Finding values in consecutive rows - r

An example of the dataframe I have is given below.
ID X
1 1
2 2
3 1
4 0
5 0
6 1
7 4
8 5
9 6
10 7
11 0
12 0
I want to apply logic to it that looks to see whether 3 or more consecutive rows have a value >0 in it. If they do I want to flag them in another column. Hence the output will look as follows.
ID X Y
1 1 1
2 2 1
3 1 1
4 0 0
5 0 0
6 1 1
7 4 1
8 5 1
9 6 1
10 7 1
11 0 0
12 0 0
EXTENSION -
How would I get the following output, givibng a different Y value for each group?
ID X Y
1 1 1
2 2 1
3 1 1
4 0 0
5 0 0
6 1 2
7 4 2
8 5 2
9 6 2
10 7 2
11 0 0
12 0 0

One option with base R. Using rle to find the adjacent values in 'X' that are greater than 0, then do the replication based on the lengths
df1$Y <- with(rle(df1$X > 0), as.integer(rep(values & lengths > 2, lengths)))
df1$Y
#[1] 1 1 1 0 0 1 1 1 1 1 0 0
For the updated case in the OP's post
df1$Y <- inverse.rle(within.list(rle(df1$X > 0), {
i1 <- values & (lengths > 2)
values[i1] <- seq_along(values[i1])}))
df1$Y
#[1] 1 1 1 0 0 2 2 2 2 2 0 0
Or using rleid from data.table
library(data.table)
setDT(df1)[, Y := as.integer((.N > 2) * (X > 0)),rleid(X > 0)]
data
df1 <- structure(list(ID = 1:12, X = c(1L, 2L, 1L, 0L, 0L, 1L, 4L, 5L,
6L, 7L, 0L, 0L)), class = "data.frame", row.names = c(NA, -12L
))

We can use rleid from data.table to create groups and use it in ave and get length of each group and assign 1 to groups which has length greater than equal to 3.
library(data.table)
df$Y <- as.integer(ave(df$X, rleid(df$X > 0), FUN = length) >= 3)
df
# ID X Y
#1 1 1 1
#2 2 2 1
#3 3 1 1
#4 4 0 0
#5 5 0 0
#6 6 1 1
#7 7 4 1
#8 8 5 1
#9 9 6 1
#10 10 7 1
#11 11 0 0
#12 12 0 0
EDIT
For updated post we could include the above data.table part with dplyr by doing
library(dplyr)
library(data.table)
df %>%
group_by(group = rleid(X > 0)) %>%
mutate(Y = ifelse(n() >= 3 & row_number() == 1, 1, 0)) %>%
ungroup() %>%
mutate(Y = cumsum(Y) * Y) %>%
group_by(group) %>%
mutate(Y = first(Y)) %>%
ungroup() %>%
select(-group)
# ID X Y
# <int> <int> <dbl>
# 1 1 1 1
# 2 2 2 1
# 3 3 1 1
# 4 4 0 0
# 5 5 0 0
# 6 6 1 2
# 7 7 4 2
# 8 8 5 2
# 9 9 6 2
#10 10 7 2
#11 11 0 0
#12 12 0 0

