Is there a way to check which value in a vector/column is nearest to a given value?
so for example I have column with number of days:
days: 50, 49, 59, 180, 170, 199, 200
I want to make a new column in the dataframe that marks an X everytime the dayscolumn has the value 183 or close to 183
It should look like this:
DAYS new column
0
12
12
14
133
140 X
0
12
14
15
178
183 X
0
15
30
72
172 X
Hope you can help me!
You're searching for local maxima, essentially. Start off by normalizing your data to your target, i.e. 183, and search for values closest to zero. Those are your local maxima. I added data with values greater than your target to demonstrate.
df <- data.frame(DAYS = c(0,12,12,14,133,140,0,12,14,15,178,183,184,190,0,15,30,72,172,172.5))
df$localmin <- abs(df$DAYS - 183)
df
> df
DAYS localmin
1 0.0 183.0
2 12.0 171.0
3 12.0 171.0
4 14.0 169.0
5 133.0 50.0
6 140.0 43.0
7 0.0 183.0
8 12.0 171.0
9 14.0 169.0
10 15.0 168.0
11 178.0 5.0
12 183.0 0.0
13 184.0 1.0
14 190.0 7.0
15 0.0 183.0
16 15.0 168.0
17 30.0 153.0
18 72.0 111.0
19 172.0 11.0
20 172.5 10.5
targets <- which(diff(sign(diff(c(df$localmin, 183)))) == 2) + 1L
df$targets <- 0
df$targets[targets] <- 1
df
> df
DAYS localmin targets
1 0.0 183.0 0
2 12.0 171.0 0
3 12.0 171.0 0
4 14.0 169.0 0
5 133.0 50.0 0
6 140.0 43.0 1
7 0.0 183.0 0
8 12.0 171.0 0
9 14.0 169.0 0
10 15.0 168.0 0
11 178.0 5.0 0
12 183.0 0.0 1
13 184.0 1.0 0
14 190.0 7.0 0
15 0.0 183.0 0
16 15.0 168.0 0
17 30.0 153.0 0
18 72.0 111.0 0
19 172.0 11.0 0
20 172.5 10.5 1
Related
I have weather dataset my data is date-dependent
I want to predict the temperature from 07 May 2008 until 18 May 2008 (which is maybe a total of 10-15 observations) my data size is around 200
I will be using decision tree/RF and SVM & NN to make my prediction
I've never handled data like this so I'm not sure how to sample non random data
I want to sample data 80% train data and 30% test data but I want to sample the data in the original order not randomly. Is that possible ?
install.packages("rattle")
install.packages("RGtk2")
library("rattle")
seed <- 42
set.seed(seed)
fname <- system.file("csv", "weather.csv", package = "rattle")
dataset <- read.csv(fname, encoding = "UTF-8")
dataset <- dataset[1:200,]
dataset <- dataset[order(dataset$Date),]
set.seed(321)
sample_data = sample(nrow(dataset), nrow(dataset)*.8)
test<-dataset[sample_data,] # 30%
train<-dataset[-sample_data,] # 80%
output
> head(dataset)
Date Location MinTemp MaxTemp Rainfall Evaporation Sunshine WindGustDir WindGustSpeed
1 2007-11-01 Canberra 8.0 24.3 0.0 3.4 6.3 NW 30
2 2007-11-02 Canberra 14.0 26.9 3.6 4.4 9.7 ENE 39
3 2007-11-03 Canberra 13.7 23.4 3.6 5.8 3.3 NW 85
4 2007-11-04 Canberra 13.3 15.5 39.8 7.2 9.1 NW 54
5 2007-11-05 Canberra 7.6 16.1 2.8 5.6 10.6 SSE 50
6 2007-11-06 Canberra 6.2 16.9 0.0 5.8 8.2 SE 44
WindDir9am WindDir3pm WindSpeed9am WindSpeed3pm Humidity9am Humidity3pm Pressure9am
1 SW NW 6 20 68 29 1019.7
2 E W 4 17 80 36 1012.4
3 N NNE 6 6 82 69 1009.5
4 WNW W 30 24 62 56 1005.5
5 SSE ESE 20 28 68 49 1018.3
6 SE E 20 24 70 57 1023.8
Pressure3pm Cloud9am Cloud3pm Temp9am Temp3pm RainToday RISK_MM RainTomorrow
1 1015.0 7 7 14.4 23.6 No 3.6 Yes
2 1008.4 5 3 17.5 25.7 Yes 3.6 Yes
