I made data scraping like table below, but I can't find solution to clean up this table vith GSUB. Namely I tried code like :
populous_table$Tax_GDP <- gsub("[:punct:]","",populous_table$Tax_GDP )
but this code can't clean brackets [] for number 7 Australia.
Can anyone help me ?
1 Afghanistan 6.4
2 Albania 22.9
3 Algeria 7.7
4 Angola 5.7
5 Argentina 37.2
6 Armenia 22.0
7 Australia 34.3 [2]
8 Austria 43.4
You may use
populous_table$Tax_GDP <- gsub("\\s*\\[\\d+]","", populous_table$Tax_GDP )
Or, if that [digits] substring is always at the end, add $:
populous_table$Tax_GDP <- gsub("\\s*\\[\\d+]$", "", populous_table$Tax_GDP )
The \s*\[\d+] pattern means
\s* - 0+ whitespaces
\[ - a [ char
\d+ - 1+ digits
] - a ] char.
See R demo:
x <- c("1 Afghanistan 6.4", "2 Albania 22.9", "3 Algeria 7.7", "4 Angola 5.7", "5 Argentina 37.2", "Armenia 22.0", "7 Australia 34.3 [2]", "8 Austria 43.4")
gsub("\\s*\\[\\d+]", "", x)
## => [1] "1 Afghanistan 6.4" "2 Albania 22.9" "3 Algeria 7.7"
[4] "4 Angola 5.7" "5 Argentina 37.2" "Armenia 22.0"
[7] "7 Australia 34.3" "8 Austria 43.4"
Related
Is it possible to construct a function, say my_mut(df, condition) such that df is a dataframe, condition is a string describing a mutation, and somewhere in the function, the mutation of df according to condition is used?
For example, if df has a foo column and you run my_mut(df, "foo = 2*foo"), then somewhere within my_mut() there would be a row that produces the same dataframe as df %>% mutate(foo = 2*foo).
I managed to do something similar with filter using eval and parse.
update_filt <- function(df,
filt,
col){
sub <- df %>%
filter(eval(parse(text = filt))) %>%
mutate("{{col}}" := 2*{{ col }})
remain <- df %>%
filter(eval(parse(
text = paste0("!(",filt,")")
))
)
return(rbind(sub, remain))
}
I am not sure the update_filt function is faultproof, but it works in some cases at least, e.g., library(gapminder) date_filt(gapminder, "year == 1952", pop) returns the expected outcome.
The same trick does not seem to work with mutate though. For example,
update_mut <- function(df, mutation){
# Evaluate mutation expression
df %>% mutate(eval(parse(text = mutation))
}
produces outcomes like
library(gapminder)
update_mut(gapminder, "year = 2*year")
# A tibble: 1,704 × 7
country continent year lifeExp pop gdpPercap `eval(parse(text = mutation))`
<fct> <fct> <int> <dbl> <int> <dbl> <dbl>
1 Afghanistan Asia 1952 28.8 8425333 779. 3904
2 Afghanistan Asia 1957 30.3 9240934 821. 3914
3 Afghanistan Asia 1962 32.0 10267083 853. 3924
4 Afghanistan Asia 1967 34.0 11537966 836. 3934
5 Afghanistan Asia 1972 36.1 13079460 740. 3944
6 Afghanistan Asia 1977 38.4 14880372 786. 3954
7 Afghanistan Asia 1982 39.9 12881816 978. 3964
8 Afghanistan Asia 1987 40.8 13867957 852. 3974
9 Afghanistan Asia 1992 41.7 16317921 649. 3984
10 Afghanistan Asia 1997 41.8 22227415 635. 3994
# … with 1,694 more rows
Instead of the expected
gapminder %>% mutate(year = 2*year)
