How is it possible with to sum up consecutive depth data with R?
For instance:
a <- data.frame(label = as.factor(c("Air","Air","Air","Air","Air","Air","Wood","Wood","Wood","Wood","Wood","Air","Air","Air","Air","Stone","Stone","Stone","Stone","Air","Air","Air","Air","Air","Wood","Wood")),
depth = as.numeric(c(1,2,3,-1,4,5,4,5,4,6,8,9,8,9,10,9,10,11,10,11,12,10,12,13,14,14)))
The given output should be something like:
Label Depth
Air 7
Wood 3
Stone 1
First the removal of negative values is done with cummax(), because depth can only increase in this special case. Hence:
label depth
1 Air 1
2 Air 2
3 Air 3
4 Air 3
5 Air 4
6 Air 5
7 Wood 5
8 Wood 5
9 Wood 5
10 Wood 6
11 Wood 8
12 Air 9
13 Air 9
14 Air 9
15 Air 10
16 Stone 10
17 Stone 10
18 Stone 11
19 Stone 11
20 Air 11
21 Air 12
22 Air 12
23 Air 12
24 Air 13
25 Wood 14
26 Wood 14
Now by max-min the increase in depth for every consecutive row you would get: (the question is how to do this step)
label depth
1 Air 4
2 Wood 3
3 Air 1
4 Stone 1
5 Air 2
5 Wood 0
And finally summing up those max-min values the output is the one presented above.
Steps tried to achieve the output:
The first obvious solution would be for instance for Air:
diff(cummax(a[a$label=="Air",]$depth))
This solution gets rid of the negative data, which is necessary due to an expected constant increase in depth.
The problem is the output also takes into account the big steps in between each consecutive subset. Hence, the sum for Air would be 12 instead of 7.
[1] 1 1 0 1 1 4 0 0 1 1 1 0 0 1
Even worse would be a solution with aggreagte, e.g.:
aggregate(depth~label, a, FUN=function(x){sum(x>0)})
Note: solutions with filtering big jumps is not what i'm looking for. Sure you could hard code a limit for instance <2 for the example of Air once again:
sum(diff(cummax(a[a$label=="Air",]$depth))[diff(cummax(a[a$label=="Air",]$depth))<2])
Gives you almost the right result but does not work as it is expected here. I'm pretty sure there is already a function for what I'm looking for because it is not a uncommon problem for many different tasks.
I guess taking the minimum and maximum value of each set of consecutive rows per material and summing those up would be one possible solution, but I'm not sure how to apply a function to only the consecutive subsets.
You can use data.table::rleid to quickly group by run, or reconstruct it with rle if you really like. After that, aggregating is fairly easy in any grammar. In dplyr,
library(dplyr)
a <- data.frame(label = c("Air","Air","Air","Air","Air","Air","Wood","Wood","Wood","Wood","Wood","Air","Air","Air","Air","Stone","Stone","Stone","Stone","Air","Air","Air","Air","Air","Wood","Wood"),
depth = c(1,2,3,-1,4,5,4,5,4,6,8,9,8,9,10,9,10,11,10,11,12,10,12,13,14,14))
a2 <- a %>%
# filter to rows where previous value is lower, equal, or NA
filter(depth >= lag(depth) | is.na(lag(depth))) %>%
# group by label and its run
group_by(label, run = data.table::rleid(label)) %>%
summarise(depth = max(depth) - min(depth)) # aggregate
a2 %>% arrange(run) # sort to make it pretty
#> # A tibble: 6 x 3
#> # Groups: label [3]
#> label run depth
#> <fctr> <int> <dbl>
#> 1 Air 1 4
#> 2 Wood 2 3
#> 3 Air 3 1
#> 4 Stone 4 1
#> 5 Air 5 2
#> 6 Wood 6 0
a3 <- a2 %>% summarise(depth = sum(depth)) # a2 is still grouped, so aggregate more
a3
#> # A tibble: 3 x 2
#> label depth
#> <fctr> <dbl>
#> 1 Air 7
#> 2 Stone 1
#> 3 Wood 3
A base R method using aggregate is
aggregate(cbind(val=cummax(a$depth)),
list(label=a$label, ID=c(0, cumsum(diff(as.integer(a$label)) != 0))),
function(x) diff(range(x)))
The first argument to aggregate calculates the cumulative maximum as the OP does above for the input vector, the use of cbind provide for the final output of the calculated vector. The second argument is the grouping argument. This uses a different method than rle, which calculates the cumulative sum of the differences. Finally, the third argument provides the function which calculates the desired output by taking a difference of the range for each group.
