Help with programming a tricky algorithm
Hi Rui, Thanks for the quick reply! It was my mistake not to notice $country in thr fourth line of your code. I went back and changed it to $name, and got the following output when I mapped the borders: http://i.imgur.com/DQ3IB.png Here is the output of your function: 1> exmpl <- sub[, c("name", "idxy", "ix", "iy")] 1> dput( head(exmpl, 30) ) structure(list(name = structure(c(1L, 6L, 6L, 1L, 5L, 6L, 1L, 1L, 1L, 1L, 1L, 6L, 1L, 1L, 1L, 5L, 1L, 6L, 1L, 5L, 6L, 1L, 1L, 3L, 1L, 6L, 1L, 6L, 1L, 3L), .Label = c("benin", "burkinafaso", "ghana", "niger", "nigeria", "togo"), class = "factor"), idxy = c(1892L, 492L, 1052L, 2732L, 3712L, 842L, 2522L, 3292L, 3012L, 3432L, 2172L, 772L, 2242L, 3152L, 1402L, 3852L, 1612L, 422L, 2662L, 3642L, 702L, 2942L, 2802L, 212L, 1682L, 1122L, 1962L, 1192L, 2452L, 142L), ix = c(27, 7, 15, 39, 53, 12, 36, 47, 43, 49, 31, 11, 32, 45, 20, 55, 23, 6, 38, 52, 10, 42, 40, 3, 24, 16, 28, 17, 35, 2), iy = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c("name", "idxy", "ix", "iy"), row.names = c(NA, 30L), class = "data.frame") I must say that I am impressed by the response of the community on this mailing list. I look forward to getting better at R such that I can be more useful to others myself! Best, Andrew
On 10/22/2012 08:33 AM, Rui Barradas wrote:
Hello,
In your original post, there was a column named 'country', it now
seems to be 'name', therefore my function shouldn't work. To see the
output of head(9 is helpfull but the better way is dput(). Try the
following:
exmpl <- sub[, c("name", "idxy", "ix", iy")]
dput( head(exmpl, 30) ) # paste the output of this in a post
And also try to change the column name like I've said above. But a
data example would really be needed.
Hope this helps,
Rui Barradas
Em 22-10-2012 16:26, Andrew Crane-Droesch escreveu:
Rui,
Thanks a lot for your help. Unfortunately this doesn't work though:
1> is.border <- function(idx, DF){
1+ i1 <- DF$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$border <- as.integer(brdr)
1> head(sub)
y x name year idxy idt rain ndvi temp pop95
1 10.05928 1.9315833 benin 2001 1892 66 112.9945 535.0 300.1728
16.81851
2 10.05928 0.4770379 togo 2001 492 66 129.3910 464.5 300.3082
24.74213
3 10.05928 1.0588560 togo 2001 1052 66 144.9040 460.5 300.2535
29.34577
4 10.05928 2.8043106 benin 2001 2732 66 153.2175 572.5 300.1122
20.39760
5 10.05928 3.8224924 nigeria 2001 3712 66 162.9701 589.5 300.0277
11.40092
6 10.05928 0.8406742 togo 2001 842 66 143.1389 510.5 300.2740
29.34577
pop00 pop05 cropland pasture rainl1 rainl2 templ1
templ2
1 19.20563 22.32065 0.04130662 0.02016802 112.05880 60.00279 303.4361
304.4833
2 29.13643 33.08628 0.51615972 0.25667389 136.98332 38.74465 303.5708
304.5079
3 34.55809 39.24327 0.25539863 0.09774620 129.25399 39.04244 303.5183
304.5009
4 23.29283 27.07122 0.10509129 0.02065746 109.87102 119.92331
303.3055 304.3630
5 13.16283 13.73924 0.12865660 0.61500187 99.13176 67.27786 303.1213
304.1993
6 34.55809 39.24327 0.20321523 0.21327476 131.74456 38.64642 303.5380
304.5035
pophat p96 p97 p98 p99 p00 p01 p02 p03 p04 p05 p06 nres ndvihat
1 19.82864 1 0 0 0 0 0 0 0 0 0 0 477.0112 576.8041
2 29.92640 1 0 0 0 0 0 0 0 0 0 0 -854.3601
486.1667
3 35.49513 1 0 0 0 0 0 0 0 0 0 0 -930.1263
487.2222
4 24.04851 1 0 0 0 0 0 0 0 0 0 0 363.3710 544.4345
5 13.27811 1 0 0 0 0 0 0 0 0 0 0 1234.5701
541.9867
6 35.49513 1 0 0 0 0 0 0 0 0 0 0 -153.9017
496.0020
res ffresid fe wvec ndviann exrate isnigeria
time D
1 32.60776 -676.2175 429.5025 261.6586 0.9947269 1.421233 0 2 0
2 41.00458 -343.6485 350.6056 251.5608 1.1106229 1.398511 0 2 0
3 38.19903 -419.1147 349.4112 245.9921 1.0967285 1.398511 0 2 0
4 113.99665 450.3076 385.6136 257.4387 1.0747216 1.421233 0 2 0
5 126.52839 778.1297 390.0818 268.2091 1.1022806 5.254477 1 2 1
6 78.50062 227.3886 359.1096 245.9921 1.0719824 1.398511 0 2 0
timexD ix iy border
1 0 27 1 0
2 0 7 1 0
3 0 15 1 0
4 0 39 1 0
5 2 53 1 0
6 0 12 1 0
1> unique(sub$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)