[R] Help with programming a tricky algorithm

Andrew Crane-Droesch andrewcd at gmail.com
Mon Oct 22 17:26:21 CEST 2012

```Rui,

Thanks a lot for your help.  Unfortunately this doesn't work though:

1> is.border <- function(idx, DF){
1+     i1 <- DF\$ix %in% (DF\$ix[idx] + c(-1, 1)) & DF\$iy == DF\$iy[idx]
1+     i2 <- DF\$iy %in% (DF\$iy[idx] + c(-1, 1)) & DF\$ix == DF\$ix[idx]
1+     any(DF\$country[idx] != DF\$country[i1 | i2])
1+ }
1>
1> brdr <- sapply(sub\$idxy, is.border, sub)
1> sub\$border <- as.integer(brdr)
y         x    name year idxy idt     rain ndvi     temp    pop95
1 10.05928 1.9315833   benin 2001 1892  66 112.9945 535.0 300.1728 16.81851
2 10.05928 0.4770379    togo 2001  492  66 129.3910 464.5 300.3082 24.74213
3 10.05928 1.0588560    togo 2001 1052  66 144.9040 460.5 300.2535 29.34577
4 10.05928 2.8043106   benin 2001 2732  66 153.2175 572.5 300.1122 20.39760
5 10.05928 3.8224924 nigeria 2001 3712  66 162.9701 589.5 300.0277 11.40092
6 10.05928 0.8406742    togo 2001  842  66 143.1389 510.5 300.2740 29.34577
pop00    pop05   cropland    pasture    rainl1 rainl2   templ1
templ2
1 19.20563 22.32065 0.04130662 0.02016802 112.05880 60.00279 303.4361
304.4833
2 29.13643 33.08628 0.51615972 0.25667389 136.98332 38.74465 303.5708
304.5079
3 34.55809 39.24327 0.25539863 0.09774620 129.25399 39.04244 303.5183
304.5009
4 23.29283 27.07122 0.10509129 0.02065746 109.87102 119.92331 303.3055
304.3630
5 13.16283 13.73924 0.12865660 0.61500187  99.13176 67.27786 303.1213
304.1993
6 34.55809 39.24327 0.20321523 0.21327476 131.74456 38.64642 303.5380
304.5035
pophat p96 p97 p98 p99 p00 p01 p02 p03 p04 p05 p06 nres  ndvihat
1 19.82864   1   0   0   0   0   0   0   0   0   0   0 477.0112 576.8041
2 29.92640   1   0   0   0   0   0   0   0   0   0   0 -854.3601 486.1667
3 35.49513   1   0   0   0   0   0   0   0   0   0   0 -930.1263 487.2222
4 24.04851   1   0   0   0   0   0   0   0   0   0   0 363.3710 544.4345
5 13.27811   1   0   0   0   0   0   0   0   0   0   0 1234.5701 541.9867
6 35.49513   1   0   0   0   0   0   0   0   0   0   0 -153.9017 496.0020
res   ffresid       fe     wvec   ndviann   exrate isnigeria time D
1  32.60776 -676.2175 429.5025 261.6586 0.9947269 1.421233         0    2 0
2  41.00458 -343.6485 350.6056 251.5608 1.1106229 1.398511         0    2 0
3  38.19903 -419.1147 349.4112 245.9921 1.0967285 1.398511         0    2 0
4 113.99665  450.3076 385.6136 257.4387 1.0747216 1.421233         0    2 0
5 126.52839  778.1297 390.0818 268.2091 1.1022806 5.254477         1    2 1
6  78.50062  227.3886 359.1096 245.9921 1.0719824 1.398511         0    2 0
timexD ix iy border
1      0 27  1      0
2      0  7  1      0
3      0 15  1      0
4      0 39  1      0
5      2 53  1      0
6      0 12  1      0
1> unique(sub\$border)
[1] 0

On 10/21/2012 09:48 AM, Rui Barradas wrote:
> is.border <- function(idx, DF){
>     i1 <- DF\$ix %in% (DF\$ix[idx] + c(-1, 1)) & DF\$iy == DF\$iy[idx]
>     i2 <- DF\$iy %in% (DF\$iy[idx] + c(-1, 1)) & DF\$ix == DF\$ix[idx]
>     any(DF\$country[idx] != DF\$country[i1 | i2])
> }
>
> brdr <- sapply(Mydata\$idxy, is.border, Mydata)
> Mydata\$border <- as.integer(brdr)

```