Sequence of arithmetic expression - r

I need to generate an arithmetic sequence but am unable to get my head around it. cf is a 20x20 matrix. I am copying the first couple of lines of the matrix. I am trying to write a program to create spot rates using bootstrapping.
head(cf)
# [3,] 4.25 4.25 104.25
# [4,] 4.50 4.50 4.50 104.50
# [5,] 5.50 5.50 5.50 5.50 105.50
# [6,] 4.75 4.75 4.75 4.75 4.75 104.75
Price is a vector of bond prices at each period.
bond_data$Price
# [1] 96.15 92.19 99.45 99.64 103.49 99.49
For the 3rd period, which is the 3rd row in cf the calculation looks something similar to this:
Z[3] = bond_data$Price[3] - CF[3,1]/(1+z[1]/2)^1 - CF[3,2]/(1+z[2]/2)^2
For 4th period which is the 4th row in the CF matrix the calculation looks something similar to this:
Z[4] = bond_data$Price[4] - CF[4,1]/(1+z[1]/2)^1 - CF[4,2]/(1+z[2]/2)^2 - CF[4,3]/(1+z[3]/2)^3
Z[1] and Z[2] are known values, I am trying to generate Z for only a couple of periods to start with and this is what I wrote:
for(for k in 3:5){
seq( from = (cf[k,1]) / (1+(z[1]/2))^1, to = (cf[k,k-1])/(1+(z[k-1]/2))^k-1 )
}
This is not working as I thought it would. Not sure where my logic is incorrect.

Related

R: Creating a sequence substracting and adding a number into a single list (or vector)

I want to create a sequence from 1 to 7, substracting and adding .25 to each number. So, 14 elements in total in a single object.
I thought this would do the trick:
1:7 + c(-.25, .25)
But I got an error because the "longer object length is not a multiple of shorter object length". Basically because that line doesn't tell R that I want to add and substract 0.25 to each number.
I tried using sapply and kinda worked:
sapply(1:7,FUN=function(x)c(x-.25,x+.25))
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 0.75 1.75 2.75 3.75 4.75 5.75 6.75
[2,] 1.25 2.25 3.25 4.25 5.25 6.25 7.25
But it creates two lists and I need one. So I finally tried with a loop and it worked:
nums = NULL
for (i in 1:7){
min = i-0.25
max = i+0.25
nums = cbind(nums,min,max)
}
nums = as.numeric(nums)
nums
[1] 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 6.75 7.25
My questions for you guys is, do you know of a better/shorter way to achieve the same result?
Thank you!

The matrix is a vector with dim attributes. So, we could change to a vector with c that drops the attributes from the matrix
c(sapply(1:7,FUN=function(x)c(x-.25,x+.25)))
#[1] 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 6.75 7.25
Or make the lengths same with rep and then +
1:7 + rep(c(-.25, .25), 7)

You just need to do it in a different order, like this:
c(1:7 - 0.25, 1:7 + 0.25)
#> [1] 0.75 1.75 2.75 3.75 4.75 5.75 6.75 1.25 2.25 3.25 4.25 5.25 6.25 7.25
If you need it sorted just wrap it in sort()
Created on 2020-11-06 by the reprex package (v0.3.0)

Ramp up/down missing time-series data in R

I have a set of time-series data (GPS speed data, specifically), which includes gaps of missing values where the signal was lost. For missing periods of short durations I am about to fill simply using a na.spline, however this is inappropriate with longer time periods. I would like to ramp the values from the last true value down to zero, based on predefined acceleration limits.
#create sample data frame
test <- as.data.frame(c(6,5.7,5.4,5.14,4.89,4.64,4.41,4.19,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,5,5.1,5.3,5.4,5.5))
names(test)[1] <- "speed"
#set rate of acceleration for ramp
ramp <- 6
#set sampling rate of receiver
Hz <- 1/10
So for missing data the ramp would use the previous value and the rate of acceleration to get the next data point, until speed reached zero (i.e. last speed [4.19] + (Hz * ramp)), yielding the following values:
3.59
2.99
2.39
1.79
1.19
0.59
0
Lastly, I need to do this in the reverse fashion, to ramp up from zero when the signal picks back up again.
Hope this is clear.
Cheers

It's not really elegant, but you can do it in a loop.
na.pos <- which(is.na(test$speed))
acc = FALSE
for (i in na.pos) {
if (acc) {
speed <- test$speed[i-1]+(Hz*ramp)
}
else {
speed <- test$speed[i-1]-(Hz*ramp)
if (round(speed,1) < 0) {
acc <- TRUE
speed <- test$speed[i-1]+(Hz*ramp)
}
}
test[i,] <- speed
}
The result is:
speed
1 6.00
2 5.70
3 5.40
4 5.14
5 4.89
6 4.64
7 4.41
8 4.19
9 3.59
10 2.99
11 2.39
12 1.79
13 1.19
14 0.59
15 -0.01
16 0.59
17 1.19
18 1.79
19 2.39
20 2.99
21 3.59
22 4.19
23 4.79
24 5.00
25 5.10
26 5.30
27 5.40
28 5.50
Note that '-0.01', because 0.59-(6*10) is -0.01, not 0. You can round it later, I decided not to.

