Hi. I'm attempting to fit a logistic/binomial model so I can determine
the influence of landscape on the probability that a box gets used by a
bird. I've looked at a few sources (MASS text, Dalgaard, Fox and
google) and the examples are almost always based on tabular predictor
variables. My data, however are not. I'm not sure if that is the
source of the problems or not because the one example that includes a
continuous predictor looks to be coded exactly the same way. Looking at
the output, I get estimates for each case when I should get a single
estimate for purbank. Any suggestions?
Many thanks,
Jeff
THE DATA: (200 boxes total, used [0 if unoccupied, 1 occupied], the rest
are landscape variables).
box use purbank purban2 purban1 pgrassk pgrass2 pgrass1 grassdist grasspatchk
1 1 0.003813435 0.02684564 0.06896552 0.3282487 0.6845638 0.7586207 0 3.73
2 1 0.04429451 0.1610738 0.1724138 0.1534174 0.3825503 0.6551724 0 1.023261
3 1 0.04458785 0.06040268 0 0.1628043 0.557047 0.7586207 0 0.9605769
4 1 0.06072162 0.2080537 0.06896552 0.01936052 0 0 323.1099 0.2284615
5 0 0.6080962 0.6979866 0.6896552 0.03168084 0.1275168 0.2413793 30 0.2627027
6 1 0.6060428 0.6107383 0.3448276 0.04077442 0.2885906 0.4482759 30 0.2978571
7 1 0.3807568 0.4362416 0.6896552 0.06864183 0.03355705 0 94.86833 0.468
8 0 0.3649164 0.3154362 0.4137931 0.06277501 0.1275168 0 120 0.4585714
THE CODE:
box.use<- read.csv("c:\\eabl\\2004\\use_logistic2.csv",
header=TRUE)
attach(box.use)
box.use <- na.omit(box.use)
use <- factor(use, levels=0:1)
levels(use) <- c("unused", "used")
glm1 <- glm(use ~ purbank, binomial)
THE OUTPUT:
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.544e-16 1.414e+00 -3.21e-16 1.000
purbank0 2.157e+01 2.923e+04 0.001 0.999
purbank0.001173365 2.157e+01 2.067e+04 0.001 0.999
purbank0.001466706 2.157e+01 2.923e+04 0.001 0.999
purbank0.001760047 6.429e-16 2.000e+00 3.21e-16 1.000
purbank0.002346729 2.157e+01 2.923e+04 0.001 0.999
purbank0.003813435 2.157e+01 2.923e+04 0.001 0.999
purbank0.004106776 2.157e+01 2.067e+04 0.001 0.999
purbank0.004693458 2.157e+01 2.067e+04 0.001 0.999
****************************************
Jeffrey A. Stratford, Ph.D.
Postdoctoral Associate
331 Funchess Hall
Department of Biological Sciences
Auburn University
Auburn, AL 36849
334-329-9198
FAX 334-844-9234
http://www.auburn.edu/~stratja
Jeffrey Stratford said the following on 10/13/2006 9:28 AM:> Hi. I'm attempting to fit a logistic/binomial model so I can determine > the influence of landscape on the probability that a box gets used by a > bird. I've looked at a few sources (MASS text, Dalgaard, Fox and > google) and the examples are almost always based on tabular predictor > variables. My data, however are not. I'm not sure if that is the > source of the problems or not because the one example that includes a > continuous predictor looks to be coded exactly the same way. Looking at > the output, I get estimates for each case when I should get a single > estimate for purbank. Any suggestions? > > Many thanks, > > Jeff > > > THE DATA: (200 boxes total, used [0 if unoccupied, 1 occupied], the rest > are landscape variables). > > box use purbank purban2 purban1 pgrassk pgrass2 pgrass1 grassdist grasspatchk > 1 1 0.003813435 0.02684564 0.06896552 0.3282487 0.6845638 0.7586207 0 3.73 > 2 1 0.04429451 0.1610738 0.1724138 0.1534174 0.3825503 