I want to get a one-way frequency table for each column in my dataframe (a count of each unique value in each column). I am following this tutorial, which suggests using the count() function from the plyr package.
for (col in mtcars[c("gear","carb")]){
freq <- count(col)
write.table(freq, file='filename.txt')
}
I would expect the output to look like this:
gear freq
1 3 15
2 4 12
3 5 5
Instead the column name is replaced with 'x':
x freq
1 3 15
2 4 12
3 5 5
Why is this happening, and how can I modify my for loop so that it prints the column name instead of 'x'?
(There is probably a better, vectorized way to do this other than using a for loop, but I'm new to R and can't quite figure out the syntax.)
In a for loop:
for (col in c("gear","carb")){
print(plyr::count(mtcars, col))
}
Using lapply():
lapply(c("gear","carb"), function(col) plyr::count(mtcars, col))
To be clear, count is not renaming anything. In your loop it receives col which is a vector. A vector does not have column names, and so count does not know what name it should use. It uses x as a place holder.
This will also work (with the names of the columns of the dataset mtcar as input, with result as a list of dataframes):
lapply(c("gear","carb"), function(x){df <- as.data.frame(table(mtcars[x])); names(df) <- c(x, 'freq'); df})
[[1]]
gear freq
1 3 15
2 4 12
3 5 5
[[2]]
carb freq
1 1 7
2 2 10
3 3 3
4 4 10
5 6 1
6 8 1
Related
I am currently try to use pre-defined strings in order to identify multiple column names in R.
To be more explicit, I am using the ave function to create identification variables for subgroups of a dataframe. The twist is that I want the identification variables to be flexible, in such a manner that I would just pass it as a generic string.
A sample code would be:
ids = with(df,ave(rep(1,nrow(df)),subcolumn1,subcolumn2,subcolumn3,FUN=seq_along))
I would like to run this code in the following fashion (code below does not work as expected):
subColumnsString = c("subcolumn1","subcolumn2","subcolumn3")
ids = with(df,ave(rep(1,nrow(df)),subColumnsString ,FUN=seq_along))
I tried something with eval, but still did not work:
subColumnsString = c("subcolumn1","subcolumn2","subcolumn3")
ids = with(df,ave(rep(1,nrow(df)),eval(parse(text=subColumnsString)),FUN=seq_along))
Any ideas?
Thanks.
EDIT: Working code example of what I want:
df = mtcars
id_names = c("vs","am")
idDF_correct = transform(df,idItem = as.numeric(interaction(vs,am)))
idDF_wrong = cbind(df,ave(rep(1,nrow(df)),df[id_names],FUN=seq_along))
Note how in idDF_correct, the unique combinations are correctly mapped into unique values of idItem. In idDF_wrong this is not the case.
I think this achieves what you requested. Here I use the mtcars dataset that ships with R:
subColumnsString <- c("cyl","gear")
ids = with(mtcars, ave(rep(1,nrow(mtcars)), mtcars[subColumnsString], FUN=seq_along))
Just index your data.frame using the sub columns which returns a list that naturally works with ave
EDIT
ids = ave(rep(1,nrow(mtcars)), mtcars[subColumnsString], FUN=seq_along)
You can omit the with and just call plain 'ol ave, as G. Grothendieck, stated and you should also use their answer as it is much more general.
This defines a function whose arguments are:
data, the input data frame
by, a character vector of column names in data
fun, a function to use in ave
Code--
Ave <- function(data, by, fun = seq_along) {
do.call(function(...) ave(rep(1, nrow(data)), ..., FUN = fun), data[by])
}
# test
Ave(CO2, c("Plant", "Treatment"), seq_along)
giving:
[1] 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3
[39] 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6
[77] 7 1 2 3 4 5 6 7
I'm trying to remove duplicate rows from a data frame, based only on the previous row. The duplicate and unique functions will remove all duplicates, leaving you only with unique rows, which is not what I want.
I've illustrated the problem here with a loop. I need to vectorize this because my actual data set is much to large to use a loop on.
x <- c(1,1,1,1,3,3,3,4)
y <- c(1,1,1,1,3,3,3,4)
z <- c(1,2,1,1,3,2,2,4)
xy <- data.frame(x,y,z)
xy
x y z
1 1 1 1
2 1 1 2
3 1 1 1
4 1 1 1 #this should be removed
5 3 3 3
6 3 3 2
7 3 3 2 #this should be removed
8 4 4 4
# loop that produces desired output
toRemove <- NULL
for (i in 2:nrow(xy)){
test <- as.vector(xy[i,] == xy[i-1,])
if (!(FALSE %in% test)){
toRemove <- c(toRemove, i) #build a vector of rows to remove
}
}
xy[-toRemove,] #exclude rows
x y z
1 1 1 1
2 1 1 2
3 1 1 1
5 3 3 3
6 3 3 2
8 4 4 4
I've tried using dplyr's lag function, but it only works on single columns, when I try to run it over all 3 columns it doesn't work.
ifelse(xy[,1:3] == lag(xy[,1:3],1), NA, xy[,1:3])
Any advice on how to accomplish this?
Looks like we want to remove if the row is same as above:
# make an index, if cols not same as above
ix <- c(TRUE, rowSums(tail(xy, -1) == head(xy, -1)) != ncol(xy))
# filter
xy[ix, ]
Why don't you just iterate the list while keeping track of the previous row to compare it to the next row?
If this is true at some point: remember that row position and remove it from the list then start iterating from the beginning of the list.
Don't delete row while iterating because you will get concurrent modification error.
