Why does R throw an error on iterative calculation - r

I'm looking at covid-19 data to calculate estimates for the reproductive number R0.
library(ggplot2)
library(dplyr)
library(tidyr)
library(stringr)
library(TTR)
# Get COVID cases, available from:
url <- "https://static.usafacts.org/public/data/covid-19/covid_confirmed_usafacts.csv"
DoubleCOV <- read.csv(url, stringsAsFactors = FALSE)
names(DoubleCOV)[1] <- "countyFIPS"
DoubleCovid <- pivot_longer(DoubleCOV, cols=starts_with("X"),
values_to="cases",
names_to=c("X","date_infected"),
names_sep="X") %>%
mutate(infected = as.Date(date_infected, format="%m.%d.%y"),
countyFIPS = str_pad(as.character(countyFIPS), 5, pad="0"))
#data is by county, summarise for the state of interest
stateData <- DoubleCovid %>% filter(State == "AL") %>% filter(cases != 0) %>%
group_by(infected) %>% summarise(sum(cases)) %>%
mutate(DaysSince = infected - min(infected))
names(stateData)[2] <- "cumCases"
#3 day moving average to smooth a little
stateData <- stateData %>% mutate(MA = runMean(cumCases,3))
#calculate doubling rate (DR) and then R0 infectious period/doubling rate
for(j in 4:nrow(stateData)){
stateData$DR[j] <- log(2)/log(stateData$MA[j]/stateData$MA[j-1])
stateData$R0[j] <- 14/stateData$DR[j]
}
CDplot <- stateData %>%
ggplot(mapping = aes(x = as.numeric(DaysSince), y = R0)) +
geom_line(color = "firebrick")
print(CDplot)
So in the above the state of interest is Alabama, hence filter(State == "AL") and this works.
But if I change the state to "NY" I get
Error in `$<-.data.frame`(`*tmp*`, "DR", value = c(NA, NA, NA, 0.733907206043719 :
replacement has 4 rows, data has 39
head(stateData) yields
infected cumCases DaysSince MA
<date> <int> <drtn> <dbl>
1 2020-03-02 1 0 days NA
2 2020-03-03 2 1 days NA
3 2020-03-04 11 2 days 4.67
4 2020-03-05 23 3 days 12
5 2020-03-06 25 4 days 19.7
6 2020-03-07 77 5 days 41.7
The moving average values in rows 3 and 4 (12 and 4.67) would yield a doubling rate of 0.734 which aligns with the value in the error message value = c(NA, NA, NA, 0.733907206043719 but why does it throw an error after that?
Bonus question: I know loops are frowned upon in R...is there a way to get the moving average and R0 calculation without one?

You have to initialise the new variables before you can access them using the j index. Due to recycling, Alabama, which has 28 rows (divisible by 4), does not return an error, only the warnings about uninitialised columns. New York, however, has 39 rows, which is not divisible by 4 so recycling fails and R returns an error. You shouldn't ignore warnings, sometimes you can, but it's not a good idea.
Try this to see what R (you) is trying to do:
stateData[4]
You should get all rows of the 4th column, not the 4th row.
Solution: initialise your DR and R0 columns first.
stateData$DR <- NA
stateData$R0 <- NA
for(j in 4:nrow(stateData)){
stateData$DR[j] <- log(2)/log(stateData$MA[j]/stateData$MA[j-1])
stateData$R0[j] <- 14/stateData$DR[j]
}
For the bonus question, you can use lag in the same mutate with MA:
stateData <- stateData %>% mutate(MA = runMean(cumCases,3),
DR = log(2)/log(MA/lag(MA)),
R0 = 14 / DR)
stateData
# A tibble: 28 x 6
infected cumCases DaysSince MA DR R0
<date> <int> <drtn> <dbl> <dbl> <dbl>
1 2020-03-13 5 0 days NA NA NA
2 2020-03-14 11 1 days NA NA NA
3 2020-03-15 22 2 days 12.7 NA NA
4 2020-03-16 29 3 days 20.7 1.42 9.89
5 2020-03-17 39 4 days 30 1.86 7.53
6 2020-03-18 51 5 days 39.7 2.48 5.64
7 2020-03-19 78 6 days 56 2.01 6.96
8 2020-03-20 106 7 days 78.3 2.07 6.78
9 2020-03-21 131 8 days 105 2.37 5.92
10 2020-03-22 167 9 days 135. 2.79 5.03
# ... with 18 more rows
I'm using Alabama's data.

