I'm working with multiple big data frames in R and I'm trying to write functions that can modify each of them (given a set of common parameters). One function is giving me trouble (shown below).
RawData <- function(x)
{
for(i in 1:nrow(x))
{
if(grep(".DERIVED", x[i,]) >= 1)
{
x <- x[-i,]
}
}
for(i in 1:ncol(x))
{
if(is.numeric(x[,i]) != TRUE)
{
x <- x[,-i]
}
}
return(x)
}
The objective of this function is twofold: first, to remove any rows that contain a ".DERIVED" string in any one of their cells (using grep), and second, to remove any columns that are non-numeric (using is.numeric). I get an error on the following condition:
if(grep(".DERIVED", x[i,]) >= 1)
The error states the "argument is of zero length", which I believe is usually associated with NULL values in a vector. However, I've used is.null on the entire data frame that is giving me errors, and it confirmed that there are no null values in the DF. I'm sure I'm missing something relatively simple here. Any advice would be greatly appreciated.
If you can use non-base-R functions, this should address your issue. df is the data.frame in question here. It will also be faster than looping over rows (generally not advised if avoidable).
library(dplyr)
library(stringr)
df %>%
filter_all(!str_detect(., '\\.DERIVED')) %>%
select_if(is.numeric)
You can make it a function just as you would anything else:
mattsFunction <- function(dat){
dat %>%
filter_all(!str_detect(., '\\.DERIVED')) %>%
select_if(is.numeric)
}
you should probably give it a better name though
The error is from the line
if(grep(".DERIVED", x[i,]) >= 1)
When grep doesn't find the term ".DERIVED", it returns something of zero length, your inequality doesn't return TRUE or FALSE, but rather returns logical(0). The error is telling you that the if statement cannot evaluate whether logical(0) >= 1
A simple example:
if(grep(".DERIVED", "1234.DERIVEDabcdefg") >= 1) {print("it works")} # Works nicely, since the inequality can be evaluated
if(grep(".DERIVED", "1234abcdefg") > 1) {print("no dice")}
You can replace that line with if(length(grep(".DERIVED", x[i,])) != 0)
There's something else you haven't noticed yet, which is that you're removing rows/columns in a loop. Say you remove the 5th column, the next loop iteration (when i = 6) will be handling what was the 7th row! (this will end in an error along the lines of Error in[.data.frame(x, , i) : undefined columns selected)
I prefer using dplyr, but if you need to use base R functions there are ways to to this without if statements.
Notice that you should consider using the regex version of "\\.DERIVED" and not ".DERIVED" which would mean "any character followed by DERIVED".
I don't have example data or output, so here's my best go...
# Made up data
test <- data.frame(a = c("data","data.DERIVED","data","data","data.DERIVED"),
b = (c(1,2,3,4,5)),
c = c("A","B","C","D","E"),
d = c(2,5,6,8,9),
stringsAsFactors = FALSE)
# Note: The following code assumes that the column class is numeric because the
# example code provided assumed that the column class was numeric. This will not
# detects if the column is full of a string of character values of only numbers.
# Using the base subset command
test2 <- subset(test,
subset = !grepl("\\.DERIVED",test$a),
select = sapply(test,is.numeric))
# > test2
# b d
# 1 1 2
# 3 3 6
# 4 4 8
# Trying to use []. Note: If only 1 column is numeric this will return a vector
# instead of a data.frame
test2 <- test[!grepl("\\.DERIVED",test$a),]
test2 <- test2[,sapply(test,is.numeric)]
# > test2
# b d
# 1 1 2
# 3 3 6
# 4 4 8
I am not sure how to handle NA within Julia DataFrames.
For example with the following DataFrame:
> import DataFrames
> a = DataFrames.#data([1, 2, 3, 4, 5]);
> b = DataFrames.#data([3, 4, 5, 6, NA]);
> ndf = DataFrames.DataFrame(a=a, b=b)
I can successfully execute the following operation on column :a
> ndf[ndf[:a] .== 4, :]
but if I try the same operation on :b I get an error NAException("cannot index an array with a DataArray containing NA values").
> ndf[ndf[:b] .== 4, :]
NAException("cannot index an array with a DataArray containing NA values")
while loading In[108], in expression starting on line 1
in to_index at /Users/abisen/.julia/v0.3/DataArrays/src/indexing.jl:85
in getindex at /Users/abisen/.julia/v0.3/DataArrays/src/indexing.jl:210
in getindex at /Users/abisen/.julia/v0.3/DataFrames/src/dataframe/dataframe.jl:268
Which is because of the presence of NA value.
