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 ")
In a dataframe I'd like to replace values in a series where they exceed a given threshold.
For example, within a group ('ID') in a series designated by 'time', if 'value' ever exceeds 3, I'd like to make all following entries also equal 3.
ID <- as.factor(c(rep("A", 3), rep("B",3), rep("C",3)))
time <- rep(1:3, 3)
value <- c(c(1,1,2), c(2,3,2), c(3,3,2))
dat <- cbind.data.frame(ID, time, value)
dat
ID time value
A 1 1
A 2 1
A 3 2
B 1 2
B 2 3
B 3 2
C 1 3
C 2 3
C 3 2
I'd like it to be:
ID time value
A 1 1
A 2 1
A 3 2
B 1 2
B 2 3
B 3 3
C 1 3
C 2 3
C 3 3
This should be easy, but I can't figure it out. Thanks!
The ave function makes this very easy by allowing you to apply a function to each of the groupings. In this case, we will adapth the cummax (cumulative maximum) to see if we've seen a 3 yet.
dat$value2<-with(dat, ave(value, ID, FUN=
function(x) ifelse(cummax(x)>=3, 3, x)))
dat;
# ID time value value2
# 1 A 1 1 1
# 2 A 2 1 1
# 3 A 3 2 2
# 4 B 1 2 2
# 5 B 2 3 3
# 6 B 3 2 3
# 7 C 1 3 3
# 8 C 2 3 3
# 9 C 3 2 3
You could also just use FUN=cummax if you want never-decreasing values. I wasn't sure about the sequence c(1,2,1) if you wanted to keep that unchanged or not.
If you can assume your data are sorted by group, then this should be fast, essentially relying on findInterval() behind the scenes:
library(IRanges)
id <- Rle(ID)
three <- which(value>=3L)
ir <- reduce(IRanges(three, end(id)[findRun(three, id)])))
dat$value[as.integer(ir)] <- 3L
This avoids looping over the groups.
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))