# A tibble: 1,704 × 6
country continent year lifeExp pop gdpPercap
<fct> <fct> <dbl> <dbl> <int> <dbl>
1 Afghanistan Asia 3904 28.8 8425333 779.
2 Afghanistan Asia 3914 30.3 9240934 821.
3 Afghanistan Asia 3924 32.0 10267083 853.
4 Afghanistan Asia 3934 34.0 11537966 836.
5 Afghanistan Asia 3944 36.1 13079460 740.
6 Afghanistan Asia 3954 38.4 14880372 786.
7 Afghanistan Asia 3964 39.9 12881816 978.
8 Afghanistan Asia 3974 40.8 13867957 852.
9 Afghanistan Asia 3984 41.7 16317921 649.
10 Afghanistan Asia 3994 41.8 22227415 635.
# … with 1,694 more rows
library(dplyr, warn.conflicts = FALSE)
my_mut <- function(df, df_filter, ...){
df %>%
filter({{ df_filter }}) %>%
mutate(newvar = 'other function stuff',
...)
}
example_df <- data.frame(a = c('zebra', 'some value'),
b = 1:2)
example_df %>%
my_mut(df_filter = a == 'some value',
b = b*5)
#> a b newvar
#> 1 some value 10 other function stuff
Created on 2021-11-11 by the reprex package (v2.0.1)
If you can't use ... because you're already using it in the function for something else, you could wrap the mutation argument in tibble when calling the function.
library(dplyr, warn.conflicts = FALSE)
my_mut <- function(df, df_filter, mutation){
df %>%
filter({{ df_filter }}) %>%
mutate(newvar = 'other function stuff',
{{ mutation }})
}
example_df <- data.frame(a = c('zebra', 'some value'),
b = 1:2)
example_df %>%
my_mut(df_filter = a == 'some value',
mutation = tibble(b = b*5))
#> a b newvar
#> 1 some value 10 other function stuff
Created on 2021-11-11 by the reprex package (v2.0.1)
If your formula is always like origianl = do_something_original(), this may helps.(for dplyr version >= 1.0)
library(dplyr)
library(stringr)
update_mut <- function(df, mutation){
xx <- word(mutation, 1)
df %>%
mutate("{xx}" := eval(parse(text = mutation)))
}
update_mut(gapminder, "year = 2*year")
country continent year lifeExp pop gdpPercap
<fct> <fct> <dbl> <dbl> <int> <dbl>
1 Afghanistan Asia 3904 28.8 8425333 779.
2 Afghanistan Asia 3914 30.3 9240934 821.
3 Afghanistan Asia 3924 32.0 10267083 853.
4 Afghanistan Asia 3934 34.0 11537966 836.
5 Afghanistan Asia 3944 36.1 13079460 740.
6 Afghanistan Asia 3954 38.4 14880372 786.
7 Afghanistan Asia 3964 39.9 12881816 978.
8 Afghanistan Asia 3974 40.8 13867957 852.
9 Afghanistan Asia 3984 41.7 16317921 649.
10 Afghanistan Asia 3994 41.8 22227415 635.
The problem is that mutate doesn't understand the assignment, because all the syntax is evaluated inside eval. So mutate simply thinks this is a nameless expression and assigns as its name the whole text of the expression.
One way to circumvent this would be to eval the whole thing, including the mutate verb, as below.
update_mut <- function(df, mutation) {
# Evaluate the mutation expression
eval(parse(text = paste0("mutate(df, ", mutation, ")")))
}
Another way would be, inside the update_mut function, to split the mutation parameter by the = character, therefore obtaining the name of the variable and the expressions. Therefore you could use a dynamic variable assingment in mutate. However this would only be more to do, since the above code simply solves the problem.
I want to create a factor variables in my dataframes based on categorical variables.
