How to resample and remodel n times by vectorization? - r

here's my for loop version of doing resample and remodel,
B <- 999
n <- nrow(butterfly)
estMat <- matrix(NA, B+1, 2)
estMat[B+1,] <- model$coef
for (i in 1:B) {
resample <- butterfly[sample(1:n, n, replace = TRUE),]
re.model <- lm(Hk ~ inv.alt, resample)
estMat[i,] <- re.model$coef
}
I tried to avoid for loop,
B <- 999
n <- nrow(butterfly)
resample <- replicate(B, butterfly[sample(1:n, replace = TRUE),], simplify = FALSE)
re.model <- lapply(resample, lm, formula = Hk ~ inv.alt)
re.model.coef <- sapply(re.model,coef)
estMat <- cbind(re.model.coef, model$coef)
It worked but didn't improve efficiency. Is there any approach I can do vectorization?
Sorry, not quite familiar with StackOverflow. Here's the dataset butterfly.
colony alt precip max.temp min.temp Hk
pd+ss 0.5 58 97 16 98
sb 0.8 20 92 32 36
wsb 0.57 28 98 26 72
jrc+jrh 0.55 28 98 26 67
sj 0.38 15 99 28 82
cr 0.93 21 99 28 72
mi 0.48 24 101 27 65
uo+lo 0.63 10 101 27 1
dp 1.5 19 99 23 40
pz 1.75 22 101 27 39
mc 2 58 100 18 9
hh 4.2 36 95 13 19
if 2.5 34 102 16 42
af 2 21 105 20 37
sl 6.5 40 83 0 16
gh 7.85 42 84 5 4
ep 8.95 57 79 -7 1
gl 10.5 50 81 -12 4

(Assuming butterfly$inv.alt <- 1/butterfly$alt)
You get the error because resample is not a list of resampled data.frames, which you can obtain with:
resample <- replicate(B, butterfly[sample(1:n, replace = TRUE),], simplify = FALSE)
The the following should work:
re.model <- lapply(resample, lm, formula = Hk ~ inv.alt)
To extract coefficients from a list of models, re.model$coef does work. The correct path to coefficients are: re.model[[1]]$coef, re.model[[2]]$coef, .... You can get all of them with the following code:
re.model.coef <- sapply(re.model, coef)
Then you can combined it with the observed coefficients:
estMat <- cbind(re.model.coef, model$coef)
In fact, you can put all of them into replicate:
re.model.coef <- replicate(B, {
bf.rs <- butterfly[sample(1:n, replace = TRUE),]
coef(lm(formula = Hk ~ inv.alt, data = bf.rs))
})
estMat <- cbind(re.model.coef, model$coef)

Related

How to use characters in variables summing in R?

