This question already has answers here:
Apply a function to groups within a data.frame in R
(4 answers)
Closed 2 years ago.
I am trying to apply a function to a dataframe to add a column which calculates the percentile rank for each record based on Weather Station ID (WSID) and Season Grouping.
## temperatures data frame:
WSID Season Date Temperature
20 Summer 24/01/2020 18
12 Summer 25/01/2020 20
20 Summer 26/01/2020 25
12 Summer 27/01/2020 17
20 Winter 18/10/2020 15
12 Winter 19/10/2020 12
12 Winter 20/10/2020 13
12 Winter 21/10/2020 14
## Code tried:
perc.rank <- function(x) trunc(rank(x))/length(x)
rank.perc = function(mdf) {
combined1 = mdf %>%
mutate(percentile = perc.rank(Temperature))
}
temperatures = temperatures %>%
split(.$WSID) %>%
map_dfr(~rank.perc(.))
## Expected Output :
WSID Season Date Temperature Percentile
20 Summer 24/01/2020 18 0.333
12 Summer 25/01/2020 20 0.444
20 Summer 26/01/2020 25 0.666
12 Summer 27/01/2020 17 0.333
20 Winter 18/10/2020 15
12 Winter 19/10/2020 12
12 Winter 20/10/2020 13
12 Winter 21/10/2020 14
Is there some elegant way to do this using functions such as group_modify, group_split, map and/or split?
I was thinking there should be as for example in case there is a 3 or more level grouping factor.
The code works for when I split the data by WSID but I cant seem to get any further when I want to group also by WSID + Season.
(Filled in Percentile values were calculated from Excel percentile rank function)
You can directly use the function with group_by instead of splitting, also function rank.perc seems unnecessary.
library(dplyr)
perc.rank <- function(x) trunc(rank(x))/length(x)
df %>%
group_by(WSID) %>%
mutate(percentile = perc.rank(Temperature))
In group_by it is easy to add more groupings later eg - group_by(WSID, Season).
Related
I am facing one problem, I calculated a monthly interest rate for a mortgage, however, I would need to sum the results in order to have it yearly (always 12 months).
H <- 2000000 # mortgage
i.m <- 0.03/12 # rate per month
year <- 15 # years
a <- (H*i.m*(1+i.m)^(12*year))/
((1+i.m)^(12*year)-1)
a # monthly payment
interest <- a*(1-(1/(1+i.m)^(0:(year*12))))
interest
cumsum(a*(1-(1/(1+i.m)^(0:(year*12))))) # first 12 values together and then next 12 values + first values and ... (I want to have for every year a value)
You may do this with tapply in base R.
monthly <- cumsum(a*(1-(1/(1+i.m)^(0:(year*12)))))
yearly <- tapply(monthly, ceiling(seq_along(monthly)/12), sum)
I think you can use the following solution:
monthly <- cumsum(a*(1-(1/(1+i.m)^(0:(year*12)))))
sapply(split(monthly, ceiling(seq_along(monthly) / 12)), function(x) x[length(x)])
1 2 3 4 5 6 7 8
2254.446 9334.668 21098.218 37406.855 58126.414 83126.695 112281.337 145467.712
9 10 11 12 13 14 15 16
182566.812 223463.138 268044.605 316202.434 367831.057 422828.023 481093.905 486093.905
I have what I think is a simple question but I can't figure it out! I have a data frame with multiple columns. Here's a general example:
colony = c('29683','25077','28695','4865','19858','2235','1948','1849','2370','23196')
age = c(21,23,4,25,7,4,12,14,9,7)
activity = c(19,45,78,33,2,49,22,21,112,61)
test.df = data.frame(colony,age,activity)
test.df
I would like for R to calculate average activity based on the age of the colony in the data frame. Specifically, I want it to only calculate the average activity of the colonies that are the same age or older than the colony in that row, not including the activity of the colony in that row. For example, colony 29683 is 21 years old. I want the average activity of colonies older than 21 for this row of my data. That would include colony 25077 and colony 4865; and the mean would be (45+33)/2 = 39. I want R to do this for each row of the data by identifying the age of the colony in the current row, then identifying the colonies that are older than that colony, and then averaging the activity of those colonies.