My question is how should DataFrames with NA should typically be handled? I can understand that > or < operation against NA would be undefined but == should work (no?).
What's your desired behavior here? If you want to do selections like this you can make the condition (not a NAN) AND (equal to 4). If the first test fails then the second one never happens.
using DataFrames
a = #data([1, 2, 3, 4, 5]);
b = #data([3, 4, 5, 6, NA]);
ndf = DataFrame(a=a, b=b)
ndf[(!isna(ndf[:b]))&(ndf[:b].==4),:]
In some cases you might just want to drop all rows with NAs in certain columns
ndf = ndf[!isna(ndf[:b]),:]
Regarding to this question I asked before, you can change this NA behavior directly in the modules sourcecode if you want. In the file indexing.jl there is a function named Base.to_index(A::DataArray) beginning at line 75, where you can alter the code to set NA's in the boolean array to false. For example you can do the following:
# Indexing with NA throws an error
function Base.to_index(A::DataArray)
A[A.na] = false
any(A.na) && throw(NAException("cannot index an array with a DataArray containing NA values"))
Base.to_index(A.data)
end
Ignoring NA's with isna() will cause a less readable sourcecode and in big formulas, a performance loss:
#timeit ndf[(!isna(ndf[:b])) & (ndf[:b] .== 4),:] #3.68 µs per loop
#timeit ndf[ndf[:b] .== 4, :] #2.32 µs per loop
## 71x179 2D Array
#timeit dm[(!isna(dm)) & (dm .< 3)] = 1 #14.55 µs per loop
#timeit dm[dm .< 3] = 1 #754.79 ns per loop
In many cases you want to treat NA as separate instances, i.e. assume that that everything that is NA is "equal" and everything else is different.
If this is the behaviour you want, current DataFrames API doesn't help you much, as both (NA == NA) and (NA == 1) returns NA instead of their expected boolean results.
This makes extremely tedious DataFrame filters using loops:
function filter(df,c)
for r in eachrow(df)
if (isna(c) && isna(r:[c])) || ( !isna(r[:c]) && r[:c] == c )
...
and breaks select-like functionalities in DataFramesMeta.jl and Query.jl when NA values are present or requested for..
One workaround is to use isequal(a,b) in place of a==b
test = #where(df, isequal.(:a,"cc"), isequal.(:b,NA) ) #from DataFramesMeta.jl
I think the new syntax in Julia is to use ismissing:
# drop NAs
df = DataFrame(col=[0,1,1,missing,0,1])
df = df[.!ismissing.(df[:col]),:]
I have three variables in a data frame and would like to swap the 4 columns around from
"dam" "piglet" "fdate" "ssire"
to
"piglet" "ssire" "dam" "tdate"
Is there any way I can do the swapping using R?
Any help would be very much appreciated.
Baz
dfrm <- dfrm[c("piglet", "ssire", "dam", "tdate")]
OR:
dfrm <- dfrm[ , c("piglet", "ssire", "dam", "tdate")]
d <- data.frame(a=1:3, b=11:13, c=21:23)
d
# a b c
#1 1 11 21
#2 2 12 22
#3 3 13 23
d2 <- d[,c("b", "c", "a")]
d2
# b c a
#1 11 21 1
#2 12 22 2
#3 13 23 3
or you can do same thing using index:
d3 <- d[,c(2, 3, 1)]
d3
# b c a
#1 11 21 1
#2 12 22 2
#3 13 23 3
To summarise the other posts, there are three ways of changing the column order, and two ways of specifying the indexing in each method.
Given a sample data frame
dfr <- data.frame(
dam = 1:5,
piglet = runif(5),
fdate = letters[1:5],
ssire = rnorm(5)
)
Kohske's answer: You can use standard matrix-like indexing using column numbers
dfr[, c(2, 4, 1, 3)]
or using column names
dfr[, c("piglet", "ssire", "dam", "fdate")]
DWin & Gavin's answer: Data frames allow you to omit the row argument when specifying the index.
dfr[c(2, 4, 1, 3)]
dfr[c("piglet", "ssire", "dam", "fdate")]
PaulHurleyuk's answer: You can also use subset.
subset(dfr, select = c(2, 4, 1, 3))
subset(dfr, select = c(c("piglet", "ssire", "dam", "fdate")))
You can use subset's 'select' argument;
#Assume df contains "dam" "piglet" "fdate" "ssire"
newdf<-subset(df, select=c("piglet", "ssire", "dam", "tdate"))
I noticed that this is almost an 8-year old question. But for people who are starting to learn R and might stumble upon this question, like I did, you can now use a much flexible select() function from dplyr package to accomplish the swapping operation as follows.
