I have this data.frame with a variable V21 in which many countries are recorded, I want to make it smaller by just specifying the continent rather then all those countries. For example 'Cuba', 'Peru', 'Argentina' rather than being separate levels of V21, I want them to become level 'South America'. Here's the code I tried to use:
recode(WaveOne.test$V21, "levels("Cuba","Colombia","Costa Rica","Argentina","Chile","Ecuador","Peru","Venezuela")= 'South America'")
levels(V21)
Can you suggest what is wrong with my code or maybe a different method?
I am a complete newbie in R and its syntax.
Thank you!
========UPDATE=========
SA_countries <- c("Cuba", "Mexico", "Argentina","Jamaica", "Haiti","West Indies", "Chile", "Ecuador", "Venezuela", "Other South America", "El Salvador", "Guatemala", "Nicaragua", "Dominican Republic", "Panama", "Costa Rica", "Peru")
Asia_countries <- c("Philippines", "Vietnam", "Laos", "Cambodia", "Hmong", "Other Asia", "China", "Hong Kong", "Taiwan", "Japan", "Korea", "India", "Pakistan")
Europe_Canada <- c("Europe/Canada")
MiddleEast_Africa <- c("Middle East/Africa")
continents <- list(`South America`= SA_countries, `Asia` = Asia_countries, `Europe_Canada` = Europe_Canada, `Middle East & Africa` = MiddleEast_Africa)
levels(WaveOne.test$V21) <- c(levels(WaveOne.test$V21), names(continents))
for(i in seq_along(continents)) WaveOne.test$V21[WaveOne.test$V21 %in% continents[[i]]] <- names(continents)[i]
levels(WaveOne.test$V21)
My output however is:
levels(WaveOne.test$V21)
1 "Cuba" "Mexico" "Nicaragua" "Colombia" "Dominican Republic" "El Salvador" "Guatemala"
[8] "Honduras" "Costa Rica" "Panama" "Argentina" "Chile" "Ecuador" "Peru"
[15] "Venezuela" "Other South America" "Haiti" "Jamaica" "West Indies" "Philippines" "Vietnam"
[22] "Laos" "Cambodia" "Hmong" "Other Asia" "China" "Hong Kong" "Taiwan"
[29] "Japan" "Korea" "India" "Pakistan" "Middle East/Africa" "Europe/Canada" "South America"
[36] "Asia" "Europe_Canada" "Middle East & Africa"
You can create a list with all of your countries and continents then reassign the values accordingly:
continents <- list(`South America`=SA_countries,
`North America` = NA_countries,
Europe=Euro_countries)
levels(df$V21) <- c(levels(df$V21), names(continents)) #necessary to add new levels
for(i in seq_along(continents)) {
df$V21[df$V21 %in% continents[[i]]] <- names(continents)[i]}
Reproducible Example
set.seed(123)
SA_countries <- c("Cuba","Colombia","Costa Rica","Argentina","Chile","Ecuador","Peru","Venezuela")
NA_countries <- c("Mexico", "USA", "Canada")
Euro_countries <- c("Germany", "France")
df <- data.frame(V21=sample(c(NA_countries,SA_countries, Europe),20,T))
df
# V21
# 1 Cuba
# 2 Venezuela
# 3 Costa Rica
# 4 Germany
# 5 France
# 6 Mexico
# 7 Argentina
# 8 Germany
# 9 Chile
# 10 Costa Rica
# 11 France
# 12 Costa Rica
# 13 Ecuador
# 14 Chile
# 15 USA
# 16 Germany
# 17 Cuba
# 18 Mexico
# 19 Colombia
# 20 France
continents <- list(`South America`=SA_countries, `North America` = NA_countries, Europe=Euro_countries)
levels(df$V21) <- c(levels(df$V21), names(continents))
for(i in seq_along(continents)) df$V21[df$V21 %in% continents[[i]]] <- names(continents)[i]
df
# V21
# 1 South America
# 2 South America
# 3 South America
# 4 Europe
# 5 Europe
# 6 North America
# 7 South America
# 8 Europe
# 9 South America
# 10 South America
# 11 Europe
# 12 South America
# 13 South America
# 14 South America
# 15 North America
# 16 Europe
# 17 South America
# 18 North America
# 19 South America
# 20 Europe
Related
I have a dataframe with data about the US States.
One of the columns in the df is "Division", which tells the location where each state belongs to ("East North Central", "East South Central", "Middle Atlantic", "Mountain", "New England", "Pacific", "South Atlantic", "West North Central", "West South Central").
