 48, 48, 48, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38,
38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38,
38, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48,
48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48), V6 = c(2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2), V7 = c(17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
17, 17, 17, 17, 17, 17, 17), V8 = c(4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), V9 = c(20, 21,
22, 23, 24, 25, 26, 27, 28, 29, 30, 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, 30, 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,
30, 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, 30, 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), V10 = c("000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V",
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"000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V000.5V000.4V000.4V000.5V000.5V000.6V000.6V000.6V000.5V",
"000.5V000.5V000.5V000.5V000.4V000.4V000.5V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.5V000.4V000.4V000.4V000.5V000.5V000.5V000.5V000.5V000.4V",
"000.4V000.4V000.3V000.3V000.3V000.3V000.4V000.4V000.6V000.7V000.6V000.5V000.5V000.5V000.5V000.5V000.4V000.4V000.5V000.5V000.6V000.7V000.8V001.1V",
"001.2V000.9V000.8V000.6V000.5V000.4V000.4V000.6V001.1V001.0V000.7V000.6V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.6V000.7V000.8V000.7V000.6V",
"000.5V000.4V000.4V000.4V000.3V000.3V000.4V000.5V000.6V000.6V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.6V000.6V000.6V000.5V000.4V",
"000.4V000.4V000.4V000.4V000.3V000.3V000.4V000.5V000.6V000.7V000.6V000.5V000.5V000.5V000.5V000.5V000.4V000.4V000.5V000.6V000.7V000.9V001.0V000.7V",
"000.5V000.4V000.4V000.4V000.3V000.3V000.4V000.5V000.7V000.8V000.6V000.6V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.7V000.7V000.7V",
"000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V000.5V000.5V000.4V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.5V000.6V000.7V000.7V",
"000.6V000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V000.5V",
"000.4V000.4V000.4V000.4V000.4V000.3V000.4V000.5V000.7V000.8V000.6V000.5V000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.7V000.9V000.7V",
"000.7V000.5V000.4V000.3V000.3V000.3V000.4V000.5V000.7V000.8V000.6V000.6V000.7V000.7V000.5V000.5V000.5V000.4V000.4V000.5V000.6V000.7V000.7V000.6V",
"000.6V000.5V000.4V000.4V000.4V000.3V000.4V000.5V000.6V000.7V000.7V000.6V000.6V000.5V000.6V000.5V000.5V000.5V000.5V000.5V000.6V000.6V000.7V000.7V",
"000.7V000.7V000.7V000.8V000.8V000.7V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V000.5V",
"000.6V000.6V000.5V000.5V000.5V000.5V000.5V000.5V000.4V000.4V000.5V000.4V000.4V000.4V000.4V000.3V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V",
"000.5V000.5V000.5V000.4V000.4V000.4V000.3V000.3V000.4V000.4V000.5V000.4V000.4V000.4V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.5V000.4V000.4V",
"000.4V000.4V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.7V000.9V",
"001.1V001.2V000.7V000.4V000.3V000.3V000.3V000.4V000.6V000.7V000.5V000.5V000.5V000.5V000.5V000.4V000.4V000.4V000.4V000.5V000.5V000.5V000.5V000.6V",
"000.4V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.5V000.6V000.5V000.5V000.5V000.5V000.4V000.4V000.4V000.3V000.4V000.5V000.6V000.7V000.5V000.4V",
"000.4V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.5V000.5V000.5V000.6V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V",
"000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.4V000.6V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V",
"000.4V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.6V000.7V000.6V000.5V002.7V000.8V000.5V000.4V000.4V000.4V000.5V000.5V000.5V000.5V000.5V000.5V",
"000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.5V000.6V000.8V000.8V",
"000.8V000.8V000.7V000.5V000.4V000.4V000.4V000.5V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.6V000.9V001.8V",
"002.0V001.1V000.6V000.4V000.4V000.4V000.4V000.5V000.7V000.9V000.7N000.5V000.5V000.4V000.5V000.5V000.5V000.5V000.5V000.6V000.6V000.6V000.6V000.7V",
"000.7V000.5V000.4V000.3V000.3V000.3V000.3V000.4V000.5V000.6V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.6V000.6V000.6V000.6V000.5V",
"000.4V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.5V000.6V000.5V000.4V000.4V000.4V000.5V000.5V000.5V000.5V000.5V000.5V000.5V000.4V000.4V000.3V",
"000.3V000.3V000.2V000.2V000.2V000.2V000.2V000.3V000.4V000.5V000.4V000.4V000.4V000.4V000.4V000.4V000.4V000.5V000.5V000.6V000.5V000.5V000.5V000.4V",
"000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.5V000.5V000.5V000.5V000.4V000.4V000.4V000.5V000.5V000.5V000.5V000.4V000.4V000.4V000.4V000.4V",
"000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.3V000.4V000.3V000.3V000.3V000.3V000.3V000.4V000.5V000.5V000.5V000.6V000.5V"
)), row.names = 50:179, class = "data.frame")

How can I use separate() to separate based on V OR N? sep = "V|N".
But I'd suggest pre-processing. Use gsub or stringr::str_replace_all to make the N entries NA before separating. Something like df$V10 = gsub("N\\d+(\\.)?\\d+", "VNA", df$V10). Replace NXXX.XX with VNA, then splitting on V should leave NAs where the Ns originally were.

