I have two data sets, one is the subset of another but the subset has additional column, with lesser observations.
Basically, I have a unique ID assigned to each participants, and then a HHID, the house id from which they were recruited (eg 15 participants recruited from 11 houses).
> Healthdata <- data.frame(ID = gl(15, 1), HHID = c(1,2,2,3,4,5,5,5,6,6,7,8,9,10,11))
> Healthdata
Now, I have a subset of data with only one participant per household, chosen who spent longer hours watching television. In this subset data, I have computed socioeconomic score (SSE) for each house.
> set.seed(1)
> Healthdata.1<- data.frame(ID=sample(1:15,11, replace=F), HHID=gl(11,1), SSE = sample(-6.5:3.5, 11, replace=TRUE))
> Healthdata.1
Now, I want to assign the SSE from the subset (Healthdata.1) to unique participants of bigger data (Healthdata) such that, participants from the same house gets the same score.
I can't merge this simply, because the data sets have different number of observations, 15 in the bigger one but only 11 in the subset.
Is there any way to do this in R? I am very new to it and I am stuck with this.
I want the required output as something like below, ie ID (participants) from same HHID (house) should have same SSE score. The following output is just meant for an example of what I need, the above seed will not give the same output.
ID HHID SSE
1 1 -6.5
2 2 -5.5
3 2 -5.5
4 3 3.3
5 4 3.0
6 5 2.58
7 5 2.58
8 5 2.58
9 6 -3.05
10 6 -3.05
11 7 -1.2
12 8 2.5
13 9 1.89
14 10 1.88
15 11 -3.02
Thanks.
You can use merge , By default it will merge by columns intersections.
merge(Healthdata,Healthdata.1,all.x=TRUE)
ID HHID SSE
1 1 1 NA
2 2 2 NA
3 3 2 NA
4 4 3 NA
5 5 4 NA
6 6 5 NA
7 7 5 NA
8 8 5 NA
9 9 6 0.7
10 10 6 NA
11 11 7 NA
12 12 8 NA
13 13 9 NA
14 14 10 NA
15 15 11 NA
Or you can choose by which column you merge :
merge(Healthdata,Healthdata.1,all.x=TRUE,by='ID')
You need to merge by HHID, not ID. Note this is somewhat confusing because the ids from the supergroup are from a different set than from the subgroup. I.e. ID.x == 4 != ID.y == 4 (in fact, in this case they are in different households). Because of that I left both ID columns here to avoid ambiguity, but you can easily subset the result to show only the ID.x one,
> merge(Healthdata, Healthdata.1, by='HHID')
HHID ID.x ID.y SSE
1 1 1 4 -5.5
2 2 2 6 0.5
3 2 3 6 0.5
4 3 4 8 -2.5
5 4 5 11 1.5
6 5 6 3 -1.5
7 5 7 3 -1.5
8 5 8 3 -1.5
9 6 9 9 0.5
10 6 10 9 0.5
11 7 11 10 3.5
12 8 12 14 -2.5
13 9 13 5 1.5
14 10 14 1 3.5
15 11 15 2 -4.5
library(plyr)
join(Healthdata, Healthdata.1)
# Inner Join
join(Healthdata, Healthdata.1, type = "inner", by = "ID")
# Left Join
# I believe this is what you are after
join(Healthdata, Healthdata.1, type = "left", by = "ID")
Related
This question already has answers here:
Summarizing by subgroup percentage in R
(2 answers)
Closed 9 months ago.
I am wrangling with a huge dataset and my R skills are very new. I am really trying to understand the terminology and processes but finding it a struggle as the R-documentation often makes no sense to me. So apologies if this is a dumb question.
I have data for plant species at different sites with different percentages of ground-cover. I want to create a new column PROP-COVER which gives the proportion of each species' cover as a percentage of the total cover of all species in a particular site. This is slightly different to calculating percentage cover by site area as it is disregards bare ground with no vegetation. This is an easy calculation with just one site, but I have over a hundred sites and need to perform the calculation on species ground-cover grouped by site. The desired column output is PROP-COVER.
SPECIES SITE COVER PROP-COVER(%)
1 1 10 7.7
2 1 20 15.4
3 1 10 7.7
4 1 20 15.4
5 1 30 23.1
6 1 40 30.8
2 2 20 22.2
3 2 50
5 2 10
6 2 10
1 3 5
2 3 25
3 3 40
5 3 10
I have looked at for loops and repeat but I can't see where the arguments should go. Every attempt I make returns a NULL.