I've tried doing this in a for loop in R. Here's the code I used:
test.avg = vector("numeric",nrow(test.df))`
for (i in 1:10){
test.avg[i] <- mean(subset(test.df$activity,test.df$age >= age[i])[-i])
}
R returns a list of values where half of them are correct and the the other half are not (I'm not even sure how it calculated those incorrect numbers..). The numbers that are correct are also out of order compared to how they're listed in the dataframe. It's clearly able to do the right thing for some iterations of the loop but not all. If anyone could help me out with my code, I would greatly appreciate it!
colony = c('29683','25077','28695','4865','19858','2235','1948','1849','2370','23196')
age = c(21,23,4,25,7,4,12,14,9,7)
activity = c(19,45,78,33,2,49,22,21,112,61)
test.df = data.frame(colony,age,activity)
library(tidyverse)
test.df %>%
mutate(result = map_dbl(age, ~mean(activity[age > .x])))
#> colony age activity result
#> 1 29683 21 19 39.00000
#> 2 25077 23 45 33.00000
#> 3 28695 4 78 39.37500
#> 4 4865 25 33 NaN
#> 5 19858 7 2 42.00000
#> 6 2235 4 49 39.37500
#> 7 1948 12 22 29.50000
#> 8 1849 14 21 32.33333
#> 9 2370 9 112 28.00000
#> 10 23196 7 61 42.00000
# base
test.df$result <- with(test.df, sapply(age, FUN = function(x) mean(activity[age > x])))
test.df
#> colony age activity result
#> 1 29683 21 19 39.00000
#> 2 25077 23 45 33.00000
#> 3 28695 4 78 39.37500
#> 4 4865 25 33 NaN
#> 5 19858 7 2 42.00000
#> 6 2235 4 49 39.37500
#> 7 1948 12 22 29.50000
#> 8 1849 14 21 32.33333
#> 9 2370 9 112 28.00000
#> 10 23196 7 61 42.00000
Created on 2021-03-22 by the reprex package (v1.0.0)
The issue in your solution is that the index would apply to the original data.frame, yet you subset that and so it does not match anymore.
Try something like this: First find minimum age, then exclude current index and calculate average activity of cases with age >= pre-calculated minimum age.
for (i in 1:10){
test.avg[i] <- {amin=age[i]; mean(subset(test.df[-i,], age >= amin)$activity)}
}
You can use map_df :
library(tidyverse)
test.df %>%
mutate(map_df(1:nrow(test.df), ~
test.df %>%
filter(age >= test.df$age[.x]) %>%
summarise(av_acti= mean(activity))))
This question already has answers here:
How to reshape data from long to wide format
(14 answers)
Closed 5 years ago.
I'm trying to change a dataframe in R to group multiple rows by a measurement. The table has a location (km), a size (mm) a count of things in that size bin, a site and year. I want to take the sizes, make a column from each one (2, 4 and 6 in this example), and place the corresponding count into each the row for that location, site and year.
It seems like a combination of transposing and grouping, but I can't figure out a way to accomplish this in R. I've looked at t(), dcast() and aggregate(), but those aren't really close at all.
So I would go from something like this:
df <- data.frame(km=c(rep(32,3),rep(50,3)), mm=rep(c(2,4,6),2), count=sample(1:25,6), site=rep("A", 6), year=rep(2013, 6))
km mm count site year
1 32 2 18 A 2013
2 32 4 2 A 2013
3 32 6 12 A 2013
4 50 2 3 A 2013
5 50 4 17 A 2013
6 50 6 21 A 2013
To this:
km site year mm_2 mm_4 mm_6
1 32 A 2013 18 2 12
2 50 A 2013 3 17 21
Edit: I tried the solution in a suggested duplicate, but I did not work for me, not really sure why. The answer below worked better.