3 1007.2 8 7 15.4 20.2 Yes 39.8 Yes
4 1007.0 2 7 13.5 14.1 Yes 2.8 Yes
5 1018.5 7 7 11.1 15.4 Yes 0.0 No
6 1021.7 7 5 10.9 14.8 No 0.2 No
> head(test)
Date Location MinTemp MaxTemp Rainfall Evaporation Sunshine WindGustDir WindGustSpeed
182 2008-04-30 Canberra -1.8 14.8 0.0 1.4 7.0 N 28
77 2008-01-16 Canberra 17.9 33.2 0.0 10.4 8.4 N 59
88 2008-01-27 Canberra 13.2 31.3 0.0 6.6 11.6 WSW 46
58 2007-12-28 Canberra 15.1 28.3 14.4 8.8 13.2 NNW 28
96 2008-02-04 Canberra 18.2 22.6 1.8 8.0 0.0 ENE 33
126 2008-03-05 Canberra 12.0 27.6 0.0 6.0 11.0 E 46
WindDir9am WindDir3pm WindSpeed9am WindSpeed3pm Humidity9am Humidity3pm Pressure9am
182 E N 2 19 80 40 1024.2
77 N NNE 15 20 58 62 1008.5
88 N WNW 4 26 71 28 1013.1
58 NNW NW 6 13 73 44 1016.8
96 SSE ENE 7 13 92 76 1014.4
126 SSE WSW 7 6 69 35 1025.5
Pressure3pm Cloud9am Cloud3pm Temp9am Temp3pm RainToday RISK_MM RainTomorrow
182 1020.5 1 7 5.3 13.9 No 0.0 No
77 1006.1 6 7 24.5 23.5 No 4.8 Yes
88 1009.5 1 4 19.7 30.7 No 0.0 No
58 1013.4 1 5 18.3 27.4 Yes 0.0 No
96 1011.5 8 8 18.5 22.1 Yes 9.0 Yes
126 1022.2 1 1 15.7 26.2 No 0.0 No
> head(train)
Date Location MinTemp MaxTemp Rainfall Evaporation Sunshine WindGustDir WindGustSpeed
7 2007-11-07 Canberra 6.1 18.2 0.2 4.2 8.4 SE 43
9 2007-11-09 Canberra 8.8 19.5 0.0 4.0 4.1 S 48
11 2007-11-11 Canberra 9.1 25.2 0.0 4.2 11.9 N 30
16 2007-11-16 Canberra 12.4 32.1 0.0 8.4 11.1 E 46
22 2007-11-22 Canberra 16.4 19.4 0.4 9.2 0.0 E 26
25 2007-11-25 Canberra 15.4 28.4 0.0 4.4 8.1 ENE 33
WindDir9am WindDir3pm WindSpeed9am WindSpeed3pm Humidity9am Humidity3pm Pressure9am
7 SE ESE 19 26 63 47 1024.6
9 E ENE 19 17 70 48 1026.1
11 SE NW 6 9 74 34 1024.4
16 SE WSW 7 9 70 22 1017.9
22 ENE E 6 11 88 72 1010.7
25 SSE NE 9 15 85 31 1022.4
Pressure3pm Cloud9am Cloud3pm Temp9am Temp3pm RainToday RISK_MM RainTomorrow
7 1022.2 4 6 12.4 17.3 No 0.0 No
9 1022.7 7 7 14.1 18.9 No 16.2 Yes
11 1021.1 1 2 14.6 24.0 No 0.2 No
16 1012.8 0 3 19.1 30.7 No 0.0 No
22 1008.9 8 8 16.5 18.3 No 25.8 Yes
25 1018.6 8 2 16.8 27.3 No 0.0 No
I use mtcars as an example. An option to non-randomly split your data in train and test is to first create a sample size based on the number of rows in your data. After that you can use split to split the data exact at the 80% of your data. You using the following code:
smp_size <- floor(0.80 * nrow(mtcars))
split <- split(mtcars, rep(1:2, each = smp_size))
With the following code you can turn the split in train and test:
train <- split$`1`
test <- split$`2`
Let's check the number of rows:
> nrow(train)
[1] 25
> nrow(test)
[1] 7
Now the data is split in train and test without losing their order.
using this function I calculate the variance of some 3d points.
centroid_3d_sq_dist <- function(
point_matrix
) {
if (nrow(point_matrix) == 1) {
return(0)
}
mean_point <- apply(point_matrix, 2, mean)
point_sq_distances <- apply(
point_matrix,
1,
function(row_point) {
sum((row_point - mean_point) ** 2)
}
)
sum_sq_distances <- sum(point_sq_distances)
return(sum_sq_distances)
}
point_3d_variance <- function(
point_matrix
) {
if (nrow(point_matrix) == 1) {
return(0)
}
dist_var <- centroid_3d_sq_dist(point_matrix) /
(nrow(point_matrix) - 1)
return(dist_var)
}
The argument of this function is a matrix (x,y,z).