Below is an example of something I tried which I am sure is totally wide of the mark, but I just can't work out where to begin with or know if it is even possible.
a<- for (i in data1$COVER) {
sum(data1$COVER[data1$SITE=="i"],na.rm = TRUE)
}
a
NULL
I have a major brain-blockage when it comes to how 'for' loops etc work, no amount of reading about it seems to help, but perhaps what I am trying to do isn't possible? :(
Many thanks for looking.
In Base R:
merge(df, prop.table(xtabs(COVER~SPECIES+SITE, df), 2)*100)
SPECIES SITE COVER Freq
1 1 1 10 7.692308
2 1 3 5 6.250000
3 2 1 20 15.384615
4 2 2 20 22.222222
5 2 3 25 31.250000
6 3 1 10 7.692308
7 3 2 50 55.555556
8 3 3 40 50.000000
9 4 1 20 15.384615
10 5 1 30 23.076923
11 5 2 10 11.111111
12 5 3 10 12.500000
13 6 1 40 30.769231
14 6 2 10 11.111111
In tidyverse you can do:
df %>%
group_by(SITE) %>%
mutate(n = proportions(COVER) * 100)
# A tibble: 14 x 4
# Groups: SITE [3]
SPECIES SITE COVER n
<int> <int> <int> <dbl>
1 1 1 10 7.69
2 2 1 20 15.4
3 3 1 10 7.69
4 4 1 20 15.4
5 5 1 30 23.1
6 6 1 40 30.8
7 2 2 20 22.2
8 3 2 50 55.6
9 5 2 10 11.1
10 6 2 10 11.1
11 1 3 5 6.25
12 2 3 25 31.2
13 3 3 40 50
14 5 3 10 12.5
The code could also be written as n = COVER/sum(COVER) or even n = prop.table(COVER)
My dataset has as features: players IDs, team, weeks and points.
I want to calculate the mean of TEAM points for previous weeks, but not all past weeks, just to the last 5 or less (if the current week is smaller than 5).
Example: For team = A, week = 7, the result will be the average of POINTS for team = A and weeks 2, 3, 4, 5 and 6.
The dataset can be created using the following code:
# set the seed for reproducibility
set.seed(123)
player_id<-c(rep(1,15),rep(2,15),rep(3,15),rep(4,15))
week<-1:15
team<-c(rep("A",30),rep("B",30))
points<-round(runif(60,1,10),0)
mydata<- data.frame(player_id=player_id,team=team,week=rep(week,4),points)
I would like to have a solution without a heavy looping, because the dataset is huge.
I have done related questions here that maybe will help, but I could not adapt to this case.
Question 1
Question 2
Thank you!
We adapt the approach from my answer to one of your other questions if you want a dplyr solution:
library(dplyr)
library(zoo)
# set the seed for reproducibility
set.seed(123)
player_id<-c(rep(1,15),rep(2,15),rep(3,15),rep(4,15))
week<-1:15
team<-c(rep("A",30),rep("B",30))
points<-round(runif(60,1,10),0)
mydata<- data.frame(player_id=player_id,team=team,week=rep(week,4),points)
roll_mean <- function(x, k) {
result <- rollapplyr(x, k, mean, partial=TRUE, na.rm=TRUE)
result[is.nan(result)] <- NA
return( result )
}
It might first be easier to aggregate by team:
team_data <- mydata %>%
select(-player_id) %>%
group_by(team, week) %>%
arrange(week) %>%
summarise(team_points = sum(points)) %>%
mutate(rolling_team_mean = roll_mean(lag(team_points), k=5)) %>%
arrange(team)
team_data
# A tibble: 30 x 4
# Groups: team [2]
team week team_points rolling_team_mean
<fctr> <int> <dbl> <dbl>
1 A 1 13 NA
2 A 2 11 13.00
3 A 3 6 12.00
4 A 4 13 10.00
5 A 5 19 10.75
6 A 6 10 12.40
7 A 7 13 11.80
8 A 8 16 12.20
9 A 9 16 14.20
10 A 10 12 14.80
# ... with 20 more rows
Then, if you like we can put everything back together:
mydata <- inner_join(mydata, team_data) %>%
arrange(week, team, player_id)
mydata[1:12, ]
player_id team week points team_points rolling_team_mean
1 1 A 1 4 13 NA
2 2 A 1 9 13 NA
3 3 B 1 10 12 NA
4 4 B 1 2 12 NA
5 1 A 2 8 11 13
6 2 A 2 3 11 13
7 3 B 2 9 12 12
8 4 B 2 3 12 12
9 1 A 3 5 6 12
10 2 A 3 1 6 12
11 3 B 3 7 12 12
12 4 B 3 5 12 12
Here's one way:
# compute points per team per week
pts <- with(mydata, tapply(points, list(team, week), sum, default = 0))
pts
# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
#A 13 11 6 13 19 10 13 16 16 12 17 11 13 10 4
#B 12 12 12 11 10 6 13 11 6 9 5 7 13 13 6
# compute the 5-week averages
sapply(setNames(seq(2, ncol(pts)), seq(2, ncol(pts))),
function(i) {
apply(pts[, seq(max(1, i - 5), i - 1), drop = FALSE], 1, mean)
})
# 2 3 4 5 6 7 8 9 10 11 12 13 14 15
#A 13 12 10 10.75 12.4 11.8 12.2 14.2 14.8 13.4 14.8 14.4 13.8 12.6
#B 12 12 12 11.75 11.4 10.2 10.4 10.2 9.2 9.0 8.8 7.6 8.0 9.4
This will give the wrong result if the week variable has gaps.