As suggested in the comment above, we can use the sep argument in spread:
library(tidyr)
spread(df, mm, count, sep = "_")
km site year mm_2 mm_4 mm_6
1 32 A 2013 4 20 1
2 50 A 2013 15 14 22
As you mentioned dcast(), here is a method using it.
set.seed(1)
df <- data.frame(km=c(rep(32,3),rep(50,3)),
mm=rep(c(2,4,6),2),
count=sample(1:25,6),
site=rep("A", 6),
year=rep(2013, 6))
library(reshape2)
dcast(df, ... ~ mm, value.var="count")
# km site year 2 4 6
# 1 32 A 2013 13 10 20
# 2 50 A 2013 3 17 1
And if you want a bit of a challenge you can try the base function reshape().
df2 <- reshape(df, v.names="count", idvar="km", timevar="mm", ids="mm", direction="wide")
colnames(df2) <- sub("count.", "mm_", colnames(df2))
df2
# km site year mm_2 mm_4 mm_6
# 1 32 A 2013 13 10 20
# 4 50 A 2013 3 17 1
I am very new to R so I am not sure how basic my question is, but I am stuck at the following point.
I have data that has a panel structure, similar to this
Country Year Outcome Country-characteristic
A 1990 10 40
A 1991 12 40
A 1992 14 40
B 1991 10 60
B 1992 12 60
For some reason I need to put this in a cross-sectional structure such I get averages over all years for each country, that is in the end, it should look like,
Country Outcome Country-Characteristic
A 12 40
B 11 60
Has anybody faced a similar problem? I was playing with lapply(table$country, table$outcome, mean) but that did not work as I wanted it.
Two tips: 1- When you ask a question, you should provide a reproducible example for the data too (as I did with read.table below). 2- It's not a good idea to use "-" in column names. You should use "_" instead.
You can get a summary using the dplyr package:
df1 <- read.table(text="Country Year Outcome Countrycharacteristic
A 1990 10 40
A 1991 12 40
A 1992 14 40
B 1991 10 60
B 1992 12 60", header=TRUE, stringsAsFactors=FALSE)
library(dplyr)
df1 %>%
group_by(Country) %>%
summarize(Outcome=mean(Outcome),Countrycharacteristic=mean(Countrycharacteristic))
# A tibble: 2 x 3
Country Outcome Countrycharacteristic
<chr> <dbl> <dbl>
1 A 12 40
2 B 11 60
We can do this in base R with aggregate
aggregate(.~Country, df1[-2], mean)
# Country Outcome Countrycharacteristic
#1 A 12 40
#2 B 11 60
I am attempting to calculate the covariance (or correlation) between the average stem counts of two species. The stem count value averages are in the "avg" column and the species are listed together in the column "Spnum", and they are assigned ID's of 2 and 18. I would like to split out these calculations by Year, Season, and Treatment.
I believe I am getting close using ddply, but I am stuck figuring out how to tell ddply that the values are in a separate column ("avg") than the species that were measured.
row.namesYear Spnum avg Season Treatment
1 1 2005 2 21.8 early delay
2 7 2005 18 18.5 early delay
3 31 2005 2 24.5 early delay
4 37 2005 18 13.2 early delay
5 60 2005 2 20.7 early ambi
6 66 2005 18 31.0 early ambi
7 89 2005 2 36.5 early ambi
...
Here are two options using dplyr and data.table. We group by 'Year', 'Season', 'Treatment' variables and then get the cor of 'avg' that corresponds to 'Spnum' value of 2 againsg the 'Spnum' value of 18 (avg[Spnum==18]).
library(dplyr)
df1 %>%
group_by(Year, Season, Treatment) %>%
summarise(Cor= cor(avg[Spnum==2], avg[Spnum==18]))
Or using data.table, we convert the 'data.frame' to 'data.table' (setDT(df1)). grouped by the variables (as described above), we get the cor.
library(data.table)
setDT(df1)[, list(Cor= cor(avg[Spnum==2], avg[Spnum==18])), by =.(Year, Season, Treatment)]