I'm relatively new to R, so I realise this type of question is asked often but I've read a lot of stack overflow posts and still can't quite get something to work on my data.
I have data on spss, in two datasets imported into R. Both of my datasets include an id (IDC), which I have been using to merge them. Before merging, I need to filter one of the datasets to select specifically the last observation of a date variable.
My dataset, d1, has a longitudinal measure in long format. There are multiple rows per IDC, representing different places of residence (neighborhood). Each row has its own "start_date", which is a variable that is NOT necessarily unique.
As it looks on spss :
IDC
neighborhood
start_date
1
22
08.07.2001
1
44
04.02.2005
1
13
21.06.2010
2
44
24.12.2014
2
3
06.03.2002
3
22
04.01.2006
4
13
08.07.2001
4
2
15.06.2011
In R, the start dates do not look the same, instead they are one long number like "13529462400". I do not understand this format but I assume it still would retain the date order...
Here are all my attempts so far to select the last date. All attempts ran, there was no error. The output just didn't give me what I want. To my perception, none of these made any change in the number of repetitions of IDC, so none of them actually selected *only the last date.
##### attempt 1 --- not working
d1$start_date_filt <- d1$start_date
d1[order(d1$IDC,d1$start_date_filt),] # Sort by ID and week
d1[!duplicated(d1$IDC, fromLast=T),] # Keep last observation per ID)
###### attempt 2--- not working
myid.uni <- unique(d1$IDC)
a<-length(myid.uni)
last <- c()
for (i in 1:a) {
temp<-subset(d1, IDC==myid.uni[i])
if (dim(temp)[1] > 1) {
last.temp<-temp[dim(temp)[1],]
}
else {
last.temp<-temp
}
last<-rbind(last, last.temp)
}
last
##### atempt 3 -- doesn't work
do.call("rbind",
by(d1,INDICES = d1$IDC,
FUN=function(DF)
DF[which.max(DF$start_date),]))
#### attempt 4 -- doesnt work
library(plyr)
ddply(d1,.(IDC), function(X)
X[which.max(X$start_date),])
### merger code -- in case something has to change with that after only the last start_date is selected
merge(d1,d2, IDC)
My goal dataset d1 would look like this:
IDC
neighborhood
start_date
1
13
21.06.2010
2
44
24.12.2014
3
22
04.01.2006
4
2
15.06.2011
I'm grateful for any help, many thanks <3
There are some problems with most approaches dealing with this data: because your dates are arbitrary strings in a format that does not sort correctly, it just-so-happens to work here because the maximum day-of-month also happens in the maximum year.
It would generally be better to work with that column as a Date object in R, so that comparisons can be better.
dat$start_date <- as.Date(dat$start_date, format = "%d.%m.%Y")
dat
# IDC neighborhood start_date
# 1 1 22 2001-07-08
# 2 1 44 2005-02-04
# 3 1 13 2010-06-21
# 4 2 44 2014-12-24
# 5 2 3 2002-03-06
# 6 3 22 2006-01-04
# 7 4 13 2001-07-08
# 8 4 2 2011-06-15
From here, things are a bit simpler:
Base R
do.call(rbind, by(dat, dat[,c("IDC"),drop=FALSE], function(z) z[which.max(z$start_date),]))
# IDC neighborhood start_date
# 1 1 13 2010-06-21
# 2 2 44 2014-12-24
# 3 3 22 2006-01-04
# 4 4 2 2011-06-15
dplyr
dat %>%
group_by(IDC) %>%
slice(which.max(start_date)) %>%
ungroup()
# # A tibble: 4 x 3
# IDC neighborhood start_date
# <int> <int> <date>
# 1 1 13 2010-06-21
# 2 2 44 2014-12-24
# 3 3 22 2006-01-04
# 4 4 2 2011-06-15
Data
dat <- structure(list(IDC = c(1L, 1L, 1L, 2L, 2L, 3L, 4L, 4L), neighborhood = c(22L, 44L, 13L, 44L, 3L, 22L, 13L, 2L), start_date = c("08.07.2001", "04.02.2005", "21.06.2010", "24.12.2014", "06.03.2002", "04.01.2006", "08.07.2001", "15.06.2011")), class = "data.frame", row.names = c(NA, -8L))
Related
My dataset contains several groups and each group can have a different number of unique observations. I carry out some calculations by group (simplified in the code below), resulting in a summary value for each group. Next, for the purpose of a bootstrap, I want to:
Randomly sample the groups with replacement (number of sampled groups = equal to number of different groups in the original dataset)
Within these sampled groups, randomly sample observations with replacement (number of sampled observations per group = equal to number of unique observations in that group in the original dataset)
A simplified version of my data set up (data1):
data1:
id group y
1001 1 10
1002 1 15
1003 1 3
3002 2 24
3003 2 15
3005 2 37
3006 2 32
3007 2 11
4001 3 12
4002 3 15
5006 4 7
5007 4 9
5009 4 22
5010 4 19
E.g. based on the dataset example above: there are 4 groups in the original dataset, so I want to sample 4 groups with replacement (e.g. groups sampled = groups 4,3,3,1), and then sample observations/rows from those 4 groups (4 ids from group 4 (e.g. 5007, 5007, 5006, 5009); 2 ids from group 3 (twice, as group 3 was sampled twice), and 3 ids from group 1, all with replacement), and return the sampled rows together in a dataframe (4+2+2+3 = 11 rows).