Related

Counting Frequencies of Sequences

Suppose there are two students - each student takes an exam multiple times (e.g.result_id = 1 is the first exam, result_id = 2 is the second exam, etc.). The student can either "pass" (1) or "fail" (0).
The data looks something like this:
library(data.table)
my_data = data.frame(id = c(1,1,1,1,1,1,2,2,2,2,2,2,2,2,2), results = c(0,1,0,1,0,0,1,1,1,0,1,1,0,1,0), result_id = c(1,2,3,4,5,6,1,2,3,4,5,6,7,8,9))
my_data = setDT(my_data)
id results result_id
1: 1 0 1
2: 1 1 2
3: 1 0 3
4: 1 1 4
5: 1 0 5
6: 1 0 6
7: 2 1 1
8: 2 1 2
9: 2 1 3
10: 2 0 4
11: 2 1 5
12: 2 1 6
13: 2 0 7
14: 2 1 8
15: 2 0 9
I am interested in counting the number of times that a student passes an exam, given that the student passed the previous two exams.
I tried to do this with the following code:
my_data$current_exam = shift(my_data$results, 0)
my_data$prev_exam = shift(my_data$results, 1)
my_data$prev_2_exam = shift(my_data$results, 2)
# Count the number of exam results for each record
out <- my_data[!is.na(prev_exam), .(tally = .N), by = .(id, current_exam, prev_exam, prev_2_exam)]
out = na.omit(out)
My code produces the following results:
> out
id current_exam prev_exam prev_2_exam tally
1: 1 0 1 0 2
2: 1 1 0 1 1
3: 1 0 0 1 1
4: 2 1 0 0 1
5: 2 1 1 0 2
6: 2 1 1 1 1
7: 2 0 1 1 2
8: 2 1 0 1 2
9: 2 0 1 0 1
However, I do not think that my code is correct.
For example, with Student_ID = 2 :
My code says that "Current_Exam = 1, Prev_Exam = 1, Prev_2_Exam = 0" happens 1 time, but looking at the actual data - this does not happen at all
Can someone please show me what I am doing wrong and how I can correct this?
Note: I think that this should be the expected output:
> expected_output
id current_exam prev_exam prev_2_exam tally
1: 1 0 1 0 2
2: 1 1 0 1 1
3: 1 0 0 1 1
4: 2 1 0 0 1
5: 2 1 1 0 1
6: 2 1 1 1 1
7: 2 0 1 1 2
8: 2 1 0 1 2
9: 2 0 1 0 0
You did not consider that you can not shift the results over id without placing NA.
. <- my_data[order(my_data$id, my_data$result_id),] #sort if needed
.$p1 <- ave(.$results, .$id, FUN = \(x) c(NA, x[-length(x)]))
.$p2 <- ave(.$p1, .$id, FUN = \(x) c(NA, x[-length(x)]))
aggregate(list(tally=.$p1), .[c("id","results", "p1", "p2")], length)
# id results p1 p2 tally
#1 1 0 1 0 2
#2 2 0 1 0 1
#3 2 1 1 0 1
#4 1 0 0 1 1
#5 1 1 0 1 1
#6 2 1 0 1 2
#7 2 0 1 1 2
#8 2 1 1 1 1
.
# id results result_id p1 p2
#1 1 0 1 NA NA
#2 1 1 2 0 NA
#3 1 0 3 1 0
#4 1 1 4 0 1
#5 1 0 5 1 0
#6 1 0 6 0 1
#7 2 1 1 NA NA
#8 2 1 2 1 NA
#9 2 1 3 1 1
#10 2 0 4 1 1
#11 2 1 5 0 1
#12 2 1 6 1 0
#13 2 0 7 1 1
#14 2 1 8 0 1
#15 2 0 9 1 0
An option would be to use filter to indicate those which had passed 3 times in a row.
cbind(., n=ave(.$results, .$id, FUN = \(x) filter(x, c(1,1,1), sides=1)))
# id results result_id n
#1 1 0 1 NA
#2 1 1 2 NA
#3 1 0 3 1
#4 1 1 4 2
#5 1 0 5 1
#6 1 0 6 1
#7 2 1 1 NA
#8 2 1 2 NA
#9 2 1 3 3
#10 2 0 4 2
#11 2 1 5 2
#12 2 1 6 2
#13 2 0 7 2
#14 2 1 8 2
#15 2 0 9 1
If olny the number of times that a student passes an exam, given that the student passed the previous two exams:
sum(ave(.$results, .$id, FUN = \(x) filter(x, c(1,1,1))==3), na.rm=TRUE)
#[1] 1
sum(ave(.$results, .$id, FUN = \(x)
x==1 & c(x[-1], 0) == 1 & c(x[-1:-2], 0, 0) == 1))
#[1] 1
When trying to count events that happen in series, cumsum() comes in quite handy. As opposed to creating multiple lagged variables, this scales well to counts across a larger number of events:
library(tidyverse)
d <- my_data |>
group_by(id) |> # group to cumulate within student only
mutate(
csum = cumsum(results), # cumulative sum of results
i = csum - lag(csum, 3, 0) # substract the cumulative sum from 3 observation before. This gives the number of exams passed in the current and previous 2 observations.
)
# Ungroup to get global count
d |>
ungroup() |>
count(i == 3) # Count the number of cases where the number of exams passes within 3 observations equals 3
#> # A tibble: 2 × 2
#> `i == 3` n
#> <lgl> <int>
#> 1 FALSE 14
#> 2 TRUE 1
# Retaining the group gives counts by student
d |>
count(i == 3) # Count the number of cases where the number of exams passes within 3 observations equals 3
#> # A tibble: 3 × 3
#> # Groups: id [2]
#> id `i == 3` n
#> <dbl> <lgl> <int>
#> 1 1 FALSE 6
#> 2 2 FALSE 8
#> 3 2 TRUE 1
Since you provided the data as data.table, here is how to do the same in that ecosystem:
my_data[ , csum := cumsum(results), .(id)]
my_data[ , i := csum - lag(csum, 3, 0), .(id)]
my_data[ , .(n_cases = sum(i ==3)), id]
#> id n_cases
#> 1: 1 0
#> 2: 2 1
Here's an approach using dplyr. It uses the lag function to look back 1 and 2 results. If the sum together with the current result is 3, then the condition is met. In the example you provided, the condition is only met once
my_data %>%
group_by(id) %>%
mutate(threex = ifelse(results + lag(results,1) + lag(results, 2) == 3, 1, 0)) %>%
filter(!is.na(threex))
id results result_id threex
<dbl> <dbl> <dbl> <dbl>
1 1 0 3 0
2 1 1 4 0
3 1 0 5 0
4 1 0 6 0
5 2 1 3 1
6 2 0 4 0
7 2 1 5 0
8 2 1 6 0
9 2 0 7 0
10 2 1 8 0
11 2 0 9 0
If you then just want to capture the cases when the condition is met, add a filter.
my_data %>%
group_by(id) %>%
mutate(threex = ifelse(results + lag(results,1) + lag(results, 2) == 3, 1, 0)) %>%
filter(threex == 1)
id results result_id threex
<dbl> <dbl> <dbl> <dbl>
1 2 1 3 1
If you are looking to understand how many times the condition is met per id, you can do this.
my_data %>%
group_by(id) %>%
mutate(threex = ifelse(results + lag(results,1) + lag(results, 2) == 3, 1, 0)) %>%
filter(threex == 1) %>%
select(id) %>%
summarize(count = n())
id count
<dbl> <int>
1 2 1