When the question says "ramp the values from the last true value down to zero" in each run of NAs I assume that that means that any remaining NAs in the run after reaching zero are also to be replaced by zero.
Now, use rleid from data.table to create a grouping vector the same length as test$speed identifying each run in is.na(test$speed) and use ave to create sequence numbers within such groups, seqno. Then calculate the declining sequences, ramp_down by combining na.locf(test$speed) and seqno. Finally replace the NAs.
library(data.table)
library(zoo)
test_speed <- test$speed
seqno <- ave(test_speed, rleid(is.na(test_speed)), FUN = seq_along)
ramp_down <- pmax(na.locf(test_speed) - seqno * ramp * Hz, 0)
result <- ifelse(is.na(test_speed), ramp_down, test_speed)
giving:
> result
[1] 6.00 5.70 5.40 5.14 4.89 4.64 4.41 4.19 3.59 2.99 2.39 1.79 1.19 0.59 0.00
[16] 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 5.00 5.10 5.30 5.40 5.50

Find where species accumulation curve reaches asymptote

I have used the specaccum() command to develop species accumulation curves for my samples.
Here is some example data:
site1<-c(0,8,9,7,0,0,0,8,0,7,8,0)
site2<-c(5,0,9,0,5,0,0,0,0,0,0,0)
site3<-c(5,0,9,0,0,0,0,0,0,6,0,0)
site4<-c(5,0,9,0,0,0,0,0,0,0,0,0)
site5<-c(5,0,9,0,0,6,6,0,0,0,0,0)
site6<-c(5,0,9,0,0,0,6,6,0,0,0,0)
site7<-c(5,0,9,0,0,0,0,0,7,0,0,3)
site8<-c(5,0,9,0,0,0,0,0,0,0,1,0)
site9<-c(5,0,9,0,0,0,0,0,0,0,1,0)
site10<-c(5,0,9,0,0,0,0,0,0,0,1,6)
site11<-c(5,0,9,0,0,0,5,0,0,0,0,0)
site12<-c(5,0,9,0,0,0,0,0,0,0,0,0)
site13<-c(5,1,9,0,0,0,0,0,0,0,0,0)
species_counts<-rbind(site1,site2,site3,site4,site5,site6,site7,site8,site9,site10,site11,site12,site13)
accum <- specaccum(species_counts, method="random", permutations=100)
plot(accum)
In order to ensure I have sampled sufficiently, I need to make sure the curve of the species accumulation plot reaches an asymptote, defined as a slope of <0.3 between the last two points (ei between sites 12 and 13).
results <- with(accum, data.frame(sites, richness, sd))
Produces this:
sites richness sd
1 1 3.46 0.9991916
2 2 4.94 1.6625403
3 3 5.94 1.7513054
4 4 7.05 1.6779918
5 5 8.03 1.6542263
6 6 8.74 1.6794660
7 7 9.32 1.5497149
8 8 9.92 1.3534841
9 9 10.51 1.0492422
10 10 11.00 0.8408750
11 11 11.35 0.7017295
12 12 11.67 0.4725816
13 13 12.00 0.0000000
I feel like I'm getting there. I could generate an lm with site vs richness and extract the exact slope (tangent?) between sites 12 and 13. Going to search a bit longer here.

Streamlining your data generation process a little bit:
species_counts <- matrix(c(0,8,9,7,0,0,0,8,0,7,8,0,
5,0,9,0,5,0,0,0,0,0,0,0, 5,0,9,0,0,0,0,0,0,6,0,0,
5,0,9,0,0,0,0,0,0,0,0,0, 5,0,9,0,0,6,6,0,0,0,0,0,
5,0,9,0,0,0,6,6,0,0,0,0, 5,0,9,0,0,0,0,0,7,0,0,3,
5,0,9,0,0,0,0,0,0,0,1,0, 5,0,9,0,0,0,0,0,0,0,1,0,
5,0,9,0,0,0,0,0,0,0,1,6, 5,0,9,0,0,0,5,0,0,0,0,0,
5,0,9,0,0,0,0,0,0,0,0,0, 5,1,9,0,0,0,0,0,0,0,0,0),
byrow=TRUE,nrow=13)
Always a good idea to set.seed() before running randomization tests (and let us know that specaccum is in the vegan package):
set.seed(101)
library(vegan)
accum <- specaccum(species_counts, method="random", permutations=100)
Extract the richness and sites components from within the returned object and compute d(richness)/d(sites) (note that the slope vector is one element shorter than the origin site/richness vectors: be careful if you're trying to match up slopes with particular numbers of sites)
(slopes <- with(accum,diff(richness)/diff(sites)))
## [1] 1.45 1.07 0.93 0.91 0.86 0.66 0.65 0.45 0.54 0.39 0.32 0.31
In this case, the slope never actually goes below 0.3, so this code for finding the first time that the slope falls below 0.3:
which(slopes<0.3)[1]
returns NA.

