I have a simple problem which can be solved in a dirty way, but I'm looking for a clean way using data.table
I have the following data.table with n columns belonging to m unequal groups. Here is an example of my data.table:
dframe <- as.data.frame(matrix(rnorm(60), ncol=30))
cletters <- rep(c("A","B","C"), times=c(10,14,6))
colnames(dframe) <- cletters
A A A A A A
1 -0.7431185 -0.06356047 -0.2247782 -0.15423889 -0.03894069 0.1165187
2 -1.5891905 -0.44468389 -0.1186977 0.02270782 -0.64950716 -0.6844163
A A A A B B B
1 -1.277307 1.8164195 -0.3957006 -0.6489105 0.3498384 -0.463272 0.8458673
2 -1.644389 0.6360258 0.5612634 0.3559574 1.9658743 1.858222 -1.4502839
B B B B B B B
1 0.3167216 -0.2919079 0.5146733 0.6628149 0.5481958 -0.01721261 -0.5986918
2 -0.8104386 1.2335948 -0.6837159 0.4735597 -0.4686109 0.02647807 0.6389771
B B B B C C
1 -1.2980799 0.3834073 -0.04559749 0.8715914 1.1619585 -1.26236232
2 -0.3551722 -0.6587208 0.44822253 -0.1943887 -0.4958392 0.09581703
C C C C
1 -0.1387091 -0.4638417 -2.3897681 0.6853864
2 0.1680119 -0.5990310 0.9779425 1.0819789
What I want to do is to take a random subset of the columns (of a sepcific size), keeping the same number of columns per group (if the chosen sample size is larger than the number of columns belonging to one group, take all of the columns of this group).
I have tried an updated version of the method mentioned in this question:
sample rows of subgroups from dataframe with dplyr
but I'm not able to map the column names to the by argument.
Can someone help me with this?
Here's another approach, IIUC:
idx <- split(seq_along(dframe), names(dframe))
keep <- unlist(Map(sample, idx, pmin(7, lengths(idx))))
dframe[, keep]
Explanation:
The first step splits the column indices according to the column names:
idx
# $A
# [1] 1 2 3 4 5 6 7 8 9 10
#
# $B
# [1] 11 12 13 14 15 16 17 18 19 20 21 22 23 24
#
# $C
# [1] 25 26 27 28 29 30
In the next step we use
pmin(7, lengths(idx))
#[1] 7 7 6
to determine the sample size in each group and apply this to each list element (group) in idx using Map. We then unlist the result to get a single vector of column indices.
Not sure if you want a solution with dplyr, but here's one with just lapply:
dframe <- as.data.frame(matrix(rnorm(60), ncol=30))
cletters <- rep(c("A","B","C"), times=c(10,14,6))
colnames(dframe) <- cletters
# Number of columns to sample per group
nc <- 8
res <- do.call(cbind,
lapply(unique(colnames(dframe)),
function(x){
dframe[,if(sum(colnames(dframe) == x) <= nc) which(colnames(dframe) == x) else sample(which(colnames(dframe) == x),nc,replace = F)]
}
))
It might look complicated, but it really just takes all columns per group if there's less than nc, and samples random nc columns if there are more than nc columns.
And to restore your original column-name scheme, gsub does the trick:
colnames(res) <- gsub('.[[:digit:]]','',colnames(res))
Related
How could i get a sample of a values of a vector but keep the order without compairing the values themself against each other?
for example:
V1 contains values (1,2,3,4,5,6,7,8,9,10,11,12,13,14)
I woule like to get a sample
sample <- (2,7,10,14)
As you can see the values are still on order but randomly selected.
But if i use a function sample or rdunif in R I get random orderd selection:
ie. (7,10,2,14)
Thank you!
With the following solution you do not compare the elements of your original vector in order to sort them; the only thing you do is shuffling a vector of logical values (TRUE or FALSE).
Let's say you want to pick n elements from the already-ordered vector v and maintain their order. Then you can do
v <- 1:14
n <- 4
set.seed(42) # for reproducibility
logi <- sample(c(rep(TRUE, n), rep(FALSE, length(v) - n)))
v[logi]
# [1] 1 6 7 14
EDIT to prove that the vector v can be any vector, and we still manage to maintain its original order.
set.seed(1)
n <- 4
v <- sample(14, replace = FALSE)
v
# [1] 9 4 7 1 2 12 3 6 10 8 5 11 13 14
set.seed(42) # for reproducibility
logi <- sample(c(rep(TRUE, n), rep(FALSE, length(v) - n)))
v[logi]
# [1] 9 12 3 14
These numbers respect indeed the original order of vector v.