I have some dataframe. Here is a small expample:
a <- rnorm(100, 5, 2)
b <- rnorm(100, 10, 3)
c <- rnorm(100, 15, 4)
df <- data.frame(a, b, c)
And I have a character variable vect <- "c('a','b')"
When I try to calculate sum of vars using command
df$d <- df[vect]
which must be an equivalent of
df$d <- df[c('a','b')]
But, as a reslut I have got an error
[.data.frame(df, vect) :undefined columns selected
You're assumption that
vect <- "c('a','b')"
df$d <- df[vect]
is equivalent to
df$d <- df[c('a','b')]
is incorrect.
As #Karthik points out, you should remove the quotation marks in the assignment to vect
However, from your question it sounds like you want to then sum the elements specified in vect and then assign to d. To do this you need to slightly change your code
vect <- c('a','b')
df$d <- apply(X = df[vect], MARGIN = 1, FUN = sum)
This does elementwise sum on the columns in df specified by vect. The MARGIN = 1 specifies that we want to apply the sum rowise rather than columnwise.
EDIT:
As #ThomasIsCoding points out below, if for some reason vect has to be a string, you can parse a string to an R expression using str2lang
vect <- "c('a','b')"
parsed_vect <- eval(str2lang(vect))
df$d <- apply(X = df[parsed_vect], MARGIN = 1, FUN = sum)
Perhaps you can try
> df[eval(str2lang(vect))]
a b
1 8.1588519 9.0617818
2 3.9361214 13.2752377
3 5.5370983 8.8739725
4 8.4542050 8.5704234
5 3.9044461 13.2642793
6 5.6679639 12.9529061
7 4.0183808 6.4746806
8 3.6415608 11.0308990
9 4.5237453 7.3255129
10 6.9379168 9.4594150
11 5.1557935 11.6776181
12 2.3829337 3.5170335
13 4.3556430 7.9706624
14 7.3274615 8.1852829
15 -0.5650641 2.8109197
16 7.1742283 6.8161200
17 3.3412044 11.6298940
18 2.5388981 10.1289533
19 3.8845686 14.1517643
20 2.4431608 6.8374837
21 4.8731053 12.7258259
22 6.9534912 6.5069513
23 4.4394807 14.5320225
24 2.0427553 12.1786148
25 7.1563978 11.9671603
26 2.4231207 6.1801862
27 6.5830372 0.9814878
28 2.5443326 9.8774632
29 1.1260322 9.4804636
30 4.0078436 12.9909014
31 9.3599808 12.2178596
32 3.5362245 8.6758910
33 4.6462337 8.6647953
34 2.0698037 7.2750532
35 7.0727970 8.9386798
36 4.8465248 8.0565347
37 5.6084462 7.5676308
38 6.7617479 9.5357666
39 5.2138482 13.6822924
40 3.6259103 13.8659939
41 5.8586547 6.5087016
42 4.3490281 9.5367522
43 7.5130701 8.1699117
44 3.7933813 9.3241308
45 4.9466813 9.4432584
46 -0.3730035 6.4695187
47 2.0646458 10.6511916
48 4.6027309 4.9207746
49 5.9919348 7.1946723
50 6.0148330 13.4702419
51 5.5354452 9.0193366
52 5.2621651 12.8856488
53 6.8580210 6.3526151
54 8.0812166 14.4659778
55 3.6039030 5.9857886
56 9.8548553 15.9081336
57 3.3675037 14.7207681
58 3.9935336 14.3186175
59 3.4308085 10.6024579
60 3.9609624 6.6595521
61 4.2358603 10.6600581
62 5.1791856 9.3241118
63 4.6976289 13.2833055
64 5.1868906 7.1323826
65 3.1810915 12.8402472
66 6.0258287 9.3805249
67 5.3768112 6.3805096
68 5.7072092 7.1130150
69 6.5789349 8.0092541
70 5.3175820 17.3377234
71 9.7706112 10.8648956
72 5.2332127 12.3418373
73 4.7626124 13.8816910
74 3.9395911 6.5270785
75 6.4394724 10.6344965
76 2.6803695 10.4501753
77 3.5577834 8.2323369
78 5.8431140 7.7932460
79 2.8596818 8.9581837
80 2.7365174 10.2902512
81 4.7560973 6.4555758
82 4.6519084 8.9786777
83 4.9467471 11.2818536
84 5.6167284 5.2641380
85 9.4700525 2.9904731
86 4.7392906 11.3572521
87 3.1221908 6.3881556
88 5.6949432 7.4518023
89 5.1435241 10.8912283
90 2.1628966 10.5080671
91 3.6380837 15.0594135
92 5.3434709 7.4034042
93 -0.1298439 0.4832707
94 7.8759390 2.7411723
95 2.0898649 9.7687250
96 4.2131549 9.3175228
97 5.0648105 11.3943350
98 7.7225193 11.4180456
99 3.1018895 12.8890257
100 4.4166832 10.4901303

