I have a table with a column "Age" that has a values from 1 to 10, and a column "Population" that has values specified for each of the "age" values. I want to generate a cumulative function for population such that resultant values start from ages at least 1 and above, 2 and above, and so on. I mean, the resultant array should be (203,180..and so on). Any help would be appreciated!
Age Population Withdrawn
1 23 3
2 12 2
3 32 2
4 33 3
5 15 4
6 10 1
7 19 2
8 18 3
9 19 1
10 22 5
You can use cumsum and rev:
df$sum_above <- rev(cumsum(rev(df$Population)))
The result:
> df
Age Population sum_above
1 1 23 203
2 2 12 180
3 3 32 168
4 4 33 136
5 5 15 103
6 6 10 88
7 7 19 78
8 8 18 59
9 9 19 41
10 10 22 22
Related
I want to use conditional statement to consecutive values in the sliding manner.
For example, I have dataset like this;
data <- data.frame(ID = rep.int(c("A","B"), times = c(24, 12)),
+ time = c(1:24,1:12),
+ visit = as.integer(runif(36, min = 0, max = 20)))
and I got table below;
> data
ID time visit
1 A 1 7
2 A 2 0
3 A 3 6
4 A 4 6
5 A 5 3
6 A 6 8
7 A 7 4
8 A 8 10
9 A 9 18
10 A 10 6
11 A 11 1
12 A 12 13
13 A 13 7
14 A 14 1
15 A 15 6
16 A 16 1
17 A 17 11
18 A 18 8
19 A 19 16
20 A 20 14
21 A 21 15
22 A 22 19
23 A 23 5
24 A 24 13
25 B 1 6
26 B 2 6
27 B 3 16
28 B 4 4
29 B 5 19
30 B 6 5
31 B 7 17
32 B 8 6
33 B 9 10
34 B 10 1
35 B 11 13
36 B 12 15
I want to flag each ID by continuous values of "visit".
If the number of "visit" continued less than 10 for 6 times consecutively, I'd attach "empty", and "busy" otherwise.
In the data above, "A" is continuously below 10 from rows 1 to 6, then "empty". On the other hand, "B" doesn't have 6 consecutive one digit, then "busy".
I want to apply the condition to next segment of 6 values if the condition weren't fulfilled in the previous segment.
I'd like achieve this using R. Any advice will be appreciated.
I have a dataframe df, consists of 2 columns: x and y coordinates.
Each row refers to a point.
I feed it into dbscan function to obtain the clusters of the points in df.
library("fpc")
db = fpc::dbscan(df, eps = 0.08, MinPts = 4)
plot(db, df, main = "DBSCAN", frame = FALSE)
By using print(db), I can see the result returned by dbscan.
> print(db)
dbscan Pts=13131 MinPts=4 eps=0.08
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
border 401 38 55 5 2 3 0 0 0 8 0 6 1 3 1 3 3 2 1 2 4 3
seed 0 2634 8186 35 24 561 99 7 22 26 5 75 17 9 9 54 1 2 74 21 3 15
total 401 2672 8241 40 26 564 99 7 22 34 5 81 18 12 10 57 4 4 75 23 7 18
22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
border 4 1 2 6 2 1 3 7 2 1 2 3 11 1 3 1 3 2 5 5 1 4 3
seed 14 9 4 48 2 4 38 111 5 11 5 14 111 6 1 5 1 8 3 15 10 15 6
total 18 10 6 54 4 5 41 118 7 12 7 17 122 7 4 6 4 10 8 20 11 19 9
45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68
border 2 4 2 1 3 2 1 1 3 1 0 2 2 3 0 3 3 3 3 0 0 2 3 1
seed 15 2 9 11 4 8 12 4 6 8 7 7 3 3 4 3 3 4 2 9 4 2 1 4
total 17 6 11 12 7 10 13 5 9 9 7 9 5 6 4 6 6 7 5 9 4 4 4 5
69 70 71
border 3 3 3
seed 1 1 1
total 4 4 4
From the above summary, I can see cluster 2 consists of 8186 seed points (core points), cluster 1 consists of 2634 seed points and cluster 5 consists of 561 points.
I define the largest cluster as the one contains the largest amount of seed points. So, in this case, the largest cluster is cluster 2. And the 1st, 2nd, 3th largest clusters are 2, 1 and 5.
Are they any direct way to return the rows (points) in the largest cluster or the k-largest cluster in general?
I can do it in an indirect way.
I can obtain the assigned cluster number of each point by
db$cluster.
Hence, I can create a new dataframe df2 with db$cluster as the
new additional column besides the original x column and y
column.
Then, I can aggregate the df2 according to the cluster numbers in
the third column and find the number of points in each cluster.
After that, I can find the k-largest groups, which are 2, 1 and 5
again.
Finally, I can select the rows in df2 with third column value equals to 2 to return the points in the largest cluster.
