I am trying to do some verification across a large dataset, cuData, that
has 23 columns.
 
Column 23 (similarity) is the outcome 0 or 1 and the other columns are
the features.
 
I do this:
 
verificationglm.model <- glm(formula = similarity ~ ., family=binomial,
data=cuData[1:1000,])
 
and produce the model:
 
> summary(verificationglm.model)
 
Call:
glm(formula = similarity ~ ., family = binomial, data = cuData[1:1000, 
    ])
 
Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.3885  -0.8943  -0.2918   0.8851   2.7025  
 
Coefficients:
                        Estimate Std. Error z value Pr(>|z|)    
(Intercept)           26.3112869 21.2229690   1.240 0.215066    
length                -0.6249415  0.1906254  -3.278 0.001044 ** 
meanPitch             -0.0110389  0.0053083  -2.080 0.037565 *  
minimumPitch           0.0002689  0.0024290   0.111 0.911845    
maximumPitch          -0.0013454  0.0038149  -0.353 0.724326    
meanF1                -0.0362153  0.0112499  -3.219 0.001286 ** 
meanF2                 0.0016765  0.0115335   0.145 0.884430    
meanF3                 0.0073960  0.0076235   0.970 0.331964    
meanF4                 0.0063015  0.0016820   3.746 0.000179 ***
meanF5                -0.0022535  0.0024885  -0.906 0.365153    
ratioF2ToF1           -1.2322825  7.0036532  -0.176 0.860334    
ratioF3ToF1           -4.9643148  4.5973552  -1.080 0.280222    
jitter                -8.7535283 14.5273818  -0.603 0.546806    
shimmer                1.6706067  2.6327972   0.635 0.525731    
percentUnvoicedFrames -0.4863219  1.1638115  -0.418 0.676042    
numberOfVoiceBreaks   -0.0335636  0.0634956  -0.529 0.597086    
percentOfVoiceBreaks  -2.9353239  0.8945600  -3.281 0.001033 ** 
meanIntensity         -0.2931293  0.3355314  -0.874 0.382321    
minimumIntensity       0.0689654  0.1531059   0.450 0.652392    
maximumIntensity       0.2186570  0.2510906   0.871 0.383848    
ratioIntensity        -8.1777871 13.1676287  -0.621 0.534565    
noSyllsIntensity       0.1714826  0.0695021   2.467 0.013614 *  
speakingRate          -0.3564808  0.1507373  -2.365 0.018034 *  
startSpeech           -1.3537348  6.7337461  -0.201 0.840669    
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 
 
(Dispersion parameter for binomial family taken to be 1)
 
    Null deviance: 1384.0  on 999  degrees of freedom
Residual deviance: 1084.7  on 976  degrees of freedom
AIC: 1132.7
 
Number of Fisher Scoring iterations: 5
 
>
 
Now I want to use the model to predict on a different part of the
dataset.
 
I try this, and get my prediction:
 
 
> pred <- predict(verificationglm.model, cuData[1001:2000,1:23])
> pred
        1001         1002         1003         1004         1005
1006         1007 
-0.495901722 -2.406349629 -0.911082179 -0.965869553 -0.488695693
-1.849622304 -1.637722247 
        1008         1009         1010         1011         1012
1013         1014 
-1.148952722 -0.191538278 -1.511895046 -2.989036645 -2.775775622
0.603852124 -0.838613048 
        1015         1016         1017         1018         1019
1020         1021 
-0.434259674 -2.004230065 -0.234829011  1.666502334  2.039631718
-0.592192326  1.667700087 
        1022         1023         1024         1025         1026
1027         1028 
 0.104644531  1.748724399  0.391461247  1.356898357  1.468154760
1.090708994  1.071487227 
        1029         1030         1031         1032         1033
1034         1035 
 0.720596788  2.378350706 -0.128248232  0.969373318  0.315142756
1.372108172 -2.399517898 
        1036         1037         1038         1039         1040
1041         1042 
-0.684530171  0.761198819 -1.298372615  1.185368711 -1.148974059
0.358234433  0.671495255 
        1043         1044         1045         1046         1047
1048         1049 
 0.683771224  0.663767266  2.009012643  0.196591464  2.063417812
0.823472345  0.696638161
 
[runs on to 2000]
 
However, I then want to check for classAgreement (an e1701 package
function).  First I want a table. I do this:
 
> t = table(pred,cuData[1001:2000,24])
> t
                      
pred                   0 1
  -8.90070098980106    0 1
  -8.0484071844879     0 1
  -7.79298548775523    1 0
  -7.18338330609013    1 0
[runs on]
 
when I expect this
 
0         1
0          ?    ?
1          ?    ?
 
with the ?’s being some count.  When I look at my slice of cuData it
looks like this:
 
> cuData[1001:2000,24]
   [1] 1 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 1 1 1
1 1 1 1
  [38] 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 1 1 1 1 0 0
0 1 1 0
  [75] 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 0 1 0 1
0 0 0 0
 [112] 0 1 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
 [149] 1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1
0 1 0 0
 [186] 0 1 0 0 0 1 0 0 0 0 1 1 0 1 1 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 1 1
1 1 0 1
 [223] 0 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 0 0 1 0 0 1 1 1 1 1
1 1 1 0
 [260] 1 0 0 0 0 1 1 1 0 1 1 1 1 0 1 0 0 1 0 0 1 1 1 0 1 1 1 1 1 1 0 1 1
1 1 1 0
 [297] 1 1 0 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 1
1 1 1 0
 [334] 0 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0
1 0 1 0
 [371] 0 0 0 0 0 0 1 1 0 0 0 0 1 1 0 1 1 0 1 0 1 0 0 0 1 0 0 1 0 0 0 1 0
1 1 0 1
 [408] 1 1 0 0 0 0 1 0 1 1 1 1
 
[etc]
 
so it looks like a different layout from my pred. Does anyone know how
to make these two compatible so table() will work?
 
Thanks.
 
Stephen

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