This returns
label ID val
1 Air 0 4
2 Wood 1 3
3 Air 2 1
4 Stone 3 1
5 Air 4 2
6 Wood 5 0
The data.table way (borrowing in part from #alistaire):
setDT(a)
a[, depth := cummax(depth)]
depth_gain <- a[,
list(
depth = max(depth) - depth[1], # Only need the starting and max values
label = label[1]
),
by = rleidv(label)
]
result <- depth_gain[, list(depth = sum(depth)), by = label]
Related
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
I'm currently reading "Practical Statistics for Data Scientists" and following along in R as they demonstrate some code. There is one chunk of code I'm particularly struggling to follow the logic of and was hoping someone could help. The code in question is creating a dataframe with 1000 rows where each observation is the mean of 5 randomly drawn income values from the dataframe loans_income. However, I'm getting confused about the logic of the code as it is fairly complicated with a tapply() function and nested rep() statements.
The code to create the dataframe in question is as follows:
samp_mean_5 <- data.frame(income = tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean),
type='mean_of_5')
In particular, I'm confused about the nested rep() statements and the 1000*5 portion of the sample() function. Any help understanding the logic of the code would be greatly appreciated!
For reference, the original dataset loans_income simply has a single column of 50,000 income values.
You have 50,000 loans_income in a single vector. Let's break your code down:
tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean)
I will replace 1000 with 10 and income with random numbers, so it's easier to explain. I also set set.seed(1) so the result can be reproduced.
sample(loans_income$income,1000*5)
We 50 random incomes from your vector without replacement. They are (temporarily) put into a vector of length 50, so the output looks like this:
> sample(runif(50000),10*5)
[1] 0.73283101 0.60329970 0.29871173 0.12637654 0.48434952 0.01058067 0.32337850
[8] 0.46873561 0.72334215 0.88515494 0.44036341 0.81386225 0.38118213 0.80978822
[15] 0.38291273 0.79795343 0.23622492 0.21318431 0.59325586 0.78340477 0.25623138
[22] 0.64621658 0.80041393 0.68511759 0.21880083 0.77455662 0.05307712 0.60320912
[29] 0.13191926 0.20816298 0.71600799 0.70328349 0.44408218 0.32696205 0.67845445
[36] 0.64438336 0.13241312 0.86589561 0.01109727 0.52627095 0.39207860 0.54643661
[43] 0.57137320 0.52743012 0.96631114 0.47151170 0.84099503 0.16511902 0.07546454
[50] 0.85970500
rep(1:1000,rep(5,1000))
Now we are creating an indexing vector of length 50:
> rep(1:10,rep(5,10))
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6
[29] 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 10 10
Those indices "group" the samples from step 1. So basically this vector tells R that the first 5 entries of your "sample vector" belong together (index 1), the next 5 entries belong together (index 2) and so on.
FUN = mean
Just apply the mean-function on the data.
tapply
So tapply takes the sampled data (sample-part) and groups them by the second argument (the rep()-part) and applies the mean-function on each group.
