From a data frame as month and a continuous number id
data.frame( id = c(1,1,2,2,3), month = c("2011-02","2011-02","2011-02","2011-03","2011-01"))
How is it possible to extract the total number of month and have the frq for every id?
Expected output:
data.frame(id = c(1,2,2,3), month = c("2011-02","2011-02","2011-03","2011-01"), frq = c(1,1,2,1)
one option is:
df$freq <- 1
aggregate(freq ~ id + month, df, sum)
id month freq
1 3 2011-01 1
2 1 2011-02 2
3 2 2011-02 1
4 2 2011-03 1
An idea via dplyr can be,
library(dplyr)
distinct(df) %>%
group_by(id) %>%
mutate(freq = seq(n()))
# A tibble: 4 x 3
# Groups: id [3]
# id month freq
# <dbl> <fct> <int>
#1 1 2011-02 1
#2 2 2011-02 1
#3 2 2011-03 2
#4 3 2011-01 1
Related
I collected some data that is different in unique in year, month, and level. I want to assign a unique code (simple numerics) to each row on these three columns alone. Any suggestions on how to proceed?
year <- c("A","J","J","S")
month <- c(2000,2001,2001,2000)
level <- c("high","low","low","low")
site <- c(1,2,3,3)
val1 <- c(1,2,3,0)
df <- data.frame(year,month,level,site,val1)
#Result desired
df$Unique.code --> 1,2,2,3
dplyr has the cur_group_id() function for this:
df %>%
group_by(year, month, level) %>%
mutate(id = cur_group_id())
# # A tibble: 4 × 6
# # Groups: year, month, level [3]
# year month level site val1 id
# <chr> <dbl> <chr> <dbl> <dbl> <int>
# 1 A 2000 high 1 1 1
# 2 J 2001 low 2 2 2
# 3 J 2001 low 3 3 2
# 4 S 2000 low 3 0 3
Or we could coerce a factor into an integer in base:
df$group_id = with(df, as.integer(factor(paste(year, month, level))))
If I had:
person_ID visit date
1 2/25/2001
1 2/27/2001
1 4/2/2001
2 3/18/2004
3 9/22/2004
3 10/27/2004
3 5/15/2008
and I wanted another column to indicate the earliest recurring observation within 90 days, grouped by patient ID, with the desired output:
person_ID visit date date
1 2/25/2001 2/27/2001
1 2/27/2001 4/2/2001
1 4/2/2001 NA
2 3/18/2004 NA
3 9/22/2004 10/27/2004
3 10/27/2004 NA
3 5/15/2008 NA
Thank you!
We convert the 'visit_date' to Date class, grouped by 'person_ID', create a binary column that returns 1 if the difference between the current and next visit_date is less than 90 or else 0, using this column, get the correponding next visit_date' where the value is 1
library(dplyr)
library(lubridate)
library(tidyr)
df1 %>%
mutate(visit_date = mdy(visit_date)) %>%
group_by(person_ID) %>%
mutate(i1 = replace_na(+(difftime(lead(visit_date),
visit_date, units = 'day') < 90), 0),
date = case_when(as.logical(i1)~ lead(visit_date)), i1 = NULL ) %>%
ungroup
-output
# A tibble: 7 x 3
# person_ID visit_date date
# <int> <date> <date>
#1 1 2001-02-25 2001-02-27
#2 1 2001-02-27 2001-04-02
#3 1 2001-04-02 NA
#4 2 2004-03-18 NA
#5 3 2004-09-22 2004-10-27
#6 3 2004-10-27 NA
#7 3 2008-05-15 NA
I have dataframe df1 containing data and groups, and df2 which stores the same groups, and one value per group.
I want to filter rows of df1 by df2 where lag by group is higher than indicated value.
