I have a long data frame with players' decisions who worked in groups.
I need to convert the data in such a way that each row (individual observation) would contain all group members decisions (so we basically can see whether they are interdependent).
Let's say the generating code is:
group_id <- c(rep(1, 3), rep(2, 3))
player_id <- c(rep(seq(1, 3), 2))
player_decision <- seq(10,60,10)
player_contribution <- seq(6,1,-1)
df <-
data.frame(group_id, player_id, player_decision, player_contribution)
So the initial data looks like:
group_id player_id player_decision player_contribution
1 1 1 10 6
2 1 2 20 5
3 1 3 30 4
4 2 1 40 3
5 2 2 50 2
6 2 3 60 1
But I need to convert it to wide per each group, but only for some of these variables, (in this example specifically for player_contribution, but in such a way that the rest of the data remains. So the head of the converted data would be:
data.frame(group_id=c(1,1),
player_id=c(1,2),
player_decision=c(10,20),
player_1_contribution=c(6,6),
player_2_contribution=c(5,5),
player_3_contribution=c(4,6)
)
group_id player_id player_decision player_1_contribution player_2_contribution player_3_contribution
1 1 1 10 6 5 4
2 1 2 20 6 5 6
I suspect I need to group_by in dplyr and then somehow gather per group but only for player_contribution (or a vector of variables). But I really have no clue how to approach it. Any hints would be welcome!
Here is solution using tidyr and dplyr.
Make a dataframe with the columns for the players contributions. Then join this dataframe back onto the columns of interest from the original Dataframe.
library(tidyr)
library(dplyr)
wide<-pivot_wider(df, id_cols= - player_decision,
names_from = player_id,
values_from = player_contribution,
names_prefix = "player_contribution_")
answer<-left_join(df[, c("group_id", "player_id", "player_decision") ], wide)
answer
group_id player_id player_decision player_contribution_1 player_contribution_2 player_contribution_3
1 1 1 10 6 5 4
2 1 2 20 6 5 4
3 1 3 30 6 5 4
4 2 1 40 3 2 1
5 2 2 50 3 2 1
6 2 3 60 3 2 1
I have the dataframe ds
CountyID ZipCode Value1 Value2 Value3 ... Value25
1 1 0 etc etc etc
2 1 3
3 1 0
4 1 1
5 2 2
6 3 3
7 4 7
8 4 2
9 5 1
10 6 0
and would like to aggregate based on ds$ZipCode and set ds$CountyID equal to the primary county based on the highest ds$Value1. For the above example, it would look like this:
CountyID ZipCode Value1 Value2 Value3 ... Value25
2 1 4 etc etc etc
5 2 2
6 3 3
7 4 9
9 5 1
10 6 0
All the ValueX columns are the sum of that column grouped by ZipCode.
I've tried a bunch of different strategies over the last couple days, but none of them work. The best I've come up with is
#initialize the dataframe
ds_temp = data.frame()
#loop through each subset based on unique zipcodes
for (zip in unique(ds$ZipCode) {
sub <- subset(ds, ds$ZipCode == zip)
len <- length(sub)
maxIndex <- which.max(sub$Value1)
#do the aggregation
row <- aggregate(sub[3:27], FUN=sum, by=list(
CountyID = rep(sub$CountyID[maxIndex], len),
ZipCode = sub$ZipCode))
rbind(ds_temp, row)
}
ds <- ds_temp
I haven't been able to test this on the real data, but with dummy datasets (such as the one above), I keep getting the error "arguments must have the same length). I've messed around with rep() and fixed vectors (eg c(1,2,3,4)) but no matter what I do, the error persists. I also occasionally get an error to the effect of
cannot subset data of type 'closure'.
Any ideas? I've also tried messing around with data.frame(), ddply(), data.table(), dcast(), etc.