0.6551724 0 1.023261 > 3 1 0.04458785 0.06040268 0 0.1628043 0.557047 0.7586207 0 0.9605769 > 4 1 0.06072162 0.2080537 0.06896552 0.01936052 0 0 323.1099 0.2284615 > 5 0 0.6080962 0.6979866 0.6896552 0.03168084 0.1275168 0.2413793 30 0.2627027 > 6 1 0.6060428 0.6107383 0.3448276 0.04077442 0.2885906 0.4482759 30 0.2978571 > 7 1 0.3807568 0.4362416 0.6896552 0.06864183 0.03355705 0 94.86833 0.468 > 8 0 0.3649164 0.3154362 0.4137931 0.06277501 0.1275168 0 120 0.4585714 > > THE CODE: > > box.use<- read.csv("c:\\eabl\\2004\\use_logistic2.csv", header=TRUE) > attach(box.use) > box.use <- na.omit(box.use) > use <- factor(use, levels=0:1) > levels(use) <- c("unused", "used") > glm1 <- glm(use ~ purbank, binomial) > > THE OUTPUT: > > Coefficients: > Estimate Std. Error z value Pr(>|z|) > (Intercept) -4.544e-16 1.414e+00 -3.21e-16 1.000 > purbank0 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001173365 2.157e+01 2.067e+04 0.001 0.999 > purbank0.001466706 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001760047 6.429e-16 2.000e+00 3.21e-16 1.000 > purbank0.002346729 2.157e+01 2.923e+04 0.001 0.999 > purbank0.003813435 2.157e+01 2.923e+04 0.001 0.999 > purbank0.004106776 2.157e+01 2.067e+04 0.001 0.999 > purbank0.004693458 2.157e+01 2.067e+04 0.001 0.999 > > > **************************************** > Jeffrey A. Stratford, Ph.D. > Postdoctoral Associate > 331 Funchess Hall > Department of Biological Sciences > Auburn University > Auburn, AL 36849 > 334-329-9198 > FAX 334-844-9234 > http://www.auburn.edu/~stratja > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.That's not what I get: lines <- "box,use,purbank,purban2,purban1,pgrassk,pgrass2,pgrass1,grassdist,grasspatchk 1,1,0.003813435,0.02684564,0.06896552,0.3282487,0.6845638,0.7586207,0,3.73 2,1,0.04429451,0.1610738,0.1724138,0.1534174,0.3825503,0.6551724,0,1.023261 3,1,0.04458785,0.06040268,0,0.1628043,0.557047,0.7586207,0,0.9605769 4,1,0.06072162,0.2080537,0.06896552,0.01936052,0,0,323.1099,0.2284615 5,0,0.6080962,0.6979866,0.6896552,0.03168084,0.1275168,0.2413793,30,0.2627027 6,1,0.6060428,0.6107383,0.3448276,0.04077442,0.2885906,0.4482759,30,0.2978571 7,1,0.3807568,0.4362416,0.6896552,0.06864183,0.03355705,0,94.86833,0.468 8,0,0.3649164,0.3154362,0.4137931,0.06277501,0.1275168,0,120,0.4585714" box.use <- read.csv(textConnection(lines)) box.use <- na.omit(box.use) box.use$use <- factor(box.use$use, levels=0:1) levels(box.use$use) <- c("unused", "used") (glm1 <- glm(use ~ purbank, binomial, box.use)) You need to check why purbank is being interpreted as a factor in your code. Also, I removed your use of "attach" because I find it dangerous (especially with no detach). Better to be explicit. HTH, --sundar
On Fri, 2006-10-13 at 09:28 -0500, Jeffrey Stratford wrote:> Hi. I'm attempting to fit a logistic/binomial model so I can determine > the influence of landscape on the probability that a box gets used by a > bird. I've looked at a few sources (MASS text, Dalgaard, Fox and > google) and the examples are almost always based on tabular predictor > variables. My data, however are not. I'm not sure if that is the > source of the problems or not because the one example that includes a > continuous predictor looks to be coded exactly the same way. Looking at > the output, I get estimates for each case when I should get a single > estimate for