Hi,
I have a character-valued vector (old_names) and want to translate
its entries whenever possible, using a dictionary (dict=data.frame).
The translation direction is dict$V3 --> dict$V2, but
some values may be undefined (NA). I suppose this is a very basic
task, but I tried in vain to make it more efficient than below.
In particular I would like to avoid the explicit (and slow) loop
Any help is very much welcome.
Thank you, TM
============================================
new_names = old_names
m = match(old_names, dict$V3)
N = length(old_names)
for (i in 1:N) {
if (is.na(m[i])) { next ; }
nn = as.vector(dict$V2)[m[i]];
if (nn == "" ) { next; }
new_names[i] = nn
}
dictionary lookup
3 messages · Duncan Murdoch, Thomas Manke
On 06/03/2008 6:45 PM, Thomas Manke wrote:
Hi,
I have a character-valued vector (old_names) and want to translate
its entries whenever possible, using a dictionary (dict=data.frame).
The translation direction is dict$V3 --> dict$V2, but
some values may be undefined (NA). I suppose this is a very basic
task, but I tried in vain to make it more efficient than below.
In particular I would like to avoid the explicit (and slow) loop
Any help is very much welcome.
Thank you, TM
============================================
new_names = old_names
m = match(old_names, dict$V3)
N = length(old_names)
for (i in 1:N) {
if (is.na(m[i])) { next ; }
nn = as.vector(dict$V2)[m[i]];
if (nn == "" ) { next; }
new_names[i] = nn
}
You can vectorize this and it should be fast. Here's a straightforward replacement for the loop. It keeps the first 2 lines, and replaces the rest with two more: new_names <- old_names m <- match(old_names, dict$V3) change <- !is.na(m) new_names[change] <- dict$V2[m[change]] Duncan Murdoch
Duncan Murdoch wrote:
On 06/03/2008 6:45 PM, Thomas Manke wrote:
Hi,
I have a character-valued vector (old_names) and want to translate
its entries whenever possible, using a dictionary (dict=data.frame).
The translation direction is dict$V3 --> dict$V2, but
some values may be undefined (NA). I suppose this is a very basic
task, but I tried in vain to make it more efficient than below.
In particular I would like to avoid the explicit (and slow) loop
Any help is very much welcome.
Thank you, TM
============================================
new_names = old_names
m = match(old_names, dict$V3)
N = length(old_names)
for (i in 1:N) {
if (is.na(m[i])) { next ; }
nn = as.vector(dict$V2)[m[i]];
if (nn == "" ) { next; }
new_names[i] = nn
}
You can vectorize this and it should be fast. Here's a straightforward replacement for the loop. It keeps the first 2 lines, and replaces the rest with two more: new_names <- old_names m <- match(old_names, dict$V3) change <- !is.na(m) new_names[change] <- dict$V2[m[change]] Duncan Murdoch
Thank you all for your responses, it certainly pointed me to the right direction. For my purposes, I only had to slightly modify Duncan's suggestion new_names <- rownames(E) m <- match(rownames(E), dict$V3) change <- ( !is.na(m) & dict$V2[m] != "") new_names[change] <- as.vector(dict$V2[m[change]] ) Best wishes, TM