I am new to R . I have a data frame(usr.query) with structure as shown below
[
Now I want to take text of each id and compare it to text of all the other id and and if there is a match, i want to append it to a new column say count of match.
A0008 with A0043,A0065,A0082,B0018,B0026
A0043 with A0008,A0065,A0082,B0018,B0026
Function to apply
count_match = length(intersect(unlist(strsplit(query1," ")),unlist(strsplit(query2," "))))
The query 1 here is text of A0008 and query 2 is text of A0043,A0065,A0082,B0018,B0026
I tried the suggested solution and here is the result.
No loops are necessary; you'll usually find that's the case in R, because it's really good at utilizing vectorized operations. In this case, you can get the necessary combinations with combn, and then make the match_count column by subsetting the original data.frame with the combinations of the new one, and testing for equality. Adding zero changes the values from Boolean to numeric (use as.integer, if you prefer).
# assemble sample data
df <- data.frame(id = 1:5, text = c('apple', 'mango', 'apple', 'apple', 'mango'))
# make combinations
df2 <- as.data.frame(t(combn(df$id, 2)))
# add names
names(df2) <- c('main_id', 'compared_to_id')
# test for match
df2$match_count <- (df[df2$main_id, 'text'] == df[df2$compared_to_id, 'text']) + 0
The result:
> df2
main_id compared_to_id match_count
1 1 2 0
2 1 3 1
3 1 4 1
4 1 5 0
5 2 3 0
6 2 4 0
7 2 5 1
8 3 4 1
9 3 5 0
10 4 5 0
I would like to get a data frame that contains only data that is within 2 SD per each numeric column.
I know how to do it for a single column but how can I do it for a bunch of columns at once?
Here is the toy data frame:
df <- read.table(text = "target birds wolfs Country
3 21 7 a
3 8 4 b
1 2 8 c
1 2 3 a
1 8 3 a
6 1 2 a
6 7 1 b
6 1 5 c",header = TRUE)
Here is the code line for getting only the data that is under 2 SD for a single column(birds).How can I do it for all numeric columns at once?
df[!(abs(df$birds - mean(df$birds))/sd(df$birds)) > 2,]
target birds wolfs Country
2 3 8 4 b
3 1 2 8 c
4 1 2 3 a
5 1 8 3 a
6 6 1 2 a
7 6 7 1 b
8 6 1 5 c
We can use lapply to loop over the dataset columns and subset the numeric vectors (by using a if/else condition) based on the mean and sd.
lapply(df, function(x) if(is.numeric(x)) x[!(abs((x-mean(x))/sd(x))>2)] else x)
EDIT:
I was under the impression that we need to remove the outliers for each column separately. But, if we need to keep only the rows that have no outliers for the numeric columns, we can loop through the columns with lapply as before, instead of returning 'x', we return the sequence of 'x' and then get the intersect of the list element with Reduce. The numeric index can be used for subsetting the rows.
lst <- lapply(df, function(x) if(is.numeric(x))
seq_along(x)[!(abs((x-mean(x))/sd(x))>2)] else seq_along(x))
df[Reduce(intersect,lst),]
I'm guessing that you are trying to filter your data set by checking that all of the numeric columns are within 2 SD (?)
In that case I would suggest to create two filters. 1 one that will indicate numeric columns, the second one that will check that all of them within 2 SD. For the second condition, we can use the built in scale function
indx <- sapply(df, is.numeric)
indx2 <- rowSums(abs(scale(df[indx])) <= 2) == sum(indx)
df[indx2,]
# target birds wolfs Country
# 2 3 8 4 b
# 3 1 2 8 c
# 4 1 2 3 a
# 5 1 8 3 a
# 6 6 1 2 a
# 7 6 7 1 b
# 8 6 1 5 c
I have a list of dataframes. I need to subset a dataframe of this list according to a criteria in one column of the dataframe.
(all dataframes of the list have the same number and names of columns, and the same number of rows)
For example, I have:
l <- list(data.frame(x=c(2,3,4,5), y = c(4,4,4,4), z=c(2,3,4,5)),
data.frame(x=c(1,4,7,3), y = c(7,7,7,7), z=c(2,5,7,8)),
data.frame(x=c(2,3,1,8), y = c(1,1,1,1), z=c(6,4,1,3)))
names(l) <- c("MH1", "MH2","MH3")
output
$MH1
x y z
1 2 4 2
2 3 4 3
3 4 4 4
4 5 4 5
$MH2
x y z
1 1 7 2
2 4 7 5
3 7 7 7
4 3 7 8
$MH3
x y z
1 2 1 6
2 3 1 4
3 1 1 1
4 8 1 3
So I want to subset the dataframe for which column "y" is the closest to a given number. For example if I say a=3, the chosen dataframe should be "MH1" (where column y=4)
If "l" was a dataframe I will do something like:
closestDF <- subset(l, abs(l$y - a) == min(abs(l$y - a))
How can I do this with the list of dataframes?
Following the answers and comments of #David Arenburg, #akrun and #shadow, here there are three possible solutions to the problem I posted:
Option 1)
library(data.table)
rbindlist(l)[abs(y - a) == min(abs(y - a))]
Option 2) (needs an R version > 3.1.2)
library(dplyr)
bind_rows(l) %>% filter(abs(y-a)==which.min(abs(y-a)))
Option 3) (also works perfectly, but computationally less faster than the first 2 options if used within a big loop or an iterative process)
l[[which.min(sapply(l, function(df) sum(abs(df$y - a))))]]