I have a wav file and I would like to calculate the Acoustic Complexity Index at each second and receive a time series output.
I understand how to modify other settings within a function like seewave::ACI() but I am unable to find out how to output a time series data frame where each row is one second of time with the corresponding ACI value.
For a reproducible example, this audio file is 20 seconds, so I'd like the output to have 20 rows, with each row printing the ACI for that 1-second of time.
library(soundecology)
data(tropicalsound)
acoustic_complexity(tropicalsound)
In fact, I'd like to achieve this is a few other indices, for example:
soundecology::ndsi(tropicalsound)
soundecology::acoustic_evenness(tropicalsound)
You can subset your wav file according to the samples it contains. Since the sampling frequency can be obtained from the wav object, we can get one-second subsets of the file and perform our calculations on each. Note that you have to set the cluster size to 1 second, since the default is 5 seconds.
library(soundecology)
data(tropicalsound)
f <- tropicalsound#samp.rate
starts <- head(seq(0, length(tropicalsound), f), -1)
aci <- sapply(starts, function(i) {
aci <- acoustic_complexity(tropicalsound[i + seq(f)], j = 1)
aci$AciTotAll_left
})
nds <- sapply(starts, function(i) {
nds <- ndsi(tropicalsound[i + seq(f)])
nds$ndsi_left
})
aei <- sapply(starts, function(i) {
aei <- acoustic_evenness(tropicalsound[i + seq(f)])
aei$aei_left
})
This allows us to create a second-by-second data frame representing a time series of each measure:
data.frame(time = 0:19, aci, nds, aei)
#> time aci nds aei
#> 1 0 152.0586 0.7752307 0.438022
#> 2 1 168.2281 0.4171902 0.459380
#> 3 2 149.2796 0.9366220 0.516602
#> 4 3 176.8324 0.8856127 0.485036
#> 5 4 162.4237 0.8848515 0.483414
#> 6 5 161.1535 0.8327568 0.511922
#> 7 6 163.8071 0.7532586 0.549262
#> 8 7 156.4818 0.7706808 0.436910
#> 9 8 156.1037 0.7520663 0.489253
#> 10 9 160.5316 0.7077717 0.491418
#> 11 10 157.4274 0.8320380 0.457856
#> 12 11 169.8831 0.8396483 0.456514
#> 13 12 165.4426 0.6871337 0.456985
#> 14 13 165.1630 0.7655454 0.497621
#> 15 14 154.9258 0.8083035 0.489896
#> 16 15 162.8614 0.7745876 0.458035
#> 17 16 148.6004 0.1393345 0.443370
#> 18 17 144.6733 0.8189469 0.458309
#> 19 18 156.3466 0.6067827 0.455578
#> 20 19 158.3413 0.7175293 0.477261
Note that this is simply a demonstration of how to achieve the desired output; you would need to check the literature to determine whether it is appropriate to use these measures over such short time periods.
the title is vague but let me explain:
I have a non-vectorized function that outputs a 15-row table of volume estimates for a tree. Each row is a different measurement unit or portion of the input tree. I have a Tables argument to help the user decide what units and measurement protocol they're looking to find, but in 99% of use case scenarios, the output for a single tree's volume estimate is a tibble with more than one row.
I've removed ~20 other arguments from the function for demonstration's sake. DBH is a tree's diameter at breast height. Vol column is arbitrary.
Est1 <- TreeVol(Tables = "All", DBH = 7)
Est1
# A tibble: 15 x 3
Tables DBH Vol
<chr> <dbl> <dbl>
1 1. Total_Above_Ground_Cubic_Volume 7 2
2 2. Gross_Inter_1/4inch_Vol 7 4
3 3. Net_Scribner_Vol 7 6
4 4. Gross_Merchantable_Vol 7 8
5 5. Net_Merchantable_Vol 7 10
6 6. Merchantable_Vol 7 12
7 7. Gross_SecondaryProduct_Vol 7 14
8 8. Net_SecondaryProduct_Vol 7 16
9 9. SecondaryProduct 7 18
10 10. Gross_Inter_1/4inch_Vol 7 20
11 11. Net_Inter_1/4inch_Vol 7 22
12 12. Gross_Scribner_SecondaryProduct 7 24
13 13. Net_Scribner_SecondaryProduct 7 26
14 14. Stump_Volume 7 28
15 15. Tip_Volume 7 30
the user can utilize the Tables argument as so:
Est2 <- TreeVol(Tables = "Scribner_BF", DBH = 7)
# A tibble: 3 x 3
Tables DBH Vol
<chr> <dbl> <dbl>
1 3. Net_Scribner_Vol 7 6
2 12. Gross_Scribner_SecondaryProduct 7 24
3 13. Net_Scribner_SecondaryProduct 7 26
The problem arises in that I'd like to write a vectorized version of this function that can calculate the volume for an entire .csv of tree inventory data. Ideally, I'd like the multi-row outputs that relate to a single tree to output as one long tibble, with each 15-row default output filtered by what the user passes to the Tables argument as so:
Est3 <- VectorizedTreeVol(Tables = "Scribner_BF", DBH = c(7, 21, 26))
# A tibble: 9 x 3
Tables DBH Vol
<chr> <dbl> <dbl>
1 3. Net_Scribner_Vol 7 6
2 12. Gross_Scribner_SecondaryProduct 7 24
3 13. Net_Scribner_SecondaryProduct 7 26
4 3. Net_Scribner_Vol 21 18
5 12. Gross_Scribner_SecondaryProduct 21 72
6 13. Net_Scribner_SecondaryProduct 21 76
7 3. Net_Scribner_Vol 26 8
8 12. Gross_Scribner_SecondaryProduct 26 78
9 13. Net_Scribner_SecondaryProduct 26 84
To achieve this, I wrote a for() loop that acts as the heart of the vectorized function. I've heard from multiple people that it's very inefficient (and I agree), but it works with the principle I'd like to achieve, in theory. Nothing I've found on this topic has suggested a better idea for application in a vectorized function like mine.
