I need to subset the columns of a dataframe taking into account the rownames of another dataframe.(in R)
Im trying to select the representative species of Brazilian Amazon subseting a great Brazilian database taking into account the percentage of representative location, information which is in another dataframe
> a <- data.frame("John" = c(2,1,1,2), "Dora" = c(1,1,3,2), "camilo" = c(1:4),"alex"=c(1,2,1,2))
> a
John Dora camilo alex
1 2 1 1 1
2 1 1 2 2
3 1 3 3 1
4 2 2 4 2
> b <- data.frame("SN" = 1:3, "Age" = c(15,31,2), "Name" = c("John","Dora","alex"))
> b
SN Age Name
1 1 15 John
2 2 31 Dora
3 3 2 alex
> result <- a[,rownames(b)[1:3]]
Error in `[.data.frame`(a, , rownames(b)[1:3]) :
undefined columns selected
I want to get this dataframe
John Dora alex
1 2 1 1
2 1 1 2
3 1 3 1
4 2 2 2
The simple a[,b$Name] does not work because b$Name is considered a factor. Be careful because it won't throw an error but you will get the wrong answer!
But this is easy to fit by using a[,as.character(b$Name)]instead!
For instance if I have this data:
ID Value
1 2
1 2
1 3
1 4
1 10
2 9
2 9
2 12
2 13
And my goal is to find the smallest value for each ID subset, and I want the number to be in the first row of the ID group while leaving the other rows blank, such that:
ID Value Start
1 2 2
1 2
1 3
1 4
1 10
2 9 9
2 9
2 12
2 13
My first instinct is to create an index for the IDs using
A <- transform(A, INDEX=ave(ID, ID, FUN=seq_along)) ## A being the name of my data
Since I am a noob, I get stuck at this point. For each ID=n, I want to find the min(A$Value) for that ID subset and place that into the cell matching condition of ID=n and INDEX=1.
Any help is much appreciated! I am sorry that I keep asking questions :(
Here's a solution:
within(A, INDEX <- "is.na<-"(ave(Value, ID, FUN = min), c(FALSE, !diff(ID))))
ID Value INDEX
1 1 2 2
2 1 2 NA
3 1 3 NA
4 1 4 NA
5 1 10 NA
6 2 9 9
7 2 9 NA
8 2 12 NA
9 2 13 NA
Update:
How it works? The command ave(Value, ID, FUN = min) applies the function min to each subset of Value along the values of ID. For the example, it returns a vector of five times 2 and four times 9. Since all values except the first in each subset should be NA, the function "is.na<-" replaces all values at the logical index defined by c(FALSE, !diff(ID)). This index is TRUE if a value is identical with the preceding one.
You're almost there. We just need to make a custom function instead of seq_along and to split value by ID (not ID by ID).
first_min <- function(x){
nas <- rep(NA, length(x))
nas[which.min(x)] <- min(x, na.rm=TRUE)
nas
}
This function makes a vector of NAs and replaces the first element with the minimum value of Value.
transform(dat, INDEX=ave(Value, ID, FUN=first_min))
## ID Value INDEX
## 1 1 2 2
## 2 1 2 NA
## 3 1 3 NA
## 4 1 4 NA
## 5 1 10 NA
## 6 2 9 9
## 7 2 9 NA
## 8 2 12 NA
## 9 2 13 NA
You can achieve this with a tapply one-liner
df$Start<-as.vector(unlist(tapply(df$Value,df$ID,FUN = function(x){ return (c(min(x),rep("",length(x)-1)))})))
I keep going back to this question and the above answers helped me greatly.
There is a basic solution for beginners too:
A$Start<-NA
A[!duplicated(A$ID),]$Start<-A[!duplicated(A$ID),]$Value
Thanks.
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))