Related

Add leading zeros to colum names

I'm surprised to find no one asked this question on Stackoverflow before. Maybe it's too stupid to ask?
So I have a dataframe that contains 48 weather variables, each representing a weather value for a month. I have drawn a simplified table shown below:
weather 1
weather 2
weather 3
weather 4
weather 5
weather 6
weather 7
weather 8
weather 9
weather 10
weather 11
weather 12
12
6
34
9
100
.01
-4
38
64
77
21
34
99
42
-3
34
34
.5
27
19
7
18
NA
20
My objective is to make the column names from "weather 1, weather 2, ..." to "weather 01, weather 02, ...." And I wrote a loop like this:
for (i in 1:9){
colnames(df) = gsub(i, 0+i, colnames(df))
}
However, instead of replacing the single-digit numbers with a leading zero, R replaced the actual letter "i" with "0+i". Can anyone let me know what's going on here and how to fix it? Or is there a better way to add leading zeros to column names?
Thank you very much!
We can use
library(stringr)
colnames(df) <- str_replace(colnames(df), "\\d+",
function(x) sprintf("%02d", as.integer(x)))
Here is another option:
library(tidyverse)
set.seed(35)
example <- tibble(`weather 1` = runif(2),
`weather 2` = runif(2),
`weather 3` = runif(2))
rename_with(example, ~str_replace(., "(weather )(\\d+)", "\\10\\2"), everything())
#> # A tibble: 2 x 3
#> `weather 01` `weather 02` `weather 03`
#> <dbl> <dbl> <dbl>
#> 1 0.857 0.553 0.486
#> 2 0.0108 0.950 0.0939
or with base R
colnames(example) <- gsub("(weather )(\\d+)", "\\10\\2", colnames(example))
example
#> # A tibble: 2 x 3
#> `weather 01` `weather 02` `weather 03`
#> <dbl> <dbl> <dbl>
#> 1 0.857 0.553 0.486
#> 2 0.0108 0.950 0.0939