The general setup for the loop looks like this:
for(i in 1:length(DBH)){
Output <- VectorizedTreeVol(Tables = Tables[[i]], DBH = DBH[[i]]) %>%
purrr::reduce(dplyr::full_join, by = NULL) %>%
SuppressWarnings()
and in functions where the non-vectorized output is always a single row, the heart of its respective vectorized function doesn't need to be encased in a for() loop and looks like this:
Output <- OtherVectorizedFunction(Tables = Tables, DBH = DBH) %>%
purrr::reduce(dplyr::full_join, by = ColumnNames) %>% #ColumnNames is a vector with all of the output's column names
SuppressWarnings()
This specific call to reduce() has worked pretty well when I've used it to vectorize the other functions in the project, but I'm open to suggestions regarding how to join the output tables. I've been stuck on this dilemma for a few months now, and any help regarding how to achieve what this for() loop is striving for in theory would be awesome. Is having a vectorized function that outputs a tibble like Est3 even possible? Any feedback/comments are much appreciated.
Given this function:
TreeVol <- function(DBH) {
data.frame(Tables = c("Tree_Vol", "Intercapillary_transfusion", "Woodiness"),
Vol = c(DBH^2, sqrt(DBH) + 3, sin(DBH)),
DBH)
}
We could put our DBH parameters into purrr::map and then bind_rows to get a data.frame.
VecTreeVol <- function(DBH) {
DBH %>%
purrr::map(TreeVol) %>%
bind_rows()
}
Result
> VecTreeVol(DBH = 1:3)
Tables Vol DBH
1 Tree_Vol 1.0000000 1
2 Intercapillary_transfusion 4.0000000 1
3 Woodiness 0.8414710 1
4 Tree_Vol 4.0000000 2
5 Intercapillary_transfusion 4.4142136 2
6 Woodiness 0.9092974 2
7 Tree_Vol 9.0000000 3
8 Intercapillary_transfusion 4.7320508 3
9 Woodiness 0.1411200 3
Simple question : in R, what's the best way to detect if there is a zero somewhere in a time series (ts class)? I run X13 (seasonal package) on hundreds of time series and I would like to identify those who contain zero values (since multiplicative models don't work when they encounter a zero). If I could detect those series, I could use a IF-THEN-ELSE statement with proper specs for the X13.
Thank you!
You can replace or delete them:
ts <- ts(0:10)
## Deleting
ts[ts != 0]
#> [1] 1 2 3 4 5 6 7 8 9 10
## Replacing
replace(ts, ts==0, 1)
#> Time Series:
#> Start = 1
#> End = 11
#> Frequency = 1
#> [1] 1 1 2 3 4 5 6 7 8 9 10
## Detecting
any(ts == 0)
#> [1] TRUE
Created on 2020-10-29 by the reprex package (v0.3.0)
I am creating 1000 random communities (vectors) from a species pool of 128 with certain operations applied to the community and stored in a new vector. For simplicity, I have been practicing writing code using 10 random communities from a species pool of 20. The problem is that there are a couple of pairs of species such that if one of the pairs is generated in the random community, I need that community to be thrown out and a new one regenerated. I have been able to code that if the pair is found in a community for that community(vector) to be labeled NA. I also know how to tell the loop to skip that vector using the "next" command. But with both of these options, I do not get all of the communities that I needing.
Here is my code using the NA option, but again that ends up shorting me communities.
C<-c(1:20)
D<-numeric(10)
X<- numeric(5)
for(i in 1:10){
X<-sample(C, size=5, replace = FALSE)
if("10" %in% X & "11" %in% X) X=NA else X=X
if("1" %in% X & "2" %in% X) X=NA else X=X
print(X)
D[i]<-sum(X)
}
print(D)
This is what my result looks like.
