[R-sig-Geo] different Moran's I results using R and matlab

Fri Mar 4 21:00:02 CET 2016

Roger,

Sorry for not replying sooner.

I check the dataset anselin.dat and I think I found the problem. xc and yc
should be latitude and longitude, but I used x coordinate and y coordinate.
After converting to latitude and longitude, I got the similar result.

Best,

Qiuhua

On Fri, Mar 4, 2016 at 5:46 AM, Roger Bivand <Roger.Bivand at nhh.no> wrote:

> On Thu, 3 Mar 2016, Roger Bivand wrote:
>
> On Thu, 3 Mar 2016, Qiuhua Ma wrote:
>>
>>  Hi,
>>>
>>>  I run the exactly same regression using R and matlab and get the same
>>>  regression results.
>>>
>>
>> Well, you need to give the exact reference to the matlab code you have
>> used - are they the functions in Spatial Econometrics Toolbox under
>> spatial/stats? I do not see the robust variants there.
>>
>> Most likely the data or the weights are different. Make your data set
>> available on a link, and I'll take a look.
>>
>
> Having not heard back, I ran moran() and lmerror() in Matlab using the
> Spatial Econometrics Toolbox on the data presented in stats/moran_d.m:
>
> load anselin.dat;
>
> y = anselin(:,1);
> n = length(y);
>
> x = [ones(n,1) anselin(:,2:3)];
>
> xc = anselin(:,4);
> yc = anselin(:,5);
> [j W j] = xy2cont(xc,yc);
>
> result = moran(y,x,W);
>
> tests$result
>>
> , , 1
>
>        [,1]
> meth   "moran"
> nobs   49
> nvar   3
> morani 0.2861962
> istat  4.019423
> imean  -0.03180435
> ivar   0.006259337
> prob   5.834084e-05
>
> (R result after moving the input data from Matlab to R with
> R.matlab::readMat())
>
> lm.morantest(lm_obj, lw, alternative="two.sided")
>>
>
>         Global Moran I for regression residuals
>
> data:
> model: lm(formula = y ~ x - 1)
> weights: lw
>
> Moran I statistic standard deviate = 4.0194, p-value = 5.834e-05
> alternative hypothesis: two.sided
> sample estimates:
> Observed Moran I      Expectation         Variance
>      0.286196209     -0.031804345      0.006259337
>
> [SAME RESULT]
>
> and similarly with lmerror()
>
> tests$result2
>>
> , , 1
>
>      [,1]
> meth "lmerror"
> lm   10.7656
> prob 0.001034042
> chi1 17.611
> nobs 49
> nvar 3
>
>
> lm.LMtests(lm_obj, lw)
>>
>
>         Lagrange multiplier diagnostics for spatial dependence
>
> data:
> model: lm(formula = y ~ x - 1)
> weights: lw
>
> LMErr = 10.766, df = 1, p-value = 0.001034
>
> [SAME RESULT]
>
> The result for lmlag() differs, probably because the Matlab code uses two
> different values for sigma and epe:
>
> sigma  = (e'*e)/(n-k);
>
> epe = (e'*e)/n;
> lm1 = (e'*W*y)/epe;
>
> t1 = trace((W+W')*W);
> D1=W*x*b;
> M=eye(n)-x*inv((x'*x))*x';
> D=(D1'*M*D1)*(1/sigma)+t1;
>
> lmlag = (lm1*lm1)*(1/D);
>
> so does not match the R code which follows Eq. 13 in Anselin et al. (1996,
> p. 84) in dividing the sum of squared errors by n, not n-k.
>
> I have not found lmlag_robust() or lmerror_robust() anywhere.
>
> Please provide the code of these functions if you need further
> clarification - the R code for the Moran's I test on OLS residuals is OK,
> as is the LM error test. The LM lag test chooses a version using ML sigma
> in harmony with the source article.
>
> Roger
>
>
>
>> Roger
>>
>>
>>>  However I got different results for Moran't I test and LM test results.
>>>
>>>  *R command:*
>>>  nb4 <- knn2nb(knearneigh(sale.sp, k = 4))
>>>  knn4listw <- nb2listw(nb4, style="W")
>>>
>>>  lm.morantest(near7_comrisk_semilog.out, knn4listw)
>>>  lm.LMtests(near7_comrisk_semilog.out, knn4listw, test=c("LMerr",
>>> "LMlag",
>>>  "RLMerr", "RLMlag"))
>>>
>>>  *R results:*
>>>  Moran I statistic standard deviate = 1.4135, p-value = 0.07875
>>>  Observed Moran's I        Expectation           Variance
>>>       8.780168e-03      -2.762542e-04       4.104967e-05
>>>
>>>  LMerr = 1.8752, df = 1, p-value = 0.1709
>>>  LMlag = 0.14335, df = 1, p-value = 0.705
>>>  RLMerr = 2.1193, df = 1, p-value = 0.1455
>>>  RLMlag = 0.38741, df = 1, p-value = 0.5337
>>>
>>>  *Matlab command:*
>>>  W_sale_4 = make_nnw(xc,yc,4);
>>>
>>>  moran4= moran(y,x, W_sale_4)
>>>  error_4 = lmerror(y,x,W_sale_4)
>>>  lag_4 = lmlag(y,x,W_sale_4)
>>>  error_4r = lmerror_robust(y,x,W_sale_4)
>>>  lag_4r = lmlag_robust(y,x,W_sale_4)
>>>
>>>  *Matlab results:*
>>>  Moran I-test for spatial correlation in residuals
>>>  Moran I                    0.13979528
>>>  Moran I-statistic         22.05018073
>>>  Marginal Probability       0.00000000
>>>  mean                      -0.00126742
>>>  standard deviation         0.00639735
>>>
>>>  error_4 =
>>>     meth: 'lmerror'
>>>       lm: 475.1839
>>>     prob: 0
>>>     chi1: 17.6110
>>>
>>>
>>>  lag_4 =
>>>     meth: 'lmlag'
>>>       lm: 465.4518
>>>     prob: 0
>>>     chi1: 17.6110
>>>
>>>  error_4r =
>>>     meth: 'lmerror_robust'
>>>       lm: 76.5845
>>>     prob: 0
>>>     chi1: 6.6400
>>>
>>>  lag_4r =
>>>
>>>     meth: 'lmlag_robust'
>>>       lm: 66.1547
>>>     prob: 4.4409e-16
>>>     chi1: 6.6400
>>>
>>>  Did I do anything wrong? Any thought on this problem?
>>>
>>>  thanks,
>>>
>>>  qiuhua
>>>
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>>>
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>>>
>>>
>>
>>
> --
> Roger Bivand
> Department of Economics, Norwegian School of Economics,
> Helleveien 30, N-5045 Bergen, Norway.
> voice: +47 55 95 93 55; fax +47 55 95 91 00
> e-mail: Roger.Bivand at nhh.no
> http://orcid.org/0000-0003-2392-6140
> https://scholar.google.no/citations?user=AWeghB0AAAAJ&hl=en
> http://depsy.org/person/434412
>

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