R data.table divide set of columns and flag using any

I am working on a data set which is large and having many columns. I am using data.table to speed up the calculations. However at certain points I am not sure how to go about and convert my data.table back to data.frame and do the calculation. This slows up the process. It would help a lot to have suggestions on how I can write the below in data.table. Below is a snap of my code on a dummy data -
library(data.table)
#### set the seed value
set.seed(9901)
#### create the sample variables for creating the data
p01 <- sample(1:100,1000,replace = T)
p02 <- sample(1:100,1000,replace = T)
p03 <- sample(1:100,1000,replace = T)
p04 <- sample(1:100,1000,replace = T)
p05 <- sample(1:100,1000,replace = T)
p06 <- sample(1:100,1000,replace = T)
p07 <- sample(1:100,1000,replace = T)
#### create the data.table
data <- data.table(cbind(p01,p02,p03,p04,p05,p06,p07))
###user input for last column
lcol <- 6
###calculate start column as last - 3
scol <- lcol-3
###calculate average for scol:lcol
data <- data[,avg:= apply(.SD,1,mean,na.rm=T),.SDcols=scol:lcol]
###converting to data.frame since do not know the solution in data.table
data <- as.data.frame(data)
###calculate the trend in percentage
data$t01 <- data[,lcol-00]/data[,"avg"]-1
data$t02 <- data[,lcol-01]/data[,"avg"]-1
data$t03 <- data[,lcol-02]/data[,"avg"]-1
data$t04 <- data[,lcol-03]/data[,"avg"]-1
data$t05 <- data[,lcol-04]/data[,"avg"]-1
###converting back to data.table
data <- as.data.table(data)
###calculate the min and max for the trend
data1 <- data[,`:=` (trend_min = apply(.SD,1,min,na.rm=T),
trend_max = apply(.SD,1,max,na.rm=T)),.SDcols=c(scol:lcol)]
###calculate flag if any of t04 OR t05 is an outlier for min and max values. This would be many columns in actual data
data1$flag1 <- ifelse(data1$t04 < data1$trend_min | data1$t04 > data1$trend_max,1,0)
data1$flag2 <- ifelse(data1$t05 < data1$trend_min | data1$t05 > data1$trend_max,1,0)
data1$flag <- ifelse(data1$flag1 == 1 | data1$flag2 == 1,1,0)
So basically, how can I -
calculate the percentages based on user input of column index. Note it is not simple divide but percentage
How can I create the flag variable....I think I need to use any function but not sure how....
Some steps can be made more efficient, i.e. instead of using the apply with MARGIN = 1, the mean, min, max can be replaced with rowMeans, pmin, pmax
library(data.table)
data[ , avg:= rowMeans(.SD, na.rm = TRUE) ,.SDcols=scol:lcol]
data[, sprintf('t%02d', 1:5) := lapply(.SD, function(x) x/avg -1),
.SDcol = patterns("^p0[1-5]")]
data[,`:=` (trend_min = do.call(pmin, c(.SD,na.rm=TRUE)),
trend_max = do.call(pmax, c(.SD,na.rm=TRUE)) ),.SDcols=c(scol:lcol)]
data
# p01 p02 p03 p04 p05 p06 p07 avg t01 t02 t03 t04 t05 trend_min trend_max
# 1: 35 53 22 82 100 59 69 65.75 -0.46768061 -0.19391635 -0.6653992 0.24714829 0.5209125 22 100
# 2: 78 75 15 65 70 69 66 54.75 0.42465753 0.36986301 -0.7260274 0.18721461 0.2785388 15 70
# 3: 15 45 27 61 63 75 99 56.50 -0.73451327 -0.20353982 -0.5221239 0.07964602 0.1150442 27 75
# 4: 41 80 13 22 63 84 17 45.50 -0.09890110 0.75824176 -0.7142857 -0.51648352 0.3846154 13 84
# 5: 53 9 75 47 25 75 66 55.50 -0.04504505 -0.83783784 0.3513514 -0.15315315 -0.5495495 25 75
# ---
# 996: 33 75 9 61 74 55 57 49.75 -0.33668342 0.50753769 -0.8190955 0.22613065 0.4874372 9 74
# 997: 24 68 74 11 43 75 37 50.75 -0.52709360 0.33990148 0.4581281 -0.78325123 -0.1527094 11 75
# 998: 62 78 82 97 56 50 74 71.25 -0.12982456 0.09473684 0.1508772 0.36140351 -0.2140351 50 97
# 999: 70 88 93 4 39 75 93 52.75 0.32701422 0.66824645 0.7630332 -0.92417062 -0.2606635 4 93
#1000: 20 50 99 94 62 66 98 80.25 -0.75077882 -0.37694704 0.2336449 0.17133956 -0.2274143 62 99
and then create the 'flag'
data[, flag := +(Reduce(`|`, lapply(.SD, function(x)
x < trend_min| x > trend_max))), .SDcols = t04:t05]