Now I have a dataset with two 3D points.
ID Trial Size PP PA FkA ciccioX ciccioY ciccioZ pinoX pinoY pinoZ
1 Gigi 1 40 39.6 1050. 31.5 521. 293. 10.6 516. 323. 6.41
2 Gigi 2 20.0 30.7 944. 9.35 525. 300. 12.6 520. 305. 7.09
3 Gigi 3 30 29.5 1056. 24.1 521. 298. 12.3 519. 321. 5.89
4 Gigi 5 60 53.0 1190. 53.0 680. 287. 64.4 699. 336. 68.6
5 Bibi 1 40 38.3 1038. 31.4 524. 289. 10.9 519. 319. 6.17
6 Bibi 2 60 64.7 1293. 47.8 516. 282. 10.4 519. 330. 6.32
7 Bibi 3 20.0 33.8 1092. 17.5 523. 300. 12.8 518. 315. 6.22
8 Bibi 4 30 35.0 1108. 26.4 525. 295. 11.7 517. 320. 5.78
9 Bibi 5 50 46.5 1199. 34.2 515. 289. 11.2 517. 323. 6.27
10 Bibi 6 30 28.7 1016. 17.1 528. 298. 12.7 524. 314. 6.36
The 3D points are:
ciccio: ciccioX ciccioY ciccioZ
pino: pinoX pinoY pinoZ
I want to calculate the variance of ciccio and the variance of pino grouped by ID and SIZE.
I tried to do:
data %>%
group_by(SubjectID, Size) %>%
summarize(as.data.frame(matrix(f4(dd[7:9],dd[10:12]), nr = 1)))
But it doesn't work.
Do you have any advice?
Your shown dataset is too small to calculate (meaningful) variations. But you could use:
library(dplyr)
df %>%
group_by(ID, Size) %>%
summarise(var_ciccio = point_3d_variance(as.matrix(across(ciccioX:ciccioZ))),
var_pino = point_3d_variance(as.matrix(across(pinoX:pinoZ))),
.groups = "drop")
This returns
# A tibble: 9 x 4
ID Size var_ciccio var_pinoo
<chr> <dbl> <dbl> <dbl>
1 Bibi 20 0 0
2 Bibi 30 9.5 42.7
3 Bibi 40 0 0
4 Bibi 50 0 0
5 Bibi 60 0 0
6 Gigi 20 0 0
7 Gigi 30 0 0
8 Gigi 40 0 0
9 Gigi 60 0 0
This question already has answers here:
Categorize numeric variable into group/ bins/ breaks
(4 answers)
Closed 1 year ago.
I am attempting to add a new column to the state sample data frame in R. I am hoping for this column to cluster the ID of states into broader categories (1-4). My code is close to what I am looking for but I am not getting it quite right.. I know I could enter each state ID line by line but is there a a quicker way? Thank you!
library(tidyverse)
#Add column to denote each state
States=state.x77
States=data.frame(States)
States <- tibble::rowid_to_column(States, "ID")
States
#Create new variable for state buckets
States <- States %>%
mutate(WAGE_BUCKET=case_when(ID <= c(1,12) ~ '1',
ID <= c(13,24) ~ '2',
ID <= c(25,37) ~ '3',
ID <= c(38,50) ~ '4',
TRUE ~ 'NA'))
View(States) #It is not grouping the states in the way I want/I am still getting some NA values but unsure why!