My data:
# A tibble: 159 x 3
name.country gpd rate_suicide
<chr> <dbl> <dbl>
1 Afghanistan 2129. 6.4
2 Albania 12003. 5.6
3 Algeria 11624. 3.3
4 Angola 7103. 8.9
5 Antigua and Barbuda 19919. 0.5
6 Argentina 20308. 9.1
7 Armenia 10704. 5.7
8 Australia 47350. 11.7
9 Austria 52633. 11.4
10 Azerbaijan 14371. 2.6
# ... with 149 more rows
I want to create factor variable region, which contains a factors as:
region <- c('Asian', 'Europe', 'South America', 'North America', 'Africa')
region = factor(region, levels = c('Asian', 'Europe', 'South America', 'North America', 'Africa'))
I want to do this with dplyr packages, that can to choose a factor levels depends on name.countrybut it doesn't work. Example:
if (new_data$name.country[new_data$name.country == "N"]) {
mutate(new_data, region_ = region[1])
}
How i can solve the problem?
I think the way I would think about your problem is
Create a reproducible problem. (see How to make a great R reproducible example. ) Since you already have the data, use dput to make it easier for people like me to recreate your data in their environment.
dput(yourdf)
structure(list(name.country = c("Afghanistan", "Albania", "Algeria"
), gpd = c(2129L, 12003L, 11624L), rate_suicide = c(6.4, 5.6,
3.3)), class = "data.frame", row.names = c(NA, -3L))
raw_data<-structure(list(name.country = c("Afghanistan", "Albania", "Algeria"
), gpd = c(2129L, 12003L, 11624L), rate_suicide = c(6.4, 5.6,
3.3)), class = "data.frame", row.names = c(NA, -3L))
Define vectors that specify your regions
Use case_when to separate countries into regions
Use as.factor to convert your character variable to a factor
asia=c("Afghanistan","India","...","Rest of countries in Asia")
europe=c("Albania","France","...","Rest of countries in Europe")
africa=c("Algeria","Egypt","...","Rest of countries in Africa")
df<-raw_data %>%
mutate(region=case_when(
name.country %in% asia ~ "asia",
name.country %in% europe ~ "europe",
name.country %in% africa ~ "africa",
TRUE ~ "other"
)) %>%
mutate(region=region %>% as.factor())
You can check that your variable region is a factor using str
str(df)
'data.frame': 3 obs. of 4 variables:
$ name.country: chr "Afghanistan" "Albania" "Algeria"
$ gpd : int 2129 12003 11624
$ rate_suicide: num 6.4 5.6 3.3
$ region : Factor w/ 3 levels "africa","asia",..: 2 3 1
Here is a working example that combines data from the question with a file of countries and region information from Github. H/T to Luke Duncalfe for maintaining the region data, which is:
...a combination of the Wikipedia ISO-3166 article for alpha and numeric country codes and the UN Statistics site for countries' regional and sub-regional codes.
regionFile <- "https://raw.githubusercontent.com/lukes/ISO-3166-Countries-with-Regional-Codes/master/all/all.csv"
regionData <- read.csv(regionFile,header=TRUE)
textFile <- "rowID|country|gdp|suicideRate
1|Afghanistan|2129.|6.4
2|Albania|12003.|5.6
3|Algeria|11624.|3.3
4|Angola|7103.|8.9
5|Antigua and Barbuda|19919.|0.5
6|Argentina|20308.|9.1
7|Armenia|10704.|5.7
8|Australia|47350.|11.7
9|Austria|52633.|11.4
10|Azerbaijan|14371.|2.6"
data <- read.csv(text=textFile,sep="|")
library(dplyr)
data %>%
left_join(.,regionData,by = c("country" = "name"))
...and the output:
rowID country gdp suicideRate alpha.2 alpha.3 country.code
1 1 Afghanistan 2129 6.4 AF AFG 4
2 2 Albania 12003 5.6 AL ALB 8
3 3 Algeria 11624 3.3 DZ DZA 12
4 4 Angola 7103 8.9 AO AGO 24
5 5 Antigua and Barbuda 19919 0.5 AG ATG 28
6 6 Argentina 20308 9.1 AR ARG 32
7 7 Armenia 10704 5.7 AM ARM 51