Dummy example:
# identify the first year of disturbance by lag by group
df1 <- data.frame(year = c(1:4, 1:4),
mort = c(5,16,40,4,5,6,10,108),
distance = rep(c("a", "b"), each = 4))
df2 = data.frame(distance = c("a", "b"),
my.median = c(12,1))
Now calculate the lag between values (creates new column) and filter df1 based on column values of df2:
# calculate lag between years
df1 %>%
group_by(distance) %>%
dplyr::mutate(yearLag = mort - lag(mort, default = 0)) %>%
filter(yearLag > df2$my.median) ##
This however does not produce expected results:
# A tibble: 3 x 4
# Groups: distance [2]
year mort distance yearLag
<int> <dbl> <fct> <dbl>
1 2 16 a 11
2 3 40 a 24
3 4 108 b 98
Instead, I expect to get:
# A tibble: 3 x 4
# Groups: distance [2]
year mort distance yearLag
<int> <dbl> <fct> <dbl>
1 3 40 a 24
2 1 5 b 5
3 3 10 b 4
The filter works great while applied to single value, but how to adapt it to vector, and especially vector of groups (as the order of elements can potentially change?)
Is this what you're trying to do?
df1 %>%
group_by(distance) %>%
dplyr::mutate(yearLag = mort - lag(mort, default = 0)) %>%
left_join(df2) %>%
filter(yearLag > my.median)
Result:
# A tibble: 4 x 5
# Groups: distance [2]
year mort distance yearLag my.median
<int> <dbl> <fct> <dbl> <dbl>
1 3 40 a 24 12
2 1 5 b 5 1
3 3 10 b 4 1
4 4 108 b 98 1
here is a data.table approach
library( data.table )
#creatae data.tables
setDT(df1);setDT(df2)
#create yearLag variable
df1[, yearLag := mort - shift( mort, type = "lag", fill = 0 ), by = .(distance) ]
#update join and filter wanted rows
df1[ df2, median.value := i.my.median, on = .(distance)][ yearLag > median.value, ][]
# year mort distance yearLag median.value
# 1: 3 40 a 24 12
# 2: 1 5 b 5 1
# 3: 3 10 b 4 1
# 4: 4 108 b 98 1
Came to the same conclusion. You should left_join the data frames.
df1 %>% left_join(df2, by="distance") %>%
group_by(distance) %>%
dplyr::mutate(yearLag = mort - lag(mort, default = 0)) %>%
filter(yearLag > my.median)
# A tibble: 4 x 5
# Groups: distance [2]
year mort distance my.median yearLag
<int> <dbl> <fct> <dbl> <dbl>
1 3 40 a 12 24
2 1 5 b 1 5
3 3 10 b 1 4
4 4 108 b 1 98
The dataframe df1 summarizes detections of individuals (ID) through the time (Date). As a short example:
df1<- data.frame(ID= c(1,2,1,2,1,2,1,2,1,2),
Date= ymd(c("2016-08-21","2016-08-24","2016-08-23","2016-08-29","2016-08-27","2016-09-02","2016-09-01","2016-09-09","2016-09-01","2016-09-10")))
df1
ID Date
1 1 2016-08-21
2 2 2016-08-24
3 1 2016-08-23
4 2 2016-08-29
5 1 2016-08-27
6 2 2016-09-02
7 1 2016-09-01
8 2 2016-09-09
9 1 2016-09-01
10 2 2016-09-10
I want to summarize either the Number of days since the first detection of the individual (Ndays) and Number of days that the individual has been detected since the first time it was detected (Ndifdays).
Additionally, I would like to include in this summary table a variable called Prop that simply divides Ndifdays between Ndays.
The summary table that I would expect would be this:
> Result
ID Ndays Ndifdays Prop
1 1 11 4 0.360 # Between 21st Aug and 01st Sept there is 11 days.
2 2 17 5 0.294 # Between 24th Aug and 10st Sept there is 17 days.
Does anyone know how to do it?
