My data have different length factors like this.
variable <- c("A,B,C","A,B","A,C","B,C")
I had used strsplit and other similar function, but I can't solve my problem
I need to get a data.frame like this
A B C
1 A B C
2 A B NA
3 A NA C
4 NA B C
Thanks
We could split the data on comma, create dataframe and assign names based on variable name. We can then bind rows by column names using bind_rows from dplyr.
dplyr::bind_rows(sapply(strsplit(variable, ","), function(x)
setNames(as.data.frame(t(x)), x)))
# A B C
#1 A B C
#2 A B <NA>
#3 A <NA> C
#4 <NA> B C
We can use rbindlist
library(data.table)
rbindlist(lapply(strsplit(variable, ","),
function(x) setNames(as.list(x), x)), fill = TRUE)
My data looks like this:
df <- data.frame(id=1:8,
f1 = c("A","B","B","C","C","C","A","A"),
f2 = c("A",NA,"B",NA,"B","A","B","A"),
f3 = c("A",NA,NA,NA,NA,"A","C","C"))
What I would like to create is a column that contains the unique values present in each row (NAs excluded). So the result would be the column "f_values":
id f1 f2 f3 f_values
1 1 A A A A
2 2 B <NA> <NA> B
3 3 B B <NA> B
4 4 C <NA> <NA> C
5 5 C B <NA> CB
6 6 C A A CA
7 7 A B C ABC
8 8 A A C AC
row1 is A b/c only A appears. row6 is CA because C and A appear uniquely. I'd describe the function as paste row-wise unique. I'm aware that it will be possible chain together a number of comparison operators and paste statements, but the real data has many more columns so I was hoping someone knew an easier way.
Given df above,
f_values<- sapply(apply(df[,-1],1, unique),function(x) paste(na.omit(x),collapse = ""))
df_new<-cbind(df,f_values)
df_new will be the desired outcome as formulated in your question.
We can also do this in data.table by grouping with 'id'.
library(data.table)
setDT(df)[, f_values := paste(na.omit(unique(unlist(.SD))), collapse="") , id]
I'm somewhat new to R programming and am in need of assistance.
I'm looking to take the sum of 4 columns in a dataframe and list these totals in a simple table.
Essentially, take the sum of 4 columns (A, B, C, D) and list the total in a table (table = column 1: A, B, C, D column 2: sum of column A, B, C, D) - something along the lines of:
A = 3
B = 4
C = 4
D = 3
Does anyone know how to get this output? Also, the less "manual" the response, the better (i.e. trying to avoid having to input several lines of code to get this output if possible).
Thank you.
If your data looks like this:
a <- c(1:4)
b <- c(2:5)
c <- c(3:6)
d <- c(4:7)
df <- data.frame(a,b,c,d)
a b c d
1 1 2 3 4
2 2 3 4 5
3 3 4 5 6
4 4 5 6 7
Use
> res <- sapply(df,sum)
to get
a b c d
10 14 18 22
in order to apply the function only on numeric columns, try
> res <- colSums(df[sapply(df,is.numeric)])
There is colSums:
colSums(Filter(is.numeric, df))
I have two data.frames that looks like these ones:
>df1
V1
a
b
c
d
e
>df2
V1 V2
1 a,k,l
2 c,m,n
3 z,b,s
4 l,m,e
5 t,r,d
I would like to match the values in df1$V1 with those from df2$V2and add a new column to df1 that corresponds to the matching and to the value of df2$V1, the desire output would be:
>df1
V1 V2
a 1
b 3
c 2
d 5
e 4
I've tried this approach but only works if df2$V2 contains just one element:
match(as.character(df1[,1]), strsplit(as.character(df2[,2], ",")) -> idx
df1$V2 <- df2[idx,1]
Many thanks
You can just use grep, which will return the position of the string found:
sapply(df1$V1, grep, x = df2$V2)
# a b c d e
# 1 3 2 5 4
If you expect repeats, you can use paste.
Let's modify your data so that there is a repeat:
df2$V2[3] <- "z,b,s,a"
And modify the solution accordingly:
sapply(df1$V1, function(z) paste(grep(z, x = df2$V2), collapse = ";"))
# a b c d e
# "1;3" "3" "2" "5" "4"
Similar to Tyler's answer, but in base using stack:
df.stack <- stack(setNames(strsplit(as.character(df2$V2), ","), df2$V1))
transform(df1, V2=df.stack$ind[match(V1, df.stack$values)])
produces:
V1 V2
1 a 1
2 b 3
3 c 2
4 d 5
5 e 4
One advantage of splitting over grep is that with grep you run the risk of searching for a and matching things like alabama, etc. (though you can be careful with the patterns to mitigate this (i.e. include word boundaries, etc.).
Note this will only find the first matching value.
Here's an approach:
library(qdap)
key <- setNames(strsplit(as.character(df2$V2), ","), df2$V1)
df1$V2 <- as.numeric(df1$V1 %l% key)
df1
## V1 V2
## 1 a 1
## 2 b 3
## 3 c 2
## 4 d 5
## 5 e 4
First we used strsplit to create a named list. Then we used qdap's lookup operator %l% to match values and create a new column (I converted to numeric though this may not be necessary).
I'm trying to implement a data.table for my relatively large datasets and I can't figure out how to operate a function over multiple columns in the same row. Specifically, I want to create a new column that contains a specifically-formatted tally of the values (i.e., a histogram) in a subset of columns. It is kind of like table() but that also includes 0 entries and is sorted--so, if you know of a better/faster method I'd appreciate that too!