# Install and load the dplyr package
install.packages("dplyr")
library("dplyr")
# Override the existing data frame with the desired column order
df <- select(df, piglet, ssire, dam, tdate)
This approach has following advantages:
You will have to type less as the select() does not require variable names to be enclosed within quotes.
In case your data frame has more than 4 variables, you can utilize select helper functions such as starts_with(), ends_with(), etc. to select multiple columns without having to name each column and rearrange them with much ease.
Relevance Note: In response to some users (myself included) that would like to swap columns without having to specify every column, I wrote this answer up.
TL;DR: A one-liner for numerical indices is provided herein and a function for swapping exactly 2 nominal and numerical indices at the end, neither using imports, that will correctly swap any two columns in a data frame of any size is provided. A function that allows the reassignment of an arbitrary number of columns that may cause unavoidable superfluous swaps if not used carefully is also made available (read more & get functions in Summary section)
Preliminary Solution
Suppose you have some huge (or not) data frame, DF, and you only know the indices of the two columns you want to swap, say 1 < n < m < length(DF). (Also important is that your columns are not adjacent, i.e. |n-m| > 1 which is very likely to be the case in our "huge" data frame but not necessarily for smaller ones; work-arounds for all degenerate cases are provided at the end).
Because it is huge, there are a ton of columns and you don't want to have to specify every other column by hand, or it isn't huge and you're just lazy someone with fine taste in coding, either way, this one-liner will do the trick:
DF <- DF[ c( 1:(n-1), m, (n+1):(m-1), n, (m+1):length(DF) ) ]
Each piece works like this:
1:(n-1) # This keeps every column before column `n` in place
m # This places column `m` where column `n` was
(n+1):(m-1) # This keeps every column between the two in place
n # This places column `n` where column `m` was
(m+1):length(DF) # This keeps every column after column `m` in place
Generalizing for Degenerates
Because of how the : operator works, i.e. allowing "backwards-ranges" like this,
> 10:0
[1] 10 9 8 7 6 5 4 3 2 1 0
we have to be careful about our choices and placements of n and m, hence our previous restrictions. For instance, n < m doesn't lose us any generality (one of the columns has to be before the other one if they are different), however, it means we do need to be careful about which goes where in our line of code. We can make it so that we don't have to check this condition with the following modification:
DF <- DF[ c( 1:(min(n,m)-1), max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m), (max(n,m)+1):length(DF) ) ]
We have replaced every instance of n and m with min(n,m) and max(n,m) respectively, meaning that the correct ordering for our code will be preserved even in the case that m > n.
In the cases where min(n,m) == 1, max(n,m) == length(DF), both of those at the same time, and |n-m| == 1, we we will make some unreadable less aesthetic modifications involving if\else to forget about having to check if these are the case. Versions for where you know that one of these are the case, (i.e. you are always swapping some interior column with the first column, swapping some interior column with the last column, swapping the first and last columns, or swapping two adjacent columns), you can actually express these actions more succinctly because they usually just require omitting parts from our restricted case:
# Swapping not the last column with the first column
# We just got rid of 1:(min(n,m)-1) because it would be invalid and not what we meant
# since min(n,m) == 1
# Now we just stick the other column right at the front
DF <- DF[ c( max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m), (max(n,m)+1):length(DF) ) ]
# Also equivalent since we know min(n,m) == 1, for the leftover index i
DF <- DF[ c( i, 2:(i-1), 1, (i+1):length(DF) ) ]
# Swapping not the first column with the last column
# Similarly, we just got rid of (max(n,m)+1):length(DF) because it would also be invalid
# and not what we meant since max(n,m) == length(DF)
# Now we just stick the other column right at the end
DF <- DF[ c( 1:(min(n,m)-1), max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m) ) ]
# Also equivalent since we know max(n,m) == length(DF), for the leftover index, say i
DF <- DF[ c( 1:(i-1), length(DF), (i+1):(length(DF)-1), i ) ]
# Swapping the first column with the last column
DF <- DF[ c( max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m) ) ]
# Also equivalent (for if you don't actually know the length beforehand, as assumed
# elsewhere)
DF <- DF[ c( length(DF), 2:(length(DF)-1), 1 ) ]
# Swapping two interior adjacent columns
# Here we drop the explicit swap on either side of our middle column segment
# This is actually enough because then the middle segment becomes a backwards range
# because we know that `min(n,m) + 1 = max(n,m)`
# The range is just an ordering of the two adjacent indices from largest to smallest
DF <- DF[ c( 1:(min(n,m)-1), (min(n,m)+1):(max(n,m)-1), (max(n,m)+1):length(DF) )]
"But!", I hear you saying, "What if more than one of these cases occur simultaneously, like did in the third version in the block above!?". Right, coding up versions for each case is an enormous waste of time if one wants to be able to "swap columns" in the most general sense.