You can try this:
data.frame(aggregate(df[,3:27], by=list(df$ZipCode), sum),
CountyID = unlist(lapply(split(df, df$ZipCode),
function(x) x$CountyID[which.max(x$Value1)])))
Fully reproducible sample data:
df<-read.table(text="
CountyID ZipCode Value1
1 1 0
2 1 3
3 1 0
4 1 1
5 2 2
6 3 3
7 4 7
8 4 2
9 5 1
10 6 0", header=TRUE)
data.frame(aggregate(df[,3], by=list(df$ZipCode), sum),
CountyID = unlist(lapply(split(df, df$ZipCode),
function(x) x$CountyID[which.max(x$Value1)])))
# Group.1 x CountyID
#1 1 4 2
#2 2 2 5
#3 3 3 6
#4 4 9 7
#5 5 1 9
#6 6 0 10
In response to your comment on Frank's answer, you can preserve the column names by using the formula method in aggregate. Using Franks's data df, this would be
> cbind(aggregate(Value1 ~ ZipCode, df, sum),
CountyID = sapply(split(df, df$ZipCode), function(x) {
with(x, CountyID[Value1 == max(Value1)]) }))
# ZipCode Value1 CountyID
# 1 1 4 2
# 2 2 2 5
# 3 3 3 6
# 4 4 9 7
# 5 5 1 9
# 6 6 0 10
Here is the data.
set.seed(23) data<-data.frame(ID=rep(1:12), group=rep(1:3,times=4), value=(rnorm(12,mean=0.5, sd=0.3)))
ID group value
1 1 1 0.4133934
2 2 2 0.6444651
3 3 3 0.1350871
4 4 1 0.5924411
5 5 2 0.3439465
6 6 3 0.3673059
7 7 1 0.3202062
8 8 2 0.8883733
9 9 3 0.7506174
10 10 1 0.3301955
11 11 2 0.7365258
12 12 3 0.1502212
I want to get z-standardized scores within each group. so I try
library(weights)
data_split<-split(data, data$group) #split the dataframe
stan<-lapply(data_split, function(x) stdz(x$value)) #compute z-scores within group
However, It looks wrong because I want to add a new variable following 'value'
How can I do that? Kindly provide some suggestions(sample code). Any help is greatly appreciated .
Use this instead:
within(data, stan <- ave(value, group, FUN=stdz))
No need to call split nor lapply.
One way using data.table package:
library(data.table)
library(weights)
set.seed(23)
data <- data.table(ID=rep(1:12), group=rep(1:3,times=4), value=(rnorm(12,mean=0.5, sd=0.3)))
setkey(data, ID)
dataNew <- data[, list(ID, stan = stdz(value)), by = 'group']
the result is:
group ID stan
1: 1 1 -0.6159312
2: 1 4 0.9538398
3: 1 7 -1.0782747
4: 1 10 0.7403661
5: 2 2 -1.2683237
6: 2 5 0.7839781
7: 2 8 0.8163844
8: 2 11 -0.3320388
9: 3 3 0.6698418
10: 3 6 0.8674548
11: 3 9 -0.2131335
12: 3 12 -1.3241632
I tried Ferdinand.Kraft's solution but it didn't work for me. I think the stdz function isn't included in the basic R install. Moreover, the within part troubled me in a large dataset with many variables. I think the easiest way is:
data$value.s <- ave(data$value, data$group, FUN=scale)
Add the new column while in your function, and have the function return the whole data frame.
stanL<-lapply(data_split, function(x) {
x$stan <- stdz(x$value)
x
})
stan <- do.call(rbind, stanL)
I have a data frame in R which is similar to the follows. Actually my real ’df’ dataframe is much bigger than this one here but I really do not want to confuse anybody so that is why I try to simplify things as much as possible.
So here’s the data frame.
id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,3)
df <-data.frame(id,a,b,c,d,e)
df
Basically what I would like to do is to get the occurrences of numbers for each column (a,b,c,d,e) and for each id group (1,2,3) (for this latter grouping see my column ’id’).
So, for column ’a’ and for id number ’1’ (for the latter see column ’id’) the code would be something like this:
as.numeric(table(df[1:10,2]))
##The results are:
[1] 3 7
Just to briefly explain my results: in column ’a’ (and regarding only those records which have number ’1’ in column ’id’) we can say that number '1' occured 3 times and number '3' occured 7 times.