You can use cut or findInterval if all of your groups will be using contiguous ID values:
findInterval(States$ID, c(0, 12, 24, 37, 51))
# [1] 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4
If you want to make it a bit more verbose, you can use dplyr::between in your case_when:
States %>%
mutate(
WAGE_BUCKET = case_when(
between(ID, 1, 12) ~ "1",
between(ID, 13, 24) ~ "2",
between(ID, 25, 37) ~ "3",
between(ID, 38, 50) ~ "4",
TRUE ~ NA_character_)
)
# ID Population Income Illiteracy Life Exp Murder HS Grad Frost Area WAGE_BUCKET
# 1 1 3615 3624 2.1 69.05 15.1 41.3 20 50708 1
# 2 2 365 6315 1.5 69.31 11.3 66.7 152 566432 1
# 3 3 2212 4530 1.8 70.55 7.8 58.1 15 113417 1
# 4 4 2110 3378 1.9 70.66 10.1 39.9 65 51945 1
# 5 5 21198 5114 1.1 71.71 10.3 62.6 20 156361 1
# 6 6 2541 4884 0.7 72.06 6.8 63.9 166 103766 1
# 7 7 3100 5348 1.1 72.48 3.1 56.0 139 4862 1
# 8 8 579 4809 0.9 70.06 6.2 54.6 103 1982 1
# 9 9 8277 4815 1.3 70.66 10.7 52.6 11 54090 1
# 10 10 4931 4091 2.0 68.54 13.9 40.6 60 58073 1
# 11 11 868 4963 1.9 73.60 6.2 61.9 0 6425 1
# 12 12 813 4119 0.6 71.87 5.3 59.5 126 82677 1
# 13 13 11197 5107 0.9 70.14 10.3 52.6 127 55748 2
# 14 14 5313 4458 0.7 70.88 7.1 52.9 122 36097 2
# 15 15 2861 4628 0.5 72.56 2.3 59.0 140 55941 2
# 16 16 2280 4669 0.6 72.58 4.5 59.9 114 81787 2
# 17 17 3387 3712 1.6 70.10 10.6 38.5 95 39650 2
# 18 18 3806 3545 2.8 68.76 13.2 42.2 12 44930 2
# 19 19 1058 3694 0.7 70.39 2.7 54.7 161 30920 2
# 20 20 4122 5299 0.9 70.22 8.5 52.3 101 9891 2
# 21 21 5814 4755 1.1 71.83 3.3 58.5 103 7826 2
# 22 22 9111 4751 0.9 70.63 11.1 52.8 125 56817 2
# 23 23 3921 4675 0.6 72.96 2.3 57.6 160 79289 2
# 24 24 2341 3098 2.4 68.09 12.5 41.0 50 47296 2
# 25 25 4767 4254 0.8 70.69 9.3 48.8 108 68995 3
# 26 26 746 4347 0.6 70.56 5.0 59.2 155 145587 3
# 27 27 1544 4508 0.6 72.60 2.9 59.3 139 76483 3
# 28 28 590 5149 0.5 69.03 11.5 65.2 188 109889 3
# 29 29 812 4281 0.7 71.23 3.3 57.6 174 9027 3
# 30 30 7333 5237 1.1 70.93 5.2 52.5 115 7521 3
# 31 31 1144 3601 2.2 70.32 9.7 55.2 120 121412 3
# 32 32 18076 4903 1.4 70.55 10.9 52.7 82 47831 3
# 33 33 5441 3875 1.8 69.21 11.1 38.5 80 48798 3
# 34 34 637 5087 0.8 72.78 1.4 50.3 186 69273 3
# 35 35 10735 4561 0.8 70.82 7.4 53.2 124 40975 3
# 36 36 2715 3983 1.1 71.42 6.4 51.6 82 68782 3
# 37 37 2284 4660 0.6 72.13 4.2 60.0 44 96184 3
# 38 38 11860 4449 1.0 70.43 6.1 50.2 126 44966 4
# 39 39 931 4558 1.3 71.90 2.4 46.4 127 1049 4
# 40 40 2816 3635 2.3 67.96 11.6 37.8 65 30225 4
# 41 41 681 4167 0.5 72.08 1.7 53.3 172 75955 4
# 42 42 4173 3821 1.7 70.11 11.0 41.8 70 41328 4
# 43 43 12237 4188 2.2 70.90 12.2 47.4 35 262134 4
# 44 44 1203 4022 0.6 72.90 4.5 67.3 137 82096 4
# 45 45 472 3907 0.6 71.64 5.5 57.1 168 9267 4
# 46 46 4981 4701 1.4 70.08 9.5 47.8 85 39780 4
# 47 47 3559 4864 0.6 71.72 4.3 63.5 32 66570 4
# 48 48 1799 3617 1.4 69.48 6.7 41.6 100 24070 4
# 49 49 4589 4468 0.7 72.48 3.0 54.5 149 54464 4
# 50 50 376 4566 0.6 70.29 6.9 62.9 173 97203 4
It is a vector of length > 1. The comparison operators works on a single vector. We could use between
library(dplyr)
States <- States %>%
mutate(WAGE_BUCKET=case_when(between(ID, 1, 12) ~ '1',
between(ID, 13,24) ~ '2',
between(ID, 25,37) ~ '3',
between(ID, 38,50) ~ '4',
TRUE ~ NA_character_))
Or another option is to use & with > and <=
States %>%
mutate(WAGE_BUCKET=case_when(ID >= 1 & ID <=12 ~ '1',
ID >= 13 & ID <= 24) ~ '2',
ID >= 25 & ID <= 37 ~ '3',
ID >= 38 & ID <= 50 ~ '4',
TRUE ~ NA_character))
Or may be the OP meant to use %in%
States %>%
mutate(WAGE_BUCKET=case_when(ID %in% c(1,12) ~ '1',
ID %in% c(13,24) ~ '2',
ID %in% c(25,37) ~ '3',
ID %in% c(38,50) ~ '4',
TRUE ~ NA_character_))
I have the following data (in a data frame), they are grouped by every 4 rows.