I am a R noob, and hope some of you can help me.
I have two data sets:
- store (containing store data, including location coordinates (x,y). The location are integer values, corresponding to GridIds)
- grid (containing all gridIDs (x,y) as well as a population variable TOT_P for each grid point)
What I want to achieve is this:
For each store I want loop over the grid date, and sum the population of the grid ids close to the store grid id.
I.e basically SUMIF the grid population variable, with the condition that
grid(x) < store(x) + 1 &
grid(x) > store(x) - 1 &
grid(y) < store(y) + 1 &
grid(y) > store(y) - 1
How can I accomplish that? My own take has been trying to use different things like merge, sapply, etc, but my R inexperience stops me from getting it right.
Thanks in advance!
Edit:
Sample data:
StoreName StoreX StoreY
Store1 3 6
Store2 5 2
TOT_P GridX GridY
8 1 1
7 2 1
3 3 1
3 4 1
22 5 1
20 6 1
9 7 1
28 1 2
8 2 2
3 3 2
12 4 2
12 5 2
15 6 2
7 7 2
3 1 3
3 2 3
3 3 3
4 4 3
13 5 3
18 6 3
3 7 3
61 1 4
25 2 4
5 3 4
20 4 4
23 5 4
72 6 4
14 7 4
178 1 5
407 2 5
26 3 5
167 4 5
58 5 5
113 6 5
73 7 5
76 1 6
3 2 6
3 3 6
3 4 6
4 5 6
13 6 6
18 7 6
3 1 7
61 2 7
25 3 7
26 4 7
167 5 7
58 6 7
113 7 7
The output I am looking for is
StoreName StoreX StoreY SUM_P
Store1 3 6 479
Store2 5 2 119
I.e for store1 it is the sum of TOT_P for Grid fields X=[2-4] and Y=[5-7]
One approach would be to use dplyr to calculate the difference between each store and all grid points and then group and sum based on these new columns.
#import library
library(dplyr)
#create example store table
StoreName<-paste0("Store",1:2)
StoreX<-c(3,5)
StoreY<-c(6,2)
df.store<-data.frame(StoreName,StoreX,StoreY)
#create example population data (copied example table from OP)
df.pop
#add dummy column to each table to enable cross join
df.store$k=1
df.pop$k=1
#dplyr to join, calculate absolute distance, filter and sum
df.store %>%
inner_join(df.pop, by='k') %>%
mutate(x.diff = abs(StoreX-GridX), y.diff=abs(StoreY-GridY)) %>%
filter(x.diff<=1, y.diff<=1) %>%
group_by(StoreName) %>%
summarise(StoreX=max(StoreX), StoreY=max(StoreY), tot.pop = sum(TOT_P) )
#output:
StoreName StoreX StoreY tot.pop
<fctr> <dbl> <dbl> <int>
1 Store1 3 6 721
2 Store2 5 2 119
I am trying to shorten a chunk of code to make it faster and easier to modify. This is a short example of my data.
order obs year var1 var2 var3
1 3 1 1 32 588 NA
2 4 1 2 33 689 2385
3 5 1 3 NA 678 2369
4 33 3 1 10 214 1274
5 34 3 2 10 237 1345
6 35 3 3 10 242 1393
7 78 6 1 5 62 NA
8 79 6 2 5 75 296
9 80 6 3 5 76 500
10 93 7 1 NA NA NA
11 94 7 2 4 86 247
12 95 7 3 3 54 207
Basically, what I want is R to find any possible and unique combination of two values (observations) in column "obs", within the same year, to create a new matrix or DF with observations being the aggregation of the originals. Order is not important, so 1+6 = 6+1. For instance, having 150 observations, I will expect 11,175 feasible combinations (each year).
