एलएम भविष्यवाणी

Question

भविष्यवाणी नहीं करेगा मेरे पास 2 डेटा फ्रेम हैं। एक प्रशिक्षण डेटा (pubs1) है, दूसरा (pubs2) परीक्षण डेटा। मैं एक रैखिक प्रतिगमन वस्तु बना सकता हूं लेकिन भविष्यवाणी करने में असमर्थ हूं। यह मेरी पहली बार ऐसा नहीं है और यह पता नहीं लगा सकता कि क्या गलत हो रहा है।

> head(pubs1) 
    id pred37 actual  weight  diff1 weightDiff1 pred1 pred2 pred3 pred4 
1 11 128.3257 128.3990 6.43482732 -0.07333650 -0.4719076922 126.3149 126.1024 126.9057 126.2718 
2 31 100.8822 100.9777 3.55520287 -0.09553741 -0.3396548680 100.7820 100.8589 100.9179 100.8903 
3 33 100.7204 100.9630 7.46413438 -0.24262409 -1.8109787866 100.8576 100.8434 100.8521 100.8914 
4 52 100.8564 100.9350 0.01299138 -0.07855588 -0.0010205495 100.8700 100.8925 100.8344 100.8714 
5 56 100.8410 100.9160 0.01299138 -0.07502125 -0.0009746298 100.8695 100.8889 100.8775 100.8871 
6 71 100.8889 100.8591 1.19266269 0.02979818 0.0355391800 100.8357 100.9205 100.8107 100.8316 
> head(pubs2) 
     id pred37  pred1 pred2  pred3  pred4 
1 762679 98.32212 97.84181 98.0776 98.03222 97.90022 
2 762680 115.79698 114.91411 115.1470 115.27129 115.45027 
3 762681 104.56418 104.81372 104.8537 104.66239 104.55240 
4 762682 106.65768 106.71011 106.6722 106.68662 106.60757 
5 762683 102.15662 103.14207 103.2035 103.31190 103.40397 
6 762684 101.96057 102.25939 102.1031 102.20659 102.04557 

> lm1 <- lm(pubs1$actual ~ pubs1$pred37 + pubs1$pred1 + pubs1$pred2 
+ + pubs1$pred3 + pubs1$pred4) 
> summary(lm1) 

Call: 
lm(formula = pubs1$actual ~ pubs1$pred37 + pubs1$pred1 + pubs1$pred2 + 
    pubs1$pred3 + pubs1$pred4) 

Residuals: 
    Min  1Q Median  3Q  Max 
-18.3415 -0.2309 0.0016 0.2236 17.8639 

Coefficients: 
       Estimate Std. Error t value Pr(>|t|)  
(Intercept) -0.122478 0.027227 -4.498 6.85e-06 *** 
pubs1$pred37 0.543270 0.005086 106.823 < 2e-16 *** 
pubs1$pred1 0.063680 0.007151 8.905 < 2e-16 *** 
pubs1$pred2 0.317768 0.010977 28.950 < 2e-16 *** 
pubs1$pred3 0.024302 0.008321 2.921 0.00349 ** 
pubs1$pred4 0.052183 0.010879 4.797 1.61e-06 *** 
--- 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 0.7298 on 99994 degrees of freedom 
Multiple R-squared: 0.9932,  Adjusted R-squared: 0.9932 
F-statistic: 2.926e+06 on 5 and 99994 DF, p-value: < 2.2e-16 

> 

> pred2 <- predict(lm1, pubs2) 
Warning message: 
'newdata' had 50000 rows but variable(s) found have 100000 rows 

> str(pubs1) 
'data.frame': 100000 obs. of 10 variables: 
$ id   : num 11 31 33 52 56 71 85 87 92 95 ... 
$ pred37  : num 128 101 101 101 101 ... 
$ actual  : num 128 101 101 101 101 ... 
$ weight  : num 6.435 3.555 7.464 0.013 0.013 ... 
$ diff1  : num -0.0733 -0.0955 -0.2426 -0.0786 -0.075 ... 
$ weightDiff1: num -0.471908 -0.339655 -1.810979 -0.001021 -0.000975 ... 
$ pred1  : num 126 101 101 101 101 ... 
$ pred2  : num 126 101 101 101 101 ... 
$ pred3  : num 127 101 101 101 101 ... 
$ pred4  : num 126 101 101 101 101 ... 
> str(pubs2) 
'data.frame': 50000 obs. of 6 variables: 
$ id : num 762679 762680 762681 762682 762683 ... 
$ pred37: num 98.3 115.8 104.6 106.7 102.2 ... 
$ pred1 : num 97.8 114.9 104.8 106.7 103.1 ... 
$ pred2 : num 98.1 115.1 104.9 106.7 103.2 ... 
$ pred3 : num 98 115 105 107 103 ... 
$ pred4 : num 97.9 115.5 104.6 106.6 103.4 ... 
> colnames(pubs1) 
[1] "id"   "pred37"  "actual"  "weight"  "diff1"  "weightDiff1" "pred1"  "pred2"  "pred3"  "pred4"  
> colnames(pubs2) 
[1] "id"  "pred37" "pred1" "pred2" "pred3" "pred4" 

क्या यहां कुछ भी है जो मुझे याद आ रही है?

Answer 1

बजाय

,

lm1 <- lm(pubs1$actual ~ pubs1$pred37 + pubs1$pred1 + pubs1$pred2 
      pubs1$pred3 + pubs1$pred4) 

कोशिश,

lm1 <- lm(actual ~ pred37 + pred1 + pred2 
      pred3 + pred4, data = pubs1) 

अन्यथा predict.lm चर अपने नए डेटा फ्रेम में pubs1$pred37 कहा जाता है की तलाश की जाएगी।