x y
1 1.495 0.0
2 1.500 30.0
3 2.500 30.0
4 2.505 0.0
5 8.495 0.0
6 8.500 30.0
7 9.500 30.0
8 9.505 0.0
9 10.495 0.0
10 10.500 30.0
11 11.500 30.0
12 11.505 0.0
13 16.495 0.0 ##From here
14 16.500 30.0
15 17.500 30.0
16 17.505 0.0
17 17.495 0.0
18 17.500 30.0
19 18.500 30.0
20 18.505 0.0 ## End here
21 19.495 0.0
22 19.500 30.0
23 20.500 30.0
24 20.505 0.0
25 23.495 0.0
26 23.500 30.0
27 24.500 30.0
28 24.505 0.0
.
.
.
I am trying to change the y-value of the rows that are overlapped (according to their x-values). For example, rows (13 to 16) are overlapped with row (17 to 20).
x-values of row 13-16: 16.495 16.500 -------- 17.500 17.505
x-values of row 17-20: ------------------ 17.495 17.500 ----------18.500 18.505
There are overlap from 17.495 to 17.505.
I would like to make the "in between" rows into something like:
13 16.495 0.0 ##From here
14 16.500 30.0
15 17.500 30.0
16 17.505 30.0
17 17.495 30.0
18 17.500 30.0
19 18.500 30.0
20 18.505 0.0 ## End here
Any idea how to do this?
Seeing the present sample data, it seems that you want to identify row(s) where a previous value in x is larger than the following value in x. In this case, row 17 is the one. Similarly, you want to identify row(s) where a value in x is larger than the following value in x. In this case, row 16 is the one. So, I tried to get row numbers for these rows in the following way. Note that your data is called mydf here.
ind <- c(which(x = lag(mydf$x) > mydf$x), which(x = lead(mydf$x) < mydf$x))
# Overwrite two specific elements in y
mydf$y[ind] <- 30
Here is the result for the part you specified. I hope this will help you.
#13 16.495 0
#14 16.500 30
#15 17.500 30
#16 17.505 30
#17 17.495 30
#18 17.500 30
#19 18.500 30
#20 18.505 0
Using a for loop, you can do the following (assuming your dataframe is called df):
# defining start and end values to process data by group of 4
start = seq(1,length(df$x),by = 4)
end = seq(4,length(df$x),by = 4)
# loop to inspect data by group of 4 and replace data in df in function of the overlap
for(i in 1:(length(start)-1))
{
if(max(df[start[i]:end[i],"x"]) > min(df[start[i+1]:end[i+1],"x"]))
{
df[end[i],"y"] = 30.0
df[start[i+1],"y"] = 30.0
}
else{}
}
And you get the following dataframe:
> df
x y
1 1.495 0
2 1.500 30
3 2.500 30
4 2.505 0
5 8.495 0
6 8.500 30
7 9.500 30
8 9.505 0
9 10.495 0
10 10.500 30
11 11.500 30
12 11.505 0
13 16.495 0
14 16.500 30
15 17.500 30
16 17.505 30
17 17.495 30
18 17.500 30
19 18.500 30
20 18.505 0
21 19.495 0
22 19.500 30
23 20.500 30
24 20.505 0
25 23.495 0
26 23.500 30
27 24.500 30
28 24.505 0
I have a function that takes in a dataframe, a percentile threshold, and the name of a given column, and computes all values that are above this threshold in the given column as a new column (0 for <, and 1 for >=). However, it won't allow me to do the df$column_name inside the quantile function because column_name is not actually a column name, but a variable storing the actual column name. Therefore df$column_name will return NULL. Is there any way to work around this and keep the code forma somewhat similar to what it is currently? Or do I have to specify the actual numerical column value instead of the name? While I can do this, it is definitely not as convenient/comprehensible as just passing in the column name.
func1 <- function(df, threshold, column_name) {
threshold_value <- quantile(df$column_name, c(threshold))
new_df <- df %>%
mutate(ifelse(column_name > threshold_value, 1, 0))
return(new_df)
}
Thank you so much for your help!
