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
Related
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))))
I have data similar to this Sample Data:
Cities Country Date Cases
1 BE A 2/12/20 12
2 BD A 2/12/20 244
3 BF A 2/12/20 1
4 V 2/12/20 13
5 Q 2/13/20 2
6 D 2/14/20 4
7 GH N 2/15/20 6
8 DA N 2/15/20 624
9 AG J 2/15/20 204
10 FS U 2/16/20 433
11 FR U 2/16/20 38
I want to organize the data by on the date and country and then sum a country's daily case. However, I try something like, it reveal the total sum:
my_data %>%
group_by(Country, Date)%>%
summarize(Cases=sum(Cases))
Your summarize function is likely being called from another package (plyr?). Try calling dplyr::sumarize like this:
my_data %>%
group_by(Country, Date)%>%
dplyr::summarize(Cases=sum(Cases))
# A tibble: 7 x 3
# Groups: Country [7]
Country Date Cases
<fct> <fct> <int>
1 A 2/12/20 257
2 D 2/14/20 4
3 J 2/15/20 204
4 N 2/15/20 630
5 Q 2/13/20 2
6 U 2/16/20 471
7 V 2/12/20 13
I sympathize with you that this is can be very frustrating. I have gotten into a habit of always using dplyr::select, dplyr::filter and dplyr::summarize. Otherwise you spend needless time frustrated about why your code isn't working.
We can also use aggregate
aggregate(Cases ~ Country + Date, my_data, sum)
I am having difficulty importing my data in the way I would like to from a .csv file to tidydata.
My data set is made up of descriptive data (age, country, etc.) and then 15 condition columns that I would like to have in just one column (long format). I have previously tried 'melting' the data in a few ways, but it does not turn out the way I intended it to. These are a few things I have tried, I know it is kind of messy. There are quite a few NAs in the data, which seem to be causing an issue. I am trying to create this specific column "Vignette" which will serve as the collective column for the 15 vignette columns I would like in long format.
head(dat)
ID Frequency Gender Country Continent Age
1 5129615189 At least weekly female France Europe 30-50 years
2 5128877943 At least daily female Spain Europe > 50 years
3 5126775994 At least weekly female Spain Europe 30-50 years
4 5126598863 At least daily male Albania Europe 30-50 years
5 5124909744 At least daily female Ireland Europe > 50 years
6 5122047758 At least weekly female Denmark Europe 30-50 years
Practice Specialty Seniority AMS
1 University public hospital centre Infectious diseases 6-10 years Yes
2 Other public hospital Infectious diseases > 10 years Yes
3 University public hospital centre Intensive care > 10 years Yes
4 University public hospital centre Infectious diseases > 10 years No
5 Private hospial/clinic Clinical microbiology > 10 years Yes
6 University public hospital centre Infectious diseases 0-5 years Yes
Durations V01 V02 V03 V04 V05 V06 V07 V08 V09 V10 V11 V12 V13 V14 V15
1 range 7 2 7 7 7 5 7 14 7 42 42 90 7 NA 5
2 range 7 10 10 5 14 5 7 14 10 42 21 42 14 14 14
3 range 7 5 5 7 14 5 5 13 10 42 42 42 5 0 7
4 range 10 7 7 5 7 10 7 5 7 28 14 42 10 10 7
5 range 7 5 7 7 14 7 7 14 10 42 42 90 10 0 7
6 fixed duration 7 3 3 7 10 10 7 14 7 90 90 90 10 7 7
dat_long %>%
gather(Days, Age, -Vignette)
dat$new_sp = NULL
names(dat) <- gsub("new_sp", "", names(dat))
dat_tidy<-melt(
data=dat,
id=0:180,
variable.name="Vignette",
value.name="Days",
na.rm=TRUE
)
dat_tidy<- mutate(dat_tidy,
Days= sub("^V", "", Days)
)
It keeps saying "Error: id variables not found in data: NA"
I have tried to get rid of NAs but it doesn't seem to do anything.
I am guessing you are loading the melt function from reshape2. I will recommend that you try tidyr which is basically the next generation of reshape2.
Your error is presumable that the argument id=0:180. This is basically asking it to keep columns 0-180 as "identifier" columns, and melt the rest (i.e. create a new row for each value in each column).
When you subset more column indices than columns in a data.frame, the non-existing columns are filled in with pure NA - you asked for them, so you get them!
I would recommend loading tidyr, as it is newer. There should be some new verbs in the package that are more intuitive, but I'll give you a solution with the older semantic:
library(tidyr)
dat_tidy <- dat %>% gather('Vignette', 'Days', starts_with('V'))
# or a bit more verbose
dat_tidy <- dat %>% gather('Vignette', 'Days', V01, V02, V03, V04)
And check out the comment #heck1 for asking even better questions.
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 have this data.frame in R.
> a <- data.frame(year = c(2001,2001,2001,2001), country = c("Japan", "Japan","US","US"), type = c("a","b","a","b"), amount = c(35,67,39,45))
> a
year country type amount
1 2001 Japan a 35
2 2001 Japan b 67
3 2001 US a 39
4 2001 US b 45
How should I transform this into a data.frame that looks like this?
year country type.a type.b
1 2001 Japan 35 67
2 2001 US 39 45
Basically I want the number of rows to be the number of (year x country) pairs, and I want to create additional columns for each type.
base solution, but requires renaming columns and rows
reshape(a, v.names="amount", timevar="type", idvar="country", direction="wide")
year country amount.a amount.b
1 2001 Japan 35 67
3 2001 US 39 45
reshape2 solution
library(reshape2)
dcast(a, year+country ~ paste("type", type, sep="."), value.var="amount")
year country type.a type.b
1 2001 Japan 35 67
2 2001 US 39 45
Another way would be to use spread in the tidyr package and rename in the dplyr package to deliver the expected outcome.
library(dplyr)
library(tidyr)
spread(a,type, amount) %>%
rename(type.a = a, type.b = b)
# year country type.a type.b
#1 2001 Japan 35 67
#2 2001 US 39 45