8 8 Australia 47350 11.7 AU AUS 36
9 9 Austria 52633 11.4 AT AUT 40
10 10 Azerbaijan 14371 2.6 AZ AZE 31
iso_3166.2 region sub.region intermediate.region
1 ISO 3166-2:AF Asia Southern Asia
2 ISO 3166-2:AL Europe Southern Europe
3 ISO 3166-2:DZ Africa Northern Africa
4 ISO 3166-2:AO Africa Sub-Saharan Africa Middle Africa
5 ISO 3166-2:AG Americas Latin America and the Caribbean Caribbean
6 ISO 3166-2:AR Americas Latin America and the Caribbean South America
7 ISO 3166-2:AM Asia Western Asia
8 ISO 3166-2:AU Oceania Australia and New Zealand
9 ISO 3166-2:AT Europe Western Europe
10 ISO 3166-2:AZ Asia Western Asia
region.code sub.region.code intermediate.region.code
1 142 34 NA
2 150 39 NA
3 2 15 NA
4 2 202 17
5 19 419 29
6 19 419 5
7 142 145 NA
8 9 53 NA
9 150 155 NA
10 142 145 NA
At this point one can decide whether to use the region, sub region, or intermediate region and convert it to a factor.
We can set region to a factor by adding a mutate() function to the dplyr pipeline:
data %>%
left_join(.,regionData,by = c("country" = "name")) %>%
mutate(region = factor(region)) -> mergedData
At this point mergedData$region is a factor.
str(mergedData$region)
table(mergedData$region)
> str(mergedData$region)
Factor w/ 5 levels "Africa","Americas",..: 3 4 1 1 2 2 3 5 4 3
> table(mergedData$region)
Africa Americas Asia Europe Oceania
2 2 3 2 1
Now the data is ready for further analysis. We will generate a table of average suicide rates by region.
library(knitr) # for kable
mergedData %>% group_by(region) %>%
summarise(suicideRate = mean(suicideRate)) %>%
kable(.)
...and the output:
|region | suicideRate|
|:--------|-----------:|
|Africa | 6.1|
|Americas | 4.8|
|Asia | 4.9|
|Europe | 8.5|
|Oceania | 11.7|
When rendered in an HTML / markdown viewer, the result looks like this:
This question already has answers here:
In R, what does a negative index do?
(3 answers)
Closed 9 years ago.
I am very new to R and at times get stuck with the codes. I came across one of this code as below. What does -7 mean in the code below?
round(cor(longley[,-7]),3)
I understand:
round for rounding,
longley as data.frame,
3: digits for rounding, but not the -7.
In the context [, -7] it means drop the 7th column from the data frame longley (or take all columns but the 7th from longley).
This is R 101 and you'd do well to read some introductory material. For example, this is covered very early on in the An Introduction to R manual that comes with R or is accessible from the R website. Or you could read ?Extract.
Here is an example
> head(longley)
GNP.deflator GNP Unemployed Armed.Forces Population Year Employed
1947 83.0 234.289 235.6 159.0 107.608 1947 60.323
1948 88.5 259.426 232.5 145.6 108.632 1948 61.122
1949 88.2 258.054 368.2 161.6 109.773 1949 60.171
1950 89.5 284.599 335.1 165.0 110.929 1950 61.187
1951 96.2 328.975 209.9 309.9 112.075 1951 63.221
1952 98.1 346.999 193.2 359.4 113.270 1952 63.639
> names(longley)
[1] "GNP.deflator" "GNP" "Unemployed" "Armed.Forces" "Population"
[6] "Year" "Employed"
> names(longley)[7]
[1] "Employed"
> head(longley[, -7])
GNP.deflator GNP Unemployed Armed.Forces Population Year
1947 83.0 234.289 235.6 159.0 107.608 1947
1948 88.5 259.426 232.5 145.6 108.632 1948
1949 88.2 258.054 368.2 161.6 109.773 1949
1950 89.5 284.599 335.1 165.0 110.929 1950
1951 96.2 328.975 209.9 309.9 112.075 1951
1952 98.1 346.999 193.2 359.4 113.270 1952
The command longley[,-7] means: All columns from longley except the 7th. This is called negative indexing.