For the above, I some have code working for these steps separately, but I cannot seem to combine them:
# Calculate group value
y.group <- tapply(data1$y,data1$group,mean)
# Step 1. Sample groups, with replacement:
sampled.group <- sample(1:length(unique(data1$group)),replace=T)
# Step 2. Sample within groups, with replacement
data2 <- data.frame(data1 %>%
group_by(group) %>% # for each group
sample_frac(1, replace = TRUE) %>%
ungroup)
Obviously, the code above in full does not do what I want, as in step 2 the sampled groups from step 1 are ignored since it just uses the original group var (I am aware of this). I have tried to solve this using step 1 and trying to generate a new dataframe containing only the sampled groups' observations (with duplicates if a group was sampled more than once, which is likely to happen), and then apply step 2 to that new dataframe, but I cannot get this to work.
I think I am just on the wrong path or overthinking things. Hopefully you can give me some advice on how to proceed.
Edit: While awaiting any potential solutions, I continued on the question myself and ended up with:
total.result <- c()
for (j in 1:length(unique(data1$group))){
sampled.group <- sample(1:length(unique(data1$group)),size=1,replace=T)
group.result <- sample_n(data1[data1$group==sampled.group,],
size=length(unique(data1$id[data1$group==sampled.group])),replace=T)
total.result <- rbind(total.result,group.result)
}
(So basically using a loop to sample the groups one at a time, creating datasets for each, and then sampling individual rows from those, and finally combining the results with rbind)
However, I think Allan Cameron's solution (see below) is more straigthforward, so I have accepted that one as the answer to my question.
I think this is what you're looking for. Let's start with your data in a reproducible format:
data1 <- structure(list(id = structure(1:14, .Label = c("1001", "1002",
"1003", "3002", "3003", "3005", "3006", "3007", "4001", "4002",
"5006", "5007", "5009", "5010"), class = "factor"), group = structure(c(1L,
1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c("1",
"2", "3", "4"), class = "factor"), y = structure(c(1L, 4L, 8L,
7L, 4L, 10L, 9L, 2L, 3L, 4L, 11L, 12L, 6L, 5L), .Label = c("10",
"11", "12", "15", "19", "22", "24", "3", "32", "37", "7", "9"
), class = "factor")), class = "data.frame", row.names = c(NA,
-14L))
And just to make sure:
data1
#> id group y
#> 1 1001 1 10
#> 2 1002 1 15
#> 3 1003 1 3
#> 4 3002 2 24
#> 5 3003 2 15
#> 6 3005 2 37
#> 7 3006 2 32
#> 8 3007 2 11
#> 9 4001 3 12
#> 10 4002 3 15
#> 11 5006 4 7
#> 12 5007 4 9
#> 13 5009 4 22
#> 14 5010 4 19
We start by splitting the data frame by group into smaller data frames, using the split function. This gives us a list with four data frames, each one containing all the members of its respective group. (The set.seed is there purely to make this example reproducible).
set.seed(69)
split_dfs <- split(data1, data1$group)
Now we can sample this list, giving us a new list of four data frames drawn with replacement from split_dfs. Each one will again contain all the members of its respective group, though of course some whole groups might be sampled more than once, and other whole groups not sampled at all.