Time difference calculated from wide data with missing rows

There is a longitudinal data set in the wide format, from which I want to compute time (in years and days) between the first observation date and the last date an individual was observed. Dates are in the format yyyy-mm-dd. The data set has four observation periods with missing dates, an example is as follows
df1<-data.frame("id"=c(1:4),
"adate"=c("2011-06-18","2011-06-18","2011-04-09","2011-05-20"),
"bdate"=c("2012-06-15","2012-06-15",NA,"2012-05-23"),
"cdate"=c("2013-06-18","2013-06-18","2013-04-09",NA),
"ddate"=c("2014-06-15",NA,"2014-04-11",NA))
Here "adate" is the first date and the last date is the date an individual was last seen. To compute the time difference (lastdate-adate), I have tried using "lubridate" package, for example
lubridate::time_length(difftime(as.Date("2012-05-23"), as.Date("2011-05-20")),"years")
However, I'm challenged by the fact that the last date is not coming from one column. I'm looking for a way to automate the calculation in R. The expected output would look like
id years days
1 1 2.99 1093
2 2 2.00 731
3 3 3.01 1098
4 4 1.01 369
Years is approximated to 2 decimal places.
Another tidyverse solution can be done by converting the data to long format, removing NA dates, and getting the time difference between last and first date for each id.
library(dplyr)
library(tidyr)
library(lubridate)
df1 %>%
pivot_longer(-id) %>%
na.omit %>%
group_by(id) %>%
mutate(value = as.Date(value)) %>%
summarise(years = time_length(difftime(last(value), first(value)),"years"),
days = as.numeric(difftime(last(value), first(value))))
#> # A tibble: 4 x 3
#> id years days
#> <int> <dbl> <dbl>
#> 1 1 2.99 1093
#> 2 2 2.00 731
#> 3 3 3.01 1098
#> 4 4 1.01 369
We could use pmap
library(dplyr)
library(purrr)
library(tidyr)
df1 %>%
mutate(out = pmap(.[-1], ~ {
dates <- as.Date(na.omit(c(...)))
tibble(years = lubridate::time_length(difftime(last(dates),
first(dates)), "years"),
days = lubridate::time_length(difftime(last(dates), first(dates)), "days"))
})) %>%
unnest_wider(out)
# A tibble: 4 x 7
# id adate bdate cdate ddate years days
# <int> <chr> <chr> <chr> <chr> <dbl> <dbl>
#1 1 2011-06-18 2012-06-15 2013-06-18 2014-06-15 2.99 1093
#2 2 2011-06-18 2012-06-15 2013-06-18 <NA> 2.00 731
#3 3 2011-04-09 <NA> 2013-04-09 2014-04-11 3.01 1098
#4 4 2011-05-20 2012-05-23 <NA> <NA> 1.01 369
Probably most of the functions introduced here might be quite complex. You should try to learn them if possible. Although will provide a Base R approach:
grp <- droplevels(interaction(df[,1],row(df[-1]))) # Create a grouping:
days <- tapply(unlist(df[-1]),grp, function(x)max(x,na.rm = TRUE) - x[1]) #Get the difference
cbind(df[1],days, years = round(days/365,2)) # Create your table
id days years
1.1 1 1093 2.99
2.2 2 731 2.00
3.3 3 1098 3.01
4.4 4 369 1.01
if comfortable with other higher functions then you could do:
dat <- aggregate(adate~id,reshape(df1,list(2:ncol(df1)), dir="long"),function(x)max(x) - x[1])
transform(dat,year = round(adate/365,2))
id adate year
1 1 1093 2.99
2 2 731 2.00
3 3 1098 3.01
4 4 369 1.01
Using base R apply :
df1[-1] <- lapply(df1[-1], as.Date)
df1[c('years', 'days')] <- t(apply(df1[-1], 1, function(x) {
x <- na.omit(x)
x1 <- difftime(x[length(x)], x[1], 'days')
c(x1/365, x1)
}))
df1[c('id', 'years', 'days')]
# id years days
#1 1 2.994521 1093
#2 2 2.002740 731
#3 3 3.008219 1098
#4 4 1.010959 369

aggregation of the region's values ​in the dataset

df <- read.csv ('https://raw.githubusercontent.com/ulklc/covid19-
timeseries/master/countryReport/raw/rawReport.csv',
stringsAsFactors = FALSE)
I processed the dataset.
Can we find the day of the least death in the Asian region?
the important thing here;
 is the sum of deaths of all countries in the asia region. Accordingly, it is to sort and find the day.
as output;
date region death
2020/02/17 asia 6300 (asia region sum)
The data in the output I created are examples. The data in the example are not real.
Since these are cumulative cases and deaths, we need to difference the data.
library(dplyr)
df %>%
mutate(day = as.Date(day)) %>%
filter(region=="Asia") %>%
group_by(day) %>%
summarise(deaths=sum(death)) %>%
mutate(d=c(first(deaths),diff(deaths))) %>%
arrange(d)
# A tibble: 107 x 3
day deaths d
<date> <int> <int>
1 2020-01-23 18 1 # <- this day saw only 1 death in the whole of Asia
2 2020-01-29 133 2
3 2020-02-21 2249 3
4 2020-02-12 1118 5
5 2020-01-24 26 8
6 2020-02-23 2465 10
7 2020-01-26 56 14
8 2020-01-25 42 16
9 2020-01-22 17 17
10 2020-01-27 82 26
# ... with 97 more rows
So the second day of records saw the least number of deaths recorded (so far).
Using the dplyr package for data treatment :
df <- read.csv ('https://raw.githubusercontent.com/ulklc/covid19-
timeseries/master/countryReport/raw/rawReport.csv',
stringsAsFactors = FALSE)
library(dplyr)
df_sum <- df %>% group_by(region,day) %>% # grouping by region and day
summarise(death=sum(death)) %>% # summing following the groups
filter(region=="Asia",death==min(death)) # keeping only minimum of Asia
Then you have :
> df_sum
# A tibble: 1 x 3
# Groups: region [1]
region day death
<fct> <fct> <int>
1 Asia 2020/01/22 17