Run a function in loop on a data frame of 3 values

Function given:
hmin = function (H,M,s) {
H - ((2*9.8*M)/1.5*s))
}
# Create a data frame to test One-at-A-Time sensitivity analysis
# Keep H and M constant, varying s
# H = 50, M = 70.5, s = 20 to 40
s1 <- numeric(length = 20)
s1 [1:21] <- c(20:40)
H1 <- c(rep(50,length(s1)))
M1 <- c(rep(70.5, length(s1)))
dataframe1 <- data.frame(H1,M1,s1)
Now with this data frame, How to run the hmin function over the variables in dataframe1 and store the output hmin in a vector to plot a barplot.
Tried using:
lapply(dataframe1, hmin (H1,M1,s1), dataframe1)
but it does not work.
Any help would be appreciated. Thank you.
My intended output is just to return the hmin values after running the data frame of variables using the function and saving it in a vector. How can I go about doing it?
Since your function is vectorized, do this:
result = hmin(H = dataframe1$H1, M = dataframe1$M1, s = dataframe1$s1)
# same thing, using the with() helper function to save typing
result = with(dataframe1, hmin(H = H1, M = M1, s = s1))
It is far more easily if you use dplyr
> library(dplyr)
> dataframe1 <- data.frame(H1,M1,s1)
> hmin <- function(H,M,s) {
+ H - ((2*9.8*M)/1.5*s)
+ }
> dataframe1 %>%
+ mutate(hminscores = hmin(H1, M1, s1))
H1 M1 s1 sensitivity
1 50 70.5 20 -18374.0
2 50 70.5 21 -19295.2
3 50 70.5 22 -20216.4
4 50 70.5 23 -21137.6
5 50 70.5 24 -22058.8
6 50 70.5 25 -22980.0
7 50 70.5 26 -23901.2
8 50 70.5 27 -24822.4
9 50 70.5 28 -25743.6
10 50 70.5 29 -26664.8
11 50 70.5 30 -27586.0
12 50 70.5 31 -28507.2
13 50 70.5 32 -29428.4
14 50 70.5 33 -30349.6
15 50 70.5 34 -31270.8
16 50 70.5 35 -32192.0
17 50 70.5 36 -33113.2
18 50 70.5 37 -34034.4
19 50 70.5 38 -34955.6
20 50 70.5 39 -35876.8
21 50 70.5 40 -36798.0
If you just want a vector, you can also use apply:
> hminscores <- apply(dataframe1, 1, function(x) {
+ hmin(x[1], x[2], x[3])
+ })
> hminscores
[1] -18374.0 -19295.2 -20216.4 -21137.6 -22058.8 -22980.0 -23901.2 -24822.4 -25743.6
[10] -26664.8 -27586.0 -28507.2 -29428.4 -30349.6 -31270.8 -32192.0 -33113.2 -34034.4
[19] -34955.6 -35876.8 -36798.0