R: Is there a way to print a data frame with some separators inserted between rows?

I bascially want to format my dataframe print. I would like to choose for each column if the text is centered or not and after some defined rows insert a 1px separator. Furthermore it would be great to define for each column the width. With what function is that possible? I would like to output that later on to a textfile, and don't want to use Latex.
EDIT:
I just want to print a dataframe to a textfile, but as a nicely formatted table. So that it looks like an Excel sheet where you hide the gridlines. After x rows I want to basically just have a separator line ("----------") filling the whole width of the table.
Example:
My Data Frame consists of the following data:
Row1: "Test Entry 1", 5, 75, 0.3
Row2: "Test 2", 0.3, 1, 0.5
Output should be
Test Entry 1 5 75 0.3
------------------------------
Test 2 0.3 1 0.5
I hope it's more clear now :)

You are probably better off using one of the table packages but if you really really want to do it you can try something like this (pretty rudimentary, but can be expanded)
df<-data.frame(Test=sample(c("Test 1","Test 2","Test 3"),10,replace=T),
D1=round(runif(10)*10,2),
D2=round(runif(10)*10,2),
D3=round(runif(10)*10,2))
sepwidth<-60
colwidth<-10
require(plyr)
ddply(df,.(Test),function(d){
print(noquote(apply(d,c(1:2),function(p)paste0(paste0(rep(" ",colwidth-length(p)),collapse=""),p,collapse=""))));
print(noquote(paste0(rep("-",sepwidth),collapse="")))
return(NULL)})
Test D1 D2 D3
[1,] Test 1 5.37 3.48 1.19
[2,] Test 1 9.49 9.51 9.44
[3,] Test 1 8.52 6.53 4.10
[4,] Test 1 0.72 0.20 0.20
[5,] Test 1 2.70 6.19 8.17
[1] ------------------------------------------------------------
Test D1 D2 D3
[1,] Test 2 0.61 0.96 2.17
[2,] Test 2 6.85 2.36 6.90
[3,] Test 2 8.99 2.86 2.32
[1] ------------------------------------------------------------
Test D1 D2 D3
[1,] Test 3 0.23 6.42 9.41
[2,] Test 3 1.53 1.84 4.60
[1] ------------------------------------------------------------

Remove indexing from matrix in R

I am trying to make a barplot for which I need my data in a matrix. I have made some really nice plots before when my matrix looked like this:
0% 20% 40% 60% 80%
C2 0.22 0.94 1.66 2.38 3.10
CC -1.38 -0.66 0.06 0.79 1.51
CCW -1.61 -0.87 -0.13 0.62 1.36
P -1.13 -0.16 0.81 1.78 2.76
PF 0.03 0.72 1.42 2.11 2.80
S2 -2.34 -1.61 -0.88 -0.16 0.57
For the rest of my data, I had to convert it from a dataframe, which I did using as.matrix(df). This matrix looks like this:
trt 2009 2010 2011
[1,] "C2" "9.0525" " 8.1400" " 8.1400"
[2,] "CC" "5.4200" " 4.7975" " 4.7975"
[3,] "CCW" "4.9675" " 4.0400" " 4.0400"
[4,] "P" "9.3150" "10.3500" "10.3500"
[5,] "PF" "9.0950" " 3.3375" " 3.3375"
[6,] "S2" "3.1725" " 3.1125" " 3.1125"
It won't work with the barplot function. I think I need to remove the index column, but haven't been able to. And what is with the quotes? I though a matrix was a matrix, so I'm not sure what is going on here.

The quotes means your matrix is in mode character. This is because matrix, as opposed to data.frame which are superficially similar, can only hold one type. Because alphanumeric characters cannot be converted to numeric, your matrix is in mode character. I would be easier to remove the first column before converting it to matrix and save yourself of converting the matrix to numeric.
m <- as.matrix(df[, -1])
#To add the row.names.
row.names(m) <- df[, 1]

Resources