This question already has an answer here:
Aggregating across list of dataframes and storing all results
(1 answer)
Closed 3 years ago.
I am working on a script where I have two lists and I am trying to combine the results so I get a new list. Each list has a date and then two numbers. The lists look like this:
date clicks impressions
1 2019-06-01 1 2
2 2019-06-02 0 0
3 2019-06-03 100 120
and
date clicks impressions
1 2019-06-01 2 14
2 2019-06-02 3 14
3 2019-06-03 11 29
I'd like a single list that is
date clicks impressions
1 2019-06-01 3 16
2 2019-06-02 3 14
3 2019-06-03 111 149
What is the best way to accomplish this. In time I will have 20 - 30 more lists that will be added to this, so I'll want to pull the first list and then combine with the second and then a third and so on. I don't know if I'll be able to assume that each date will be in each list.
Assuming your list is called list_df, you can bind them all together using bind_rows, group_by date and then sum all the other columns.
library(dplyr)
list_df %>%
bind_rows() %>%
group_by(date) %>%
summarise_all(sum)
# A tibble: 3 x 3
# date clicks impressions
# <fct> <int> <int>
#1 2019-06-01 3 16
#2 2019-06-02 3 14
#3 2019-06-03 111 149
which in base R could be achieved using Reduce
aggregate(.~date, Reduce(rbind, list_df), sum)
We can use data.table
library(data.table)
rbindlist(list_df)[, lapply(.SD, sum), date]
# date clicks impressions
#1: 2019-06-01 3 16
#2: 2019-06-02 3 14
#3: 2019-06-03 111 149
data
list_df <- mget(paste0("df", 1:2))
We can do:
cbind(date=df1[,1],do.call(`+`, list(df1[,-1],df2[,-1])),
row.names = NULL)
date clicks impressions
1 2019-06-01 3 16
2 2019-06-02 3 14
3 2019-06-03 111 149
If you are not sure about the presence of dates(can then cbind as above):
do.call(`+`,lapply(list(df1,df2), function(x) x[,-1]))
clicks impressions
1 3 16
2 3 14
3 111 149
This assumes that the data sets will have the same structure always.
Related
This question already has an answer here:
Aggregating across list of dataframes and storing all results
(1 answer)
Closed 3 years ago.
I am working on a script where I have two lists and I am trying to combine the results so I get a new list. Each list has a date and then two numbers. The lists look like this:
date clicks impressions
1 2019-06-01 1 2
2 2019-06-02 0 0
3 2019-06-03 100 120
and
date clicks impressions
1 2019-06-01 2 14
2 2019-06-02 3 14
3 2019-06-03 11 29
I'd like a single list that is
date clicks impressions
1 2019-06-01 3 16
2 2019-06-02 3 14
3 2019-06-03 111 149
What is the best way to accomplish this. In time I will have 20 - 30 more lists that will be added to this, so I'll want to pull the first list and then combine with the second and then a third and so on. I don't know if I'll be able to assume that each date will be in each list.
Assuming your list is called list_df, you can bind them all together using bind_rows, group_by date and then sum all the other columns.
library(dplyr)
list_df %>%
bind_rows() %>%
group_by(date) %>%
summarise_all(sum)
# A tibble: 3 x 3
# date clicks impressions
# <fct> <int> <int>
#1 2019-06-01 3 16
#2 2019-06-02 3 14
#3 2019-06-03 111 149
which in base R could be achieved using Reduce
aggregate(.~date, Reduce(rbind, list_df), sum)
We can use data.table
library(data.table)
rbindlist(list_df)[, lapply(.SD, sum), date]
# date clicks impressions
#1: 2019-06-01 3 16
#2: 2019-06-02 3 14
#3: 2019-06-03 111 149
data
list_df <- mget(paste0("df", 1:2))
We can do:
cbind(date=df1[,1],do.call(`+`, list(df1[,-1],df2[,-1])),
row.names = NULL)
date clicks impressions
1 2019-06-01 3 16
2 2019-06-02 3 14
3 2019-06-03 111 149
If you are not sure about the presence of dates(can then cbind as above):
do.call(`+`,lapply(list(df1,df2), function(x) x[,-1]))
clicks impressions
1 3 16
2 3 14
3 111 149
This assumes that the data sets will have the same structure always.
