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 a data frame in which each individual (row) has two data points per variable.
Example data:
df1 <- read.table(text = "IID L1.1 L1.2 L2.1 L2.2
1 1 38V1 38V1 48V1 52V1
2 2 36V1 38V2 50V1 48Y1
3 3 37Y1 36V1 50V2 48V1
4 4 38V2 36V2 52V1 50V2",
stringsAsFactor = FALSE, header = TRUE)
I have many more columns than this in the full dataset and would like to recode these values to label unique identifiers across the two columns. I know how to get identifiers and relabel a single column from previous questions (Creating a unique ID and How to assign a unique ID number to each group of identical values in a column) but I don't know how to include the information for two columns, as R identifies and labels factors per column.
Ultimately I want something that would look like this for the above data:
(df2)
IID L1.1 L1.2 L2.1 L2.2
1 1 1 1 1 4
2 2 2 4 2 5
3 3 3 2 3 1
4 4 1 5 4 3
It doesn't really matter what the numbers are, as long as they indicate unique values across both columns. I've tried creating a function based on the output from:
unique(df1[,1:2])
but am struggling as this still looks at unique entries per column, not across the two.
Something like this would work...
pairs <- (ncol(df1)-1)/2
for(i in 1:pairs){
refs <- unique(c(df1[,2*i],df1[,2*i+1]))
df1[,2*i] <- match(df1[,2*i],refs)
df1[,2*i+1] <- match(df1[,2*i+1],refs)
}
df1
IID L1.1 L1.2 L2.1 L2.2
1 1 1 1 1 4
2 2 2 4 2 5
3 3 3 2 3 1
4 4 4 5 4 3
You could reshape it to long format, assign the groups and then recast it to wide:
library(data.table)
df_m <- melt(df, id.vars = "IID")
setDT(df_m)[, id := .GRP, by = .(gsub("(.*).","\\1", df_m$variable), value)]
dcast(df_m, IID ~ variable, value.var = "id")
# IID L1.1 L1.2 L2.1 L2.2
#1 1 1 1 6 9
#2 2 2 4 7 10
#3 3 3 2 8 6
#4 4 1 5 9 8
This should also be easily expandable to multiple groups of columns. I.e. if you have L3. it should work with that as well.
I have a table like the one below with 100's of rows of data.
ID RANK
1 2
1 3
1 3
2 4
2 8
3 3
3 3
3 3
4 6
4 7
4 7
4 7
4 7
4 7
4 6
I want to try to find a way to group the data by ID so that I can ReRank each group separately. The ReRank column is based on the Rank column and basically renumbering it starting at 1 from least to greatest, but it's important to note that the the number in the ReRank column can be put in more than once depending on the numbers in the Rank column .
In other words, the output needs to look like this
ID Rank ReRANK
1 3 2
1 2 1
1 3 2
2 4 1
2 8 2
3 3 1
3 3 1
3 3 1
For the life of me, I can't figure out how to be able to ReRank the the columns by the grouped columns and the value of the Rank columns.
This has been my best guess so far, but it definitely is not doing what I need it to do
ReRANK = mat.or.vec(length(RANK),1)
ReRANK[1] = counter = 1
for(i in 2:length(RANK)) {
if (RANK[i] != RANK[i-1]) { counter = counter + 1 }
ReRANK[i] = counter
}
Thank you in advance for the help!!
Here is a base R method using ave and rank:
df$ReRank <- ave(df$Rank, df$ID, FUN=function(i) rank(i, ties.method="min"))
The min argument in rank assures that the minimum ranking will occur when there are ties. the default is to take the mean of the ranks.
In the case that you have ties lower down in the groups, rank will count those lower values and then add continue with the next lowest value as the count of the lower values + 1. These values wil still be ordered and distinct. If you really want to have the count be 1, 2, 3, and so on rather than 1, 3, 6 or whatever depending on the number of duplicate values, here is a little hack using factor:
df$ReRank <- ave(df$Rank, df$ID, FUN=function(i) {
as.integer(factor(rank(i, ties.method="min"))))
Here, we use factor to build values counting from upward for each level. We then coerce it to be an integer.
For example,
temp <- c(rep(1, 3), 2,5,1,4,3,7)
[1] 2.5 2.5 2.5 5.0 8.0 2.5 7.0 6.0 9.0
rank(temp, ties.method="min")
[1] 1 1 1 5 8 1 7 6 9
as.integer(factor(rank(temp, ties.method="min")))
[1] 1 1 1 2 5 1 4 3 6
data
df <- read.table(header=T, text="ID Rank
1 2
1 3
1 3
2 4
2 8
3 3
3 3
3 3 ")
I need a sequence of repeated numbers, i.e. 1 1 ... 1 2 2 ... 2 3 3 ... 3 etc. The way I implemented this was:
nyear <- 20
names <- c(rep(1,nyear),rep(2,nyear),rep(3,nyear),rep(4,nyear),
rep(5,nyear),rep(6,nyear),rep(7,nyear),rep(8,nyear))
which works, but is clumsy, and obviously doesn't scale well.
How do I repeat the N integers M times each in sequence?
I tried nesting seq() and rep() but that didn't quite do what I wanted.
I can obviously write a for-loop to do this, but there should be an intrinsic way to do this!
You missed the each= argument to rep():
R> n <- 3
R> rep(1:5, each=n)
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
R>
so your example can be done with a simple
R> rep(1:8, each=20)
Another base R option could be gl():
gl(5, 3)
Where the output is a factor:
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
Levels: 1 2 3 4 5
If integers are needed, you can convert it:
as.numeric(gl(5, 3))
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
For your example, Dirk's answer is perfect. If you instead had a data frame and wanted to add that sort of sequence as a column, you could also use group from groupdata2 (disclaimer: my package) to greedily divide the datapoints into groups.
# Attach groupdata2
library(groupdata2)
# Create a random data frame
df <- data.frame("x" = rnorm(27))
# Create groups with 5 members each (except last group)
group(df, n = 5, method = "greedy")
x .groups
<dbl> <fct>
1 0.891 1
2 -1.13 1
3 -0.500 1
4 -1.12 1
5 -0.0187 1
6 0.420 2
7 -0.449 2
8 0.365 2
9 0.526 2
10 0.466 2
# … with 17 more rows
There's a whole range of methods for creating this kind of grouping factor. E.g. by number of groups, a list of group sizes, or by having groups start when the value in some column differs from the value in the previous row (e.g. if a column is c("x","x","y","z","z") the grouping factor would be c(1,1,2,3,3).
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))