sampled_group_dfs <- split_dfs[sample(length(split_dfs), replace = TRUE)]
Now we can sample within each group by sampling with replacement from the rows of each data frame in our new list. We do this for all our data frames in our list by using lapply
all_sampled <- lapply(sampled_group_dfs, function(x) x[sample(nrow(x), replace = TRUE), ])
All that remains is to stick all the resultant dataframes in this list back together to get our result:
result <- do.call(rbind, all_sampled)
As you can see from the final result, it just so happens that each of the four groups was sampled once (this is just by chance - alter set.seed to get different results). However, within the groups there have clearly been some duplicates drawn. In fact, since R mandates unique row names in a data frame, these are easy to pick out by the .1 that has been appended to the duplicate row names. If you don't like this, you can reset the row names with rownames(result) <- seq(nrow(result))
result
#> id group y
#> 4.14 5010 4 19
#> 4.14.1 5010 4 19
#> 4.11 5006 4 7
#> 4.13 5009 4 22
#> 1.3 1003 1 3
#> 1.3.1 1003 1 3
#> 1.2 1002 1 15
#> 3.9 4001 3 12
#> 3.9.1 4001 3 12
#> 2.5 3003 2 15
#> 2.5.1 3003 2 15
#> 2.6 3005 2 37
#> 2.7 3006 2 32
#> 2.5.2 3003 2 15
Created on 2020-02-15 by the reprex package (v0.3.0)
I need to embed a condition in a remove duplicates function. I am working with large student database from South Africa, a highly multilingual country. Last week you guys gave me the code to remove duplicates caused by retakes, but I now realise my language exam data shows some students offering more than 2 different languages.
The source data, simplified looks like this
STUDID MATSUBJ SCORE
101 AFRIKAANSB 1
101 AFRIKAANSB 4
102 ENGLISHB 2
102 ISIZULUB 7
102 ENGLISHB 5
The result file I need is
STUDID MATSUBJ SCORE flagextra
101 AFRIKAANS 4
102 ENGLISH 5
102 ISIZULUB 7 1
I need to flag the extra language so that I can see what languages they are and make new category for this
Two stage procedure works better for me as a newbie to R:
1- remove the duplicates caused by subject retakes:
df <- LANGSEC%>%
group_by(STUDID,MATRICSUBJ) %>%
top_n(1,SUBJSCORE)
2- Then flag one of the two subjects causing the remaining duplicates:
LANGSEC$flagextra <- as.integer(duplicated(LANGSEC$STUDID),LANGSEC$MATRICSUBJ
Then filter for this third language and make new file:
LANG3<-LANGSEC%>% filter(flagextra==1)
Then remove these from the other file:
LANG2<-LANGSEC %>% filter(!flagextra==1)
May be this helps
library(tidyverse)
df1 %>%
group_by(STUDID, MATSUBJ) %>%
summarise(SCORE = max(SCORE),
flagextra = as.integer(!sum(duplicated(MATSUBJ))))
# A tibble: 3 x 4
# Groups: STUDID [?]
# STUDID MATSUBJ SCORE flagextra
# <int> <chr> <dbl> <int>
#1 101 AFRIKAANSB 4 0
#2 102 ENGLISHB 5 0
#3 102 ISIZULUB 7 1
Or with base R
i1 <- !(duplicated(df1[1:2])|duplicated(df1[1:2], fromLast = TRUE))
transform(aggregate(SCORE ~ ., df1, max),
flagextra = as.integer(MATSUBJ %in% df1$MATSUBJ[i1]))
data
df1 <- structure(list(STUDID = c(101L, 101L, 102L, 102L, 102L), MATSUBJ
= c("AFRIKAANSB",
"AFRIKAANSB", "ENGLISHB", "ISIZULUB", "ENGLISHB"), SCORE = c(1L,
4L, 2L, 7L, 5L)), class = "data.frame", row.names = c(NA, -5L
))
I'd like to split a dataset made of character strings into columns specified by start and end.
My dataset looks something like this:
>head(templines,3)
[1] "201801 1 78"
[2] "201801 2 67"
[3] "201801 1 13"
and i'd like to split it by specifying my columns using the data dictionary:
>dictionary
col_name col_start col_end
year 1 4
week 5 6
gender 8 8
age 11 12
so it becomes:
year week gender age
2018 01 1 78
2018 01 2 67
2018 01 1 13
In reality the data comes from a long running survey and the white spaces between some columns represent variables that are no longer collected. It has many variables so i need a solution that would scale.