Some context first:
I'm working with a data set which includes health related data. It includes questionnaire scores pre and post treatment. However, some clients reappear within the data for further treatment. I've provided a mock example of the data in the code section.
I have tried to come up with a solution on dplyr as this is package I'm most familiar with, but I didn't achieve what I've wanted.
#Example/mock data
ClientNumber<-c("4355", "2231", "8894", "9002", "4355", "2231", "8894", "9002", "4355", "2231")
Pre_Post<-c(1,1,1,1,2,2,2,2,1,1)
QuestionnaireScore<-c(62,76,88,56,22,30, 35,40,70,71)
df<-data.frame(ClientNumber, Pre_Post, QuestionnaireScore)
df$ClientNumber<-as.character(df$ClientNumber)
df$Pre_Post<-as.factor(df$Pre_Post)
View(df)
#tried solution
df2<-df%>%
group_by(ClientNumber)%>%
filter( Pre_Post==1|Pre_Post==2)
#this doesn't work, or needs more code to it
As you can see, the first four client numbers both have a pre and post treatment score. This is good. However, client numbers 4355 and 2231 appear again at the end (you could say they have relapsed and started new treatment). These two clients do not have a post treatment score.
I only want to analyse clients that have a pre and post score, therefore I need to filter clients which have completed treatment, while excluding ones that do not have a post treatment score if they have appeared in the data again. In relation to the example I've provided, I want to include the first 8 for analysis while excluding the last two, as they do not have a post treatment score.
If these cases are to be kept in order, you could try:
library(dplyr)
df %>%
group_by(ClientNumber) %>%
filter(!duplicated(Pre_Post) & n_distinct(Pre_Post) == 2)
ClientNumber Pre_Post QuestionnaireScore
<fct> <dbl> <dbl>
1 4355 1 62
2 2231 1 76
3 8894 1 88
4 9002 1 56
5 4355 2 22
6 2231 2 30
7 8894 2 35
8 9002 2 40
I don't know if you actually need to use n_distinct() but it won't hurt to keep it. This will remove cases who have a pre score but no post score if they exist in the data.
First arrange ClientNumbers then group_by and finally filter using dplyr::lead and dplyr::lag
library(dplyr)
df %>% arrange(ClientNumber) %>% group_by(ClientNumber) %>%
filter(Pre_Post==1 & lead(Pre_Post)==2 | Pre_Post==2 & lag(Pre_Post)==1)
# A tibble: 8 x 3
# Groups: ClientNumber [4]
ClientNumber Pre_Post QuestionnaireScore
<fct> <dbl> <dbl>
1 2231 1 76
2 2231 2 30
3 4355 1 62
4 4355 2 22
5 8894 1 88
6 8894 2 35
7 9002 1 56
8 9002 2 40
Another option is to create groups of 2 for every ClientNumber and select only those groups which have 2 rows in them.
library(dplyr)
df %>%
arrange(ClientNumber) %>%
group_by(ClientNumber, group = cumsum(Pre_Post == 1)) %>%
filter(n() == 2) %>%
ungroup() %>%
select(-group)
# ClientNumber Pre_Post QuestionnaireScore
# <chr> <fct> <dbl>
#1 2231 1 76
#2 2231 2 30
#3 4355 1 62
#4 4355 2 22
#5 8894 1 88
#6 8894 2 35
#7 9002 1 56
#8 9002 2 40
The same can be translated in base R using ave
new_df <- df[order(df$ClientNumber), ]
subset(new_df, ave(Pre_Post,ClientNumber,cumsum(Pre_Post == 1),FUN = length) == 2)
I have a two data.frames (call them dataset.new and dataset.old) that both contain information about some individuals. These individuals all have a identification number (a variable we can call ”individual”) that occurs in both of the data.frames and each frame has information on when the data was collected, stored in a column that we can call ”some.date”.
The second of these two data.frames (dataset.old) contains historical data for the individuals, i.e. values of some other variables measured at other times and thus each individual appears many times in dataset.old.
What I wish to do is the following. For each individual in dataset.new, find the rows from dataset.old that are the newest but still older than the observations in dataset.new. For the individuals that have no such date present in dataset.old, I want it to return NA.