Moving average of 4 rows of data.table with multiple columns

I have a data.table xSet with multiple columns. I need a new table with a moving 4 row average for each column individually.

We could use rollapplyr from zoo
library(zoo)
library(dplyr)
df1 %>%
mutate_all(funs(New = rollapplyr(., FUN = mean, width = 4, partial = TRUE)))
Or similar option with data.table
library(data.table)
setDT(df1)[, paste0("New", names(df1)) := lapply(.SD,
function(x) rollapplyr(x, FUN = mean, width = 4, partial = TRUE))]
data
set.seed(24)
df1 <- as.data.frame(matrix(sample(0:9, 3 * 15, replace = TRUE),
ncol = 3, dimnames = list(NULL, paste0("Col", 1:3))))

The answers by akrun and G. Grothendieck call the rollapplr() function which uses a right aligned window by default.
But this is in contrast to the definition the OP has shown in the image.
This can be visualised by creating some suitable input data and by using toString() instead of mean() as aggregation function:
library(data.table)
# create suitable input data
DT <- data.table(col1 = 1:15, col2 = 21:35, col3 = 41:55)
DT[, cbind(.SD, New = zoo::rollapplyr(.SD, 4, toString, partial = TRUE))]
col1 col2 col3 New.col1 New.col2 New.col3
1: 1 21 41 1 21 41
2: 2 22 42 1, 2 21, 22 41, 42
3: 3 23 43 1, 2, 3 21, 22, 23 41, 42, 43
4: 4 24 44 1, 2, 3, 4 21, 22, 23, 24 41, 42, 43, 44
5: 5 25 45 2, 3, 4, 5 22, 23, 24, 25 42, 43, 44, 45
6: 6 26 46 3, 4, 5, 6 23, 24, 25, 26 43, 44, 45, 46
7: 7 27 47 4, 5, 6, 7 24, 25, 26, 27 44, 45, 46, 47
8: 8 28 48 5, 6, 7, 8 25, 26, 27, 28 45, 46, 47, 48
9: 9 29 49 6, 7, 8, 9 26, 27, 28, 29 46, 47, 48, 49
10: 10 30 50 7, 8, 9, 10 27, 28, 29, 30 47, 48, 49, 50
11: 11 31 51 8, 9, 10, 11 28, 29, 30, 31 48, 49, 50, 51
12: 12 32 52 9, 10, 11, 12 29, 30, 31, 32 49, 50, 51, 52
13: 13 33 53 10, 11, 12, 13 30, 31, 32, 33 50, 51, 52, 53
14: 14 34 54 11, 12, 13, 14 31, 32, 33, 34 51, 52, 53, 54
15: 15 35 55 12, 13, 14, 15 32, 33, 34, 35 52, 53, 54, 55
col1 is equal to the row numbers, New.col1 shows the row indices which are being involved in computing rollapplyr().
Compared to OP's image, only rows 1 and 2 do match. Apparently, a right aligned window does not meet OP's definition.
We can compare OP's requirement with the other alignment options for rolling windows:
DT <- data.table(col1 = 1:15, col2 = 21:35, col3 = 41:55)
align_window <- c("center", "left", "right")
DT[, (align_window) := lapply(align_window,
function(x) zoo::rollapply(
col1, 4, toString, partial = TRUE, align = x))]
# add OP's definition from image
DT[1:2, OP := right][3, OP := toString(2:4)][4:15, OP := center][]
col1 col2 col3 center left right OP
1: 1 21 41 1, 2, 3 1, 2, 3, 4 1 1
2: 2 22 42 1, 2, 3, 4 2, 3, 4, 5 1, 2 1, 2
3: 3 23 43 2, 3, 4, 5 3, 4, 5, 6 1, 2, 3 2, 3, 4
4: 4 24 44 3, 4, 5, 6 4, 5, 6, 7 1, 2, 3, 4 3, 4, 5, 6
5: 5 25 45 4, 5, 6, 7 5, 6, 7, 8 2, 3, 4, 5 4, 5, 6, 7
6: 6 26 46 5, 6, 7, 8 6, 7, 8, 9 3, 4, 5, 6 5, 6, 7, 8
7: 7 27 47 6, 7, 8, 9 7, 8, 9, 10 4, 5, 6, 7 6, 7, 8, 9
8: 8 28 48 7, 8, 9, 10 8, 9, 10, 11 5, 6, 7, 8 7, 8, 9, 10
9: 9 29 49 8, 9, 10, 11 9, 10, 11, 12 6, 7, 8, 9 8, 9, 10, 11
10: 10 30 50 9, 10, 11, 12 10, 11, 12, 13 7, 8, 9, 10 9, 10, 11, 12
11: 11 31 51 10, 11, 12, 13 11, 12, 13, 14 8, 9, 10, 11 10, 11, 12, 13
12: 12 32 52 11, 12, 13, 14 12, 13, 14, 15 9, 10, 11, 12 11, 12, 13, 14
13: 13 33 53 12, 13, 14, 15 13, 14, 15 10, 11, 12, 13 12, 13, 14, 15
14: 14 34 54 13, 14, 15 14, 15 11, 12, 13, 14 13, 14, 15
15: 15 35 55 14, 15 15 12, 13, 14, 15 14, 15
None of the alignment options does completely meet OP's definition. "center" is the best match except for the first 3 rows.