Have a look at ?Extract for further information.
Here is first 4 rows of my data;
X...Country.Name Country.Code Indicator.Name
1 Turkey TUR Inflation, GDP deflator (annual %)
2 Turkey TUR Unemployment, total (% of total labor force)
3 Afghanistan AFG Inflation, GDP deflator (annual %)
4 Afghanistan AFG Unemployment, total (% of total labor force)
Indicator.Code X2010
1 NY.GDP.DEFL.KD.ZG 5.675740
2 SL.UEM.TOTL.ZS 11.900000
3 NY.GDP.DEFL.KD.ZG 9.437322
4 SL.UEM.TOTL.ZS NA
I want my data reshaped into two colums, one of each Indicator code, and I want each row correspond to a country, something like this;
Country Name NY.GDP.DEFL.KD.ZG SL.UEM.TOTL.ZS
Turkey 5.6 11.9
Afghanistan 9.43 NA
I think I could do this with Excel, but I want to learn the R way, so that I don't need to rely on excel everytime I have a problem. Here is dput of data if you need it.
Edit: I actually want 3 colums, one for each indicator and one for the country's name.
Sticking with base R, use reshape. I took the liberty of cleaning up the column names. Here, I'm only showing you a few rows of the output. Remove head to see the full output. This assumes your data.frame is named "mydata".
names(mydata) <- c("CountryName", "CountryCode",
"IndicatorName", "IndicatorCode", "X2010")
head(reshape(mydata[-c(2:3)],
direction = "wide",
idvar = "CountryName",
timevar = "IndicatorCode"))
# CountryName X2010.NY.GDP.DEFL.KD.ZG X2010.SL.UEM.TOTL.ZS
# 1 Turkey 5.675740 11.9
# 3 Afghanistan 9.437322 NA
# 5 Albania 3.459343 NA
# 7 Algeria 16.245617 11.4
# 9 American Samoa NA NA
# 11 Andorra NA NA
Another option in base R is xtabs, but NA gets replaced with 0:
head(xtabs(X2010 ~ CountryName + IndicatorCode, mydata))
# IndicatorCode
# CountryName NY.GDP.DEFL.KD.ZG SL.UEM.TOTL.ZS
# Afghanistan 9.437322 0.0
# Albania 3.459343 0.0
# Algeria 16.245617 11.4
# American Samoa 0.000000 0.0
# Andorra 0.000000 0.0
# Angola 22.393924 0.0
The result of xtabs is a matrix, so if you want a data.frame, wrap the output with as.data.frame.matrix.
I have a balanced panel by country from 1951 to 2007 in a data frame. I'd like to transform it into a new data frame of five year averages of my other variables. When I sat down to do this I realized the only way I could think to do this involved a for loop and then decided that it was time to come to stackoverflow for help.
So, is there an easy way to turn data that looks like this:
country country.isocode year POP ci grgdpch
Argentina ARG 1951 17517.34 18.445022145 3.4602044759
Argentina ARG 1952 17876.96 17.76066507 -7.887407586
Argentina ARG 1953 18230.82 18.365255769 2.3118720688
Argentina ARG 1954 18580.56 16.982113434 1.5693778844
Argentina ARG 1955 18927.82 17.488907008 5.3690276523
Argentina ARG 1956 19271.51 15.907756547 0.3125559183
Argentina ARG 1957 19610.54 17.028450999 2.4896639667
Argentina ARG 1958 19946.54 17.541597134 5.0025894968
Argentina ARG 1959 20281.15 16.137310492 -6.763501447
Argentina ARG 1960 20616.01 20.519539628 8.481742144
...
