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. ???
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