purbank. Any suggestions? > > Many thanks, > > Jeff > > > THE DATA: (200 boxes total, used [0 if unoccupied, 1 occupied], the rest > are landscape variables). > > box use purbank purban2 purban1 pgrassk pgrass2 pgrass1 grassdist grasspatchk > 1 1 0.003813435 0.02684564 0.06896552 0.3282487 0.6845638 0.7586207 0 3.73 > 2 1 0.04429451 0.1610738 0.1724138 0.1534174 0.3825503 0.6551724 0 1.023261 > 3 1 0.04458785 0.06040268 0 0.1628043 0.557047 0.7586207 0 0.9605769 > 4 1 0.06072162 0.2080537 0.06896552 0.01936052 0 0 323.1099 0.2284615 > 5 0 0.6080962 0.6979866 0.6896552 0.03168084 0.1275168 0.2413793 30 0.2627027 > 6 1 0.6060428 0.6107383 0.3448276 0.04077442 0.2885906 0.4482759 30 0.2978571 > 7 1 0.3807568 0.4362416 0.6896552 0.06864183 0.03355705 0 94.86833 0.468 > 8 0 0.3649164 0.3154362 0.4137931 0.06277501 0.1275168 0 120 0.4585714 > > THE CODE: > > box.use<- read.csv("c:\\eabl\\2004\\use_logistic2.csv", header=TRUE) > attach(box.use) > box.use <- na.omit(box.use) > use <- factor(use, levels=0:1) > levels(use) <- c("unused", "used") > glm1 <- glm(use ~ purbank, binomial) > > THE OUTPUT: > > Coefficients: > Estimate Std. Error z value Pr(>|z|) > (Intercept) -4.544e-16 1.414e+00 -3.21e-16 1.000 > purbank0 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001173365 2.157e+01 2.067e+04 0.001 0.999 > purbank0.001466706 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001760047 6.429e-16 2.000e+00 3.21e-16 1.000 > purbank0.002346729 2.157e+01 2.923e+04 0.001 0.999 > purbank0.003813435 2.157e+01 2.923e+04 0.001 0.999 > purbank0.004106776 2.157e+01 2.067e+04 0.001 0.999 > purbank0.004693458 2.157e+01 2.067e+04 0.001 0.999It appears that the 'purbank' variable is being imported as a factor, hence the multiple levels indicated in the left hand column. Check: str(box.use) right after the read.csv() step and see what it shows.>From the sample data above, it _should_ be along the lines of:> str(box.use)'data.frame': 8 obs. of 10 variables: $ box : int 1 2 3 4 5 6 7 8 $ use : int 1 1 1 1 0 1 1 0 $ purbank : num 0.00381 0.04429 0.04459 0.06072 0.60810 ... $ purban2 : num 0.0268 0.1611 0.0604 0.2081 0.6980 ... $ purban1 : num 0.069 0.172 0.000 0.069 0.690 ... $ pgrassk : num 0.3282 0.1534 0.1628 0.0194 0.0317 ... $ pgrass2 : num 0.685 0.383 0.557 0.000 0.128 ... $ pgrass1 : num 0.759 0.655 0.759 0.000 0.241 ... $ grassdist : num 0 0 0 323 30 ... $ grasspatchk: num 3.730 1.023 0.961 0.228 0.263 ... Hence, you should be able to use:> model <- glm(use ~ purbank, data = box.use, family = binomial)> summary(model)Call: glm(formula = use ~ purbank, family = binomial, data = box.use) Deviance Residuals: Min 1Q Median 3Q Max -1.61450 -0.03098 0.31935 0.45888 1.39194 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 3.223 2.225 1.448 0.147 purbank -6.129 4.773 -1.284 0.199 (Dispersion parameter for binomial family taken to be 1) Null deviance: 8.9974 on 7 degrees of freedom Residual deviance: 6.5741 on 6 degrees of freedom AIC: 10.574 Number of Fisher Scoring iterations: 5 Note that na.omit() is the default operation for most R models, so is redundant. Also, I would not attach the data frame, as you can use the 'data' argument in model related functions. This avoids the confusion of having multiple copies of the source data set and wondering why changes made can become confusing and problematic. HTH, Marc Schwartz