I created an array with the average expectancy life for each division, using an existing column called "Life Exp:
avg.life.exp = tapply(df[["Life Exp"]], df$Division, mean, na.rm=TRUE)
Which returns the following:
East North Central East South Central Middle Atlantic
70.99000 69.33750 70.63667
Mountain New England Pacific
70.94750 71.57833 71.69400
South Atlantic West North Central West South Central
69.52625 72.32143 70.43500
Now I would like to add a new column to the df, with the average life expectancy of each Division. So basically I would like to do a Left Join, where if the state belonged to the East Noth Central, it would return 70.99000, and so on.
I need to do this without using packages.
Thank you in advance for any help you can provide!
One option would be to use merge:
merge(df, data.frame(Division = names(avg.life.exp), avg.life.exp), all.x = TRUE)
A second option would be to use match
df$avg.life.exp <- avg.life.exp[match(df$Division, names(avg.life.exp))]
Using the gapminder dataset as example data:
library(gapminder)
# Example data
df <- gapminder[gapminder$year == 2007, c("country", "continent", "lifeExp")]
avg.life.exp <- tapply(df[["lifeExp"]], df$continent, mean, na.rm=TRUE)
avg.life.exp
#> Africa Americas Asia Europe Oceania
#> 54.80604 73.60812 70.72848 77.64860 80.71950
# Using merge
df1 <- merge(df, data.frame(continent = names(avg.life.exp), avg.life.exp), all.x = TRUE)
head(df1)
#> continent country lifeExp avg.life.exp
#> 1 Africa Reunion 76.442 54.80604
#> 2 Africa Eritrea 58.040 54.80604
#> 3 Africa Algeria 72.301 54.80604
#> 4 Africa Congo, Rep. 55.322 54.80604
#> 5 Africa Equatorial Guinea 51.579 54.80604
#> 6 Africa Malawi 48.303 54.80604
# Using match
df$avg.life.exp <- avg.life.exp[match(df$continent, names(avg.life.exp))]
head(df)
#> # A tibble: 6 × 4
#> country continent lifeExp avg.life.exp
#> <fct> <fct> <dbl> <dbl>
#> 1 Afghanistan Asia 43.8 70.7
#> 2 Albania Europe 76.4 77.6
#> 3 Algeria Africa 72.3 54.8
#> 4 Angola Africa 42.7 54.8
#> 5 Argentina Americas 75.3 73.6
#> 6 Australia Oceania 81.2 80.7
I have a large dataset with various currencies. As for visualisation purpose I would like to display the full name of the currencies and not the ISO-4217-Code (currency code).
Let me take the following excerpt of the data:
# dataframe
id = rep(1:8)
country = c("USA", "Canada", "UK", "Switzerland", "USA", "Sweden", "Switzerland", "Canada")
currency_code = c("USD", "CAD", "GBP", "CHF", "USD", "SEK", "CHF", "CAD")
df1 = data.frame(id, country, currency_code)
I could now with a left join manually assigning the currency name to the ISO-4217-Code which would work. But I'm looking personally for a more elegant way to do it than below shown:
country = c("USA", "Canada", "UK", "Switzerland", "Sweden")
currency_name = c("US Dollar", "Canadian Dollar", "Great Britain Pound", "Swiss franc", "Swedish krona")
df2 = data.frame(country, currency_name)
#left join
merge(df1, df2, by = "country", all.x = TRUE)
So my desired data would look like this
country id currency_code currency_name
Canada 2 CAD Canadian Dollar
Canada 8 CAD Canadian Dollar
Sweden 6 SEK Swedish krona
Switzerland 4 CHF Swiss franc
Switzerland 7 CHF Swiss franc
UK 3 GBP Great Britain Pound
USA 1 USD US Dollar
USA 5 USD US Dollar
With the countrycode package it is possible to assign country names to the country code (vice versa). Also it is possible to assign the currency code to a country, which is sadly not what my goal is.
I am aware that a left join would also give me a solution, but I would appreciate it if someone knows a better approach.