This is perhaps easiest illustrated through some example data, presented below.
dataset.new
individual some.date
1 1 2016-05-01
2 2 2016-01-28
3 7 2016-03-03
dataset.old
individual some.date
1 1 2016-01-12
2 1 2015-12-30
3 1 2016-04-27
4 1 2016-05-02
5 2 2015-11-15
6 2 2012-01-27
7 2 2016-02-06
8 3 2016-04-30
9 3 2016-01-27
10 4 2016-03-01
11 4 2011-01-16
In this example, I am looking for a way get the following output:
individual row.nr
1 1 3
2 2 5
3 7 NA
since those rows correspond to the newest data in dataset.old that still is older than the data in dataset.new.
I have a code that solves the problem, but it is too slow for the data that I have in mind (which has well over 20 000 rows in dataset.new and many, many more in dataset.old). My solution is basically a loop over all individuals, subsetting the data at each stage.
find.previous <- function(dataset.old, individual, some.new.date){
subsetted.dataset <- dataset.old[dataset.old[, "individual"] == individual, ] # We only look at the individual in question.
subsetted.dataset <- subsetted.dataset[subsetted.dataset[, "some.date"] < some.new.date, ]# Here we get all the rows that have data that are measured BEFORE timepoint.
row.index <- which.min(some.new.date - subsetted.dataset[, "some.date"]) # This can be done, since we have already made sure that fromdatum < timepoint.
ifelse(length(row.index)!= 0, as.integer(rownames(subsetted.dataset[row.index,])), NA) # Then we output the row that had that information.
}
output <- matrix(ncol=2, nrow=0)
for(i in 1:nrow(dataset.new)){
output <- rbind(output, cbind(dataset.new[, "individual"][i], find.previous(dataset.old, dataset.new[, "individual"][i], dataset.new[, "some.date"][i])))
}
colnames(output) <- c("individual", "row.nr")
output
Any help on how to solve this problem would be greatly appreciated. I have tried using my Google skills as well as reading other posts on here stackoverflow, but without success.
The example data can be replicated by copying the following lines of code:
dataset.new <- data.frame(individual=c(1, 2, 7), some.date=as.Date(c("2016-05-01", "2016-01-28", "2016-03-03")))
dataset.old <- data.frame(individual=c(1,1,1,1,2,2,2,3,3,4,4), some.date=as.Date(c("2016-01-12", "2015-12-30", "2016-04-27", "2016-05-02", "2015-11-15", "2012-01-27", "2016-02-06", "2016-04-30", "2016-01-27", "2016-03-01", "2011-01-16")))
You can solve this efficiently with a merge.
First make the rownumber variable you want in dataset.old. Then merge dataset.new with dataset.old on individual (left join, or merge(lhs, rhs, all.x = TRUE)). This can get you:
dataset.old
individual new.date old.date old.rownumber
1 1 2016-05-01 2016-01-12 1
2 1 2016-05-01 2015-12-30 2
3 1 2016-05-01 2016-04-27 3
4 1 2016-05-01 2016-05-02 4
5 2 2016-01-28 2015-11-15 5
6 2 2016-01-28 2012-01-27 6
7 2 2016-01-28 2016-02-06 7
8 7 2016-03-03 NA NA
Subset to new.date > old.date or is.na(old.date):
dataset.old
individual new.date old.date old.rownumber
1 1 2016-05-01 2016-01-12 1
2 1 2016-05-01 2015-12-30 2
3 1 2016-05-01 2016-04-27 3
5 2 2016-01-28 2015-11-15 5
6 2 2016-01-28 2012-01-27 6
8 7 2016-03-03 NA NA
Subset to old.date == max(old.date) or is.na(old.date) grouped by individual.