You could achieve using various summarising functions in dplyr
library(dplyr)
df1 %>%
group_by(ID) %>%
summarise(Ndays = as.integer(max(Date) - min(Date)),
Ndifdays = n_distinct(Date),
Prop = Ndifdays/Ndays)
# ID Ndays Ndifdays Prop
# <dbl> <int> <int> <dbl>
#1 1 11 4 0.364
#2 2 17 5 0.294
The data.table version of this would be
library(data.table)
df12 <- setDT(df1)[, .(Ndays = as.integer(max(Date) - min(Date)),
Ndifdays = uniqueN(Date)), by = ID]
df12$Prop <- df12$Ndifdays/df12$Ndays
and base R with aggregate
df12 <- aggregate(Date~ID, df1, function(x) c(max(x) - min(x), length(unique(x))))
df12$Prop <- df1$Ndifdays/df1$Ndays
After grouping by 'ID', get the diff or range of 'Date' to create 'Ndays', and then get the unique number of 'Date' with n_distinct, divide by the number of distinct by the Ndays to get the 'Prop'
library(dplyr)
df1 %>%
group_by(ID) %>%
summarise(Ndays = as.integer(diff(range(Date))),
Ndifdays = n_distinct(Date),
Prop = Ndifdays/Ndays)
# A tibble: 2 x 4
# ID Ndays Ndifdays Prop
# <dbl> <int> <int> <dbl>
#1 1 11 4 0.364
#2 2 17 5 0.294
I have a data in which I have 2 fields in a table sf -> Customer id and Buy_date. Buy_date is unique but for each customer, but there can be more than 3 different values of Buy_dates for each customer. I want to calculate difference in consecutive Buy_date for each Customer and its mean value. How can I do this.
Example
Customer Buy_date
1 2018/03/01
1 2018/03/19
1 2018/04/3
1 2018/05/10
2 2018/01/02
2 2018/02/10
2 2018/04/13
I want the results for each customer in the format
Customer mean
Here's a dplyr solution.
Your data:
df <- data.frame(Customer = c(1,1,1,1,2,2,2), Buy_date = c("2018/03/01", "2018/03/19", "2018/04/3", "2018/05/10", "2018/01/02", "2018/02/10", "2018/04/13"))
Grouping, mean Buy_date calculation and summarising:
library(dplyr)
df %>% group_by(Customer) %>% mutate(mean = mean(as.POSIXct(Buy_date))) %>% group_by(Customer, mean) %>% summarise()
Output:
# A tibble: 2 x 2
# Groups: Customer [?]
Customer mean
<dbl> <dttm>
1 1 2018-03-31 06:30:00
2 2 2018-02-17 15:40:00
Or as #r2evans points out in his comment for the consecutive days between Buy_dates:
df %>% group_by(Customer) %>% mutate(mean = mean(diff(as.POSIXct(Buy_date)))) %>% group_by(Customer, mean) %>% summarise()
Output:
# A tibble: 2 x 2
# Groups: Customer [?]
Customer mean
<dbl> <time>
1 1 23.3194444444444
2 2 50.4791666666667
I am not exactly sure of the desired output but this what I think you want.
library(dplyr)
library(zoo)
dat <- read.table(text =
"Customer Buy_date
1 2018/03/01
1 2018/03/19
1 2018/04/3
1 2018/05/10
2 2018/01/02
2 2018/02/10
2 2018/04/13", header = T, stringsAsFactors = F)
dat$Buy_date <- as.Date(dat$Buy_date)
dat %>% group_by(Customer) %>% mutate(diff_between = as.vector(diff(zoo(Buy_date), na.pad=TRUE)),
mean_days = mean(diff_between, na.rm = TRUE))
This produces:
Customer Buy_date diff_between mean_days
<int> <date> <dbl> <dbl>
1 1 2018-03-01 NA 23.3
2 1 2018-03-19 18 23.3
3 1 2018-04-03 15 23.3
4 1 2018-05-10 37 23.3
5 2 2018-01-02 NA 50.5
6 2 2018-02-10 39 50.5
7 2 2018-04-13 62 50.5
EDITED BASED ON USER COMMENTS:
Because you said that you have factors and not characters just convert them by doing the following:
dat$Buy_date <- as.Date(as.character(dat$Buy_date))
dat$Customer <- as.character(dat$Customer)