Let's see if we can't do this when the original V1 is not in numerical order.
set.seed(42)
v <- sample(1:14,14,rep=FALSE)
# [1] 1 5 14 9 10 4 2 8 12 11 6 13 7 3
n <- 4
foo <- sample(v,length(v)-n,rep=FALSE)
match(foo,v)
v[-match(foo,v)]
# [1] 1 13 7 3
Now the output sample values are in the same order they are in the original vector.
V1 <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14)
sample_V1 <- sample(V1, 4)
sort(sample_V1)
I am trying to polishing my R skills and sort of hit my limit.
The issue I am trying to find the solution as follows.
Suppose my dataframe is as below ,
n = c(2, 15, 31 , 33)
n2 = c( 10 , 9, 10 , 40)
n3 = c( 11 , 10 , 11 , 42)
df = data.frame(n , n2 , n3)
> df
n n2 n3
1 2 10 11
2 15 9 10
3 31 10 11
4 33 40 42
if I would like to go through each row , and generate a random pair :eg 2,10 and go through each of the rest of the rows to find a repeated pair and print out the common pairs and number of occurrence , how can I do that?
In the above example , the only pair that repeat is 10 and 11 at rows 1 and 3.
So far I have thought about the pseudo code as follows
for(each row in the dataframe)
{
for (each of the values in the row)
{
for every pair
}
find a repeated pair
if found store in a dataframe
}
and to find the random pair using combn function.
But I am a little lost at the part on iteration through out the dataframe rows.
Pls help.
Thanks a lot!
I think this is what you want. Instead of thinking about selecting every combination of two values for each row, we'll get every combination of two column numbers - which will be the same for every row. Then we use plyr::count as a convenience function to count rows with the same values for an entire data frame at once. This way we can loop over the combinations of column indices rather than over rows. I use apply, but you could write it as a for loop instead.
pairs = combn(ncol(df), m = 2)
result = apply(pairs, MAR = 2, FUN = function(p) {
plyr::count(df[p])
})
names(result) = apply(pairs, MAR = 2, FUN = paste, collapse = "_")
The result is a list where each item is is a data frame with two columns and a freq column giving the number of rows in the original data each value-pair occurred.
result
# $`1_2`
# n n2 freq
# 1 2 10 1
# 2 15 9 1
# 3 31 10 1
# 4 33 40 1
#
# $`1_3`
# n n3 freq
# 1 2 11 1
# 2 15 10 1
# 3 31 11 1
# 4 33 42 1
#
# $`2_3`
# n2 n3 freq
# 1 9 10 1
# 2 10 11 2
# 3 40 42 1
If you want to omit the values that aren't repeated, we can just subset them out:
lapply(result, subset, freq > 1)
# $`1_2`
# [1] n n2 freq
# <0 rows> (or 0-length row.names)
#
# $`1_3`
# [1] n n3 freq
# <0 rows> (or 0-length row.names)
#
# $`2_3`
# n2 n3 freq
# 2 10 11 2
Slightly different method
n = c(2, 15, 31 , 15) # changed the dataset to have some common pairs in n and n2 too
n2 = c( 10 , 9, 10 , 9)
n3 = c( 11 , 10 , 11 , 42)
df = data.frame(n , n2 , n3)
library(dplyr)
library(rlang)
library(utils)
cols<-colnames(df) # define the columns that you want to do the pair checking for
combinations<- as.data.frame(combn(cols,2),stringsAsFactors = FALSE)
# picks up all combinations of columns
#iterates over each pair of columns
all_combs<- lapply(names(combinations[cols1]), function(x){
df %>%
group_by(!! sym( combinations[[x]][1]),!! sym( combinations[[x]][2])) %>%
filter(n()>1) # groups by the two columns, and filters out pairs that occur more than once. You can add a distinct command below if you
#dont want them repeated
})
all_combs_df <- do.call("rbind", all_combs)# all_combs is in a list format, use rbind to convert into a dataframe
all_combs_df
the output is this
n n2 n3
<dbl> <dbl> <dbl>
1 15. 9. 10.
2 15. 9. 42.
3 2. 10. 11.
4 31. 10. 11.
