I have been trying to make a procedure in R. I want to ADD a new column base on several categories of another column.
I put an example :
Column New Column
A 1
B 2
C 3
D 4
D 4
A 1
My question is how to add this new column with a particular value base on the values (in characters) of the first column.
It is really similar using the function of MUTATE and CASE_WHEN. The problem is that this function just takes into consideration numeric values and in this case I want take characters (categories) and base on this give a specific value to the new column.
Assuming you have a column of categories (not only letters), you can convert it to "ordered factors" to order the categories and then convert to integers.
x <- c("A", "B", "C", "D", "D", "A")
# make the dataframe
v <- data.frame(x, as.integer(as.ordered(x)))
#
colnames(v) <- c("Column", "New Column")
v
# output
> v
Column New Column
1 A 1
2 B 2
3 C 3
4 D 4
5 D 4
6 A 1
If I understand you correctly you want to create a new column that has numbers corresponding to letters, with 1 corresponding to the first letter of the alphabet A, 2 corresponding to B, 3 to C, and so on. If that premise is correct, then this code will work for you:
ILLUSTRATIVE DATA
set.seed(12)
df <- data.frame(
Column = sample(LETTERS[1:5],10, replace = T)
)
df
Column
1 A
2 E
3 E
4 B
5 A
6 A
7 A
8 D
9 A
10 A
SOLUTION:
Assign the indices of LETTERS, which is an ordered sequence of integers starting with 1, to the letters in df$COlumn where they match the letters in LETTERS:
df$Newcolumn <- seq(LETTERS)[match(df$Column, LETTERS)]
RESULt:
df
Column Newcolumn
1 A 1
2 E 5
3 E 5
4 B 2
5 A 1
6 A 1
7 A 1
8 D 4
9 A 1
10 A 1
Related
So, I'm trying to read a excel files. What happens is that some of the rows are empty for some of the columns but not for all of them. I want to skip all the rows that are not complete, i.e., that don't have information in all of the columns. For example:
In this case I would like to skip the lines 1,5,6,7,8 and so on.
There is probably more elegant way of doing it, but a possible solution is to count the number of elements per rows that are not NA and keep only rows with the number of elements equal to the number of columns.
Using this dummy example:
df <- data.frame(A = LETTERS[1:6],
B = c(sample(1:10,5),NA),
C = letters[1:6])
A B C
1 A 5 a
2 B 9 b
3 C 1 c
4 D 3 d
5 E 4 e
6 F NA f
Using apply, you can for each rows count the number of elements without NA:
v <- apply(df,1, function(x) length(na.omit(x)))
[1] 3 3 3 3 3 2
And then, keep only rows with the number of elements equal to the number of columns (which correspond to complete rows):
df1 <- df[v == ncol(df),]
A B C
1 A 5 a
2 B 9 b
3 C 1 c
4 D 3 d
5 E 4 e
Does it answer your question ?
I have a data frame that contains duplicate column names. I'm aware that it's non-standard to use duplicated column names, but these names are actually being reassigned downstream using user inputs. For now, I'm attempting to positionally subset a data frame, but the column names become deduplicated. Here's an example.
> df <- data.frame(x = 1:4, y = 2:5, y = LETTERS[2:5], y = (2+(2:5)), check.names = F)
> df
x y y y
1 1 2 B 4
2 2 3 C 5
3 3 4 D 6
4 4 5 E 7
However, when I attempt to subset, the names change...
> df[, 1:3]
x y y.1
1 1 2 B
2 2 3 C
3 3 4 D
4 4 5 E
Is there any way to prevent this from happening? It only occurs when I subset on columns, not rows.
> df[1:3,]
x y y y
1 1 2 B 4
2 2 3 C 5
3 3 4 D 6
Edit for others noticing this behavior:
I've done some digging into the behavior and this relevant section from the help page for extract.data.frame (type ?'[')
The relevant section states:
If [ returns a data frame it will have unique (and non-missing) row
names, if necessary transforming the row names using make.unique.
Similarly, if columns are selected column names will be transformed to
be unique if necessary (e.g., if columns are selected more than once,
or if more than one column of a given name is selected if the data
frame has duplicate column names).
This explains the why, appreciate the comments so far on addressing how to best navigate this.
