[R] Assignment problems

[Rui Barradas]

Hello,


phillip03 wrote
> 
> The text below is a part of, some work I have to do, which is due in 2
> days and I am strung up with a lot of other stuff, so I was hoping someone
> would take 5 mins and help me ??
> 
> Here is a part of my data.frame:
> 
>      year country1 country2 contig comlang        pop1        gdp1       
> pop2         gdp2 rta    dist      avgflow
> 1    1992      AUS      AUT      0       0  17.4950008  321708.281  
> 7.7825189   194684.078   0 15608.4 1.075999e+02
> 2    1992      AUS      BEL      0       0  17.4950008  321708.281 
> 10.0450001   231762.094   0 16319.2 4.767162e+02
> 3    1992      AUS      CAN      0       1  17.4950008  321708.281 
> 28.5195980   570291.188   0 15391.1 7.456945e+02
> 4    1992      AUS      CHE      0       0  17.4950008  321708.281  
> 6.8750000   249471.422   0 16170.1 4.625214e+02
> 5    1992      AUS      DEU      0       0  17.4950008  321708.281 
> 80.6240005  2062141.500   0 15935.1 2.047573e+03
> 6    1992      AUS      DNK      0       0  17.4950008  321708.281  
> 5.1700001   150195.484   0 15725.5 1.453406e+02
> 7    1992      AUS      ESP      0       0  17.4950008  321708.281 
> 39.0677490   612585.250   0 17072.9 2.106880e+02
> 8    1992      AUS      FIN      0       0  17.4950008  321708.281  
> 5.0419998   109859.438   0 14849.5 2.025125e+02
> 9    1992      AUS      FRA      0       0  17.4950008  321708.281 
> 57.2422981  1371706.000   0 16513.0 1.070802e+03
> 10   1992      AUS      GBR      0       1  17.4950008  321708.281 
> 57.9023476  1071537.375   0 16602.3 2.279130e+03
> 11   1992      AUS      GRC      0       0  17.4950008  321708.281 
> 10.3699999   102022.352   0 14845.6 4.164985e+01
> 12   1992      AUS      IRL      0       1  17.4950008  321708.281  
> 3.5490999    54272.410   0 16895.0 1.076323e+02
> 13   1992      AUS      ISL      0       0  17.4950008  321708.281  
> 0.2611000     6976.168   0 16443.6 2.190602e+01
> 14   1992      AUS      ITA      0       0  17.4950008  321708.281 
> 56.7976494  1265800.125   0 15855.4 9.683720e+02
> 15   1992      AUS      JPN      0       0  17.4950008  321708.281
> 124.2289963  3766884.000   0  7827.1 1.026065e+04
> 16   1992      AUS      NLD      0       0  17.4950008  321708.281 
> 15.1780005   348224.562   0 16227.5 6.510009e+02
> 17   1992      AUS      NOR      0       0  17.4950008  321708.281  
> 4.2863998   127170.328   0 15646.2 9.357240e+01
> 18   1992      AUS      NZL      0       1  17.4950008  321708.281  
> 3.5316999    40706.199   1  2736.4 2.267670e+03
> 19   1992      AUS      PRT      0       0  17.4950008  321708.281  
> 9.9630003   102890.258   0 17625.3 2.611476e+02
> 20   1992      AUS      SWE      0       0  17.4950008  321708.281  
> 8.6680002   264822.875   0 15385.4 4.653388e+02
> 
> 
> there is 3400 observations.
> 
> 3.1.1. Construct a dummy variable, EMU, that in any given year takes the
> value 1 if both countries are members of the EMU and 0 otherwise. How big
> a proportion of the observations are among EMU member countries?
> 
> This problem is solved with:
> 
> euro<-c("AUT","BEL","DEU","ESP","FIN","FRA","GRC","IRL","ITA","NLD","PRT")
>  countries<-data.frame(country1,country2,stringsAsFactors=FALSE)
>  data1<-cbind(data,EMU=Reduce(`&`, lapply(countries, function(x) x %in%
> euro)))
>  
>  data1[EMU==TRUE,13]
> 
>  a<-table(EMU)
> 
> 
> 3.1.2. Are the member and non-member country-pairs alike? 
> 
> What I need here is:
> I want to find the mean of avgflow, but only for the data where 2
> countries are in the euro vector/if EMU=TRUE ?
> I have tried with:
>>avgflowONLY<-cbind(avgflow,EMU)
> 
>> NEWavgflow<-rep(0,nrow(avgflowONLY))
> 
>> for (i in 1:nrow(avgflowONLY)){if
>> (EMU==1){NEWavgflow[i]<-mean(avgflow[i])}}
> 
> BUT it gives me: 
> Warning messages:
> 1: In if (EMU == 1) { ... :
>   the condition has length > 1 and only the first element will be used
> etc. ???
> 

You're forgeting the index in the conditon, EMU[i] == 1.
Note that since EMU is a logical vector, you don't need the explicit
comparison.

If you just want the mean of avgflow where EMU == TRUE, this is much
simpler, but
returns one value, not a vector.

mean(avgflow[ EMU ])


Hope this helps,

Rui Barradas


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