Swapping any Two Columns
It will be easiest to generalize our code to cover all of the cases at the same time, because they all employ essentially the same strategy. We will use if\else to keep our code a one-liner:
DF <- DF[ if (n==m) 1:length(DF) else c( (if (min(n,m)==1) c() else 1:(min(n,m)-1) ), (if (min(n,m)+1 == max(n,m)) (min(n,m)+1):(max(n,m)-1) else c( max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m))), (if (max(n,m)==length(DF)) c() else (max(n,m)+1):length(DF) ) ) ]
That's totally unreadable and probably pretty unfriendly to anyone who might try to understand or recreate your code (including yourself), so better to box it up in a function.
# A function that swaps the `n` column and `m` column in the data frame DF
swap <- function(DF, n, m)
{
return (DF[ if (n==m) 1:length(DF) else c( (if (min(n,m)==1) c() else 1:(min(n,m)-1) ), (if (min(n,m)+1 == max(n,m)) (min(n,m)+1):(max(n,m)-1) else c( max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m))), (if (max(n,m)==length(DF)) c() else (max(n,m)+1):length(DF) ) ) ])
}
A more robust version that can also swap on column names and has semi-explanatory comments:
# Returns data frame object with columns `n` and `m` swapped
# `n` and `m` can be column names, numerical indices, or a heterogeneous pair of both
swap <- function(DF, n, m)
{
# Of course, first, we want to make sure that n != m,
# because if they do, we don't need to do anything
if (n==m) return(DF)
# Next, if either n or m is a column name, we want to get its index
# We assume that if they aren't column names, they are indices (integers)
n <- if (class(n)=="character" & is.na(suppressWarnings(as.integer(n)))) which(colnames(DF)==n) else as.integer(n)
m <- if (class(m)=="character" & is.na(supressWarnings(as.integer(m)))) which(colnames(DF)==m) else as.integer(m)
# Make sure each index is actually valid
if (!(1<=n & n<=length(DF))) stop( "`n` represents invalid index!" )
if (!(1<=m & m<=length(DF))) stop( "`m` represents invalid index!" )
# Also, for readability, lets go ahead and set which column is earlier, and which is later
earlier <- min(n,m)
later <- max(n,m)
# This constructs the first third of the indices
# These are the columns that, if any, come before the earlier column you are swapping
firstThird <- if ( earlier==1 ) c() else 1:(earlier-1)
# This constructs the last third of the the indices
# These are the columns, if any, that come after the later column you are swapping
lastThird <- if ( later==length(DF) ) c() else (later+1):length(DF)
# This checks if the columns to be swapped are adjacent and then constructs the
# secondThird accordingly
if ( earlier+1 == later )
{
# Here; the second third is a list of the two columns ordered from later to earlier
secondThird <- (earlier+1):(later-1)
}
else
{
# Here; the second third is a list of
# the later column you want to swap
# the columns in between
# and then the earlier column you want to swap
secondThird <- c( later, (earlier+1):(later-1), earlier)
}
# Now we assemble our indices and return our permutation of DF
return (DF[ c( firstThird, secondThird, lastThird ) ])
}
And, for ease of repatriation with less of the spatial cost, a comment-less version that checks index validity and can handle column names, i.e. does everything in pretty close to the smallest space it can (yes, you could vectorize, using ifelse(...), the two checks that get performed, but then you'd have to unpack the vector back into n,m or change how the final line is written):
swap <- function(DF, n, m)
{
n <- if (class(n)=="character" & is.na(suppressWarnings(as.integer(n)))) which(colnames(DF)==n) else as.integer(n)
m <- if (class(m)=="character" & is.na(suppressWarnings(as.integer(m)))) which(colnames(DF)==m) else as.integer(m)
if (!(1<=n & n<=length(DF))) stop( "`n` represents invalid index!" )
if (!(1<=m & m<=length(DF))) stop( "`m` represents invalid index!" )
return (DF[ if (n==m) 1:length(DF) else c( (if (min(n,m)==1) c() else 1:(min(n,m)-1) ), (if (min(n,m)+1 == max(n,m)) (min(n,m)+1):(max(n,m)-1) else c( max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m))), (if (max(n,m)==length(DF)) c() else (max(n,m)+1):length(DF) ) ) ])
}
Permutations (or How to Do Specifically What the Question Asked and More!)
