In R have a table containing a set of insect species and an empty column "habitat specifity". Additionally, a vector specifies those species considerated habitat specialists: Species B and C are habitat specialists, species A, D and E are habitat generalists.
example.species <- data.frame (species = c("A","B","C","D","E"), habitat.specifity=NA)
example.species
species habitat.specifity
1 A NA
2 B NA
3 C NA
4 D NA
5 E NA
example.specialists <- c("B","C")
I simply want to fill column two ("habitat specifity") with "s" for specialist and "g" for generalist. The table should then look like this:
species habitat.specifity
1 A g
2 B s
3 C s
4 D g
5 E g
I think it must be a simple task to accomplish, but I cannot figure out how. Any help is appreciated!
Here's a straightforward way in base R:
example.species <- data.frame (species = c("A","B","C","D","E"), habitat.specifity=NA)
example.species$habitat.specifity <- "g" # default value
example.species$habitat.specifity[example.species$species %in% c("B","C")] <- "s"
# species habitat.specifity
# 1 A g
# 2 B s
# 3 C s
# 4 D g
# 5 E g
Example with dplyr:
library(dplyr)
# Your data
example.species <- data.frame(species = c("A","B","C","D","E"),habitat.specifity=NA)
# Simple if_else with dplyr and pipes
example.species %>%
mutate(habitat.specifity = if_else(species %in% c("B","C"), "s", "g"))
# Result
species habitat.specifity
1 A g
2 B s
3 C s
4 D g
5 E g
Related
I have a data frame. I want to filter out some issues only in the case they are associated with a specific group.
For a dummy example, suppose I have the following:
> mydf
Group Issue
1 A G
2 A H
3 A L
4 B V
5 B M
6 C G
7 C H
8 C L
9 C X
10 D G
11 D H
12 D I
I want to filter out rows with a "G" or "H" or "L" issue if there is also an "L" issue in that Group.
So in this case, I want to filter out rows 1, 2, 3, 6,7,8 but leave rows 4,5,9, 10,11, and 12. Thus the result would be:
> mydf
Group Issue
4 B V
5 B M
9 C X
10 D G
11 D H
12 D I
I think I first need to group_by(Group) but then I'm wondering what's the best way to do this.
Thanks!
If the rule is
When a group contains L, drop L, G & H.
then
mydf %>%
group_by(Group) %>%
filter( if (any(Issue=="L")) !(Issue %in% c("G","H","L")) else TRUE )
# Group Issue
# 1 B V
# 2 B M
# 3 C X
# 4 D G
# 5 D H
# 6 D I
I Would like to extract the next 'n' rows after I find a string in R.
For example, let's say I have the following data frame:
df<-as.data.frame(rep(c("a","b","c","d","e","f"),10))
I would like to extract every row that includes "b", as well as the next two rows (in this example, I would like to extract rows with "b", or "c", or "d")
BUT, please, I don't want to specify "c" and "d", I just want the next two rows after "b" as well (in my real data the next two rows are not consistent).
I've tried many things, but no success.. Thanks in advance! Nick
You can find the indices of rows with b and then use those and the next two of each, something like this:
df <- data.frame(col1=rep(c("a","b","c","d","e","f"),3), col2=letters[1:18], stringsAsFactors = FALSE)
df
col1 col2
1 a a
2 b b
3 c c
4 d d
5 e e
6 f f
7 a g
8 b h
9 c i
10 d j
11 e k
12 f l
13 a m
14 b n
15 c o
16 d p
17 e q
18 f r
bs <- which(df$col1=="b")
df[sort(bs+rep(0:2, each=length(bs)),] #2 is the number of rows you want after your desired match (b).
col1 col2
2 b b
3 c c
4 d d
8 b h
9 c i
10 d j
14 b n
15 c o
16 d p
I added a second column to illustrate the dataframe better, otherwise a vector would be returned.
My "SOfun" package has a function called getMyRows which does what you ask for, with the exception of returning a list instead of a data.frame.