I modified your function as follows. Now the function can take a data frame, a threshold, and a column name. This function only needs the base R.
# Modified function
func1 <- function(df, threshold, column_name) {
threshold_value <- quantile(df[[column_name]], threshold)
new_df <- df
new_df[["new_col"]] <- ifelse(df[[column_name]] > threshold_value, 1, 0)
return(new_df)
}
# Take the trees data frame as an example
head(trees)
# Girth Height Volume
# 1 8.3 70 10.3
# 2 8.6 65 10.3
# 3 8.8 63 10.2
# 4 10.5 72 16.4
# 5 10.7 81 18.8
# 6 10.8 83 19.7
# Apply the function
func1(trees, 0.5, "Volume")
# Girth Height Volume new_col
# 1 8.3 70 10.3 0
# 2 8.6 65 10.3 0
# 3 8.8 63 10.2 0
# 4 10.5 72 16.4 0
# 5 10.7 81 18.8 0
# 6 10.8 83 19.7 0
# 7 11.0 66 15.6 0
# 8 11.0 75 18.2 0
# 9 11.1 80 22.6 0
# 10 11.2 75 19.9 0
# 11 11.3 79 24.2 0
# 12 11.4 76 21.0 0
# 13 11.4 76 21.4 0
# 14 11.7 69 21.3 0
# 15 12.0 75 19.1 0
# 16 12.9 74 22.2 0
# 17 12.9 85 33.8 1
# 18 13.3 86 27.4 1
# 19 13.7 71 25.7 1
# 20 13.8 64 24.9 1
# 21 14.0 78 34.5 1
# 22 14.2 80 31.7 1
# 23 14.5 74 36.3 1
# 24 16.0 72 38.3 1
# 25 16.3 77 42.6 1
# 26 17.3 81 55.4 1
# 27 17.5 82 55.7 1
# 28 17.9 80 58.3 1
# 29 18.0 80 51.5 1
# 30 18.0 80 51.0 1
# 31 20.6 87 77.0 1
If you still want to use dplyr, it is essential to learn how to deal with non-standard evaluation. Please see this to learn more (https://cran.r-project.org/web/packages/dplyr/vignettes/programming.html). The following code will also works.
library(dplyr)
func2 <- function(df, threshold, column_name) {
col_en <- enquo(column_name)
threshold_value <- quantile(df %>% pull(!!col_en), threshold)
new_df <- df %>%
mutate(new_col := ifelse(!!col_en >= threshold_value, 1, 0))
return(new_df)
}
func2(trees, 0.5, Volume)
# Girth Height Volume new_col
# 1 8.3 70 10.3 0
# 2 8.6 65 10.3 0
# 3 8.8 63 10.2 0
# 4 10.5 72 16.4 0
# 5 10.7 81 18.8 0
# 6 10.8 83 19.7 0
# 7 11.0 66 15.6 0
# 8 11.0 75 18.2 0
# 9 11.1 80 22.6 0
# 10 11.2 75 19.9 0
# 11 11.3 79 24.2 1
# 12 11.4 76 21.0 0
# 13 11.4 76 21.4 0
# 14 11.7 69 21.3 0
# 15 12.0 75 19.1 0
# 16 12.9 74 22.2 0
# 17 12.9 85 33.8 1
# 18 13.3 86 27.4 1
# 19 13.7 71 25.7 1
# 20 13.8 64 24.9 1
# 21 14.0 78 34.5 1
# 22 14.2 80 31.7 1
# 23 14.5 74 36.3 1
# 24 16.0 72 38.3 1
# 25 16.3 77 42.6 1
# 26 17.3 81 55.4 1
# 27 17.5 82 55.7 1
# 28 17.9 80 58.3 1
# 29 18.0 80 51.5 1
# 30 18.0 80 51.0 1
# 31 20.6 87 77.0 1