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zom$country.code is int.
zom$Country.Code<- c(1,14,30,37,94,148,162,166,184,189,191,208,214,215,216)

r <-c(India,Australia,Brazil,Canada,Indonesia,NewZealand,Phillipines,Qatar,Singapore,southAfrica,SriLanka,Turkey,UAE,UnitedKingdom,UnitedStates)

I want a output like:

zom$Country.Code <- c(India,Australia,Brazil,Canada,Indonesia,NewZealand,Phillipines,Qatar,Singapore,southAfrica,SriLanka,Turkey,UAE,UnitedKingdom,UnitedStates)

How can I solve this problem in R.

vidhi amin
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    Welcome to SO. Users here expect you to (1) have demonstrated research, (2) present a code attempt including, (3) sample data, and (4) your expected output. Please review [how to ask](https://stackoverflow.com/help/how-to-ask) questions and how to provide a [minimal reproducible example/attempt](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example). – Maurits Evers Apr 21 '18 at 01:30
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    In particular, what do you mean by assign, what is the logic for deciding what country gets what number, what is the format your data is in, how many numbers there are – Calum You Apr 21 '18 at 01:32
  • Is `data$e` supposed to be a vector of row-indices into `r`, or factor levels, or what? Please explain. Also, use `str()` and `dput()` on your dataframes to understand their structure, then tell us what we need to know to understand your issue. – smci Apr 21 '18 at 01:41
  • data$e is an int type. – vidhi amin Apr 21 '18 at 01:49
  • @vidhiamin - The vector assignment in the last line of code in your 5th edit generates an error in R because the country name strings should be within quotes, as I illustrate in my updated answer to your question. – Len Greski Apr 21 '18 at 08:03

1 Answers1

1

The factor() function can be used to associate a vector of numbers with a set of labels. For example:

x <- c(1,1,1,2,3,3,2,3,4,4)

theLabels <- c("India","Canada","United States","Mexico")

y <- factor(x,1:4,theLabels)
y

produces the following output:

> y <- factor(x,1:4,theLabels)
> y
 [1] India         India         India         Canada        United States
 [6] United States Canada        United States Mexico        Mexico

Levels: India Canada United States Mexico

To demonstrate that this answer works with the data provided in the fifth edit of the OP:

r <-c("India","Australia","Brazil","Canada","Indonesia","NewZealand",
      "Phillipines","Qatar","Singapore","southAfrica","SriLanka","Turkey","UAE","UnitedKingdom","UnitedStates")
zom<- data.frame(Country.Code=c(1,14,30,37,94,148,162,166,184,189,191,208,214,215,216))

zom$Country.Code <- factor(zom$Country.Code,
                           levels = c(1,14,30,37,94,148,162,166,184,189,191,208,214,215,216),
                           labels = r)

zom$Country.Code

...and the output:

> zom$Country.Code
 [1] India         Australia     Brazil        Canada        Indonesia     NewZealand    Phillipines   Qatar        
 [9] Singapore     southAfrica   SriLanka      Turkey        UAE           UnitedKingdom UnitedStates 
15 Levels: India Australia Brazil Canada Indonesia NewZealand Phillipines Qatar Singapore southAfrica SriLanka Turkey ... UnitedStates

NOTE: Once the original codes are converted to a factor, the underlying codes are lost because a side effect of factor is that the factor levels become an ordered list from 1 to the number of unique labels associated with the factor.

An alternative to the factor() approach is to create a lookup table of country names and codes, and to merge this with the original data. This approach preserves the original values of Country.Code.

To illustrate, we'll create a data frame containing multiple rows of Country.Code from the OP, and merge it with a lookup table via dplyr::inner_join(). We'll then generate a cross-tab of Country.Name and Country.Code to illustrate accuracy of the join process. 

library(dplyr)
# first, build a data frame containg multiple rows with same country code
zom<- data.frame(Country.Code=c(1,14,30,37,94,148,162,166,184,189,191,208,214,215,216,
                                1,14,30,37,94,148,162,166,184,189,191,208,214,215,216,
                                1,14,30,37,94,148,162,166,184,189,191,208,214,215,216))
# second, create lookup table of codes and names, one row per country
countryNames <- data.frame(Country.Code=c(1,14,30,37,94,148,162,166,184,189,191,208,214,215,216),
                           Country.Name= c("India","Australia","Brazil","Canada","Indonesia","NewZealand",
                                           "Phillipines","Qatar","Singapore","southAfrica","SriLanka","Turkey","UAE","UnitedKingdom","UnitedStates"),
     stringsAsFactors=FALSE)

# use dplyr::inner_join() to join country names 
mergedData <- zom %>% inner_join(countryNames)
table(mergedData$Country.Name,mergedData$Country.Code)

...and the output:

> table(mergedData$Country.Name,mergedData$Country.Code)

                1 14 30 37 94 148 162 166 184 189 191 208 214 215 216
  Australia     0  3  0  0  0   0   0   0   0   0   0   0   0   0   0
  Brazil        0  0  3  0  0   0   0   0   0   0   0   0   0   0   0
  Canada        0  0  0  3  0   0   0   0   0   0   0   0   0   0   0
  India         3  0  0  0  0   0   0   0   0   0   0   0   0   0   0
  Indonesia     0  0  0  0  3   0   0   0   0   0   0   0   0   0   0
  NewZealand    0  0  0  0  0   3   0   0   0   0   0   0   0   0   0
  Phillipines   0  0  0  0  0   0   3   0   0   0   0   0   0   0   0
  Qatar         0  0  0  0  0   0   0   3   0   0   0   0   0   0   0
  Singapore     0  0  0  0  0   0   0   0   3   0   0   0   0   0   0
  southAfrica   0  0  0  0  0   0   0   0   0   3   0   0   0   0   0
  SriLanka      0  0  0  0  0   0   0   0   0   0   3   0   0   0   0
  Turkey        0  0  0  0  0   0   0   0   0   0   0   3   0   0   0
  UAE           0  0  0  0  0   0   0   0   0   0   0   0   3   0   0
  UnitedKingdom 0  0  0  0  0   0   0   0   0   0   0   0   0   3   0
  UnitedStates  0  0  0  0  0   0   0   0   0   0   0   0   0   0   3
>
Len Greski
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