But the above approach re-computes many known results as stated in the summary of print(db).
The dbscan function doesn't appear to retain the data.
library(fpc)
set.seed(665544)
n <- 600
df <- data.frame(x=runif(10, 0, 10)+rnorm(n, sd=0.2), y=runif(10, 0, 10)+rnorm(n,sd=0.2))
(dbs <- dbscan(df, 0.2))
#dbscan Pts=600 MinPts=5 eps=0.2
# 0 1 2 3 4 5 6 7 8 9 10 11
#border 28 4 4 8 5 3 3 4 3 4 6 4
#seed 0 50 53 51 52 51 54 54 54 53 51 1
#total 28 54 57 59 57 54 57 58 57 57 57 5
attributes(dbs)
#$names
#[1] "cluster" "eps" "MinPts" "isseed"
#$class
#[1] "dbscan"
Your indirect steps are not that indirect (only two lines needed), and these commands won't recalculate the clusters. So just run those commands, or put them in a function and then call the function in one command.
cluster_k <- function(dbs, data, k){
kth <- names(rev(sort(table(dbs$cluster)))[k])
data[dbs$cluster == kth,]
}
cluster_k(dbs=dbs, data=df, k=1)
## x y
## 3 6.580695 8.715245
## 13 6.704379 8.528486
## 23 6.809558 8.160721
## 33 6.375842 8.756433
## 43 6.603195 8.640206
## 53 6.728533 8.425067
## a data frame with 59 rows
The following randomly splits a data frame into halves.
df <- read.csv("https://raw.githubusercontent.com/HirokiYamamoto2531/data/master/data.csv")
head(df, 3)
# dv iv subject item
#1 562 -0.5 1 7
#2 790 0.5 1 21
#3 NA -0.5 1 19
r <- seq_len(nrow(df))
first <- sample(r, 240)
second <- r[!r %in% first]
df_1 <- df[first, ]
df_2 <- df[second, ]
However, in this way, each data frame (df_1 and df_2) is not balanced on subject and item: e.g.,
table(df_1$subject)
# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
# 7 8 3 5 5 3 8 1 5 7 7 6 7 7 9 8 8 9 6 7 8 5 4 4 5 2 7 6 9
# 30 31 32 33 34 35 36 37 38 39 40
# 7 5 7 7 7 3 5 7 5 3 8
table(df_1$item)
# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
# 12 11 12 12 9 11 11 8 11 12 10 8 14 7 14 10 8 7 9 9 7 11 9 8
# There are 40 subjects and 24 items, and each subject is assigned to 12 items and each item to 20 subjects.
I would like to know how to split the data frame into halves that are balanced on subject and item (i.e., exactly 6 data points from each subject and 10 data points from each item).
You can use the createDataPartition function from the caret package to create a balanced partition of one variable.
The code below creates a balanced partition of the dataset according to the variable subject:
df <- read.csv("https://raw.githubusercontent.com/HirokiYamamoto2531/data/master/data.csv")
partition <- caret::createDataPartition(df$subject, p = 0.5, list = FALSE)
first.half <- df[partition, ]
second.half <- df[-partition, ]
table(first.half$subject)
table(second.half$subject)
I'm not sure whether it's possible to balance two variables at once. You can try balancing for one variable and checking if you're happy with the partition of the second variable.
ID MON in out
2 1 23 12
3 1 23 12
7 1 33 22
1 2 22 11
2 2 111 100
1 3 21 10
2 3 22 11
2 4 111 100
7 4 21 10
2 5 31 20
7 2046 41 30
I have a large data set in this format. I want to extract column four for the value of column 1==2 and column 2 smaller then 5.
It's basic R.
df[,4][df[,1]==2 & df[,2]<5]
I have a dataframe with 20 classrooms [1 to 20] indexes and 20 different number of students in each class, how to obtain all sub-samples of size n = 8 and store them because i want to use them later for calculations. I used combn() but that takes only one vector, can i use it with a dataframe and how? (sorry but i'm new in R),
dataframe below:
classrooms students
1 1 29
2 2 30
3 3 35
4 4 28
5 5 32
6 6 20
7 7 25
8 8 22
9 9 32
10 10 26
11 11 27
12 12 34
13 13 27
14 14 28
15 15 33
16 16 21
17 17 36
18 18 24
19 19 19
20 20 32
It is as simple as passing a function to combn. simplify = FALSE means that a list will be returned.
Assuming you want all possible combinations of 8 classrooms from the dataset classrooms
combinations <- combn(nrow(classrooms), 8, function(x,data) data[x,],
simplify = FALSE, data =classrooms )
head(combinations, n = 2)
[[1]]
classrooms students
1 1 29
2 2 30
3 3 35
4 4 28
5 5 32
6 6 20
7 7 25
8 8 22
[[2]]
classrooms students
1 1 29
2 2 30
3 3 35
4 4 28
5 5 32
6 6 20
7 7 25
9 9 32