Iteration for time series data, using purrr

I have a bunch of time series data stacked on top of one another in a data frame; one series for each region in a country. I'd like to apply the seas() function (from the seasonal package) to each series, iteratively, to make the series seasonally adjusted. To do this, I first have to convert the series to a ts class. I'm struggling to do all this using purrr.
Here's a minimum worked example:
library(seasonal)
library(tidyverse)
set.seed(1234)
df <- data.frame(region = rep(1:10, each = 20),
quarter = rep(1:20, 10),
var = sample(5:200, 200, replace = T))
For each region (indexed by a number) I'd like to perform the following operations. Here's the first region as an example:
tem1 <- df %>% filter(region==1)
tem2 <- ts(data = tem1$var, frequency = 4, start=c(1990,1))
tem3 <- seas(tem2)
tem4 <- as.data.frame(tem3$data)
I'd then like to stack the output (ie. the multiple tem4 data frames, one for each region), along with the region and quarter identifiers.
So, the start of the output for region 1 would be this:
final seasonaladj trend irregular region quarter
1 27 27 96.95 -67.97279 1 1
2 126 126 96.95 27.87381 1 2
3 124 124 96.95 27.10823 1 3
4 127 127 96.95 30.55075 1 4
5 173 173 96.95 75.01355 1 5
6 130 130 96.95 32.10672 1 6
The data for region 2 would be below this etc.
I started with the following but without luck so far. Basically, I'm struggling to get the time series into the tibble:
seas.adjusted <- df %>%
group_by(region) %>%
mutate(data.ts = map(.x = data$var,
.f = as.ts,
start = 1990,
freq = 4))
I don't know much about the seasonal adjustment part, so there may be things I missed, but I can help with moving your calculations into a map-friendly function.
After grouping by region, you can nest the data so there's a nested data frame for each region. Then you can run essentially the same code as you had, but inside a function in map. Unnesting the resulting column gives you a long-shaped data frame of adjustments.
Like I said, I don't have the expertise to know whether those last two columns having NAs is expected or not.
Edit: Based on #wibeasley's question about retaining the quarter column, I'm adding a mutate that adds a column of the quarters listed in the nested data frame.
library(seasonal)
library(tidyverse)
set.seed(1234)
df <- data.frame(region = rep(1:10, each = 20),
quarter = rep(1:20, 10),
var = sample(5:200, 200, replace = T))
df %>%
group_by(region) %>%
nest() %>%
mutate(data.ts = map(data, function(x) {
tem2 <- ts(x$var, frequency = 4, start = c(1990, 1))
tem3 <- seas(tem2)
as.data.frame(tem3$data) %>%
mutate(quarter = x$quarter)
})) %>%
unnest(data.ts)
#> # A tibble: 200 x 8
#> region final seasonaladj trend irregular quarter seasonal adjustfac
#> <int> <dbl> <dbl> <dbl> <dbl> <int> <dbl> <dbl>