If I have the following data table
m = matrix(1:12, ncol=4)
colnames(m) = c('A1','A2','B1','B2')
d = data.table(m)
is it possible to execute a function on sets of columns?
For example the following would be the sum of A1,A2 and B1,B2.
A B
1: 5 17
2: 7 19
3: 9 21
The solution would preferably work with a 500k x 100 matrix
Solution
A trick would be to split the column into groups.
Then you can use rowSums as Frank suggests (see comments on question):
# using your data example
m <- matrix(1:12, ncol = 4)
colnames(m) <- c('A1', 'A2', 'B1', 'B2')
d <- data.table(m)
# 1) group columns
groups <- split(colnames(d), substr(colnames(d), 1, 1))
# 2) group wise row sums
d[,lapply(groups, function(i) {rowSums(d[, i, with = FALSE])})]
Result
This will return the data.table:
A B
1: 5 17
2: 7 19
3: 9 21
Explanation
split creates a list of column names for each group, defined by a (something coercable to a) factor.
substr(colnames(m), 1, 1) takes the first letter as group id, use a different approach (e.g. sub("([A-Z]).*", "\\1", colnames(m)) for variable number of letters).
lapply is commonly used to apply functions over multiple columns in a data.table. Here we create a list output, named as the groups, containing the rowSums. with = FALSE is important to use the value of i to get the respective columns from d.
Definitely possible...
d[, ":=" (A = A1 + A2, B = B1 + B2)]
d
A1 A2 B1 B2 A B
1: 1 4 7 10 5 17
2: 2 5 8 11 7 19
3: 3 6 9 12 9 21
# Want to drop the old columns?
set(d, j = which(names(d) %in% c("A1", "B1", "A2", "B2")), value = NULL)
d
A B
1: 5 17
2: 7 19
3: 9 21
Whether it is desirable I shall not tell. Probably better to follow Frank's advice (see comments).
I've noticed that aggregate() appears to return its result ordered by the grouping column(s). Is this a guarantee? Can this be relied upon in surrounding logic?
A couple of examples:
set.seed(1); df <- data.frame(group=sample(letters[1:3],10,replace=T),value=1:10);
aggregate(value~group,df,sum);
## group value
## 1 a 16
## 2 b 22
## 3 c 17
And with two groups (notice the second group is ordered first, then the first group to break ties):
set.seed(1); df <- data.frame(group1=sample(letters[1:3],10,replace=T),group2=sample(letters[4:6],10,replace=T),value=1:10);
aggregate(value~group1+group2,df,sum);
## group1 group2 value
## 1 a d 1
## 2 b d 2
## 3 b e 9
## 4 c e 10
## 5 a f 15
## 6 b f 11
## 7 c f 7
Note: I'm asking because I just came up with an answer for Aggregating while merging two dataframes in R which, at least in its current form at the time of writing, depends on aggregate() returning its result ordered by the grouping column.
Yes, as long as you understand the natural ordering of factors to be by their integer keys. You can see this in the code:
y <- as.data.frame(by, stringsAsFactors = FALSE)
... # y becomes the "integerized" dataframe of index vectors
grp <- rank(do.call(paste, c(lapply(rev(y), ident), list(sep = "."))),
ties.method = "min")
y <- y[match(sort(unique(grp)), grp, 0L), , drop = FALSE]
...
I would like to aggregate an R data.frame by equal amounts of the cumulative sum of one of the variables in the data.frame. I googled quite a lot, but probably I don't know the correct terminology to find anything useful.
Suppose I have this data.frame:
> x <- data.frame(cbind(p=rnorm(100, 10, 0.1), v=round(runif(100, 1, 10))))
> head(x)
p v
1 10.002904 4
2 10.132200 2
3 10.026105 6
4 10.001146 2
5 9.990267 2
6 10.115907 6
7 10.199895 9
8 9.949996 8
9 10.165848 8
10 9.953283 6
11 10.072947 10
12 10.020379 2
13 10.084002 3
14 9.949108 8
15 10.065247 6
16 9.801699 3
17 10.014612 8
18 9.954638 5
19 9.958256 9
20 10.031041 7
I would like to reduce the x to a smaller data.frame where each line contains the weighted average of p, weighted by v, corresponding to an amount of n units of v. Something of this sort:
> n <- 100
> cum.v <- cumsum(x$v)
> f <- cum.v %/% n
> x.agg <- aggregate(cbind(v*p, v) ~ f, data=x, FUN=sum)
> x.agg$'v * p' <- x.agg$'v * p' / x.agg$v
> x.agg
f v * p v
1 0 10.039369 98
2 1 9.952049 94
3 2 10.015058 104
4 3 9.938271 103
5 4 9.967244 100
6 5 9.995071 69
First question, I was wondering if there is a better (more efficient approach) to the code above. The second, more important, question is how to correct the code above in order to obtain more precise bucketing. Namely, each row in x.agg should contain exacly 100 units of v, not just approximately as it is the case above. For example, the first row contains the aggregate of the first 17 rows of x which correspond to 98 units of v. The next row (18th) contains 5 units of v and is fully included in the next bucket. What I would like to achieve instead would be attribute 2 units of row 18th to the first bucket and the remaining 3 units to the following one.
