I'd like to add rows to a dataframe based on a vector within the dataframe. Here are the dataframes (df2 is the one I'd like to add rows to; df1 is the one I'd like to take the rows from):
ID=c(1:5)
x=c(rep("a",3),rep("b",2))
y=c(rep(0,5))
df1=data.frame(ID,x,y)
df2=df1[2:4,1:2]
df2$y=c(5,2,3)
df1
ID x y
1 1 a 0
2 2 a 0
3 3 a 0
4 4 b 0
5 5 b 0
df2
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
I'd like to add to df2 any rows that aren't in df1, based on the ID vector. so my output dataframe would look like this:
ID x y
1 b 0
5 b 0
2 a 5
3 a 2
4 b 3
Can anyone see a way of doing this neatly, please? I need to do it for a lot of dataframes, all with different numbers of rows. I've tried using merge or rbind but I haven't been able to work out how to do it based on the vector.
Thank you!
A solution with dplyr:
bind_rows(df2,anti_join(df1,df2,by="ID"))
# ID x y
#1 2 a 5
#2 3 a 2
#3 4 b 3
#4 1 a 0
#5 5 b 0
You can do the following:
missingIDs <- which(!df1$ID %in% df2$ID) #check which df1 ID's are not in df2, see function is.element()
df.toadd <- df1[missingIDs,] #define the data frame to add to df2
result <- rbind(df.toadd, df2) #use rbind to add it
result
ID x y
1 1 a 0
5 5 b 0
2 2 a 5
3 3 a 2
4 4 b 3
What about this one-liner?
rbind(df2, df1[!df1$ID %in% df2$ID,])
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
1 1 a 0
5 5 b 0
I need to create (with R) a rolling index of pairs from a data set that includes groups. Consider the following data set:
times <- c(4,3,2)
V1 <- unlist(lapply(times, function(x) seq(1, x)))
df <- data.frame(group = rep(1:length(times), times = times),
V1 = V1,
rolling_index = c(1,1,2,2,3,3,4,5,5))
df
group V1 rolling_index
1 1 1 1
2 1 2 1
3 1 3 2
4 1 4 2
5 2 1 3
6 2 2 3
7 2 3 4
8 3 1 5
9 3 2 5
The data frame I have includes the variables group and V1. Within each group V1 designates a running index (that may or may not start at 1).
I want to create a new indexing variable that looks like rolling_index. This variable groups rows within the same group and consecutive V1 value, thus creating a new rolling index. This new index must be consecutive over groups. If there is an uneven amount of rows within a group (e.g. group 2), then the last, single row gets its own rolling index value.
You can try
library(data.table)
setDT(df)[, gr:=as.numeric(gl(.N, 2, .N)), group][,
rollindex:=cumsum(c(TRUE,abs(diff(gr))>0))][,gr:= NULL]
# group V1 rolling_index rollindex
#1: 1 1 1 1
#2: 1 2 1 1
#3: 1 3 2 2
#4: 1 4 2 2
#5: 2 1 3 3
#6: 2 2 3 3
#7: 2 3 4 4
#8: 3 1 5 5
#9: 3 2 5 5
Or using base R
indx1 <- !duplicated(df$group)
indx2 <- with(df, ave(group, group, FUN=function(x)
gl(length(x), 2, length(x))))
cumsum(c(TRUE,diff(indx2)>0)|indx1)
#[1] 1 1 2 2 3 3 4 5 5
Update
The above methods are based on the 'group' column. Suppose you already have a sequence column ('V1') by group as showed in the example, creation of rolling index is easier
cumsum(!!df$V1 %%2)
#[1] 1 1 2 2 3 3 4 5 5
As mentioned in the post, if the 'V1' column do not start at '1' for some groups, we can get the sequence from the 'group' and then do the cumsum as above
cumsum(!!with(df, ave(seq_along(group), group, FUN=seq_along))%%2)
#[1] 1 1 2 2 3 3 4 5 5
There is probably a simpler way but you can do:
rep_each <- unlist(mapply(function(q,r) {c(rep(2, q),rep(1, r))},
q=table(df$group)%/%2,
r=table(df$group)%%2))
df$rolling_index <- inverse.rle(x=list(lengths=rep_each, values=seq(rep_each)))
df$rolling_index
#[1] 1 1 2 2 3 3 4 5 5
I would like to break a dataset into two frames - one for which the original dataset has duplicate observations based on a condition and one for which the original dataset does not have duplicate observations based on a condition. In the following example, I would like to break the frame into one for which there is only one coder for an observation and one for which there are two coders::
frame <- data.frame(id = c(1,1,1,2,2,3), coder = c("A", "A", "B", "A", "B", "A"), y = c(4,5,4,1,1,2))
frame
For this, I would like to produce, such that:
frame1:
id coder y
1 1 A 4
2 1 A 5
3 1 B 4
4 2 A 1
5 2 B 1
frame2:
6 3 A 2
You can use aggregate to determine the ids you want in each data frame:
cts <- aggregate(coder~id, frame, function(x) length(unique(x)))
cts
# id coder
# 1 1 2
# 2 2 2
# 3 3 1
Then you can subset as appropriate based on this:
subset(frame, id %in% cts$id[cts$coder >= 2])
# id coder y
# 1 1 A 4
# 2 1 A 5
# 3 1 B 4
# 4 2 A 1
# 5 2 B 1
subset(frame, id %in% cts$id[cts$coder < 2])
# id coder y
# 6 3 A 2
You may also try:
indx <- !colSums(!table(frame$coder, frame$id))
frame[frame$id %in% names(indx)[indx],]
# id coder y
#1 1 A 4
#2 1 A 5
#3 1 B 4
#4 2 A 1
#5 2 B 1
frame[frame$id %in% names(indx)[!indx],]
# id coder y
#6 3 A 2
Explanation
table(frame$coder, frame$id)
# 1 2 3
# A 2 1 1
# B 1 1 0 #Here for id 3, B==0
If we Negate that, the result would be a logical index
!table(frame$coder, frame$id).
Do the colSums of the above, which results
# 1 2 3
# 0 0 1
Negate again and get the index for ids and subset those ids which are TRUE
From this you can subset by matching with the names of the ids
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))