If you are familiar with data.frames and the dplyr package, take a look at this (only the first 10 rows are displayed):
set.seed(1)
df <- data.frame(income=sample(runif(5000),10*5), index=rep(1:10,rep(5,10)))
income index
1 0.42585569 1
2 0.16931091 1
3 0.48127444 1
4 0.68357403 1
5 0.99374923 1
6 0.53227877 2
7 0.07109499 2
8 0.20754511 2
9 0.35839481 2
10 0.95615917 2
I attached the an index to the random numbers (your income). Now we calculate the mean per group:
df %>%
group_by(index) %>%
summarise(mean=mean(income))
which gives us
# A tibble: 10 x 2
index mean
<int> <dbl>
1 1 0.551
2 2 0.425
3 3 0.827
4 4 0.391
5 5 0.590
6 6 0.373
7 7 0.514
8 8 0.451
9 9 0.566
10 10 0.435
Compare it to
set.seed(1)
tapply(sample(runif(5000),10*5),
rep(1:10,rep(5,10)),
mean)
which yields basically the same result:
1 2 3 4 5 6 7 8 9
0.5507529 0.4250946 0.8273149 0.3905850 0.5902823 0.3730092 0.5143829 0.4512932 0.5658460
10
0.4352546
I have a set of observations that measure time delay from an initial event, such as the elapsed time from when an email is sent to when it is opened.
Given a set of 100 observations, how can I tell at what point in time 90 percent of the opens took place. I want to be able to say "90 % of the opens took place within 4 hours of send time."
I can generate a histogram of delays, which shows that most opens happen early, but I do not know how to get a cumulative measure for all counts in the bins. (I'm not explaining myself very well, not a stats wonk)
So with this sample data I have 10 observations with a delay of 1 hour, 5 with a delay of 2 hours, 3 with a delay of 3 hours and 2 with a delay of 4 hours. This means that 90% of the opens came within less than 4 hours. How do I determine that 90% limit for a real set of observations?
Edited with more compact sample data creation and added first cut at plot of cumulative percentage. Would welcome better solutions.
library(tidyverse)
library(ggplot2)
all_delays <- tibble(delay = rep(1:4, c(10, 5, 3, 2)))
all_delays
#> # A tibble: 20 x 1
#> delay
#> <int>
#> 1 1
#> 2 1
#> 3 1
#> 4 1
#> 5 1
#> 6 1
#> 7 1
#> 8 1
#> 9 1
#> 10 1
#> 11 2
#> 12 2
#> 13 2
#> 14 2
#> 15 2
#> 16 3
#> 17 3
#> 18 3
#> 19 4
#> 20 4
# histogram of data
ggplot(all_delays) + aes(delay) +
geom_histogram() +
scale_y_continuous(breaks = seq(0,10,1))
#> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
# newbie incorrect way to get plot of cumulative percentage.
# would welcome better way to do this.
all_delays <- all_delays %>% mutate(cnt = 1) %>%
arrange(delay) %>%
mutate(cs = cumsum(cnt))
ggplot(all_delays) + aes(cs/nrow(all_delays),delay) +
geom_line() +
scale_x_continuous(breaks = c(0,.25,.50,.75,.90,1),
labels=c("0","25%","50%","75%","90%","100%")) +
geom_vline(xintercept =.9) +
xlab("Cumulative Percentage of opens") +
ylab("hours since open")
Created on 2019-04-27 by the reprex package (v0.2.1)
I guess my expected results are something that would say "90% limit = 3", or some kind of cumulative curve that would start at the shortest open delay and then increase in value until 100 % was reached with a tick at 90 %.
Thanks for the quantile() answer!
Email open rates typically have a long tail where mot activity happens within a day or two of the email send, and then a very long tail as people browse their email inboxes weeks or even months after the email was sent.
What you describe is called a quantile. The code below removes all delays above the 90th percentile; i.e., the remaining delays give you the points by which 90% of the events have occurred.
> all_delays %>% filter(delay <= quantile(delay, 0.9))
# A tibble: 18 x 1
delay
<dbl>
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 2
12 2
13 2
14 2
15 2
16 3
17 3
18 3
I have some data in the form:
Person.ID Household.ID Composition
1 4593 1A_0C
2 4992 2A_1C
3 9843 1A_1C
4 8385 2A_2C
5 9823 8A_1C
6 3458 1C_9C
7 7485 2C_0C
: : :
We can think of the composition variable as a count of adults/children i.e. 2A_1C would equate to two adults and two children.
What I want to do is reduce the amount of possible levels of composition. For person 5 we have composition of 8A_1C, I am looking for a way to reduce this to 4+A_0C. So for example we would have 4+ for any composition value with greater than 4A.