I had left the result as a list to make it easier to handle some edge cases, like where the requests for rows would overlap. For example, in the following sample data, there are two consecutive "b" values. There's also a "b" value in the final row.
df <- data.frame(col1 = c("a", "b", "b",
rep(c("a", "b", "c", "d", "e", "f"), 3), "b"),
col2 = letters[1:22])
library(SOfun)
getMyRows(df, which(df$col1 == "b"), 0:2, TRUE)
# [[1]]
# col1 col2
# 2 b b
# 3 b c
# 4 a d
#
# [[2]]
# col1 col2
# 3 b c
# 4 a d
# 5 b e
#
# [[3]]
# col1 col2
# 5 b e
# 6 c f
# 7 d g
#
# [[4]]
# col1 col2
# 11 b k
# 12 c l
# 13 d m
#
# [[5]]
# col1 col2
# 17 b q
# 18 c r
# 19 d s
#
# [[6]]
# col1 col2
# 22 b v
The usage is essentially:
Specify the data.frame.
Specify the index positions to use as the base. Here, we want all rows where "col1" equals "b" to be our base index position.
Specify the range of rows interested in. -1:3, for example, would give you one row before to three rows after the base.
TRUE means that you are specifying the starting points by their numeric indices.
Given:
df <- data.frame(rep = letters[sample(4, 30, replace=TRUE)], loc = LETTERS[sample(5:8, 30, replace=TRUE)], y= rnorm(30))
lookup <- data.frame(rep=letters[1:4], loc=LETTERS[5:8])
This will give me the rows in df that have rep,loc combinations that occur in lookup:
mdply(lookup, function(rep,loc){
r=rep
l=loc
subset(df, rep==r & loc==l)
})
But I've read that using subset() inside a function is poor practice due to scoping issues. So how do I get the desired result using index notation?
In this particular case, merge seems to make the most sense to me:
merge(df, lookup)
# rep loc y
# 1 a E 1.6612394
# 2 a E 1.1050825
# 3 a E -0.7016759
# 4 b F 0.4364568
# 5 d H 1.3246636
# 6 d H -2.2573545
# 7 d H 0.5061980
# 8 d H 0.1397326
A simple alternative might be to paste together the "rep" and "loc" columns from df and from lookup and subset based on that:
df[do.call(paste, df[c("rep", "loc")]) %in% do.call(paste, lookup), ]
# rep loc y
# 4 d H 1.3246636
# 10 b F 0.4364568
# 14 a E -0.7016759
# 15 a E 1.6612394
# 19 d H 0.5061980
# 20 a E 1.1050825
# 22 d H -2.2573545
# 28 d H 0.1397326
I have a data frame like this:
n = c(2, 2, 3, 3, 4, 4)
n <- as.factor(n)
s = c("a", "b", "c", "d", "e", "f")
df = data.frame(n, s)
df
n s
1 2 a
2 2 b
3 3 c
4 3 d
5 4 e
6 4 f
and I want to access the first element of each level of my factor (and have in this example a vector containing a, c, e).
It is possible to reach the first element of one level, with
df$s[df$n == 2][1]
but it does not work for all levels:
df$s[df$n == levels(n)]
[1] a f
How would you do that?
And to go further, I’d like to modify my data frame to see which is the first element for each level at every occurrence. In my example, a new column should be:
n s rep firstelement
1 2 a a a
2 2 b c a
3 3 c e c
4 3 d a c
5 4 e c e
6 4 f e e
Edit. The first part of my answer addresses the original question, i.e. before "And to go further" (which was added by OP in an edit).
Another possibility, using duplicated. From ?duplicated: "duplicated() determines which elements of a vector or data frame are duplicates of elements with smaller subscripts."