#> 1 1 27 27 97.0 -68.0 1 NA NA
#> 2 1 126 126 97.0 27.9 2 NA NA
#> 3 1 124 124 97.0 27.1 3 NA NA
#> 4 1 127 127 97.0 30.6 4 NA NA
#> 5 1 173 173 97.0 75.0 5 NA NA
#> 6 1 130 130 97.0 32.1 6 NA NA
#> 7 1 6 6 97.0 -89.0 7 NA NA
#> 8 1 50 50 97.0 -46.5 8 NA NA
#> 9 1 135 135 97.0 36.7 9 NA NA
#> 10 1 105 105 97.0 8.81 10 NA NA
#> # ... with 190 more rows
I also gave a bit more thought to doing this without nesting, and instead tried doing it with a split. Passing that list of data frames into imap_dfr let me take each split piece of the data frame and its name (in this case, the value of region), then return everything rbinded back together into one data frame. I sometimes shy away from nested data just because I have trouble seeing what's going on, so this is an alternative that is maybe more transparent.
df %>%
split(.$region) %>%
imap_dfr(function(x, reg) {
tem2 <- ts(x$var, frequency = 4, start = c(1990, 1))
tem3 <- seas(tem2)
as.data.frame(tem3$data) %>%
mutate(region = reg, quarter = x$quarter)
}) %>%
select(region, quarter, everything()) %>%
head()
#> region quarter final seasonaladj trend irregular seasonal adjustfac
#> 1 1 1 27 27 96.95 -67.97274 NA NA
#> 2 1 2 126 126 96.95 27.87378 NA NA
#> 3 1 3 124 124 96.95 27.10823 NA NA
#> 4 1 4 127 127 96.95 30.55077 NA NA
#> 5 1 5 173 173 96.95 75.01353 NA NA
#> 6 1 6 130 130 96.95 32.10669 NA NA
Created on 2018-08-12 by the reprex package (v0.2.0).
I put all the action inside of f(), and then called it with purrr::map_df(). The re-inclusion of quarter is a hack.
f <- function( .region ) {
d <- df %>%
dplyr::filter(region == .region)
y <- d %>%
dplyr::pull(var) %>%
ts(frequency = 4, start=c(1990,1)) %>%
seas()
y$data %>%
as.data.frame() %>%
# dplyr::select(-seasonal, -adjustfac) %>%
dplyr::mutate(
quarter = d$quarter
)
}
purrr::map_df(1:10, f, .id = "region")
results:
region final seasonaladj trend irregular quarter seasonal adjustfac
1 1 27.00000 27.00000 96.95000 -6.797279e+01 1 NA NA
2 1 126.00000 126.00000 96.95000 2.787381e+01 2 NA NA
3 1 124.00000 124.00000 96.95000 2.710823e+01 3 NA NA
4 1 127.00000 127.00000 96.95000 3.055075e+01 4 NA NA
5 1 173.00000 173.00000 96.95000 7.501355e+01 5 NA NA
6 1 130.00000 130.00000 96.95000 3.210672e+01 6 NA NA
7 1 6.00000 6.00000 96.95000 -8.899356e+01 7 NA NA
8 1 50.00000 50.00000 96.95000 -4.647254e+01 8 NA NA
9 1 135.00000 135.00000 96.95000 3.671077e+01 9 NA NA
10 1 105.00000 105.00000 96.95000 8.806955e+00 10 NA NA
...
96 5 55.01724 55.01724 60.25848 9.130207e-01 16 1.9084928 1.9084928
97 5 60.21549 60.21549 59.43828 1.013076e+00 17 1.0462424 1.0462424
98 5 58.30626 58.30626 58.87065 9.904130e-01 18 0.1715082 0.1715082
99 5 61.68175 61.68175 58.07827 1.062045e+00 19 1.0537962 1.0537962
100 5 59.30138 59.30138 56.70798 1.045733e+00 20 2.5294523 2.5294523
...