[1] 5 1 7 3 14
[1] 20 8 3 18 17
[1] NA
[1] NA
[1] 4 7 1 5 3
[1] 16 1 11 3 12
[1] 14 3 8 10 15
[1] 7 6 18 3 17
[1] 6 5 7 3 20
[1] 16 14 17 7 9
> print(D)
[1] 30 66 NA NA 20 43 50 51 41 63
Thanks so much!
I have the following data set:
dat<-as.data.frame(rbind(10,8,2,7,10,10,1,10,14,9,2,6,10,8,10,8,10,10,7,11,10))
colnames(dat)<-"Score"
print(dat)
Score
10
8
2
7
10
10
1
10
14
9
2
6
10
8
10
8
10
10
7
11
10
these are the test scores which students obtained, a student could get a maximum of 15 or a minimum of 0 in this test (by the way, nobody got the max or the min), however the lowest score obtained in this test was 1 and the highest was 14.
Now, I want to normalize/scale this data to the scale of 0 to 20.
How to achieve this in excel? or in R?
My final goal is to normalize the scores in this test to the above scale and to compare them with another set of data for which the max and min is 5 and 0 respectively.
How to compare these two different scaled data sets correctly against each other?
What I tried:
I went through many stuff on the internet, and came up with this:
which I got it from the wikipedia.
Is this method reliable?
In your case I would use the feature scale formula you posted on your question. The (x - min(x)) / (max(x) - min(x)) will essentially convert your test marks to the range between 0-1.
Since your edges are indeed 0 and 15 and not 2 and 14, your min(x)=0 and your max(x)=15. Once you have your marks between 0-1 using the above, you just multiply by 20.
i.e.
tests <- read.table(header=T, file='clipboard')
tests2 <- (tests - 0) / (15 - 0) #or equally tests / 15
And multiply by 20 to get marks between 0-20:
> tests2 * 20
Score
1 13.333333
2 10.666667
3 2.666667
4 9.333333
5 13.333333
6 13.333333
7 1.333333
8 13.333333
9 18.666667
10 12.000000
11 2.666667
12 8.000000
13 13.333333
14 10.666667
15 13.333333
16 10.666667
17 13.333333
18 13.333333
19 9.333333
20 14.666667
21 13.333333
The results are intuitive and the function is reliable. For example the person who scored 14/15 should get the highest mark (and very close to 20) which is the case here (after the transformation they scored 18.6666).
In Excel, if you want the normalized data to have a min of 0 and and max of 20, then we need to solve:
y = A * x + b
for two points.
Put the max of the raw data in C1:
=MAX(A:A)
Put the min of the raw data in C2:
=MIN(A:A)
Put the desired max in D1 and the desired min in D2. Put the formula for the A-coefficient in C3:
=($D$1-$D$2)/($C$1-$C$2)
and the formula for the B-coefficient in C4:
=$D$1-$C$3*$C$1
Finally put the scaling formula in B1:
=A1*$C$3+$C$4
and copy down:
Naturally, if you want the scaling to be independent of the raw max or min, you would use 15 in C1 and 0 in C2.
You can scale between 0 to 20 with this command in R:
newvalue <- 20/(max(score)-min(score))*(score-min(score))
The math way is fairly straightforward if the floor for all scales is 0.
new_value = new_ceiling * old_value / old_ceiling
The next formula will account for different floors on each scale:
new_value = new_floor + (new_ceiling - old_ceiling) * ((old_value-old_floor)/(old_ceiling-old_floor)) which is actually the formula you posted from Wikipedia. ;)
Hope this helps!
That is very simple. Due to the fact that both of those grades are linear, that a simple multiple ratio will do the work. Or in other word each grade in your set needs to be *20/15.
Here's a little r function which can help you run this if you need to repeat the operation and give you some flexibility on what you rescale to. Also one must be careful of NA values because min() and max() do not drop them by default which will then return NA. Therefore I provided an option on to handle NA values (drops them by default).
# function rescales data from 0 to 1 and optionally multiplies by new max
rescale <- function(x, new_max = 1, na.rm = T) {
as.vector(new_max * scale(x,
center = min(x, na.rm = na.rm),
scale = (max(x, na.rm = na.rm) - min(x, na.rm = na.rm))))
}
# old scores
scores <- c(10,8,2,7,10,10,1,10,14,9,2,6,10,8,10,8,10,10,7,11,10)
# new scores
data.frame(old = scores,
new = rescale(scores, new_max = 20))
#> old new
#> 1 10 13.846154
#> 2 8 10.769231
#> 3 2 1.538462
#> 4 7 9.230769
#> 5 10 13.846154
#> 6 10 13.846154
#> 7 1 0.000000
#> 8 10 13.846154
#> 9 14 20.000000
#> 10 9 12.307692
#> 11 2 1.538462
#> 12 6 7.692308
#> 13 10 13.846154
#> 14 8 10.769231
#> 15 10 13.846154
#> 16 8 10.769231
#> 17 10 13.846154
#> 18 10 13.846154
#> 19 7 9.230769
#> 20 11 15.384615
#> 21 10 13.846154
Created on 2022-03-10 by the reprex package (v2.0.1)