On Fri, 2006-10-13 at 09:28 -0500, Jeffrey Stratford wrote:> Hi. I'm attempting to fit a logistic/binomial model so I can determine > the influence of landscape on the probability that a box gets used by a > bird. I've looked at a few sources (MASS text, Dalgaard, Fox and > google) and the examples are almost always based on tabular predictor > variables. My data, however are not. I'm not sure if that is the > source of the problems or not because the one example that includes a > continuous predictor looks to be coded exactly the same way. Looking at > the output, I get estimates for each case when I should get a single > estimate for purbank. Any suggestions? > > Many thanks, > > JeffHi Jeff, using the snippet of data you provided (copy/paste into a text file and read in with read.table) worked fine: box.use <- read.table("~/tmp/tmp.txt", header = TRUE) box.use str(box.use) 'data.frame': 8 obs. of 10 variables: $ box : int 1 2 3 4 5 6 7 8 $ use : int 1 1 1 1 0 1 1 0 $ purbank : num 0.00381 0.04429 0.04459 0.06072 0.60810 ... $ purban2 : num 0.0268 0.1611 0.0604 0.2081 0.6980 ... $ purban1 : num 0.069 0.172 0.000 0.069 0.690 ... $ pgrassk : num 0.3282 0.1534 0.1628 0.0194 0.0317 ... $ pgrass2 : num 0.685 0.383 0.557 0.000 0.128 ... $ pgrass1 : num 0.759 0.655 0.759 0.000 0.241 ... $ grassdist : num 0 0 0 323 30 ... $ grasspatchk: num 3.730 1.023 0.961 0.228 0.263 ... Now I don't like attach, and you just don't need it so I deviate a little now. Replace box.use$use directly and make use of the data argument in glm. Also, your data didn't have any missing data so I'm not sure whether the response or predictor is missing and whether your na.omit is needed or not - I don't use it below. box.use$use <- factor(box.use$use, levels=0:1) levels(box.use$use) <- c("unused", "used") box.use str(box.use) glm1 <- glm(use ~ purbank, data = box.use, family = binomial()) summary(glm1) Call: glm(formula = use ~ purbank, family = binomial(), data = box.use) Deviance Residuals: Min 1Q Median 3Q Max -1.61450 -0.03098 0.31935 0.45888 1.39194 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 3.223 2.225 1.448 0.147 purbank -6.129 4.773 -1.284 0.199 (Dispersion parameter for binomial family taken to be 1) Null deviance: 8.9974 on 7 degrees of freedom Residual deviance: 6.5741 on 6 degrees of freedom AIC: 10.574 Number of Fisher Scoring iterations: 5 I suspect something got messed up in your reading of the data and R thought purbank was a factor or character. Always check your data after reading in, and str() is a your friend here as printed representations are not always what they seem. HTH G> > > THE DATA: (200 boxes total, used [0 if unoccupied, 1 occupied], the rest > are landscape variables). > > box use purbank purban2 purban1 pgrassk pgrass2 pgrass1 grassdist grasspatchk > 1 1 0.003813435 0.02684564 0.06896552 0.3282487 0.6845638 0.7586207 0 3.73 > 2 1 0.04429451 0.1610738 0.1724138 0.1534174 0.3825503 0.6551724 0 1.023261 > 3 1 0.04458785 0.06040268 0 0.1628043 0.557047 0.7586207 0 0.9605769 > 4 1 0.06072162 0.2080537 0.06896552 0.01936052 0 0 323.1099 0.2284615 > 5 0 0.6080962 0.6979866 0.6896552 0.03168084 0.1275168 0.2413793 30 0.2627027 > 6 1 0.6060428 0.6107383 0.3448276 0.04077442 0.2885906 0.4482759 30 0.2978571 > 7 1 0.3807568 0.4362416 0.6896552 0.06864183 0.03355705 0 94.86833 0.468 > 8 0 0.3649164 