Here is an option, although I think it is not a good idea to have duplicated column names.
as.data.frame(as.list(df)[1:3], check.names = F)
# x y y
# 1 1 2 B
# 2 2 3 C
# 3 3 4 D
# 4 4 5 E
We have a data frame as below :
raw<-data.frame(v1=c("A","B","C","D"),v2=c(NA,"B","C","A"),v3=c(NA,"A",NA,"D"),v4=c(NA,"D",NA,NA))
I need a result data frame in the following format :
result<-data.frame(v1=c("A","B","C","D"), v2=c(3,2,2,3))
Used the following code to get the count across one particular column :
count_raw<-sqldf("SELECT DISTINCT(v1) AS V1, COUNT(v1) AS count FROM raw GROUP BY v1")
This would return count of unique values across an individual column.
Any help would be highly appreciated.
Use this
table(unlist(raw))
Output
A B C D
3 2 2 3
For data frame type output wrap this with as.data.frame.table
as.data.frame.table(table(unlist(raw)))
Output
Var1 Freq
1 A 3
2 B 2
3 C 2
4 D 3
If you want a total count,
sapply(unique(raw[!is.na(raw)]), function(i) length(which(raw == i)))
#A B C D
#3 2 2 3
We can use apply with MARGIN = 1
cbind(raw[1], v2=apply(raw, 1, function(x) length(unique(x[!is.na(x)]))))
If it is for each column
sapply(raw, function(x) length(unique(x[!is.na(x)])))
Or if we need the count based on all the columns, convert to matrix and use the table
table(as.matrix(raw))
# A B C D
# 3 2 2 3
If you have only character values in your dataframe as you've provided, you can unlist it and use unique or to count the freq, use count
> library(plyr)
> raw<-data.frame(v1=c("A","B","C","D"),v2=c(NA,"B","C","A"),v3=c(NA,"A",NA,"D"),v4=c(NA,"D",NA,NA))
> unique(unlist(raw))
[1] A B C D <NA>
Levels: A B C D
> count(unlist(raw))
x freq
1 A 3
2 B 2
3 C 2
4 D 3
5 <NA> 6
How can I find the 5 highest values of a column in a data frame
I tried the order() function but it gives me only the indices of the rows, wherease I need the actual data from the column. Here's what I have so far:
tail(order(DF$column, decreasing=TRUE),5)
You need to pass the result of order back to DF:
DF <- data.frame( column = 1:10,
names = letters[1:10])
order(DF$column)
# 1 2 3 4 5 6 7 8 9 10
head(DF[order(DF$column),],5)
# column names
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
# 5 5 e
You're correct that order just gives the indices. You then need to pass those indices to the data frame, to pick out the rows at those indices.
Also, as mentioned in the comments, you can use head instead of tail with decreasing = TRUE if you'd like, but that's a matter of taste.
I am trying to simulate n times the measuring order and see how measuring order effects my study subject. To do this I am trying to generate integer random numbers to a new column in a dataframe. I have a big dataframe and i would like to add a column into the dataframe that consists a random number according to the number of observations in a block.
Example of data(each row is an observation):
df <- data.frame(A=c(1,1,1,2,2,3,3,3,3),
B=c("x","b","c","g","h","g","g","u","l"),
C=c(1,2,4,1,5,7,1,2,5))
A B C
1 1 x 1
2 1 b 2
3 1 c 4
4 2 g 1
5 2 h 5
6 3 g 7
7 3 g 1
8 3 u 2
9 3 l 5
What I'd like to do is add a D column and generate random integer numbers according to the length of each block. Blocks are defined in column A.
Result should look something like this:
df <- data.frame(A=c(1,1,1,2,2,3,3,3,3),
B=c("x","b","c","g","h","g","g","u","l"),
C=c(1,2,4,1,5,7,1,2,5),
D=c(2,1,3,2,1,4,3,1,2))
> df
A B C D
1 1 x 1 2
2 1 b 2 1
3 1 c 4 3
4 2 g 1 2
5 2 h 5 1
6 3 g 7 4
7 3 g 1 3
8 3 u 2 1
9 3 l 5 2
I have tried to use R:s sample() function to generate random numbers but my problem is splitting the data according to block length and adding the new column. Any help is greatly appreciated.
It can be done easily with ave
df$D <- ave( df$A, df$A, FUN = function(x) sample(length(x)) )
(you could replace length() with max(), or whatever, but length will work even if A is not numbers matching the length of their blocks)
This is really easy with ddply from plyr.
ddply(df, .(A), transform, D = sample(length(A)))
The longer manual version is:
Use split to split the data frame by the first column.
split_df <- split(df, df$A)
Then call sample on each member of the list.
split_df <- lapply(split_df, function(df)
{
df$D <- sample(nrow(df))
df
})
Then recombine with
df <- do.call(rbind, split_df)
One simple way:
df$D = 0
counts = table(df$A)
for (i in 1:length(counts)){
df$D[df$A == names(counts)[i]] = sample(counts[i])
}