In tidyr::separate it looks like you can only split by specifying the position to split at, rather than the start and end positions. Is there a way to use start / end?
I thought of doing this with read_fwf but I can't seem to be able to use it on my already loaded dataset. I only managed to get it to work by first exporting as a txt and then reading from this .txt:
write_lines(templines,"t1.txt")
read_fwf("t1.txt",
fwf_positions(start = dictionary$col_start,
end = dictionary$col_end,
col_names = dictionary$col_name)
is it possible to use read_fwf on an already loaded dataset?
Answering your question directly: yes, it is possible to use read_fwf with already loaded data. The relevant part of the docs is the part about the argument file:
Either a path to a file, a connection, or literal data (either a single string or a raw vector).
...
Literal data is most useful for examples and tests.
It must contain at least one new line to be recognised as data (instead of a path).
Thus, you can simply collapse your data and then use read_fwf:
templines %>%
paste(collapse = "\n") %>%
read_fwf(., fwf_positions(start = dictionary$col_start,
end = dictionary$col_end,
col_names = dictionary$col_name))
This should scale to multiple columns, and is fast for many rows (on my machine for 1 million rows and four columns about half a second).
There are a few warnings regarding parsing failures, but they stem from your dictionary. If you change the last line to age, 11, 12 it works as expected.
A solution with substring:
library(data.table)
x <- transpose(lapply(templines, substring, dictionary$col_start, dictionary$col_end))
setDT(x)
setnames(x, dictionary$col_name)
# > x
# year week gender age
# 1: 2018 01 1 78
# 2: 2018 01 2 67
# 3: 2018 01 1 13
How about this?
data.frame(year=substr(templines,1,4),
week=substr(templines,5,6),
gender=substr(templines,7,8),
age=substr(templines,11,13))
Using base R:
m = list(`attr<-`(dat$col_start,"match.length",dat$col_end-dat$col_start+1))
d = do.call(rbind,regmatches(x,rep(m,length(x))))
setNames(data.frame(d),dat$col_name)
year week gender age
1 2018 01 1 78
2 2018 01 2 67
3 2018 01 1 13
DATA USED:
x = c("201801 1 78", "201801 2 67", "201801 1 13")
dat=read.table(text="col_name col_start col_end
year 1 4
week 5 6
gender 8 8
age 11 13 ",h=T)
We could use separate from tidyverse
library(tidyverse)
data.frame(Col = templines) %>%
separate(Col, into = dictionary$col_name, sep= head(dictionary$col_end, -1))
# year week gender age
#1 2018 01 1 78
#2 2018 01 2 67
#3 2018 01 1 13
The convert = TRUE argument can also be used with separate to have numeric columns as output
tibble(Col = templines) %>%
separate(Col, into = dictionary$col_name,
sep= head(dictionary$col_end, -1), convert = TRUE)
# A tibble: 3 x 4
# year week gender age
# <int> <int> <int> <int>
#1 2018 1 1 78
#2 2018 1 2 67
#3 2018 1 1 13
data
dictionary <- structure(list(col_name = c("year", "week", "gender", "age"),
col_start = c(1L, 5L, 8L, 11L), col_end = c(4L, 6L, 8L, 13L
)), .Names = c("col_name", "col_start", "col_end"),
class = "data.frame", row.names = c(NA, -4L))
templines <- c("201801 1 78", "201801 2 67", "201801 1 13")
This is an explicit function which seems to be working the way you wanted.
split_func<-function(char,ref,name,start,end){
res<-data.table("ID" = 1:length(char))
for(i in 1:nrow(ref)){
res[,ref[[name]][i] := substr(x = char,start = ref[[start]][i],stop = ref[[end]][i])]
}
return(res)
}
I have created the same input files as you:
templines<-c("201801 1 78","201801 2 67","201801 1 13")
dictionary<-data.table("col_name" = c("year","week","gender","age"),"col_start" = c(1,5,8,11),
"col_end" = c(4,6,8,13))
# col_name col_start col_end
#1: year 1 4
#2: week 5 6
#3: gender 8 8
#4: age 11 13
As for the arguments,
char - The character vector with the values you want to split
ref - The reference table or dictionary
name - The column number in the reference table containing the column names you want
start - The column number in the reference table containing the start points
end - The column number in the reference table containing the stop points
If I use this function with these inputs, I get the following result:
out<-split_func(char = templines,ref = dictionary,name = 1,start = 2,end = 3)
#>out
# ID year week gender age
#1: 1 2018 01 1 78
#2: 2 2018 01 2 67
#3: 3 2018 01 1 13
I had to include an "ID" column to initiate the data table and make this easier. In case you want to drop it later you can just use:
out[,ID := NULL]
Hope this is closer to the solution you were looking for.