Simplified test case:
DF<-data.frame("A"=c("a","d","a"),"B"=c("b","a","a"),"C"=c("c","a","a"),"D"=c("a","b","c"),"E"=c("a","a","c"))
DT<-as.data.table(DF)
> DT
A B C D E
1: a b c a a
2: d a a b a
3: a a a c c
my klunky histogram function:
histo<-function(vec){
foo<-c("a"=0,"b"=0,"c"=0,"d"=0)
for(i in vec){foo[i]=foo[i]+1}
return(foo)}
>histo(unname(unlist(DF[1,])))
a b c d
3 1 1 0
>histo(unname(unlist(DF[2,])))
a b c d
3 1 0 1
>histo(unname(unlist(DF[3,])))
a b c d
3 0 2 0
pseduocode of desired function and output
>DT[,his:=some_func_with_histo(A:E)]
>DT
A B C D E his
1: a b c a a (3,1,1,0)
2: d a a b a (3,1,0,1)
3: a a a c c (3,0,2,0)
df <- data.table(DF)
df$hist <- unlist(apply(df, 1, function(x) {
list(
sapply(letters[1:4], function(d) {
b <- sum(!is.na(grep(d,x)))
assign(d, b)
}))
}), recursive=FALSE)
Your df$hist column is a list, with each value named:
> df
A B C D E hist
1: a b c a a 3,1,2,0
2: d a a b a 3,1,1,1
3: a a a c c 3,0,3,0
> df$hist
[[1]]
a b c d
3 1 2 0
[[2]]
a b c d
3 1 1 1
[[3]]
a b c d
3 0 3 0
NOTE: Answer has been updated to to OP's request and mnel's comment
OK, how do you like that solution:
library(data.table)
DT <- data.table(A=c("a","d","a"),
B=c("b","a","a"),
C=c("c","a","a"),
D=c("a","b","c"),
E=c("a","a","c"))
fun <- function(vec, char) {
sum(vec==char)
}
DT[, Vec_Nr:= paste(Vectorize(fun, 'char')(.SD, letters[1:4]), collapse=","),
by=1:nrow(DT),
.SDcols=LETTERS[1:5]]
A B C D E Vec_Nr
1: a b c a a 3,1,1,0
2: d a a b a 3,1,0,1
3: a a a c c 3,0,2,0
I basically split up your problem into several steps:
First, I define a function fun that gives me the number of occurrences for one character. To see how
that function works, just call
fun(c("a", "a", "b"), "b")
[1] 1
Next, I vectorize this function because you don't want to know that for only one character "b", but for many. To pass a vector of arguments to a function,
use Vectorize. To see how that works, just type
Vectorize(fun, "char")(c("a", "a", "b"), c("a", "b"))
a b
2 1
Next, I collapse the results into one string and save that as a new column. Note that I deliberatly used the letters and LETTERS here to show you how make this more dynamic.
EDIT (also see below): Provided you first convert column classes to character, e.g., with DT <- DT[,lapply(.SD,as.character)]...
By using factor, you can convert vec and pass the values (a,b,c,d) in one step:
histo2 <- function(x) table(factor(x,levels=letters[1:4]))
Then you can iterate over rows by passing by=1:nrow(DT).
DT[,as.list(histo2(.SD)),by=1:nrow(DT)]
This gives...
nrow a b c d
1: 1 3 1 1 0
2: 2 3 1 0 1
3: 3 3 0 2 0
Also, this iterates over columns. This works because .SD is a special variable holding the subset of data associated with the call to by. In this case, that subset is the data.table consisting of one of the rows. histo2(DT[1]) works the same way.
EDIT (responding to OP's comment): Oh, sorry, I instinctively replaced your first line with
DF<-data.frame("A"=c("a","d","a"),"B"=c("b","a","a"),"C"=c("c","a","a"),"D"=c("a","b","c"),"E"=c("a","a","c")
,stringsAsFactors=FALSE)
since I dislike using factors except when making tables. If you do not want to convert your factor columns to character columns in this way, this will work:
histo3 <- function(x) table(factor(sapply(x,as.character),levels=letters[1:4]))
To put the output into a single column, you use := as you suggested...
DT[,hist:=list(list(histo3(.SD))),by=1:nrow(DT)]
The list(list()) part is key; I always figure this out by trial-and-error. Now DT looks like this:
A B C D E hist
1: a b c a a 3,1,1,0
2: d a a b a 3,1,0,1
3: a a a c c 3,0,2,0
You might find that it's a pain to access the information directly from your new column. For example, to access the "a" column of the "histogram", I think the fastest route is...
DT[,hist[[1]][["a"]],by=1:nrow(DT)]
My initial suggestion created an auxiliary data.table with just the counts. I think it's cleaner to do whatever you want to do with the counts in that data.table and then cbind it back. If you choose to store it in a column, you can always create the auxiliary data.table later with
DT[,as.list(hist[[1]]),by=1:nrow(DT)]
You are correct about using .SDcols. For your example, ...
cols = c("A","C")
histname = paste(c("hist",cols),collapse="")
DT[,(histname):=list(list(histo3(.SD))),by=1:nrow(DT),.SDcols=cols]
This gives
A B C D E hist histAC
1: a b c a a 3,1,1,0 1,0,1,0
2: d a a b a 3,1,0,1 1,0,0,1
3: a a a c c 3,0,2,0 2,0,0,0