Create a new dataframe with 1's and 0's from summarized data?

I have the below dataset that I am working with in R:
df <- data.frame(day=seq(1,3,1), tot.infected=c(1,2,4), tot.ind=5)
df
And I would like to transform the tot.infected column into a binomial variable with 1's and 0's, such as the following dataframe:
df2 <- data.frame(year = c(rep(1,5), rep(2,5), rep(3,5)), infected = c(rep(1,1), rep(0,4), rep(1,2), rep(0,3), rep(1,4), rep(0,1)))
Is there a more elegant way to do this in R?
Thank you for your help!
I tried hard-coding a dataframe using rep(), but this is extremely time-consuming for large datasets and I was looking for a more elegant way to achieve this.
base R
tmp <- do.call(Map, c(list(f = function(y, inf, ind) data.frame(year = y, infected = replace(integer(ind), seq(ind) <= inf, 1L))), unname(df)))
do.call(rbind, tmp)
# year infected
# 1 1 1
# 2 1 0
# 3 1 0
# 4 1 0
# 5 1 0
# 6 2 1
# 7 2 1
# 8 2 0
# 9 2 0
# 10 2 0
# 11 3 1
# 12 3 1
# 13 3 1
# 14 3 1
# 15 3 0
dplyr
library(dplyr)
df %>%
rowwise() %>%
summarize(tibble(year = day, infected = replace(integer(tot.ind), seq(tot.ind) <= tot.infected, 1L)))
# # A tibble: 15 x 2
# year infected
# <dbl> <int>
# 1 1 1
# 2 1 0
# 3 1 0
# 4 1 0
# 5 1 0
# 6 2 1
# 7 2 1
# 8 2 0
# 9 2 0
# 10 2 0
# 11 3 1
# 12 3 1
# 13 3 1
# 14 3 1
# 15 3 0
We can do it this way:
library(dplyr)
df %>%
group_by(day) %>%
summarise(cur_data()[seq(unique(tot.ind)),]) %>%
#mutate(x = row_number())
mutate(tot.infected = ifelse(row_number() <= first(tot.infected),
first(tot.infected)/first(tot.infected), 0), .keep="used")
day tot.infected
<dbl> <dbl>
1 1 1
2 1 0
3 1 0
4 1 0
5 1 0
6 2 1
7 2 1
8 2 0
9 2 0
10 2 0
11 3 1
12 3 1
13 3 1
14 3 1
15 3 0
Using rep.int and replace, basically.
with(df, data.frame(
year=do.call(rep.int, unname(df[c(1, 3)])),
infected=unlist(Map(replace, Map(rep.int, 0, tot.ind), lapply(tot.infected, seq), 1))
))
# year infected
# 1 1 1
# 2 1 0
# 3 1 0
# 4 1 0
# 5 1 0
# 6 2 1
# 7 2 1
# 8 2 0
# 9 2 0
# 10 2 0
# 11 3 1
# 12 3 1
# 13 3 1
# 14 3 1
# 15 3 0
Data:
df <- structure(list(day = c(1, 2, 3), tot.infected = c(1, 2, 4), tot.ind = c(5,
5, 5)), class = "data.frame", row.names = c(NA, -3L))