I am trying write a function or use cut to assign a grouping variable to some date data when those dates are close (user definition of close). For example, I would like to create a common grouping variable for some samples that were collected on consecutive dates. I was thinking cut would work here but then I realized cut doesn't group variables when they are close and rather creates a series of groups based on a sequence.
So take this dataframe for example:
df <- structure(list(Num = c(0.888401849195361, 0.185766335576773,
0.493163562379777, 0.13070688676089, 0.484760325402021, 0.603240836178884,
0.893201333936304, 0.641203448642045, 0.16957180458121, 0.0101411847863346
), Date = structure(c(10592, 10597, 10598, 10605, 10606, 10608,
10609, 10616, 10617, 10618), class = "Date"), day = c(1L, 6L,
7L, 14L, 15L, 17L, 18L, 25L, 26L, 27L)), .Names = c("Num", "Date",
"day"), row.names = c(NA, -10L), class = "data.frame")
If was to apply a cut function as I understand its usage like this:
df$cutVar <- cut(df$day, breaks= seq(0, 31, by = 1), right=TRUE)
I would be left with a range that went right through values that I'd prefer to be grouped together. For example, the 6th and 7th should be grouped together based on their proximity to each other. Similar to 14th and 15th and so on.
> df
Num Date day cutVar
1 0.88840185 1999-01-01 1 (0,1]
2 0.18576634 1999-01-06 6 (5,6]
3 0.49316356 1999-01-07 7 (6,7]
4 0.13070689 1999-01-14 14 (13,14]
5 0.48476033 1999-01-15 15 (14,15]
6 0.60324084 1999-01-17 17 (16,17]
7 0.89320133 1999-01-18 18 (17,18]
8 0.64120345 1999-01-25 25 (24,25]
9 0.16957180 1999-01-26 26 (25,26]
10 0.01014118 1999-01-27 27 (26,27]
So the basic question here is how to group a continuous variable (a date in this instance) such that close (defined by the user) numbers are grouped together in a factor range?
Is this something you'd like? where 3 is a threshold I chose for convenience. It can be any number you prefer:
df$group <- cumsum(c(1, diff.Date(df$Date)) >= 3)
df
Num Date day group
1 0.88840185 1999-01-01 1 0
2 0.18576634 1999-01-06 6 1
3 0.49316356 1999-01-07 7 1
4 0.13070689 1999-01-14 14 2
5 0.48476033 1999-01-15 15 2
6 0.60324084 1999-01-17 17 2
7 0.89320133 1999-01-18 18 2
8 0.64120345 1999-01-25 25 3
9 0.16957180 1999-01-26 26 3
10 0.01014118 1999-01-27 27 3
I really hate to ask two questions in a row but this is something that I can’t wrap my head around. So let’s say I have a data frame, as follows:
df
Row# User Morning Evening Measure Date
1 1 NA NA 2/18/11
2 1 50 115 2/19/11
3 1 85 128 2/20/11
4 1 62 NA 2/25/11
5 1 48 100.8 3/8/11
6 1 19 71 3/9/11
7 1 25 98 3/10/11
8 1 NA 105 3/11/11
9 2 48 105 2/18/11
10 2 28 203 2/19/11
11 2 35 80.99 2/21/11
12 2 91 78.25 2/22/11
Is it possible in R to take the difference between the previous consecutive day (and only the previous day, not the previous result) evening value of 1 row and the morning value of a different row for each user group? So my desired results would be this.
df
Row# User Morning Evening Date Difference
1 1 NA NA 2/18/11 NA
2 1 50 115 2/19/11 NA
3 1 85 129 2/20/11 30
4 1 62 NA 2/25/11 NA
5 1 48 100.8 3/8/11 NA
6 1 19 71 3/9/11 81.8
7 1 25 98 3/10/11 46
8 1 10 105 3/11/11 88
9 2 48 105 2/18/11 NA
10 2 28 203 2/19/11 77
11 2 35 80.99 2/21/11 NA
12 2 91 78.25 2/22/11 -10.01
All I want this to do is to take the morning value and subtract it from the evening value of the previous consecutive day for each user group. As you can see, some parts of my data frame contain NA values in the morning and evening columns, in addition, not all of the dates are in consecutive order for each different user, so naturally, NA should be assigned.