0.3154362 0.4137931 0.06277501 0.1275168 0 120 0.4585714 > > THE CODE: > > box.use<- read.csv("c:\\eabl\\2004\\use_logistic2.csv", header=TRUE) > attach(box.use) > box.use <- na.omit(box.use) > use <- factor(use, levels=0:1) > levels(use) <- c("unused", "used") > glm1 <- glm(use ~ purbank, binomial) > > THE OUTPUT: > > Coefficients: > Estimate Std. Error z value Pr(>|z|) > (Intercept) -4.544e-16 1.414e+00 -3.21e-16 1.000 > purbank0 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001173365 2.157e+01 2.067e+04 0.001 0.999 > purbank0.001466706 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001760047 6.429e-16 2.000e+00 3.21e-16 1.000 > purbank0.002346729 2.157e+01 2.923e+04 0.001 0.999 > purbank0.003813435 2.157e+01 2.923e+04 0.001 0.999 > purbank0.004106776 2.157e+01 2.067e+04 0.001 0.999 > purbank0.004693458 2.157e+01 2.067e+04 0.001 0.999 > > > **************************************** > Jeffrey A. Stratford, Ph.D. > Postdoctoral Associate > 331 Funchess Hall > Department of Biological Sciences > Auburn University > Auburn, AL 36849 > 334-329-9198 > FAX 334-844-9234 > http://www.auburn.edu/~stratja > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.-- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Gavin Simpson [t] +44 (0)20 7679 0522 ECRC & ENSIS, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/cv/ London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/ %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Gavin, That worked! I went through and I found a few missing cases where I had "." instead of "NA" - I'm still in SAS mode. Many thanks! **************************************** Jeffrey A. Stratford, Ph.D. Postdoctoral Associate 331 Funchess Hall Department of Biological Sciences Auburn University Auburn, AL 36849 334-329-9198 FAX 334-844-9234 http://www.auburn.edu/~stratja ****************************************>>> Gavin Simpson <gavin.simpson at ucl.ac.uk> 10/13/06 11:23 AM >>>On Fri, 2006-10-13 at 09:28 -0500, Jeffrey Stratford wrote:> Hi. I'm attempting to fit a logistic/binomial model so I candetermine> the influence of landscape on the probability that a box gets used bya> bird. I've looked at a few sources (MASS text, Dalgaard, Fox and > google) and the examples are almost always based on tabular predictor > variables. My data, however are not. I'm not sure if that is the > source of the problems or not because the one example that includes a > continuous predictor looks to be coded exactly the same way. Lookingat> the output, I get estimates for each case when I should get a single > estimate for purbank. Any suggestions? > > Many thanks, > > JeffHi Jeff, using the snippet of data you provided (copy/paste into a text file and read in with read.table) worked fine: box.use <- read.table("~/tmp/tmp.txt", header = TRUE) box.use str(box.use) 'data.frame': 8 obs. of 10 variables: $ box : int 1 2 3 4 5 6 7 8 $ use : int 1 1 1 1 0 1 1 0 $ purbank : num 0.00381 0.04429 0.04459 0.06072 0.60810 ... $ purban2 : num 0.0268 0.1611 0.0604 0.2081 0.6980 ... $ purban1 : num 0.069 0.172 0.000 0.069 0.690 ... $ pgrassk : num 0.3282 0.1534 0.1628 0.0194 0.0317 ... $ pgrass2 : num 0.685 0.383 0.557 0.000 0.128 ... $ pgrass1 : num 0.759 0.655 0.759 0.000 0.241 ... $ grassdist : num 0 0 0 323 30 ... $ grasspatchk: num 3.730 1.023 0.961 0.228 0.263 ... Now I don't like attach, and you