Venezuela VEN 1997 22361.80 21.923577413 5.603872759
Venezuela VEN 1998 22751.36 24.451736863 -0.781844721
Venezuela VEN 1999 23128.64 21.585034168 -8.728234466
Venezuela VEN 2000 23492.75 20.224310777 2.6828641218
Venezuela VEN 2001 23843.87 23.480311721 0.2476965412
Venezuela VEN 2002 24191.77 16.290691319 -8.02535946
Venezuela VEN 2003 24545.43 10.972153646 -8.341989049
Venezuela VEN 2004 24904.62 17.147693312 14.644028806
Venezuela VEN 2005 25269.18 18.805970212 7.3156977879
Venezuela VEN 2006 25641.46 22.191098769 5.2737381326
Venezuela VEN 2007 26023.53 26.518210052 4.1367897561
into something like this:
country country.isocode period AvPOP Avci Avgrgdpch
Argentina ARG 1 18230 17.38474 1.423454
...
Venezuela VEN 12 25274 21.45343 5.454334
Do I need to transform this data frame using a specific panel data package? Or is there another easy way to do this that I'm missing?
This is the stuff aggregate is made for. :
Df <- data.frame(
year=rep(1951:1970,2),
country=rep(c("Arg","Ven"),each=20),
var1 = c(1:20,51:70),
var2 = c(20:1,70:51)
)
Level <-cut(Df$year,seq(1951,1971,by=5),right=F)
id <- c("var1","var2")
> aggregate(Df[id],list(Df$country,Level),mean)
Group.1 Group.2 var1 var2
1 Arg [1951,1956) 3 18
2 Ven [1951,1956) 53 68
3 Arg [1956,1961) 8 13
4 Ven [1956,1961) 58 63
5 Arg [1961,1966) 13 8
6 Ven [1961,1966) 63 58
7 Arg [1966,1971) 18 3
8 Ven [1966,1971) 68 53
The only thing you might want to do, is to rename the categories and the variable names.
For this type of problem, the plyr package is truely phenomenal. Here is some code that gives you what you want in essentially a single line of code plus a small helper function.
library(plyr)
library(zoo)
library(pwt)
# First recreate dataset, using package pwt
data(pwt6.3)
pwt <- pwt6.3[
pwt6.3$country %in% c("Argentina", "Venezuela"),
c("country", "isocode", "year", "pop", "ci", "rgdpch")
]
# Use rollmean() in zoo as basis for defining a rolling 5-period rolling mean
rollmean5 <- function(x){
rollmean(x, 5)
}
# Use ddply() in plyr package to create rolling average per country
pwt.ma <- ddply(pwt, .(country), numcolwise(rollmean5))
Here is the output from this:
> head(pwt, 10)
country isocode year pop ci rgdpch
ARG-1950 Argentina ARG 1950 17150.34 13.29214 7736.338
ARG-1951 Argentina ARG 1951 17517.34 18.44502 8004.031
ARG-1952 Argentina ARG 1952 17876.96 17.76067 7372.721
ARG-1953 Argentina ARG 1953 18230.82 18.36526 7543.169
ARG-1954 Argentina ARG 1954 18580.56 16.98211 7661.550
ARG-1955 Argentina ARG 1955 18927.82 17.48891 8072.900
ARG-1956 Argentina ARG 1956 19271.51 15.90776 8098.133
ARG-1957 Argentina ARG 1957 19610.54 17.02845 8299.749
ARG-1958 Argentina ARG 1958 19946.54 17.54160 8714.951
ARG-1959 Argentina ARG 1959 20281.15 16.13731 8125.515
> head(pwt.ma)
country year pop ci rgdpch
1 Argentina 1952 17871.20 16.96904 7663.562
2 Argentina 1953 18226.70 17.80839 7730.874
3 Argentina 1954 18577.53 17.30094 7749.694
4 Argentina 1955 18924.25 17.15450 7935.100
5 Argentina 1956 19267.39 16.98977 8169.456
6 Argentina 1957 19607.51 16.82080 8262.250
Note that rollmean(), by default, calculates the centred moving mean. You can modify this behaviour to get the left or right moving mean by passing this parameter to the helper function.