I've tried searching google but there wasn't much information on being able to apply functions to different rows for each group of rows on different columns (if that makes any sense).
My attempts include many variations of this.
df$Difference<-ave((df$Morning,df$Evening),
df$User,
FUN=function(x){
c('NA',diff(df$Evening-df$Morning)),na.rm=T
})
Again, any help would be greatly appreciated. Thanks.
Note: The input data you show and the output data are not the same. There is a NA which is replaced by 10 in output and the last date is 2/14/11 in input and 2/22/11 in output.
I've assumed the output to be the original data to create this answer to match your result.
df$Diff <- c(NA, head(df$Evening, -1) - tail(df$Morning, -1))
df$Diff[which(c(0, diff(as.Date(as.character(df$Measure_Date),
format="%m/%d/%Y"))) != 1)] <- NA
> df
# Row User Morning Evening Measure_Date Diff
# 1 1 1 NA NA 2/18/11 NA
# 2 2 1 50 115.00 2/19/11 NA
# 3 3 1 85 128.00 2/20/11 30.00
# 4 4 1 62 NA 2/25/11 NA
# 5 5 1 48 100.80 3/8/11 NA
# 6 6 1 19 71.00 3/9/11 81.80
# 7 7 1 25 98.00 3/10/11 46.00
# 8 8 1 10 105.00 3/11/11 88.00
# 9 9 2 48 105.00 2/18/11 NA
# 10 10 2 28 203.00 2/19/11 77.00
# 11 11 2 35 80.99 2/21/11 NA
# 12 12 2 91 78.25 2/22/11 -10.01
#user1342086's edit (that got rejected, but was right indeed):
df$Diff[which(diff(df$User) != 0)] <- NA
seems to take care of the grouping by "User".
A blind first shot (untested). Relies on the data frame being already sorted by User and Date.
#if necessary, transform your dates from factor to Date
df$Date <- as.Date(levels(df$Date)[df$Date],format="%m/%d/%y")
df <- within(df,
Difference <- ifelse(c(NA,diff(Measure_Date)) == 1 & diff(User) == 0,
c(NA,head(Evening,-1)) - Morning, NA
)
)
I used plyr, so be sure you have it installed. This solution should work even if user data are mixed (i.e. not in consecutive rows) and dates are not in chronological order.
# Your example data, as you should post it for us to use
df <-
structure(list(User = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L), Morning = c(NA, 50L, 85L, 62L, 48L, 19L, 25L, NA, 48L,
28L, 35L, 91L), Evening = c(NA, 115, 128, NA, 100.8, 71, 98,
105, 105, 203, 80.99, 78.25), Measure_Date = structure(c(1L,
2L, 3L, 5L, 9L, 10L, 6L, 7L, 1L, 2L, 4L, 8L), .Label = c("2/18/11",
"2/19/11", "2/20/11", "2/21/11", "2/25/11", "3/10/11", "3/11/11",
"3/14/11", "3/8/11", "3/9/11"), class = "factor")), .Names = c("User",
"Morning", "Evening", "Measure_Date"), class = "data.frame", row.names = c(NA,
-12L))
# As already stated by Arun, you need the date as class Date
df$Measure_Date <- as.Date(df$Measure_Date, format='%m/%d/%y')
# Use plyr to procces the dataframe by user
library(package=plyr)
ddply(.data=df, .variables='User',
.fun=function(x){
# Complete sequence of dates for each user
tdf <- data.frame(Measure_Date=seq(from=min(x$Measure_Date),
to=max(x$Measure_Date),
by='1 day'))
# Merge to fill in NAs for unused dates
tdf <- merge(tdf, x, all=TRUE)
# Put desired values side by side
tdf$Evening <- c(NA, tdf$Evening[-length(tdf$Evening)])
# Diference
tdf$Difference <- tdf$Evening - tdf$Morning
# Return desired value to original data
tdf <- tdf[,c('Measure_Date', 'Difference')]
x <- merge(x, tdf)
x
})