How to flag non-sequential numbers in R

I was wondering if there's a way to flag non-consequential numbers? For instance below.
Number
3
4
5
6
10
11
12
16
Is there a way to flag the number before the non-sequential number like so?
Number Flag
3 0
4 0
5 0
6 1
10 0
11 0
12 1
16
etc..
Thank you!
dat$Flag <- +c(diff(dat$Number) != 1, NA)
dat
# Number Flag
# 1 3 0
# 2 4 0
# 3 5 0
# 4 6 1
# 5 10 0
# 6 11 0
# 7 12 1
# 8 16 NA
I would use lead() here:
df$Flag <- as.numeric(lead(df$Number) != df$Number + 1)
df
Number Flag
1 3 0
2 4 0
3 5 0
4 6 1
5 10 0
6 11 0
7 12 1
8 16 NA
Here's another option with data.table:
library(data.table)
setDT(df)[, Flag := +(shift(Number, type = "lead") - (Number + 1) != 0)]
Output
Number Flag
<int> <int>
1: 3 0
2: 4 0
3: 5 0
4: 6 1
5: 10 0
6: 11 0
7: 12 1
8: 16 NA
Or another option with dplyr:
library(dplyr)
df %>%
mutate(Flag = +(lead(Number) - (Number + 1) != 0))
Data
df <- structure(list(Number = c(3L, 4L, 5L, 6L, 10L, 11L, 12L, 16L)), class = "data.frame", row.names = c(NA,
-8L))