Here we use !, the logical negation (NOT), to select not duplicated elements of 'n', i.e. first elements of each level of 'n'.
df[!duplicated(df$n), ]
# n s
# 1 2 a
# 3 3 c
# 5 4 e
Update Didn't see your "And to go further" edit until now. My first suggestion would definitely be to use ave, as already proposed by #thelatemail and #sparrow. But just to dig around in the R toolbox and show you an alternative, here's a dplyr way:
Group the data by n, use the mutate function to create a new variable 'first', with the value 'first element of s' (s[1]),
library(dplyr)
df %.%
group_by(n) %.%
mutate(
first = s[1])
# n s first
# 1 2 a a
# 2 2 b a
# 3 3 c c
# 4 3 d c
# 5 4 e e
# 6 4 f e
Or go all in with dplyr convenience functions and use first instead of [1]:
df %.%
group_by(n) %.%
mutate(
first = first(s))
A dplyr solution for your original question would be to use summarise:
df %.%
group_by(n) %.%
summarise(
first = first(s))
# n first
# 1 2 a
# 2 3 c
# 3 4 e
Here is an approach using match:
df$s[match(levels(n), df$n)]
EDIT: Maybe this looks a bit confusing ...
To get a column which lists the first elements you could use match twice (but with x and table arguments swapped):
df$firstelement <- df$s[match(levels(n), df$n)[match(df$n, levels(n))]]
df$firstelement
# [1] a a c c e e
# Levels: a b c d e f
Lets look at this in detail:
## this returns the first matching elements
match(levels(n), df$n)
# [1] 1 3 5
## when we swap the x and table argument in match we get the level index
## for each df$n (the duplicated indices are important)
match(df$n, levels(n))
# [1] 1 1 2 2 3 3
## results in
c(1, 3, 5)[c(1, 1, 2, 2, 3, 3)]
# [1] 1 1 3 3 5 5
df$s[c(1, 1, 3, 3, 5, 5)]
# [1] a a c c e e
# Levels: a b c d e f
the function ave is useful in these cases:
df$firstelement = ave(df$s, df$n, FUN = function(x) x[1])
df
n s firstelement
1 2 a a
2 2 b a
3 3 c c
4 3 d c
5 4 e e
6 4 f e
In this case I prefer plyr package, it gives further freedom to manipulate the data.
library(plyr)
ddply(df,.(n),function(subdf){return(subdf[1,])})
n s
1 2 a
2 3 c
3 4 e
You could also use data.table
library(data.table)
dt = as.data.table(df)
dt[, list(firstelement = s[1]), by=n]
which would get you:
n firstelement
1: 2 a
2: 3 c
3: 4 e
The by=n bit groups everything by each value of n so s[1] is getting the first element of each of those groups.
To get this as an extra column you could do:
dt[, newcol := s[1], by=n]
dt
# n s newcol
#1: 2 a a
#2: 2 b a
#3: 3 c c
#4: 3 d c
#5: 4 e e
#6: 4 f e
So this just takes the value of s from the first row of each group and assigns it to a new column.
df$s[sapply(levels(n), function(particular.level) { which(df$n == particular.level)[1]})]
I believe your problem is that you are comparing two vectors df$n is a vector and levels(n) is a vector. vector == vector only happens to work for you since df$n is a multiple length of levels(n)
Surprised not to see this classic in the answer stream yet.
> do.call(rbind, lapply(split(df, df$n), function(x) x[1,]))
## n s
## 2 2 a
## 3 3 c
## 4 4 e
I have two dataframe in R.
dataframe 1
A B C D E F G
1 2 a a a a a
2 3 b b b c c
4 1 e e f f e
dataframe 2
X Y Z
1 2 g
2 1 h
3 4 i
1 4 j
I want to match dataframe1's column A and B with dataframe2's column X and Y. It is NOT a pairwise comparsions, i.e. row 1 (A=1 B=2) are considered to be same as row 1 (X=1, Y=2) and row 2 (X=2, Y=1) of dataframe 2.
When matching can be found, I would like to add columns C, D, E, F of dataframe1 back to the matched row of dataframe2, as follows: with no matching as na.
Final dataframe
X Y Z C D E F G
1 2 g a a a a a
2 1 h a a a a a
3 4 i na na na na na
1 4 j e e f f e
I can only know how to do matching for single column, however, how to do matching for two exchangable columns and merging two dataframes based on the matching results is difficult for me. Pls kindly help to offer smart way of doing this.
