I have 1000+ rows and I want to calculate the CV for each row that has the same condition.
The data look like this:
Condition Y
0.5 25
0.5 26
0.5 27
1 43
1 45
1 75
5 210
5 124
5 20
10 54
10 78
10 10
and then I did:
CV <- function(x){
(sd(x)/mean(x))*100
}
CV.for every row. <- aggregate(y ~ Condition,
data = df,
FUN = CV)
I have the feeling that what I did, uses the mean of the whole column, cause the results are a bit whatever.
Related
I have a two-column dataframe contataining thousands of IDs where each ID has hundreds of data rows, in other words a data frame of about 6 million rows. I am grouping (using either dplyr or data.table) this data frame by IDs and performing a "tso" (outlier detection) function on grouped data frame. The problem is after hours of computation it returns me an error related to ARIMA specification of one of the IDs. Question is how can I identify the ID (or the row number) where my function is returning error?? (if I detect it then I can remove that ID from dataframe)
I tried to manually perform my function on subgroups of this dataframe however I cannot reach the erroneous ID because there are thousands of IDs so it takes me weeks to find them this way.
outlier.detection <- function(x,iter) {
y <- as.ts(x)
out2 <- tso(y,maxit.iloop = iter,tsmethod = "auto.arima",remove.method = "bottom-up",cval=3)
y[out2$outliers$ind] <- NA
return(y)}
df <- data.table(outlying1);setkey(df,id)
test <- df[,list(new.weight = outlier.detection(weight,iter=1)),by=id]
the above function finds the annomalies and replace them with NAs. here is an example,
ID weight
1 a 50
2 a 50
3 a 51
4 a 51.5
5 a 52
6 b 80
7 b 81
8 b 81.5
9 b 90
10 b 82
it will look like the following,
ID weight
1 a 50
2 a 50
3 a 51
4 a 51.5
5 a 52
6 b 80
7 b 81
8 b 81.5
9 b NA
10 b 82
I am relatively new to R from Stata. I have a data frame that has 100+ columns and thousands of rows. Each row has a start value, stop value, and 100+ columns of numerical values. The goal is to get the sum of each row from the column that corresponds to the start value to the column that corresponds to the stop value. This is direct enough to do in a loop, that looks like this (data.frame is df, start is the start column, stop is the stop column):
for(i in 1:nrow(df)) {
df$out[i] <- rowSums(df[i,df$start[i]:df$stop[i]])
}
This works great, but it is taking 15 minutes or so. Does anyone have any suggestions on a faster way to do this?
You can do this using some algebra (if you have a sufficient amount of memory):
DF <- data.frame(start=3:7, end=4:8)
DF <- cbind(DF, matrix(1:50, nrow=5, ncol=10))
# start end 1 2 3 4 5 6 7 8 9 10
#1 3 4 1 6 11 16 21 26 31 36 41 46
#2 4 5 2 7 12 17 22 27 32 37 42 47
#3 5 6 3 8 13 18 23 28 33 38 43 48
#4 6 7 4 9 14 19 24 29 34 39 44 49
#5 7 8 5 10 15 20 25 30 35 40 45 50
take <- outer(seq_len(ncol(DF)-2)+2, DF$start-1, ">") &
outer(seq_len(ncol(DF)-2)+2, DF$end+1, "<")
diag(as.matrix(DF[,-(1:2)]) %*% take)
#[1] 7 19 31 43 55
If you are dealing with values of all the same types, you typically want to do things in matrices. Here is a solution in matrix form:
rows <- 10^3
cols <- 10^2
start <- sample(1:cols, rows, replace=T)
end <- pmin(cols, start + sample(1:(cols/2), rows, replace=T))
# first 2 cols of matrix are start and end, the rest are
# random data
mx <- matrix(c(start, end, runif(rows * cols)), nrow=rows)
# use `apply` to apply a function to each row, here the
# function sums each row excluding the first two values
# from the value in the start column to the value in the
# end column
apply(mx, 1, function(x) sum(x[-(1:2)][x[[1]]:x[[2]]]))
# df version
df <- as.data.frame(mx)
df$out <- apply(df, 1, function(x) sum(x[-(1:2)][x[[1]]:x[[2]]]))
You can convert your data.frame to a matrix with as.matrix. You can also run the apply directly on your data.frame as shown, which should still be reasonably fast. The real problem with your code is that your are modifying a data frame nrow times, and modifying data frames is very slow. By using apply you get around that by generating your answer (the $out column), which you can then cbind back to your data frame (and that means you modify your data frame just once).