EDIT:
#Joris Meys gently pointed out that you might in fact be after the average for five-year periods.
Here is the modified code to do this:
pwt$period <- cut(pwt$year, seq(1900, 2100, 5))
pwt.ma <- ddply(pwt, .(country, period), numcolwise(mean))
pwt.ma
And the output:
> pwt.ma
country period year pop ci rgdpch
1 Argentina (1945,1950] 1950.0 17150.336 13.29214 7736.338
2 Argentina (1950,1955] 1953.0 18226.699 17.80839 7730.874
3 Argentina (1955,1960] 1958.0 19945.149 17.42693 8410.610
4 Argentina (1960,1965] 1963.0 21616.623 19.09067 9000.918
5 Argentina (1965,1970] 1968.0 23273.736 18.89005 10202.665
6 Argentina (1970,1975] 1973.0 25216.339 19.70203 11348.321
7 Argentina (1975,1980] 1978.0 27445.430 23.34439 11907.939
8 Argentina (1980,1985] 1983.0 29774.778 17.58909 10987.538
9 Argentina (1985,1990] 1988.0 32095.227 15.17531 10313.375
10 Argentina (1990,1995] 1993.0 34399.829 17.96758 11221.807
11 Argentina (1995,2000] 1998.0 36512.422 19.03551 12652.849
12 Argentina (2000,2005] 2003.0 38390.719 15.22084 12308.493
13 Argentina (2005,2010] 2006.5 39831.625 21.11783 14885.227
14 Venezuela (1945,1950] 1950.0 5009.006 41.07972 7067.947
15 Venezuela (1950,1955] 1953.0 5684.009 44.60849 8132.041
16 Venezuela (1955,1960] 1958.0 6988.078 37.87946 9468.001
17 Venezuela (1960,1965] 1963.0 8451.073 26.93877 9958.935
18 Venezuela (1965,1970] 1968.0 10056.910 28.66512 11083.242
19 Venezuela (1970,1975] 1973.0 11903.185 32.02671 12862.966
20 Venezuela (1975,1980] 1978.0 13927.882 36.35687 13530.556
21 Venezuela (1980,1985] 1983.0 16082.694 22.21093 10762.718
22 Venezuela (1985,1990] 1988.0 18382.964 19.48447 10376.123
23 Venezuela (1990,1995] 1993.0 20680.645 19.82371 10988.096
24 Venezuela (1995,2000] 1998.0 22739.062 20.93509 10837.580
25 Venezuela (2000,2005] 2003.0 24550.973 17.33936 10085.322
26 Venezuela (2005,2010] 2006.5 25832.495 24.35465 11790.497
Use cut on your year variable to make the period variable, then use melt and cast from the reshape package to get the averages. There's a lot of other answers that can show you how; see https://stackoverflow.com/questions/tagged/r+reshape
There is a base stats and a plyr answer, so for completeness, here is a dplyr based answer. Using the toy data given by Joris, we have
Df <- data.frame(
year=rep(1951:1970,2),
country=rep(c("Arg","Ven"),each=20),
var1 = c(1:20,51:70),
var2 = c(20:1,70:51)
)
Now, using cut to create the periods, we can then group on them and get the means:
Df %>% mutate(period = cut(Df$year,seq(1951,1971,by=5),right=F)) %>%
group_by(country, period) %>% summarise(V1 = mean(var1), V2 = mean(var2))
Source: local data frame [8 x 4]
Groups: country
country period V1 V2
1 Arg [1951,1956) 3 18
2 Arg [1956,1961) 8 13
3 Arg [1961,1966) 13 8
4 Arg [1966,1971) 18 3
5 Ven [1951,1956) 53 68
6 Ven [1956,1961) 58 63
7 Ven [1961,1966) 63 58
8 Ven [1966,1971) 68 53