For the ease of discussion (thanks for the comments by Vincent and DWin (my previous quesiton) that I should test the quote.) There are the quota for loading dataframe 1 and 2 to R.
df1 <- data.frame(A = c(1,2,4), B=c(2,3,1), C=c('a','b','e'),
D=c('a','b','e'), E=c('a','b','f'),
F=c('a','c','f'), G=c('a','c', 'e'))
df2 <- data.frame(X = c(1,2,3,1), Y=c(2,1,4,4), Z=letters[7:10])
The following works, but no doubt can be improved.
I first create a little helper function that performs a row-wise sort on A and B (and renames it to V1 and V2).
replace_index <- function(dat){
x <- as.data.frame(t(sapply(seq_len(nrow(dat)),
function(i)sort(unlist(dat[i, 1:2])))))
names(x) <- paste("V", seq_len(ncol(x)), sep="")
data.frame(x, dat[, -(1:2), drop=FALSE])
}
replace_index(df1)
V1 V2 C D E F G
1 1 2 a a a a a
2 2 3 b b b c c
3 1 4 e e f f e
This means you can use a straight-forward merge to combine the data.
merge(replace_index(df1), replace_index(df2), all.y=TRUE)
V1 V2 C D E F G Z
1 1 2 a a a a a g
2 1 2 a a a a a h
3 1 4 e e f f e j
4 3 4 <NA> <NA> <NA> <NA> <NA> i
This is slightly clunky, and has some potential collision and order issues but works with your example
df1a <- df1; df1a$A <- df1$B; df1a$B <- df1$A #reverse A and B
merge(df2, rbind(df1,df1a), by.x=c("X","Y"), by.y=c("A","B"), all.x=TRUE)
to produce
X Y Z C D E F G
1 1 2 g a a a a a
2 1 4 j e e f f e
3 2 1 h a a a a a
4 3 4 i <NA> <NA> <NA> <NA> <NA>
One approach would be to create an id key for matching that is order invariant.
# create id key to match
require(plyr)
df1 = adply(df1, 1, transform, id = paste(min(A, B), "-", max(A, B)))
df2 = adply(df2, 1, transform, id = paste(min(X, Y), "-", max(X, Y)))
# combine data frames using `match`
cbind(df2, df1[match(df2$id, df1$id),3:7])
This produces the output
X Y Z id C D E F G
1 1 2 g 1 - 2 a a a a a
1.1 2 1 h 1 - 2 a a a a a
NA 3 4 i 3 - 4 <NA> <NA> <NA> <NA> <NA>
3 1 4 j 1 - 4 e e f f e
You could also join the tables both ways (X == A and Y == B, then X == B and Y == A) and rbind them. This will produce duplicate pairs where one way yielded a match and the other yielded NA, so you would then reduce duplicates by slicing only a single row for each X-Y combination, the one without NA if one exists.
library(dplyr)
m <- left_join(df2,df1,by = c("X" = "A","Y" = "B"))
n <- left_join(df2,df1,by = c("Y" = "A","X" = "B"))
rbind(m,n) %>%
group_by(X,Y) %>%
arrange(C,D,E,F,G) %>% # sort to put NA rows on bottom of pairs
slice(1) # take top row from combination
Produces:
Source: local data frame [4 x 8]
Groups: X, Y
X Y Z C D E F G
1 1 2 g a a a a a
2 1 4 j e e f f e
3 2 1 h a a a a a
4 3 4 i NA NA NA NA NA
Here's another possible solution in base R. This solution cbind()s new key columns (K1 and K2) to both data.frames using the vectorized pmin() and pmax() functions to derive the canonical order of the key columns, and merges on those:
merge(cbind(df2,K1=pmin(df2$X,df2$Y),K2=pmax(df2$X,df2$Y)),cbind(df1,K1=pmin(df1$A,df1$B),K2=pmax(df1$A,df1$B)),all.x=T)[,-c(1:2,6:7)];
## X Y Z C D E F G
## 1 1 2 g a a a a a
## 2 2 1 h a a a a a
## 3 1 4 j e e f f e
## 4 3 4 i <NA> <NA> <NA> <NA> <NA>
Note that the use of pmin() and pmax() is only possible for this problem because you only have two key columns; if you had more, then you'd have to use some kind of apply+sort solution to achieve the canonical key order for merging, similar to what #Andrie does in his helper function, which would work for any number of key columns, but would be less performant.