We can use the currency_list dataset from currencycode and join the 'currency_code' column with the input dataset 'df1'
library(dplyr)
# remotes::install_github("KKulma/currencycode")
library(currencycode)
data(currency_list)
currency_list %>%
select(currency_code, currency_name) %>%
filter(complete.cases(currency_code)) %>%
right_join(df1)
-output
# currency_code currency_name id country
#1 CAD Canadian Dollar 2 Canada
#2 CAD Canadian Dollar 8 Canada
#3 CHF Swiss Franc 4 Switzerland
#4 CHF Swiss Franc 7 Switzerland
#5 GBP Pound Sterling 3 UK
#6 SEK Swedish Krona 6 Sweden
#7 USD US Dollar 1 USA
#8 USD US Dollar 5 USA
The countrycode package also includes currency names...
library(dplyr)
library(countrycode)
df1 %>%
mutate(currency_name = countrycode(country, "country.name", "currency"))
#> id country currency_code currency_name
#> 1 1 USA USD US Dollar
#> 2 2 Canada CAD Canadian Dollar
#> 3 3 UK GBP Pound Sterling
#> 4 4 Switzerland CHF Swiss Franc
#> 5 5 USA USD US Dollar
#> 6 6 Sweden SEK Swedish Krona
#> 7 7 Switzerland CHF Swiss Franc
#> 8 8 Canada CAD Canadian Dollar
all of countrycode's currency options...
library(countrycode)
country = c("USA", "Canada", "UK", "Switzerland", "USA", "Sweden", "Switzerland", "Canada")
countrycode(country, "country.name", "iso4217c")
# [1] "USD" "CAD" "GBP" "CHF" "USD" "SEK" "CHF" "CAD"
countrycode(country, "country.name", "iso4217n")
# [1] 840 124 826 756 840 752 756 124
countrycode(country, "country.name", "currency")
# [1] "US Dollar" "Canadian Dollar" "Pound Sterling" "Swiss Franc"
# [5] "US Dollar" "Swedish Krona" "Swiss Franc" "Canadian Dollar"
I am using this excel sheet that I have currently read into R: https://www.knomad.org/sites/default/files/2018-04/bilateralmigrationmatrix20170_Apr2018.xlsx
dput(head(remittance, 5))
The output is:
structure(list(`Remittance-receiving country (across) - Remittance-sending country (down)` = c("Australia",
"Brazil", "Canada"), Brazil = c("27.868809286999106", "0", "31.284184411144214"
), Canada = c("46.827693406219382", "1.5806325278762619", "0"
), `Czech Republic` = c("104.79905129342241", "3.0488843262423089",
"176.79676736179096"), Finland = c("26.823089572300752", "1.3451674211686246",
"37.781150857376964"), France = c("424.37048861305249", "123.9763417712491",
"1296.7352242506483"), Germany = c("556.4140279523856", "66.518143815367239",
"809.9621650533453"), Hungary = c("200.08597014449356", "11.953328254521287",
"436.0811601171776"), Indonesia = c("172.0021287331823", "1.3701340430259537",
"33.545925908780198"), Italy = c("733.51652291459231", "116.74264895322995",
"1072.1119887588022"), `Korea, Rep.` = c("259.97044386689589",
"20.467939414361016", "326.94157937864327"), Netherlands = c("133.48932759488602",
"4.7378343766684532", "181.28828076733771"), Philippines = c("1002.3593555086774",
"1.5863355979877207", "2369.5223195675494"), Poland = c("109.73486651698796",
"5.8313637459523129", "341.10408952685464"), `Russian Federation` = c("19.082541158574934",
"1.0136604494838692", "58.760989426089431"), `Saudi Arabia` = c("13.578431465294949",
"0.32506772760873404", "15.511213677040857"), Sweden = c("91.887827513176489",
"5.1132733094740352", "65.860232580192786"), Thailand = c("383.08245004577498",
"2.7410805494977684", "79.370683058792849"), `United Kingdom` = c("1084.0742194994727",
"4.2050614573174592", "568.62605950140266"), `United States` = c("188.06242727403128",
"49.814372612310521", "661.98049661387927"), WORLD = c("5578.0296723604206",
"422.37127035334271", "8563.264510816849")), row.names = c(NA,
-3L), class = c("tbl_df", "tbl", "data.frame"))
I currently have a dataframe of two columns "Source" and "Destination" where each row is a pair of countries which I created by doing:
countries = c("Australia","Brazil", "Canada", "Czech Republic", "Germany", "Finland", "United Kingdom", "Italy", "Poland", "Russian Federation", "Sweden", "United States", "Philippines", "France", "Netherlands", "Hungary", "Saudi Arabia", "Thailand", "Korea, Rep.", "Indonesia")
pairs = t(combn(countries, 2))
I would like to use each pair to extract its corresponding value from the excel sheet above. (In the Excel sheet "Source" is the first column of countries-down and "Destination is the first row countries-across)
For example a sample of the df that I have looks as follows (it currently contains 190 pairs):
pairs = data.frame(Source = c("Australia", "Australia", "Australia"), Destination = c("Brazil", "Canada", "Czech Republic"))
Where the first pair in my df is (Australia, Brazil) which corresponds to a value of 27.868809286999106 from the excel sheet that I reproduced above. Is there a built-in R function that would match the pairs from my df to extract its corresponding value? Thanks
Perhaps what you need is dplyr::pivot_longer?