just don't need it so I deviate a little now. Replace box.use$use directly and make use of the data argument in glm. Also, your data didn't have any missing data so I'm not sure whether the response or predictor is missing and whether your na.omit is needed or not - I don't use it below. box.use$use <- factor(box.use$use, levels=0:1) levels(box.use$use) <- c("unused", "used") box.use str(box.use) glm1 <- glm(use ~ purbank, data = box.use, family = binomial()) summary(glm1) Call: glm(formula = use ~ purbank, family = binomial(), data = box.use) Deviance Residuals: Min 1Q Median 3Q Max -1.61450 -0.03098 0.31935 0.45888 1.39194 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 3.223 2.225 1.448 0.147 purbank -6.129 4.773 -1.284 0.199 (Dispersion parameter for binomial family taken to be 1) Null deviance: 8.9974 on 7 degrees of freedom Residual deviance: 6.5741 on 6 degrees of freedom AIC: 10.574 Number of Fisher Scoring iterations: 5 I suspect something got messed up in your reading of the data and R thought purbank was a factor or character. Always check your data after reading in, and str() is a your friend here as printed representations are not always what they seem. HTH G> > > THE DATA: (200 boxes total, used [0 if unoccupied, 1 occupied], therest> are landscape variables). > >box use purbank purban2 purban1 pgrassk pgrass2 pgrass1 grassdist grasspatchk>1 1 0.003813435 0.02684564 0.06896552 0.3282487 0.6845638 0.7586207 0 3.73>2 1 0.04429451 0.1610738 0.1724138 0.1534174 0.3825503 0.6551724 0 1.023261>3 1 0.04458785 0.06040268 0 0.1628043 0.557047 0.7586207 0 0.9605769>4 1 0.06072162 0.2080537 0.06896552 0.01936052 0 0 323.1099 0.2284615>5 0 0.6080962 0.6979866 0.6896552 0.03168084 0.1275168 0.2413793 30 0.2627027>6 1 0.6060428 0.6107383 0.3448276 0.04077442 0.2885906 0.4482759 30 0.2978571>7 1 0.3807568 0.4362416 0.6896552 0.06864183 0.03355705 0 94.86833 0.468>8 0 0.3649164 0.3154362 0.4137931 0.06277501 0.1275168 0 120 0.4585714> > THE CODE: > > box.use<- read.csv("c:\\eabl\\2004\\use_logistic2.csv", header=TRUE) > attach(box.use) > box.use <- na.omit(box.use) > use <- factor(use, levels=0:1) > levels(use) <- c("unused", "used") > glm1 <- glm(use ~ purbank, binomial) > > THE OUTPUT: > > Coefficients: > Estimate Std. Error z value Pr(>|z|) > (Intercept) -4.544e-16 1.414e+00 -3.21e-16 1.000 > purbank0 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001173365 2.157e+01 2.067e+04 0.001 0.999 > purbank0.001466706 2.157e+01 2.923e+04 0.001 0.999 > purbank0.001760047 6.429e-16 2.000e+00 3.21e-16 1.000 > purbank0.002346729 2.157e+01 2.923e+04 0.001 0.999 > purbank0.003813435 2.157e+01 2.923e+04 0.001 0.999 > purbank0.004106776 2.157e+01 2.067e+04 0.001 0.999 > purbank0.004693458 2.157e+01 2.067e+04 0.001 0.999 > > > **************************************** > Jeffrey A. Stratford, Ph.D. > Postdoctoral Associate > 331 Funchess Hall > Department of Biological Sciences > Auburn University > Auburn, AL 36849 > 334-329-9198 > FAX 334-844-9234 > http://www.auburn.edu/~stratja > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.-- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Gavin Simpson [t] +44 (0)20 7679 0522 ECRC & ENSIS, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/cv/ London, UK. WC1E 6BT. [w] http://www.ucl.ac.uk/~ucfagls/ %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%