I searched SO, but I could not seem to find the right code that is applicable to my question. It is similar to this question: Linear Regression calculation several times in one dataframe
I got a dataframe of LR coefficients following Andrie's code:
Cddply <- ddply(test, .(sumtest), function(test)coef(lm(Area~Conc, data=test)))
sumtest (Intercept) Conc
1 -108589.2726 846.0713372
2 -49653.18701 811.3982918
3 -102598.6252 832.6419926
4 -72607.4017 727.0765558
5 54224.28878 391.256075
6 -42357.45407 357.0845661
7 -34171.92228 367.3962888
8 -9332.569856 289.8631555
9 -7376.448899 335.7047756
10 -37704.92277 359.1457617
My question is how to apply each of these LR models (1-10) to specific row intervals in another dataframe in order to get x, the independent variable, into a 3rd column. For example, I would like to apply sumtest1 to Samples 6:29, sumtest2 to samples 35:50, sumtest3 to samples 56:79, etc.. in intervals of 24 and 16 samples. The sample numbers repeats after 200, so sumtest9 will be for Samples 6:29 again.
Sample Area
6 236211
7 724919
8 1259814
9 1574722
10 268836
11 863818
12 1261768
13 1591845
14 220322
15 608396
16 980182
17 1415859
18 276276
19 724532
20 1130024
21 1147840
22 252051
23 544870
24 832512
25 899457
26 285093
27 4291007
28 825922
29 865491
35 246707
36 538092
37 767269
38 852410
39 269152
40 971471
41 1573989
42 1897208
43 261321
44 481486
45 598617
46 769240
47 229695
48 782691
49 1380597
50 1725419
The resulting dataframe would look like this:
Sample Area Calc
6 236211 407.5312917
7 724919 985.1525288
8 1259814 1617.363812
9 1574722 1989.564693
10 268836 446.0919309
...
35 246707 365.2452551
36 538092 724.3591324
37 767269 1006.805521
38 852410 1111.736505
39 269152 392.9073207
Thank you for your assistance.
Is this what you want? I made up a slightly larger dummy data set of 'area' to make it easier to see how the code worked when I tried it out.
# create 400 rows of area data
set.seed(123)
df <- data.frame(area = round(rnorm(400, mean = 1000000, sd = 100000)))
# "sample numbers repeats after 200" -> add a sample nr 1-200, 1-200
df$sample_nr <- 1:200
# create a factor which cuts the vector of sample_nr into pieces of length 16, 24, 16, 24...
# repeat to a total length of the pieces is 200
# i.e. 5 repeats of (16, 24)
grp <- cut(df$sample_nr, breaks = c(-Inf, cumsum(rep(c(16, 24), 5))))
# add a numeric version of the chunks to data frame
# this number indicates the model from which coefficients will be used
# row 1-16 (16 rows): model 1; row 17-40 (24 rows): model 2;
# row 41-56 (16 rows): model 3; and so on.
df$mod <- as.numeric(grp)
# read coefficients
coefs <- read.table(text = "intercept beta_conc
1 -108589.2726 846.0713372
2 -49653.18701 811.3982918
3 -102598.6252 832.6419926
4 -72607.4017 727.0765558
5 54224.28878 391.256075
6 -42357.45407 357.0845661
7 -34171.92228 367.3962888
8 -9332.569856 289.8631555
9 -7376.448899 335.7047756
10 -37704.92277 359.1457617", header = TRUE)
# add model number
coefs$mod <- rownames(coefs)
head(df)
head(coefs)
# join area data and coefficients by model number
# (use 'join' instead of merge to avoid sorting)
library(plyr)
df2 <- join(df, coefs)
# calculate conc from area and model coefficients
# area = intercept + beta_conc * conc
# conc = (area - intercept) / beta_conc
df2$conc <- (df2$area - df2$intercept) / df2$beta_conc
head(df2, 41)
I have two matrices, one is a 10x1 double matrix, that can be expanded to any user preset number, eg. 100.
View(min_matrx)
V1
1 27
2 46
3 30
4 59
5 46
6 45
7 34
8 31
9 52
10 46
The other matrix looks like this, there are more rows not shown:
View(main_matrx)
row.names sum_value
s17 45
s7469 213
s20984 24
s17309 214
s7432369 43
s221320984 12
s17556 34
s741269 11
s20132984 35
For each row name in main_matrx I want to count the number of times that a value more than the sum_value in main_matrx appears in min_matrx. Then I want to divide that by the number of rows in min_matrx and add that value as a new column in main_matrx.
For example, in row 1 of main_matrx for s17, the number of times a value appears that is more than 45 in min_matrx =5 times.
Now divide that 5 by 10 rows of min_matrx=> 5/10 =0.5 would be the value I'd like to have as a new column in main_matrx for s17. Then the same formula for all the s_ids in the row names.