Use R to find values for which a condition is first met

Consider the following sample dataset. Id is an individual identifier.
rm(list=ls()); set.seed(1)
n<-100
X<-rbinom(n, 1, 0.5) #binary covariate
j<-rep (1:n)
dat<-data.frame(id=1:n, X)
ntp<- rep(4, n)
mat<-matrix(ncol=3,nrow=1)
m=0; w <- mat
for(l in ntp)
{
m=m+1
ft<- seq(from = 2, to = 8, length.out = l)
# ft<- seq(from = 1, to = 9, length.out = l)
ft<-sort(ft)
seq<-rep(ft,each=2)
seq<-c(0,seq,10)
matid<-cbind( matrix(seq,ncol=2,nrow=l+1,byrow=T ) ,m)
w<-rbind(w,matid)
}
d<-data.frame(w[-1,])
colnames(d)<-c("time1","time2","id")
D <- round( merge(d,dat,by="id") ,2) #merging dataset
nr<-nrow(D)
D$Survival_time<-round(rexp(nr, 0.1)+1,3)
head(D,15)
id time1 time2 X Survival_time
1 1 0 2 0 21.341
2 1 2 4 0 18.987
3 1 4 6 0 4.740
4 1 6 8 0 13.296
5 1 8 10 0 6.397
6 2 0 2 0 10.566
7 2 2 4 0 2.470
8 2 4 6 0 14.907
9 2 6 8 0 8.620
10 2 8 10 0 13.376
11 3 0 2 1 45.239
12 3 2 4 1 11.545
13 3 4 6 1 11.352
14 3 6 8 1 19.760
15 3 8 10 1 7.547
How can I obtain the value at which Survival_time is less that time2 for the very first time per individual. I should end up with the following values
id Survival_time
1 4.740
2 2.470
3 7.547
Also, how can I subset the data to stop individualwise when this condition occurs. i.e obtain
id time1 time2 X Survival_time
1 1 0 2 0 21.341
2 1 2 4 0 18.987
3 1 4 6 0 4.740
6 2 0 2 0 10.566
7 2 2 4 0 2.470
11 3 0 2 1 45.239
12 3 2 4 1 11.545
13 3 4 6 1 11.352
14 3 6 8 1 19.760
15 3 8 10 1 7.547
Using data.table
library(data.table)
setDT(D)[, .SD[seq_len(.N) <= which(Survival_time < time2)[1]], id]
-output
id time1 time2 X Survival_time
1: 1 0 2 0 21.341
2: 1 2 4 0 18.987
3: 1 4 6 0 4.740
4: 2 0 2 0 10.566
5: 2 2 4 0 2.470
6: 3 0 2 1 45.239
7: 3 2 4 1 11.545
8: 3 4 6 1 11.352
9: 3 6 8 1 19.760
10: 3 8 10 1 7.547
Slight variation:
library(dplyr)
D %>% # Take D, and then
group_by(id) %>% # group by id, and then
filter(Survival_time < time2) %>% # keep Survival times < time2, and then
slice(1) %>% # keep the first row per id, and then
ungroup() # ungroup
You can use -
library(dplyr)
D %>%
group_by(id) %>%
summarise(Survival_time = Survival_time[match(TRUE, Survival_time < time2)])
#Also using which.max
#summarise(Survival_time = Survival_time[which.max(Survival_time < time2)])
# id Survival_time
# <int> <dbl>
#1 1 4.74
#2 2 2.47
#3 3 7.55
To select the rows you may till that point you may use -
D %>%
group_by(id) %>%
filter(row_number() <= match(TRUE, Survival_time < time2)) %>%
ungroup
# id time1 time2 X Survival_time
# <int> <int> <int> <int> <dbl>
# 1 1 0 2 0 21.3
# 2 1 2 4 0 19.0
# 3 1 4 6 0 4.74
# 4 2 0 2 0 10.6
# 5 2 2 4 0 2.47
# 6 3 0 2 1 45.2
# 7 3 2 4 1 11.5
# 8 3 4 6 1 11.4
# 9 3 6 8 1 19.8
#10 3 8 10 1 7.55