dataset.old
individual new.date old.date old.rownumber
3 1 2016-05-01 2016-04-27 3
6 2 2016-01-28 2012-01-27 5
8 7 2016-03-03 NA NA
Edit:
I'm partial to data.table. The code would look something like:
dataset.old[, old.rownumber := 1:.N]
setnames(dataset.old, "some.date", "old.date")
setnames(dataset.new, "some.date", "new.date")
dataset.merge <- merge(dataset.old, dataset.new, by = "individual", all.x = TRUE)
dataset.merge <- dataset.merge[, new.date > old.date]
dataset.merge[old.date == max(old.date) | is.na(old.date), by = individual]
We can skip the NA search by finding the minimum square root. The negative values will be coerced to missing for us:
dataset.old$rn <- 1:nrow(dataset.old)
minp <- function(x) if(!length(m <- which.min(as.numeric(x)^.5))) NA else m
mrg <- merge(dataset.new, dataset.old, by="individual", all.x=TRUE)
mrg %>% group_by(individual) %>%
summarise(row.nr=rn[minp(some.date.x - some.date.y)])
# A tibble: 3 x 2
# individual row.nr
# <int> <int>
# 1 1 3
# 2 2 5
# 3 7 NA
Have a look at the simplified table below. I want for each product a vector containing the quantities sold within each delivery time. A delivery time is defined as 4 days. So if we look at product A, we see that it starts at 03/12/15 and within the first delivery term (until 07/12/15) it has sold a quantity of 4. The second delivery term starts at 08/12/15 and ends at 12/12/15. So for this period there is 1 quantity sold. The following delivery term starts at 13/12/15 and ends at 17/12/15. During these period there are no quantities sold and thus for this period the vector must have a value of 0. In the last period, finally, 2 products are sold. So basically the problem here is that information regarding the periods were no products are sold is missing.
Any ideas on how the vector I want can be created using R? I've been thinking of for or while loops, but these do not seem to give the requested results. Note that the code must be applicable on a real dataset containing over 1000 product categories, so it has to be 'automatized' in one way.
I would be very gratefull if somebody could point me in the right direction.
Product Quantity Date
A 1 03/12/15
A 2 04/12/15
A 1 05/12/15
A 1 08/12/15
A 1 17/12/16
A 1 18/12/16
B 1 19/12/15
B 2 10/05/15
B 2 11/05/15
C 1 01/06/15
C 1 02/06/15
C 1 12/06/15
Assume that dt is the dataset you provided. You'll get a better understanding of the process if you run it step by step (and maybe with an even simpler dataset).
library(lubridate)
library(dplyr)
# create date time columns
dt$Date = dmy(dt$Date)
dt %>%
group_by(Product) %>%
do(data.frame(days = seq(min(.$Date), max(.$Date), by="1 day"))) %>% # create all combinations between product and days
mutate(dist = as.numeric(difftime(days,min(days), units="days"))) %>% # create distance of each day with min date
ungroup() %>%
left_join(dt, by=c("Product"="Product","days"="Date")) %>% # join info to get quantities for each day
mutate(Quantity = ifelse(is.na(Quantity), 0, Quantity), # replace NAs with 0s
id = floor(dist/5 + 1)) %>% # create the 4 period id
group_by(Product, id) %>%
summarise(Sum = sum(Quantity),
min_date = min(days),
max_date = max(days)) %>%
ungroup
# Product id Sum min_date max_date
# 1 A 1 4 2015-12-03 2015-12-07
# 2 A 2 1 2015-12-08 2015-12-12
# 3 A 3 0 2015-12-13 2015-12-17
# 4 A 4 0 2015-12-18 2015-12-22
# 5 A 5 0 2015-12-23 2015-12-27
# 6 A 6 0 2015-12-28 2016-01-01
# 7 A 7 0 2016-01-02 2016-01-06
# 8 A 8 0 2016-01-07 2016-01-11
# 9 A 9 0 2016-01-12 2016-01-16
# 10 A 10 0 2016-01-17 2016-01-21
# .. ... .. ... ... ...
First row of the output tells you that for product A in the first 4 days period (id = 1) you had 4 quantities in total and the period is from 3/12 to 7/12.
I would suggest {dplyr}'s summarise(),mutate() and group_by() functions. group_by() groups your data by desired variables (in your case - product and delivery term),mutate() allows operations on grouped columns, and summarise() applies a summarising function over these groups (in your case sum(Quantity)).