I sort of got what I want with basic coding but, as you will see, is way too long (I have built this way 66 different new data sets so it does not really make a sense) and I am wondering how to shorten it. I did some trials (plyr,...) with no real success. Here what I did:
# For the 1st year, groups of 2 obs
newmatrix <- data.frame(t(combn(unique(data$obs[data$year==1]), 2)))
colnames(newmatrix) <- c("obs1", "obs2")
newmatrix$name <- do.call(paste, c(newmatrix[c("obs1", "obs2")], sep = "_"))
# and the aggregation of var. using indexes, which I will skip here to save your time :)
To ilustrate, here the result, considering above sample, of what I would get for the 1st year. NA is because I only computed those where the 2 values were valid. And only for variables 1 and 3. More, I did the sum but it could be any other possible Function:
order obs1 obs2 year var1 var3
1 1 1 3 1_3 42 NA
2 2 1 6 1_6 37 NA
3 3 1 7 1_7 NA NA
4 4 3 6 3_6 15 NA
5 5 3 7 3_7 NA NA
6 6 6 7 6_7 NA NA
As for the 2 first lines in the 3rd year, same type of matrix:
order obs1 obs2 year var1 var3
1 1 1 3 1_3 NA 3762
2 2 1 6 1_6 NA 2868
.......... etc ............
I hope I explained myself. Thank you in advance for your hints on how to do this more efficient.
I would use split-apply-combine to split by year, find all the combinations, and then combine back together:
do.call(rbind, lapply(split(data, data$year), function(x) {
p <- combn(nrow(x), 2)
data.frame(order=paste(x$order[p[1,]], x$order[p[2,]], sep="_"),
obs1=x$obs[p[1,]],
obs2=x$obs[p[2,]],
year=x$year[1],
var1=x$var1[p[1,]] + x$var1[p[2,]],
var2=x$var2[p[1,]] + x$var2[p[2,]],
var3=x$var3[p[1,]] + x$var3[p[2,]])
}))
# order obs1 obs2 year var1 var2 var3
# 1.1 3_33 1 3 1 42 802 NA
# 1.2 3_78 1 6 1 37 650 NA
# 1.3 3_93 1 7 1 NA NA NA
# 1.4 33_78 3 6 1 15 276 NA
# 1.5 33_93 3 7 1 NA NA NA
# 1.6 78_93 6 7 1 NA NA NA
# 2.1 4_34 1 3 2 43 926 3730
# 2.2 4_79 1 6 2 38 764 2681
# 2.3 4_94 1 7 2 37 775 2632
# 2.4 34_79 3 6 2 15 312 1641
# 2.5 34_94 3 7 2 14 323 1592
# 2.6 79_94 6 7 2 9 161 543
# 3.1 5_35 1 3 3 NA 920 3762
# 3.2 5_80 1 6 3 NA 754 2869
# 3.3 5_95 1 7 3 NA 732 2576
# 3.4 35_80 3 6 3 15 318 1893
# 3.5 35_95 3 7 3 13 296 1600
# 3.6 80_95 6 7 3 8 130 707
This enables you to be very flexible in how you combine data pairs of observations within a year --- x[p[1,],] represents the year-specific data for the first element in each pair and x[p[2,],] represents the year-specific data for the second element in each pair. You can return a year-specific data frame with any combination of data for the pairs, and the year-specific data frames are combined into a single final data frame with do.call and rbind.
I want to make new column in my data set with the values determined by values in another data set, but it's not as simple as the values in one column being a function of the values in the other. Here's an example:
>df1
chromosome position
1 1 1
2 1 2
3 1 4
4 1 5
5 1 7
6 1 12
7 1 13
8 1 15
9 1 21
10 1 23
11 1 24
12 2 1
13 2 5
14 2 7
15 2 8
16 2 12
17 2 15
18 2 18
19 2 21
20 2 22
and
>df2
chromosome segment_start segment_end segment.number
1 1 1 5 1.1
2 1 6 20 1.2
3 1 21 25 1.3
4 2 1 7 2.1
5 2 8 16 2.2
6 2 18 22 2.3
I want to make a new column in df1 called 'segment', and the value in segment is to be determined by which segment (as determined by 'segment_start', 'segment_end', and 'chromosome' from df2) the value in 'position' belongs to. For example, in df1, row 7, position=13, and chromosome=1. Because 13 is between 6 and 20, the entry in my hypothetical 'segment' column would be 1.2, from row 2 of df2, because 13 falls between segment_start and segment_end from that row (6 and 20, respectively), and the 'chromosome' value from df1 row 7 is 1, just as 'chromosome' in df2 row 2 is 1.