grouping continuous data with specific pattern

I have a data frame with a column like this (I am not posting other columns)
Value
1
1
1
0
0
1
0
0
1
1
2
2
0
0
1
0
0
1
1
1
0
0
2
2
1
1
2
0
0
1
0
I am trying to group it based on a specific condition. Grouping has to be done when I have 1 and 2. But conditions like these are one group :
1 1 0 0 1 1 0 0
Basically I need to group occurrences of 1 but in between 0s are allowed
Expected output:
Value Group
1 1
1 1
1 1
0 1
0 1
1 1
0 1
0 1
1 1
1 1
2 2
2 2
0 2
0 2
1 3
0 3
0 3
1 3
1 3
1 3
0 3
0 3
2 4
2 4
1 5
1 5
2 6
0 6
0 6
1 7
0 7
2 8
0 8
2 8
1 9
Here is another option using data.table:
DT[, Group := .GRP, .(date, rleid(nafill(replace(Value, Value==0L, NA_integer_), "locf")))]
Here is another base approach that uses ave() to count the changes between 1 and 2 and then uses cummax() on the result to give the final groupings.
dat$Group <- cummax(ave(dat$Value, dat$Value == 0, FUN = function(x) cumsum(c(x[1], diff(x) != 0))))
dat
Value Group
1 1 1
2 1 1
3 1 1
4 0 1
5 0 1
6 1 1
7 0 1
8 0 1
9 1 1
10 1 1
11 2 2
12 2 2
13 0 2
14 0 2
15 1 3
16 0 3
17 0 3
18 1 3
19 1 3
20 1 3
21 0 3
22 0 3
23 2 4
24 2 4
25 1 5
26 1 5
27 2 6
28 0 6
29 0 6
30 1 7
31 0 7
In response to your comment, if you want the result by grouped by date, you can use a nested ave():
ave(ave(dat$Value, dat$Value == 0, dat$date, FUN = function(x) cumsum(c(x[1], diff(x) != 0))), dat$date, FUN = cummax)
This loop in Base-R does the trick
group <- 0
lastgroupvalue <- NA
data$Group <- NA
for(i in 1:nrow(data)){
if(!data$Value[i] %in% c(lastgroupvalue, 0)){
group <- group + 1
lastgroupvalue <- data$Value[i]
}
data$Group[i] <- group
}
> data
Value Group
1 1 1
2 1 1
3 1 1
4 0 1
5 0 1
6 1 1
7 0 1
8 0 1
9 1 1
10 1 1
11 2 2
12 2 2
13 0 2
14 0 2
15 1 3
16 0 3
17 0 3
18 1 3
19 1 3
20 1 3
21 0 3
22 0 3
23 2 4
24 2 4
25 1 5
26 1 5
27 2 6
28 0 6
29 0 6
30 1 7
31 0 7
Data:
data <- structure(list(Value = c(1L, 1L, 1L, 0L, 0L, 1L, 0L, 0L, 1L,
1L, 2L, 2L, 0L, 0L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 2L, 2L, 1L,
1L, 2L, 0L, 0L, 1L, 0L)), class = "data.frame", row.names = c(NA,
-31L))
Another solution that avoids loops, that works similar to Limey's solution but uses cumsum to create the groups.
df$Group <- dplyr::na_if(df$Value, 0)
df <- tidyr::fill(df, Group, .direction = "down")
df$Group <- cumsum(df$Group != dplyr::lag(df$Group, default = -1))
> df
Value Group
1 1 1
2 1 1
3 1 1
4 0 1
5 0 1
6 1 1
7 0 1
8 0 1
9 1 1
10 1 1
11 2 2
12 2 2
13 0 2
14 0 2
15 1 3
16 0 3
17 0 3
18 1 3
19 1 3
20 1 3
21 0 3
22 0 3
23 2 4
24 2 4
25 1 5
26 1 5
27 2 6
28 0 6
29 0 6
30 1 7
31 0 7
Or a tidyverse solution that avoids loops:
x <- tibble(Value=c(1,1,1,0,0,1,0,0,1,1,2,2,0,0,1,0,0,1,1,1,
0,0,2,2,1,1,2,0,0,1,0,2,0,2,1)) %>%
mutate(ModValue=ifelse(Value == 0, NA, Value)) %>%
fill(ModValue, .direction="down")
runLengths <- rle(x$ModValue)
groupIndex <- unlist(lapply(1:length(runLengths$lengths),
function(x) rep(x, runLengths$lengths[x]))
)
x <- x %>% add_column(Group=groupIndex) %>% select(-ModValue)
Your input data has a different length to your expected output. Took me a while to work that out... :)
** Edit **
And an inelegant solution to account for changing days (or other super-groupings...
x <- tibble(
RowNumber=1:35,
Date=lubridate::ymd(c(rep("2020-05-31", 20), rep("2020-06-01", 15))),
Value=c(1,1,1,0,0,1,0,0,1,1,2,2,0,0,1,0,0,1,1,1,0,0,2,2,1,1,2,0,0,1,0,2,0,2,1))
# Check we have a change of date mid-sequence
x %>% filter(row_number() > 15 & row_number() < 25)
x <- x %>%
mutate(ModValue=ifelse(Value == 0, NA, Value)) %>%
fill(ModValue, .direction="down")
# Inelegantly compute the groups
make_groups <- function(x) {
runs <- rle(x)
return(tibble(GroupWithinDay=unlist(
lapply(1:length(runs$lengths),
function(x) rep(x, runs$lengths[x])))))
}
y <- x %>% group_by(Date) %>% do(make_groups(.$ModValue))
x <- x %>% add_column(GroupWithinDay=y$GroupWithinDay) %>% select(-ModValue)
# Check the change of date is handled correctly
x %>% filter(row_number() > 15 & row_number() < 25)
Giving
# A tibble: 9 x 4
RowNumber Date Value GroupWithinDay
<int> <date> <dbl> <int>
1 16 2020-05-31 0 3
2 17 2020-05-31 0 3
3 18 2020-05-31 1 3
4 19 2020-05-31 1 3
5 20 2020-05-31 1 3
6 21 2020-06-01 0 1
7 22 2020-06-01 0 1
8 23 2020-06-01 2 2
9 24 2020-06-01 2 2

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