R - Keep reading line if 7 or more numbers are => 10

I have a file foo.txt that looks like this:
7, 3, 5, 7, 3, 3, 3, 3, 3, 3, 3, 6, 7, 5, 5, 22, 18, 14, 23, 16, 18, 5, 13, 34, 24, 17, 50, 30, 42, 35, 29, 27, 52, 35, 44, 52, 36, 39, 25, 40, 50, 52, 40, 2, 52, 52, 31, 35, 30, 19, 32, 46, 50, 43, 36, 15, 21, 16, 36, 25, 7, 3, 5, 7, 3, 3, 3, 3, 3, 3, 3, 6
I want to read the numbers in sets of 15, moving to the right one number at the time:
7, 3, 5, 7, 3, 3, 3, 3, 3, 3, 3, 6, 7, 5, 5
then
3, 5, 7, 3, 3, 3, 3, 3, 3, 3, 6, 7, 5, 5, 22
and so on.
If 7 or more of those 15 numbers are =>10 then keep them in a growing object that ends when the condition isn't met. So the first one to keep would be
3, 3, 3, 6, 7, 5, 5, 22, 18, 14, 23, 16, 18, 5, 13
because 7 out of those 15 numbers are => 10 (those numbers are 22, 18, 14, 23, 16, 18 and 13
The output file would look like this:
3, 3, 3, 6, 7, 5, 5, 22, 18, 14, 23, 16, 18, 5, 13, 34, 24, 17, 50, 30, 42, 35, 29, 27, 52, 35, 44, 52, 36, 39, 25, 40, 50, 52, 40, 2, 52, 52, 31, 35, 30, 19, 32, 46, 50, 43, 36, 15, 21, 16, 36, 25, 7, 3, 5, 7, 3, 3, 3, 3
So far I'm stuck at getting sets of 15 digits but I don't know how to make the condition "7 or more must be => 10"
qual <- readLines("foo.txt", 1)
separados <- unlist(strsplit(qual, ", "))
for (i in 1:length(qual)) {
separados[(i):(i + 14)] -> numbers
I don't mind the language as long as it does the work

I've added two ='s to Vlo's solutions and made this for you. Does this answer your question?
foo.txt <- c(7, 3, 5, 7, 3, 3, 3, 3, 3, 3, 3, 6, 7, 5, 5, 22, 18, 14, 23, 16, 18, 5,
13, 34, 24, 17, 50, 30, 42, 35, 29, 27, 52, 35, 44, 52, 36, 39, 25, 40,
50, 52, 40, 2, 52, 52, 31, 35, 30, 19, 32, 46, 50, 43, 36, 15, 21, 16,
36, 25, 7, 3, 5, 7, 3, 3, 3, 3, 3, 3, 3, 6)
# install.packages(c("zoo"), dependencies = TRUE)
require(zoo)
bar <- rollapply(foo.txt, 15, function(x) sum(x >= 10 ) >= 7)
(product <- foo.txt[bar])
[1] 3 3 3 6 7 5 5 22 18 14 23 16 18 5 13 34 24 17 50 30 42 35 29 27
[25] 52 35 44 52 36 39 25 40 50 52 40 2 52 52 31 35 30 19 32 46 50 43 3 3
[49] 3 3 3 6

I would do it in Python (you said you don't mind the language):
array = []
with open("foo.txt","r") as f:
for line in f:
for num in line.strip().split(', '):
array.append(int(num))
result = []
growing = False
while len(array) >= 15:
if sum(1 for e in filter(lambda x: x>=10, array[:15])) >= 7:
if growing:
result.append(array[15])
else:
result.extend(array[:15])
growing = True
else:
growing = False
del(array[0])
print(str(result)[1:-1])
Short explanation: first while simply reads the lines in the file, strips end of line, separates every number between ", " characters and appends each number to array.
Second while checks the first 15 numbers in array; if they have at least 7 numbers >= 0, it appends all the numbers, or just the last one (depending if the last iteration), to result. At the end of the loop, it removes the first number in array so that the loop can continue with the next 15 numbers.

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