Thanks in advance for any help provided.
Here's another method that does this with out repeating each p v times. And the way I understand it is, the place where it crosses 100 (see below)
18 9.954638 5 98
19 9.958256 9 107
should be changed to:
18 9.954638 5 98
19.1 9.958256 2 100 # ---> 2 units will be considered with previous group
19.2 9.958256 7 107 # ----> remaining 7 units will be split for next group
The code:
n <- 100
# get cumulative sum, an id column (for retrace) and current group id
x <- transform(x, cv = cumsum(x$v), id = seq_len(nrow(x)), grp = cumsum(x$v) %/% n)
# Paste these two lines in R to install IRanges
source("http://bioconductor.org/biocLite.R")
biocLite("IRanges")
require(IRanges)
ir1 <- successiveIRanges(x$v)
ir2 <- IRanges(seq(n, max(x$cv), by=n), width=1)
o <- findOverlaps(ir1, ir2)
# gets position where multiple of n(=100) occurs
# (where we'll have to do something about it)
pos <- queryHits(o)
# how much do the values differ from multiple of 100?
val <- start(ir2)[subjectHits(o)] - start(ir1)[queryHits(o)] + 1
# we need "pos" new rows of "pos" indices
x1 <- x[pos, ]
x1$v <- val # corresponding values
# reduce the group by 1, so that multiples of 100 will
# belong to the previous row
x1$grp <- x1$grp - 1
# subtract val in the original data x
x$v[pos] <- x$v[pos] - val
# bind and order them
x <- rbind(x1,x)
x <- x[with(x, order(id)), ]
# remove unnecessary entries
x <- x[!(duplicated(x$id) & x$v == 0), ]
x$cv <- cumsum(x$v) # updated cumsum
x$id <- NULL
require(data.table)
x.dt <- data.table(x, key="grp")
x.dt[, list(res = sum(p*v)/sum(v), cv = tail(cv, 1)), by=grp]
Running on your data:
# grp res cv
# 1: 0 10.037747 100
# 2: 1 9.994648 114
Running on #geektrader's data:
# grp res cv
# 1: 0 9.999680 100
# 2: 1 10.040139 200
# 3: 2 9.976425 300
# 4: 3 10.026622 400
# 5: 4 10.068623 500
# 6: 5 9.982733 562
Here's a benchmark on a relatively big data:
set.seed(12345)
x <- data.frame(cbind(p=rnorm(1e5, 10, 0.1), v=round(runif(1e5, 1, 10))))
require(rbenchmark)
benchmark(out <- FN1(x), replications=10)
# test replications elapsed relative user.self
# 1 out <- FN1(x) 10 13.817 1 12.586
It takes about 1.4 seconds on 1e5 rows.
If you are looking for precise bucketing, I am assuming value of p is same for 2 "split" v
i.e. in your example, value of p for 2 units of row 18th that go in first bucket is 9.954638
With above assumption, you can do following for not super large datasets..
> set.seed(12345)
> x <- data.frame(cbind(p=rnorm(100, 10, 0.1), v=round(runif(100, 1, 10))))
> z <- unlist(mapply(function(x,y) rep(x,y), x$p, x$v, SIMPLIFY=T))
this creates a vector with each value of p repeated v times for each row and result is combined into single vector using unlist.
After this aggregation is trivial using aggregate function
> aggregate(z, by=list((1:length(z)-0.5)%/%100), FUN=mean)
Group.1 x
1 0 9.999680
2 1 10.040139
3 2 9.976425
4 3 10.026622
5 4 10.068623
6 5 9.982733