So this is how it will look:
convert date into proper format:
library(dplyr)
df=tbl_df(df)
df$Date=as.Date(df$Date,format="%d/%m/%y")
calculating delivery terms
df=group_by(df,Product) %>% arrange(Date)
df=mutate(df,term=1+unclass((Date-min(Date)))%/%4)
group by product and terms and calculate sum of quantity:
df=group_by(df,Product,term)
summarise(df,sum=sum(Quantity))
Here's a base R way:
df$groups <- ave(as.numeric(df$Date), df$Product, FUN=function(x) {
intrvl <- findInterval(x, seq(min(x), max(x),4))
as.numeric(factor(intrvl))
})
df
# Product Quantity Date groups
# 1 A 1 2015-12-03 1
# 2 A 2 2015-12-04 1
# 3 A 1 2015-12-05 1
# 4 A 1 2015-12-08 2
# 5 A 1 2016-12-17 3
# 6 A 1 2016-12-18 3
# 7 B 1 2015-12-19 2
# 8 B 2 2015-05-10 1
# 9 B 2 2015-05-11 1
# 10 C 1 2015-06-01 1
# 11 C 1 2015-06-02 1
# 12 C 1 2015-06-12 2
The dates should be converted to one of the date classes. I chose as.Date. When it converts to numeric, the output will be the number of days from a specified date. From there, we are able to group by 4 day increments.
Data
df$Date <- as.Date(df$Date, format="%d/%m/%y")
Having the following table which comprises some key columns which are: customer ID | order ID | product ID | Quantity | Amount | Order Date.
All this data is in LONG Format, in that you will get multi line items for the 1 Customer ID.
I can get the first date last date using R DateDiff but converting the file to WIDE format using Plyr, still end up with the same problem of getting multiple orders by customer, just less rows and more columns.
Is there an R function that extends R DateDiff to work out how to get the time interval between purchases by Customer ID? That is, time between order 1 and 2, order 2 and 3, and so on assuming these orders exists.
CID Order.Date Order.DateMY Order.No_ Amount Quantity Category.Name Locality
1 26/02/13 Feb-13 zzzzz 1 r MOSMAN
1 26/05/13 May-13 qqqqq 1 x CHULLORA
1 28/05/13 May-13 wwwww 1 r MOSMAN
1 28/05/13 May-13 wwwww 1 x MOSMAN
2 19/08/13 Aug-13 wwwwww 1 o OAKLEIGH SOUTH
3 3/01/13 Jan-13 wwwwww 1 x CURRENCY CREEK
4 28/08/13 Aug-13 eeeeeee 1 t BRISBANE
4 10/09/13 Sep-13 rrrrrrrrr 1 y BRISBANE
4 25/09/13 Sep-13 tttttttt 2 e BRISBANE
It is not clear what do you want to do since you don't give the expected result. But I guess you want to the the intervals between 2 orders.
library(data.table)
DT <- as.data.table(DF)
DT[, list(Order.Date,
diff = c(0,diff(sort(as.Date(Order.Date,'%d/%m/%y')))) ),CID]
CID Order.Date diff
1: 1 26/02/13 0
2: 1 26/05/13 89
3: 1 28/05/13 2
4: 1 28/05/13 0
5: 2 19/08/13 0
6: 3 3/01/13 0
7: 4 28/08/13 0
8: 4 10/09/13 13
9: 4 25/09/13 15
Split the data frame and find the intervals for each Customer ID.
df <- data.frame(customerID=as.factor(c(rep("A",3),rep("B",4))),
OrderDate=as.Date(c("2013-07-01","2013-07-02","2013-07-03","2013-06-01","2013-06-02",
"2013-06-03","2013-07-01")))
dfs <- split(df,df$customerID)
lapply(dfs,function(x){
tmp <-diff(x$OrderDate)
tmp
})
Or use plyr
library(plyr)
dfs <- dlply(df,.(customerID),function(x)return(diff(x$OrderDate)))
I know this question is very old, but I just figured out another way to do it and wanted to record it:
> library(dplyr)
> library(lubridate)
> df %>% group_by(customerID) %>%
mutate(SinceLast=(interval(ymd(lag(OrderDate)),ymd(OrderDate)))/86400)
# A tibble: 7 x 3
# Groups: customerID [2]
customerID OrderDate SinceLast
<fct> <date> <dbl>
1 A 2013-07-01 NA
2 A 2013-07-02 1.
3 A 2013-07-03 1.
4 B 2013-06-01 NA
5 B 2013-06-02 1.
6 B 2013-06-03 1.
7 B 2013-07-01 28.