Again, just to show you another example. For column ’a’ and for id number ’2’ (for the latter grouping see again column ’id’):
as.numeric(table(df[11:20,2]))
##After running the codes the results are:
[1] 4 3 3
Let me explain a little again: in column ’a’ and regarding only those observations which have number ’2’ in column ’id’) we can say that number '1' occured 4 times, number '2' occured 3 times and number '3' occured 3 times.
So this is what I would like to do. Calculating the occurrences of numbers for each custom-defined subsets (and then collecting these values into a data frame). I know it is not a difficult task but the PROBLEM is that I’m gonna have to change the input ’df’ dataframe on a regular basis and hence both the overall number of rows and columns might change over time…
What I have done so far is that I have separated the ’df’ dataframe by columns, like this:
for (z in (2:ncol(df))) assign(paste("df",z,sep="."),df[,z])
So df.2 will refer to df$a, df.3 will equal df$b, df.4 will equal df$c etc. But I’m really stuck now and I don’t know how to move forward…
Is there a proper, ”automatic” way to solve this problem?
How about -
> library(reshape)
> dftab <- table(melt(df,'id'))
> dftab
, , value = 1
variable
id a b c d e
1 3 8 2 2 4
2 4 6 3 2 4
3 4 2 1 5 1
, , value = 2
variable
id a b c d e
1 0 1 4 3 3
2 3 3 3 6 2
3 1 4 5 3 4
, , value = 3
variable
id a b c d e
1 7 1 4 5 3
2 3 1 4 2 4
3 5 4 4 2 5
So to get the number of '3's in column 'a' and group '1'
you could just do
> dftab[3,'a',1]
[1] 4
A combination of tapply and apply can create the data you want:
tapply(df$id,df$id,function(x) apply(df[id==x,-1],2,table))
However, when a grouping doesn't have all the elements in it, as in 1a, the result will be a list for that id group rather than a nice table (matrix).
$`1`
$`1`$a
1 3
3 7
$`1`$b
1 2 3
8 1 1
$`1`$c
1 2 3
2 4 4
$`1`$d
1 2 3
2 3 5
$`1`$e
1 2 3
4 3 3
$`2`
a b c d e
1 4 6 3 2 4
2 3 3 3 6 2
3 3 1 4 2 4
$`3`
a b c d e
1 4 2 1 5 1
2 1 4 5 3 4
3 5 4 4 2 5
I'm sure someone will have a more elegant solution than this, but you can cobble it together with a simple function and dlply from the plyr package.
ColTables <- function(df) {
counts <- list()
for(a in names(df)[names(df) != "id"]) {
counts[[a]] <- table(df[a])
}
return(counts)
}
results <- dlply(df, "id", ColTables)
This gets you back a list - the first "layer" of the list will be the id variable; the second the table results for each column for that id variable. For example:
> results[['2']]['a']
$a
1 2 3
4 3 3
For id variable = 2, column = a, per your above example.
A way to do it is using the aggregate function, but you have to add a column to your dataframe
> df$freq <- 0
> aggregate(freq~a+id,df,length)
a id freq
1 1 1 3
2 3 1 7
3 1 2 4
4 2 2 3
5 3 2 3
6 1 3 4
7 2 3 1
8 3 3 5
Of course you can write a function to do it, so it's easier to do it frequently, and you don't have to add a column to your actual data frame
> frequency <- function(df,groups) {
+ relevant <- df[,groups]
+ relevant$freq <- 0
+ aggregate(freq~.,relevant,length)
+ }
> frequency(df,c("b","id"))
b id freq
1 1 1 8
2 2 1 1
3 3 1 1
4 1 2 6
5 2 2 3
6 3 2 1
7 1 3 2
8 2 3 4
9 3 3 4
You didn't say how you'd like the data. The by function might give you the output you like.
by(df, df$id, function(x) lapply(x[,-1], table))