Extend data frame column with inflation in R

I'm trying to extend some code to be able to:
1) read in a vector of prices
2) left join that vector of prices to a data frame of years (or years and months)
3) append/fill the prices for missing years with interpolated data based on the last year of available prices plus a specified inflation rate. Consider an example like this one:
prices <- data.frame(year=2018:2022,
wti=c(75,80,90,NA,NA),
brent=c(80,85,94,93,NA))
What I need is something that will fill the missing rows of each column with the last price plus inflation (suppose 2%). I can do this in a pretty brute force way as:
i_rate<-0.02
for(i in c(1:nrow(prices))){
if(is.na(prices$wti[i]))
prices$wti[i]<-prices$wti[i-1]*(1+i_rate)
if(is.na(prices$brent[i]))
prices$brent[i]<-prices$brent[i-1]*(1+i_rate)
}
It seems to me there should be a way to do this using some combination of apply() and/or fill() but I can't seem to make it work.
Any help would be much appreciated.
As noted by #camille, the problem with dplyr::lag is that it doesn't work here with consecutive NAs because it uses the "original" ith element of a vector instead of the "revised" ith element. We'd have to first create a version of lag that will do this by creating a new function:
impute_inflation <- function(x, rate) {
output <- x
y <- rep(NA, length = length(x)) #Creating an empty vector to fill in with the loop. This makes R faster to run for vectors with a large number of elements.
for (i in seq_len(length(output))) {
if (i == 1) {
y[i] <- output[i] #To avoid an error attempting to use the 0th element.
} else {
y[i] <- output[i - 1]
}
if (is.na(output[i])) {
output[i] <- y[i] * (1 + rate)
} else {
output[i]
}
}
output
}
Then it's a pinch to apply this across a bunch of variables with dplyr::mutate_at():
library(dplyr)
mutate_at(prices, vars(wti, brent), impute_inflation, 0.02)
year wti brent
1 2018 75.000 80.00
2 2019 80.000 85.00
3 2020 90.000 94.00
4 2021 91.800 93.00
5 2022 93.636 94.86
You can use dplyr::lag to get the previous value in a given column. Your lagged values look like this:
library(dplyr)
inflation_factor <- 1.02
prices <- data_frame(year=2018:2022,
wti=c(75,80,90,NA,NA),
brent=c(80,85,94,93,NA)) %>%
mutate_at(vars(wti, brent), as.numeric)
prices %>%
mutate(prev_wti = lag(wti))
#> # A tibble: 5 x 4
#> year wti brent prev_wti
#> <int> <dbl> <dbl> <dbl>
#> 1 2018 75 80 NA
#> 2 2019 80 85 75
#> 3 2020 90 94 80
#> 4 2021 NA 93 90
#> 5 2022 NA NA NA
When a value is NA, multiply the lagged value by the inflation factor. As you can see, that doesn't handle consecutive NAs, however.
prices %>%
mutate(wti = ifelse(is.na(wti), lag(wti) * inflation_factor, wti),
brent = ifelse(is.na(brent), lag(brent) * inflation_factor, brent))
#> # A tibble: 5 x 3
#> year wti brent
#> <int> <dbl> <dbl>
#> 1 2018 75 80
#> 2 2019 80 85
#> 3 2020 90 94
#> 4 2021 91.8 93
#> 5 2022 NA 94.9
Or to scale this and avoid doing the same multiplication over and over, gather the data into a long format, get lags within each group (wti, brent, or any others you may have), and adjust values as needed. Then you can spread back to the original shape:
prices %>%
tidyr::gather(key = key, value = value, wti, brent) %>%
group_by(key) %>%
mutate(value = ifelse(is.na(value), lag(value) * inflation_factor, value)) %>%
tidyr::spread(key = key, value = value)
#> # A tibble: 5 x 3
#> year brent wti
#> <int> <dbl> <dbl>
#> 1 2018 80 75
#> 2 2019 85 80
#> 3 2020 94 90
#> 4 2021 93 91.8
#> 5 2022 94.9 NA
Created on 2018-07-12 by the reprex package (v0.2.0).

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