So far I have fiddled with:
for(s in 1:length(main_matrx)) {
new<-sum(main_matrx[s,]>min_CPRS_set)/length(min_matrx)
}
and I tried using apply() but I'm still not getting results.
apply(main_matrx,1:length(main_matrx), function(x) sum(main_matrx>min_CPRS_set)/length(min_matrx)))
Now, I'm just stuck because it's not working. I'm still new to R so my code isn't particularly efficient. Any suggestions?
Lots of ways to approach this. Here's one that came to my head (I think I understand what you're after; again it's much easier to understand an example than with words alone. In the future I'd suggest an example to accompany the text question.)
Where x is an element, y is a vector
FUN <- function(x, y = min_matrix[, 1]) {
sum(y > x)/length(y)
}
main_matrx$new <- sapply(main_matrx[, 2], FUN)
## > main_matrx
## row.names sum_value new
## 1 s17 45 0.5
## 2 s7469 213 0.0
## 3 s20984 24 1.0
## 4 s17309 214 0.0
## 5 s7432369 43 0.6
## 6 s221320984 12 1.0
## 7 s17556 34 0.6
## 8 s741269 11 1.0
## 9 s20132984 35 0.6
I have two vectors, a and b. See attached.
a is the signal and is a probability.
b is the absolute percentage change the next period.
Signalt <- seq(0, 1, 0.05)
I would like to find the maximum absolute return occuring within each intermediate 5%-tile (Signalt) of the a vector. So if it is
0.01, 0.02, 0.03, 0.06 0.07
then it should calculate the maximum return between
0.01 and 0.02,
0.01 and 0.03,
0.02 and 0.03.
Then move on to
0.06 and 0.07 do it over etc.
Output would then be combined in a matrix or table when the entire sequence has run.
It should follow the index from vector a and b.
i is an index that is updated by one every time that a crosses into a new percentile. t(i) is the bucket associated with the ith cross.
a is the probability vector which has length tao. This vector should be analyzed in its 5% tiles, with the maximum intermediate absolute return being the output. The price change of next period is the vector b. This would be represented by P in the equation below.
l and m are indexes.
Every time Signal moves from one 5% tile to another, we compute the
largest absolute return that occurs between any two intermediate
buckets, until Signal moves to another 5% tile. For example, suppose
that Signal moves into the 85th percentile and 4 volume buckets later
moves into the 90th percentile. We would then calculate absolute
returns between buckets 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3
and 4. We are interested in the maximum absolute return. We would then
calculate the max return in the following percentile bucket, move on
to the next, which could be an 85th percentile and so on. So we let i
be an index that is updated by 1 every time that Signal moves from one
percentile into another, and τ(i) the bucket associated with the ith
cross.
This is the equation I am using. The notation might vary slightly.
Now my question is how to go about this. Perhaps someone has an intuitive solution to this.
I hope my question is clear.
"a","b"
0,0.013013698630137
0,0.0013522650439487
0,0.00135409614082593
0,0.00203389830508471
0.27804813511593,0.00135317997293627
0.300237801284318,0
0.495965075167796,0.00405405405405412
0.523741892051237,0.000672947510094168
0.558753750296458,0.00202020202020203
0.665762829019002,0.000672043010752743
0.493106479913899,0.000671591672263272
0.344592579573497,0.000672043010752854
0.336263897823707,0.00201748486886366
0.35884763774257,0.00536912751677865
0.23662807979007,0.00133511348464632
0.212636893966841,0.00267379679144386
0.362212830513403,0.000666666666666593
0.319216408413927,0.00333555703802535
0.277670854167344,0
0.310143323100971,0
0.374104373036218,0.00267737617135211
0.190943075221511,0.00268456375838921
0.165770070508112,0.00200803212851386
0.240310208616952,0.00133600534402145
0.212418038918236,0.00200133422281523
0.204282022136019,0.00200534759358306
0.363725074298064,0.000667111407605114
0.451807761954326,0.000666666666666593
0.369296011692801,0.000666222518321047
0.37503495989363,0.0026666666666666
0.323386355686901,0.00132978723404265
0.189216171830472,0.00266311584553924
0.185252052821193,0.00199203187250996
0.174882909380997,0.000662690523525522
0.149291525540782,0.00132625994694946
0.196824215268048,0.00264900662251666