Person.ID Household.ID Composition
5 9823 4+A_1C
6 3458 1A_4+C
: : :
I am unsure of how to do this in R, I am thinking of using filter() or select() from dyplyr. Otherwise I would need to use some sort of regular expression.
Any help would be appreciated. Thanks
Data:
Person.ID <- c(1,2,3,4,5,6,7,8)
Household.ID <- c(4593,4992,9843,8385,9823,3458,7485)
Composition <- c("1A_0C","2A_1C","1A_1C","2A_2C","8A_1C","1A_9C","2A_0C")
dat <- tibble(Person.ID, Household.ID, Composition)
Function:
above4 <- function(f){
ff <- gsub("[^0-9]","",f)
if(ff>4){return("4+")}
if(ff<=4){return(ff)}
}
Apply function (done on separated data, but can recombine after):
dat_ <- dat %>% tidyr::separate(., col=Composition,
into=c("Adults", "Children"),
sep="_") %>%
dplyr::mutate(Adults_ = unlist(lapply(Adults,above4)),
Children_ = unlist(lapply(Children,above4)))
You might then use select, filter to get your required dataset.
dat_ %>% dplyr::mutate(Composition_ = paste0(Adults_, "A_", Children_, "C")) %>%
dplyr::select(Person.ID, Household.ID, Composition=Composition_)
# A tibble: 7 x 3
Person.ID Household.ID Composition
<dbl> <dbl> <chr>
1 1. 4593. 1A_0C
2 2. 4992. 2A_1C
3 3. 9843. 1A_1C
4 4. 8385. 2A_2C
5 5. 9823. 4+A_1C
6 6. 3458. 1A_4+C
7 7. 7485. 2A_0C
We can use gsub:
df$Composition <- gsub("(?<!\\d)([5-9]|\\d{2,})(?=[AC])", "4+", df$Composition, perl = TRUE)
This assumes that 2 or more consecutive digits represent a number that's always greater than 4 (i.e. no 01, 02, or 001).
Output:
Person.ID Household.ID Composition
1 1 4593 1A_0C
2 2 4992 2A_1C
3 3 9843 1A_1C
4 4 8385 2A_2C
5 5 9823 4+A_1C
6 6 3458 1C_4+C
7 7 7485 2C_0C
I have a dataset of Ages for the customer and I wanted to make a frequency distribution by 9 years of a gap of age.
Ages=c(83,51,66,61,82,65,54,56,92,60,65,87,68,64,51,
70,75,66,74,68,44,55,78,69,98,67,82,77,79,62,38,88,76,99,
84,47,60,42,66,74,91,71,83,80,68,65,51,56,73,55)
My desired outcome would be similar to below-shared table, variable names can be differed(as you wish)
Could I use binCounts code into it ? if yes could you help me out using the code as not sure of bx and idxs in this code?
binCounts(x, idxs = NULL, bx, right = FALSE) ??
Age Count
38-46 3
47-55 7
56-64 7
65-73 14
74-82 10
83-91 6
92-100 3
Much Appreciated!
I don't know about the binCounts or even the package it is in but i have a bare r function:
data.frame(table(cut(Ages,0:7*9+37)))
Var1 Freq
1 (37,46] 3
2 (46,55] 7
3 (55,64] 7
4 (64,73] 14
5 (73,82] 10
6 (82,91] 6
7 (91,100] 3
To exactly duplicate your results:
lowerlimit=c(37,46,55,64,73,82,91,101)
Labels=paste(head(lowerlimit,-1)+1,lowerlimit[-1],sep="-")#I add one to have 38 47 etc
group=cut(Ages,lowerlimit,Labels)#Determine which group the ages belong to
tab=table(group)#Form a frequency table
as.data.frame(tab)# transform the table into a dataframe
group Freq
1 38-46 3
2 47-55 7
3 56-64 7
4 65-73 14
5 74-82 10
6 83-91 6
7 92-100 3
All this can be combined as:
data.frame(table(cut(Ages,s<-0:7*9+37,paste(head(s+1,-1),s[-1],sep="-"))))