Extracting chunks from a matrix by columns

Say I have a matrix with 1000 columns. I want to create a new matrix with every other n columns from the original matrix, starting from column i.
So let say that n=3 and i=5, then the columns I need from the old matrix are 5,6,7,11,12,13,17,18,19 and so on.
Using two seq()s to create the start and stop bounds, then using a mapply() on those to build your true column index intervals. Then just normal bracket notation to extract from your matrix.
set.seed(1)
# using 67342343's test case
M <- matrix(runif(100^2), ncol = 100)
n <- 3
i <- 5
starts <- seq(i, ncol(M), n*2)
stops <- seq(i+(n-1), ncol(M), n*2)
col_index <- c(mapply(seq, starts, stops)) # thanks Jaap and Sotos
col_index
[1] 5 6 7 11 12 13 17 18 19 23 24 25 29 30 31 35 36 37 41 42 43 47 48 49 53 54 55 59 60 61 65 66 67 71 72 73 77 78
[39] 79 83 84 85 89 90 91 95 96 97
M[, col_index]
Another solution is based on the fact that R uses index recycling:
i <- 5; n <- 3
M <- matrix(runif(100^2), ncol = 100)
id <- seq(i, ncol(M), by = 1)[rep(c(TRUE, FALSE), each = n)]
M_sub <- M[, id]
I would write a function that determines the indices of the columns you want, and then call that function as needed.
col_indexes <- function(mat, start = 1, by = 1){
n <- ncol(mat)
inx <- seq(start, n, by = 2*by)
inx <- c(sapply(inx, function(i) i:(i + by -1)))
inx[inx <= n]
}
m <- matrix(0, nrow = 1, ncol = 20)
icol <- col_indexes(m, 5, 3)
icol
[1] 5 6 7 11 12 13 17 18 19
Here is a method using outer.
c(outer(5:7, seq(0L, 95L, 6L), "+"))
[1] 5 6 7 11 12 13 17 18 19 23 24 25 29 30 31 35 36 37 41 42 43 47 48 49 53
[26] 54 55 59 60 61 65 66 67 71 72 73 77 78 79 83 84 85 89 90 91 95 96 97
To generalize this, you could do
idx <- c(outer(seq(i, i + n), seq(0L, ncol(M) - i, 2 * n), "+"))
The idea is to construct the initial set of columns (5:7 or seq(i, i + n)), calculate the starting points for every subsequent set (seq(0L, 95L, 6L) or seq(0L, ncol(M) - i, 2 * n)) then use outer to calculate the sum of every combination of these two vectors.
you can subset the matrix using [ like M[, idx].

Given data points and y value, give x value

Given a set of (x,y) coordinates, how can I solve for x, from y. If you were to plot the coordinates, they would be non-linear, but pretty close to exponential. I tried approx(), but it is way off. Here is example data. In this scenario, how could I solve for y == 50?
V1 V3
1 5.35 11.7906
2 10.70 15.0451
3 16.05 19.4243
4 21.40 20.7885
5 26.75 22.0584
6 32.10 25.4367
7 37.45 28.6701
8 42.80 30.7500
9 48.15 34.5084
10 53.50 37.0096
11 58.85 39.3423
12 64.20 41.5023
13 69.55 43.4599
14 74.90 44.7299
15 80.25 46.5738
16 85.60 47.7548
17 90.95 49.9749
18 96.30 51.0331
19 101.65 52.0207
20 107.00 52.9781
21 112.35 53.8730
22 117.70 54.2907
23 123.05 56.3025
24 128.40 56.6949
25 133.75 57.0830
26 139.10 58.5051
27 144.45 59.1440
28 149.80 60.0687
29 155.15 60.6627
30 160.50 61.2313
31 165.85 61.7748
32 171.20 62.5587
33 176.55 63.2684
34 181.90 63.7085
35 187.25 64.0788
36 192.60 64.5807
37 197.95 65.2233
38 203.30 65.5331
39 208.65 66.1200
40 214.00 66.6208
41 219.35 67.1952
42 224.70 67.5270
43 230.05 68.0175
44 235.40 68.3869
45 240.75 68.7485
46 246.10 69.1878
47 251.45 69.3980
48 256.80 69.5899
49 262.15 69.7382
50 267.50 69.7693
51 272.85 69.7693
52 278.20 69.7693
53 283.55 69.7693
54 288.90 69.7693
I suppose the problem you have is that approx solves for y given x, while you are talking about solving for x given y. So you need to switch your variables x and y when using approx:
df <- read.table(textConnection("
V1 V3
85.60 47.7548
90.95 49.9749
96.30 51.0331
101.65 52.0207
"), header = TRUE)
approx(x = df$V3, y = df$V1, xout = 50)
# $x
# [1] 50
#
# $y
# [1] 91.0769
Also, if y is exponential with respect to x, then you have a linear relationship between x and log(y), so it makes more sense to use a linear interpolator between x and log(y), then take the exponential to get back to y:
exp(approx(x = df$V3, y = log(df$V1), xout = 50)$y)
# [1] 91.07339

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