library(dplyr)
colnames(remittance)[1] <- 'source'
remittance %>% pivot_longer(-source, names_to = 'destination')
#----
# A tibble: 60 x 3
source destination value
<chr> <chr> <chr>
1 Australia Brazil 27.868809286999106
2 Australia Canada 46.827693406219382
3 Australia Czech Republic 104.79905129342241
4 Australia Finland 26.823089572300752
Note remittance is the dataframe in the OP dput.
Probably you are interested in keeping the flexibility of your nice combn approach.
To loop over your pairs data frame (it's actually a matrix though) you may use apply with MARGIN=1 for row-wise. In the FUN= argument we create data frames of one row each with source corresponding to column 1 of pairs and destination to column 2. The distance (or whatever this value is) we get by subsetting at the corresponding rows and columns of remittance (for brevity I shortend to rem).
Since we will get a list of single-line data frames, we want to rbind, and because we have multiple objects we need do.call.
res <- do.call(rbind,
apply(pairs, MARGIN=1, FUN=function(x)
data.frame(source=x[1], destination=x[2],
dist=as.integer(rem[rem[, 1] == x[1], rem[1, ] == x[2]])))
)
Since the .xlsx has zeros where actually should be NAs we should declare them as such in the result.
res[res == 0] <- NA
Result
head(res, 25)
# source destination dist
# 1 Australia Brazil 721
# 2 Australia Canada 24721
# 3 Australia Czech Republic 1074
# 4 Australia Germany 13938
# 5 Australia Finland 1121
# 6 Australia United Kingdom 135000
# 7 Australia Italy 19350
# 8 Australia Poland 974
# 9 Australia Russian Federation 543
# 10 Australia Sweden 3988
# 11 Australia United States 93179
# 12 Australia Philippines 4118
# 13 Australia France 8475
# 14 Australia Netherlands 10697
# 15 Australia Hungary 997
# 16 Australia Saudi Arabia NA
# 17 Australia Thailand 11298
# 18 Australia Korea, Rep. 5381
# 19 Australia Indonesia 11094
# 20 Brazil Canada 26647
# 21 Brazil Czech Republic 742
# 22 Brazil Germany 44000
# 23 Brazil Finland 1378
# 24 Brazil United Kingdom 55772
# 25 Brazil Italy 104779
Data:
u <- "https://www.knomad.org/sites/default/files/2018-04/bilateralmigrationmatrix20170_Apr2018.xlsx"
rem <- openxlsx::read.xlsx(u)
countries <- c("Australia", "Brazil", "Canada", "Czech Republic", "Germany",
"Finland", "United Kingdom", "Italy", "Poland", "Russian Federation",
"Sweden", "United States", "Philippines", "France", "Netherlands",
"Hungary", "Saudi Arabia", "Thailand", "Korea, Rep.", "Indonesia")
pairs <- t(combn(countries, 2))
I have a dataset (survey) and a column of birth_country, where people have written their country of birth. An example of it:
1 america
2 usa
3 american
4 us of a
5 united states
6 england
7 english
8 great britain
9 uk
10 united kingdom
how I would like it to look:
1 america
2 america
3 america
4 america
5 america
6 uk
7 uk
8 uk
9 uk
10 uk
I have tried using str_replace to manually insert the different spellings, to replace them with 'america' but when I look at my dataset, nothing has changed
e.g.