With our swap function in tow, we can try to actually do what the original question asked. The easiest way to do this, is to build a function that utilizes the really cool power that comes with a choice of heterogeneous arguments. Create a mapping:
mapping <- data.frame( "piglet" = 1, "ssire" = 2, "dam" = 3, "tdate" = 4)
In the case of the original question, these are all of the columns in our original data frame, but we will build a function where this doesn't have to be the case:
# A function that takes two data frames, one with actual data: DF, and the other with a
# rearrangement of the columns: R
# R must be structured so that colnames(R) is a subset of colnames(DF)
# Alternatively, R can be structured so that 1 <= as.integer(colnames(R)) <= length(DF)
# Further, 1 <= R$column <= length(DF), and length(R$column) == 1
# These structural requirements on R are not checked
# This is for brevity and because most likely R has been created specifically for use with
# this function
rearrange <- function(DF, R)
{
for (col in colnames(R))
{
DF <- swap(DF, col, R[col])
}
return (DF)
}
Wait, that's it? Yup. This will swap every column name to the appropriate placement. The power for such simplicity comes from swap taking heterogeneous arguments meaning we can specify the moving column name that we want to put somewhere, and so long as we only ever try to put one column in each position (which we should), once we put that column where it belongs, it won't move again. This means that even though it seems like later swaps could undo previous placements, the heterogeneous arguments make certain that won't happen, and so additionally, the order of the columns in our mapping also doesn't matter. This is a really nice quality because it means that we aren't kicking this whole "organizing the data" issue down the road too much. You only have to be able to determine which placement you want to send each column you want to move to.
Ok, ok, there is a catch. If you don't reassign the entire data frame when you do this, then you have superfluous swaps that occur, meaning that if you re-arrange over a subset of columns that isn't "closed", i.e. not every column name has an index that is represented in the rearrangement, then other columns that you didn't explicitly say to move may get moved to other places they don't exactly belong. This can be handled by creating your mapping very carefully, or simply using numerical indices mapping to other numerical indices. In the latter case, this doesn't solve the issue, but it makes more explicit what swaps are taking place and in what order so planning the rearrangement is more explicit and thus less likely to lead to problematic superfluous swaps.
Summary
You can use the swap function that we built to successfully swap exactly two columns or the rearrange function with a "rearrangement" data frame specifying where to send each column name you want to move. In the case of the rearrange function, if any of the placements chosen for each column name are not already occupied by one of the specified columns (i.e. not in colnames(R)), then superfluous swaps can and are very likely to occur (The only instance they won't is when every superfluous swap has a partner superfluous swap that undoes it before the end. This is, as stated, very unlikely to happen by accident, but the mapping can be structured to accomplish this outcome in practice).
swap <- function(DF, n, m)
{
n <- if (class(n)=="character" & is.na(suppressWarnings(as.integer(n)))) which(colnames(DF)==n) else as.integer(n)
m <- if (class(m)=="character" & is.na(suppressWarnings(as.integer(m)))) which(colnames(DF)==m) else as.integer(m)
if (!(1<=n & n<=length(DF))) stop( "`n` represents invalid index!" )
if (!(1<=m & m<=length(DF))) stop( "`m` represents invalid index!" )
return (DF[ if (n==m) 1:length(DF) else c( (if (min(n,m)==1) c() else 1:(min(n,m)-1) ), (if (min(n,m)+1 == max(n,m)) (min(n,m)+1):(max(n,m)-1) else c( max(n,m), (min(n,m)+1):(max(n,m)-1), min(n,m))), (if (max(n,m)==length(DF)) c() else (max(n,m)+1):length(DF) ) ) ])
}
rearrange <- function(DF, R)
{
for (col in colnames(R))
{
DF <- swap(DF, col, R[col])
}
return (DF)
}
I quickly wrote a function that takes a vector v and column indexes a and b which you want to swap.