0.164611993131396,0.000660501981505912
0.125470998266484,0.00132187706543285
0.179999532586703,0.00264026402640272
0.368749638521621,0.000658327847267826
0.427799340926225,0
My interpretation of the question
I hope I understand your question correctly. Here is what I understood:
For each row you compute which 5% percentile it belongs to
Whenever that percentile changes, you start a new bucket
All rows from the same bucket result in a single resulting value
If there is only a single row in a bucket, the b value from that row is the resulting value
Otherwise, you compute all abs(b[l]/b[m]-1) where m<l and both belong to the same bucket
Basic answer
Code
This code here does what I describe above:
# read the data (shortened, full data in OP)
d <- read.table(textConnection("a,b
0,0.013013698630137
[…]
0.427799340926225,0
"), sep=",", header=TRUE)
# compute percentile number for each line
d$percentile <- floor(d$a/0.05)*5 + 5
# start a new bucket whenever the percentile changes
d$bucket <- cumsum(c(1, diff(d$percentile) != 0))
# compute a single number for all rows of the same bucket
aggregate(b ~ percentile + bucket, d, function(b) {
if(length(b) == 1) return(b); # special case of only a single row
m <- outer(b, b, function(pm, pl) abs(pl/pm - 1)) # compare all pairs
return(max(m[upper.tri(m)])) # only return pairs with m < l
})
Output
The result will look like this:
percentile bucket b
1 5 1 0.8960891071
2 30 2 0.0013531800
3 35 3 0.0000000000
4 50 4 0.0040540541
5 55 5 0.0006729475
6 60 6 0.0020202020
7 70 7 0.0006720430
8 50 8 0.0006715917
9 35 9 2.0020174849
10 40 10 0.0053691275
11 25 11 1.0026737968
12 40 12 0.0006666667
13 35 13 0.0033355570
14 30 14 0.0000000000
15 35 15 0.0000000000
16 40 16 0.0026773762
17 20 17 0.2520080321
18 25 18 0.5010026738
19 40 19 0.0006671114
20 50 20 0.0006666667
21 40 21 3.0026666667
22 35 22 0.0013297872
23 20 23 0.7511597084
24 15 24 0.0013262599
25 20 25 0.7506605020
26 15 26 0.0013218771
27 20 27 0.0026402640
28 40 28 0.0006583278
29 45 29 0.0000000000
Additional columns
Code
If you also want to know the number of items in each group, then I suggest you use the plyr library:
library(plyr)
aggB <- function(b) {
if(length(b) == 1) return(b)
m <- outer(b, b, function(pm, pl) abs(pl/pm - 1))
return(max(m[upper.tri(m)]))
}
ddply(d, .(bucket), summarise,
percentile = percentile[1], n = length(b), maxr = aggB(b))
Output
This will give you the following result:
bucket percentile n maxr
1 1 5 4 0.8960891071
2 2 30 1 0.0013531800
3 3 35 1 0.0000000000
4 4 50 1 0.0040540541
5 5 55 1 0.0006729475
6 6 60 1 0.0020202020
7 7 70 1 0.0006720430
8 8 50 1 0.0006715917
9 9 35 2 2.0020174849
10 10 40 1 0.0053691275
11 11 25 2 1.0026737968
12 12 40 1 0.0006666667
13 13 35 1 0.0033355570
14 14 30 1 0.0000000000
15 15 35 1 0.0000000000
16 16 40 1 0.0026773762
17 17 20 2 0.2520080321
18 18 25 3 0.5010026738
19 19 40 1 0.0006671114
20 20 50 1 0.0006666667
21 21 40 2 3.0026666667
22 22 35 1 0.0013297872
23 23 20 3 0.7511597084
24 24 15 1 0.0013262599
25 25 20 2 0.7506605020
26 26 15 1 0.0013218771
27 27 20 1 0.0026402640
28 28 40 1 0.0006583278
29 29 45 1 0.0000000000
I am not sure to understand but here an attempt. My idea is to group data by centiles than do calculation on each group using by
To group data I create a new variable split
##dat$split <- cut(dat$a,seq(0, 1, 0.05),include.lowest=T)
dat$split <- c(0,cumsum(diff(dat$a) > 0.05))
Using by, I can performs my function en each group. I remove the singular cases of NULL prob values or one values.
by(dat,dat$split,FUN =function(x){
P <- x$b
if( is.null(P)||length(P) ==1) return(0)
nn <- length(P)
ind <- expand.grid(1:nn,1:nn) ## I generate indexes here
ret <- abs(P[ind[,1]]/P[ind[,2]]-1) ## perfom P_l/P_m-1 (vectorized)
list(P=P,
ret.max = max(ret),
ret.ind = ind[which.max(ret),])
})
Here the result list. For each interval I show ,
P ( Prob values),
The maximum return
The indexes from which this maximum is computed.
For example:
dat$split: 0
$P
[1] 0.0130 0.0014 0.0014 0.0020
$ret.max
[1] 8.6236
$ret.ind
Var1 Var2
5 1 2
---------------------------------------------------------------------------------------------------------------
dat$split: 1
$P
[1] 0.0014 0.0000
$ret.max
[1] 1
$ret.ind
Var1 Var2
2 2 1