survey <- structure(list(birth_country = c("america", "usa", "american", "us of a", "united states", "england", "english", "great britain", "uk", "united kingdom")), row.names = c(NA, -10L), class = "data.frame")
survey$birth_country <- str_replace(survey$birth_country, ' "united state"|"united statea"|"united states of america"', "america")
thank you in advance
Come up with some patterns that only match for each country and basically loop over what you are already doing (you can change the replacement below with your favorite function)
survey <- structure(list(birth_country = c("america", "usa", "american", "us of a", "united states", "england", "english", "great britain", "uk", "united kingdom")), row.names = c(NA, -10L), class = "data.frame")
## use a _named_ list of regular expressions
## the name will be the replacement string
l <- list(
america = 'amer|us|states',
uk = 'eng|brit|king|uk',
'another country' = 'ano|an co',
chaz = 'chaz|chop'
)
f <- function(x, list) {
for (ii in seq_along(list)) {
x[grepl(list[[ii]], x, ignore.case = TRUE)] <- names(list)[ii]
}
x
}
## test it
f(survey$birth_country, l)
# [1] "america" "america" "america" "america" "america" "uk" "uk" "uk" "uk" "uk"
within(survey, {
clean <- f(birth_country, l)
})
# birth_country clean
# 1 america america
# 2 usa america
# 3 american america
# 4 us of a america
# 5 united states america
# 6 england uk
# 7 english uk
# 8 great britain uk
# 9 uk uk
# 10 united kingdom uk
Note that 1) if you don't give a pattern that matches, nothing will change, but 2) if you give a pattern that matches both countries (e.g., "united"), the first in the list will be used (unless the replacement itself is also matched)
Looks like the problem is in how you specified your regular expression. Try this (updated based on #Gabriella 's comment, and another tidyverse approach, similar to #MarBIo ):
library(tidyverse)
survey <- survey %>%
mutate(birth_country = if_else(
str_detect(birth_country,
"(united state)|(united statea)|(united states of america)"), #If your regular expression matches any in birth_country
"america", #Change it to "america"
birth_country #Otherwise, keep as is.
) #end of if_else
) #end of mutate
Other people are suggesting you come up with a more complex regular expression, which you can certainly do as well. Consecutive "or" (i.e. "|") statements in your regular expression works though.
In case you allow tidyverse`s mutate you can do:
library(tidyverse)
survey <- structure(list(birth_country = c("america", "usa", "american", "us of a", "united states", "england", "english", "great britain", "uk", "united kingdom")), row.names = c(NA, -10L), class = "data.frame")
americas <- c("america", "usa", "american", "us of a", "united states")
englands <- c("england", "english", "great britain")
survey %>%
mutate(birth_country = ifelse(birth_country %in% americas, 'america', 'UK'))
#> birth_country
#> 1 america
#> 2 america
#> 3 america
#> 4 america
#> 5 america
#> 6 UK
#> 7 UK
#> 8 UK
#> 9 UK
#> 10 UK
I have a data frame (df) that lists the countries associated with every site
Site Country
Site1 USA
Site2 Vietnam
Site3 Spain
Site4 Germany
Site5 China
I want to attach a column, where for each country I associate its corresponding continent. I wrote a simple if loop to do this:
df$Continent <- NA
if(df$Country == "USA" |df$Country == "Canada" |df$Country == "Mexico")
{df$Continent <- "North America"}
if(df$Country == "Spain" |df$Country == "France" |df$Country == "Germany")
{df$Continent <- "Europe"}
## .. etc
summary(df)
However, each time I run it the df, I find that it assigns North America to all the countries. I understand that this may sound trivial, but does it make a difference if I use if statments everywhere and not else or if else? Any suggestions for correcting this?
Build a lookup table and merge() it with the data.
For example:
lookup <- data.frame(Country = c("USA", "Canada", "Mexico",
"Spain", "France", "Germany",
"Vietnam", "China"),
Continent = rep(c("North America", "Europe", "Asia"),
times = c(3,3,2)))
Using your snippet of data as data frame df, we can add Continent via merge() (a join in database terminology):
> merge(df, lookup, sort = FALSE, all.x = TRUE)
Country Site Continent
1 USA Site1 North America
2 Vietnam Site2 Asia
3 Spain Site3 Europe
4 Germany Site4 Europe
5 China Site5 Asia
If you're working with a factor you can also do some nonsense with levels, or levels<- to be exact:
`levels<-`(dat$Country, list(
`North America` = c("USA","Canada","Mexico"),
`Europe` = c("Spain","France","Germany"),
`Asia` = c("Vietnam","China")
))
#[1] North America Asia Europe Europe Asia
#Levels: North America Europe Asia
I like ifelse() for things like this. You could use it with the %in% operator like this:
df$Continent <- ifelse(df$Country %in% c("USA", "Canada", "Mexico"),
"North America", df$Continent)
df$Continent <- ifelse(df$Country %in% c("Spain", "France", "Germany"),
"Europe", df$Continent)
df
Site Country Continent
1 Site1 USA North America
2 Site2 Vietnam <NA>
3 Site3 Spain Europe
4 Site4 Germany Europe
5 Site5 China <NA>