Each row in df1 belongs to one of the segments described in df2; that is, it lies on the same chromosome as one of the segments, and its 'position' is >=segment_start and <=segment_end. And I want to get that information into df1, so it says what segment each position belongs to.
I was thinking of using an if function, and started with:
if(df1$position>=df2$segment_start & df1$position<=df2$segment_end & df1$chromosome==df2$chromosome) df1$segment<-df2$segment.number
But am not sure that way will be feasible. If nothing else maybe the code can help illustrate what it is I'm trying to do. Basically, I want match each row by its position and chromosome to a segment in df2. Thanks.
This appears to be a rolling join. You can use data.table for this
require(data.table)
DT1 <- data.table(df1, key = c('chromosome','position'))
DT2 <- data.table(df2, key = c('chromosome','section_start'))
# this will perform the join you want (but retain all the
# columns with names names of DT2)
# DT2[DT1, roll=TRUE]
# which is why I have renamed and subset here)
DT2[DT1, roll=TRUE][ ,list(chromosome,position = segment_start,segment.number)]
# chromosome position segment.number
# 1: 1 1 1.1
# 2: 1 2 1.1
# 3: 1 4 1.1
# 4: 1 5 1.1
# 5: 1 7 1.2
# 6: 1 12 1.2
# 7: 1 13 1.2
# 8: 1 15 1.2
# 9: 1 21 1.3
# 10: 1 23 1.3
# 11: 1 24 1.3
# 12: 2 1 2.1
# 13: 2 5 2.1
# 14: 2 7 2.1
# 15: 2 8 2.2
# 16: 2 12 2.2
# 17: 2 15 2.2
# 18: 2 18 2.3
# 19: 2 21 2.3
# 20: 2 22 2.3
You really need to check out the GenomicRanges package from Bioconductor. It provides the data structures that are appropriate for your use case.
First, we create the GRanges objects:
gr1 <- with(df1, GRanges(chromosome, IRanges(position, width=1L)))
gr2 <- with(df2, GRanges(chromosome, IRanges(segment_start, segment_end),
segment.number=segment.number))
Then we find the overlaps and do the merge:
hits <- findOverlaps(gr1, gr2)
gr1$segment[queryHits(hits)] <- gr2$segment.number[subjectHits(hits)]
I'm going to assume that the regions in df2 are non-overlapping, continuous and complete (not missing any positions from df1). I seem to do this differently every time I try, so here's my latest idea.
First, make sure chromosome is a factor in both data sets
df1$chromosome<-factor(df1$chromosome)
df2$chromosome<-factor(df2$chromosome)
Now I want to unwrap, chr/pos into one over all generic position, i'll do that with
ends<-with(df2, tapply(segment_end, chromosome, max))
offset<-head(c(0,cumsum(ends)),-1)
names(offset)<-names(ends)
This assigns unique position values to all positions across all chromosomes and it tracks the offset to the beginning of each chromosome in this new system. Now we will build a translation function from the data in df2
seglookup <- approxfun(with(df2, offset[chromosome]+segment_start), 1:nrow(df2),
method="constant", rule=2)
We use approxfun to find the right interval in the genetic position space for each segment. Now we use this function on df1
segid <- with(df1, seglookup(offset[chromosome]+position))
Now we have the correct ID for each position. We can verify this by merging the data and looking at the results
cbind(df1, df2[segid,-1])
chromosome position segment_start segment_end segment.number
1 1 1 1 5 1.1
2 1 2 1 5 1.1
3 1 4 1 5 1.1
4 1 5 1 5 1.1
5 1 7 6 20 1.2
6 1 12 6 20 1.2
7 1 13 6 20 1.2
8 1 15 6 20 1.2
9 1 21 21 25 1.3
10 1 23 21 25 1.3
11 1 24 21 25 1.3
12 2 1 1 7 2.1
13 2 5 1 7 2.1
14 2 7 1 7 2.1
15 2 8 8 16 2.2
16 2 12 8 16 2.2
17 2 15 8 16 2.2
18 2 18 18 22 2.3
19 2 21 18 22 2.3
20 2 22 18 22 2.3
So it looks like we did alright.