swappy = function(v,a,b){ # where v is a dataframe, a and b are the columns indexes to swap
name = deparse(substitute(v))
helpy = v[,a]
v[,a] = v[,b]
v[,b] = helpy
name1 = colnames(v)[a]
name2 = colnames(v)[b]
colnames(v)[a] = name2
colnames(v)[b] = name1
assign(name,value = v , envir =.GlobalEnv)
}
I was using the function by Khôra Willis, which is helpful. But I encountered an error. I tried to make corrections. Here is R code that finally works. The arguments n and m could either be column names or column numbers in data frame DF.
require(tidyverse)
swap <- function(DF, n, m)
{
if (class(n)=="character") n <- which(colnames(DF)==n)
if (class(m)=="character") m <- which(colnames(DF)==m)
p <- NCOL(DF)
if (!(1<=n & n<=p)) stop("`n` represents invalid index!")
if (!(1<=m & m<=p)) stop("`m` represents invalid index!")
index <- 1:p
index[n] <- m; index[m] <- n
DF0 <- DF %>% select(all_of(index))
return(DF0)
}
I have a dataframe with millions of rows and three columns labeled Keywords, Impressions, Clicks. I'd like to add a column with values depending on the evaluation of this function:
isType <- function(Impressions, Clicks)
{
if (Impressions >= 1 & Clicks >= 1){return("HasClicks")} else if (Impressions >=1 & Clicks == 0){return("NoClicks")} else {return("ZeroImp")}
}
so far so good. I then try this to create the column but 1) it takes for ever and 2) it marks all the rows has "HasClicks" even the ones where it shouldn't.
# Creates a dataframe
Type <- data.frame()
# Loops until last row and store it in data.frame
for (i in c(1:dim(Mydf)[1])) {Type <- rbind(Type,isType(Mydf$Impressions[i], Mydf$Clicks[i]))}
# Add the column to Mydf
Mydf <- transform(Mydf, Type = Type)
input data:
Keywords,Impressions,Clicks
"Hello",0,0
"World",1,0
"R",34,23
Wanted output:
Keywords,Impressions,Clicks,Type
"Hello",0,0,"ZeroImp"
"World",1,0,"NoClicks"
"R",34,23,"HasClicks"
Building on Joshua's solution, I find it cleaner to generate Type in a single shot (note however that this presumes Clicks >= 0...)
Mydf$Type = ifelse(Mydf$Impressions >= 1,
ifelse(Mydf$Clicks >= 1, 'HasClicks', 'NoClicks'), 'ZeroImp')
First, the if/else block in your function will return the warning:
Warning message:
In if (1:2 > 2:3) TRUE else FALSE :
the condition has length > 1 and only the first element will be used
which explains why it all the rows are the same.
Second, you should allocate your data.frame and fill in the elements rather than repeatedly combining objects together. I imagine this is causing your long run-times.
EDIT: My shared code. I'd love for someone to provide a more elegant solution.
Mydf <- data.frame(
Keywords = sample(c("Hello","World","R"),20,TRUE),
Impressions = sample(0:3,20,TRUE),
Clicks = sample(0:3,20,TRUE) )
Mydf$Type <- "ZeroImp"
Mydf$Type <- ifelse(Mydf$Impressions >= 1 & Mydf$Clicks >= 1,
"HasClicks", Mydf$Type)
Mydf$Type <- ifelse(Mydf$Impressions >= 1 & Mydf$Clicks == 0,
"NoClicks", Mydf$Type)
This is a case where arithmetic can be cleaner and most likely faster than nested ifelse statements.
Again building on Joshua's solution:
Mydf$Type <- factor(with(Mydf, (Impressions>=1)*2 + (Clicks>=1)*1),
levels=1:3, labels=c("ZeroImp","NoClicks","HasClicks"))