我正在做一些计数数据分析。数据在此链接中：[1]：https ://www.dropbox.com/s/q7fwqicw3ebvwlg/stackquestion.csv?dl=0
A列是计数数据，其他列是自变量。一开始我是用泊松回归分析的：
m0<-glm(A~.,data=d,family="poisson")
summary(m0)
#We see that the residual deviance is greater than the degrees of freedom so that we have over-dispersion.
Call:
glm(formula = A ~ ., family = "poisson", data = d)
Deviance Residuals:
Min 1Q Median 3Q Max
-28.8979 -4.5110 0.0384 5.4327 20.3809
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 8.7054842 0.9100882 9.566 < 2e-16 ***
B -0.1173783 0.0172330 -6.811 9.68e-12 ***
C 0.0864118 0.0182549 4.734 2.21e-06 ***
D 0.1169891 0.0301960 3.874 0.000107 ***
E 0.0738377 0.0098131 7.524 5.30e-14 ***
F 0.3814588 0.0093793 40.670 < 2e-16 ***
G -0.3712263 0.0274347 -13.531 < 2e-16 ***
H -0.0694672 0.0022137 -31.380 < 2e-16 ***
I -0.0634488 0.0034316 -18.490 < 2e-16 ***
J -0.0098852 0.0064538 -1.532 0.125602
K -0.1105270 0.0128016 -8.634 < 2e-16 ***
L -0.3304606 0.0155454 -21.258 < 2e-16 ***
M 0.2274175 0.0259872 8.751 < 2e-16 ***
N 0.2922063 0.0174406 16.754 < 2e-16 ***
O 0.1179708 0.0119332 9.886 < 2e-16 ***
P 0.0618776 0.0260646 2.374 0.017596 *
Q -0.0303909 0.0060060 -5.060 4.19e-07 ***
R -0.0018939 0.0037642 -0.503 0.614864
S 0.0383040 0.0065841 5.818 5.97e-09 ***
T 0.0318111 0.0116611 2.728 0.006373 **
U 0.2421129 0.0145502 16.640 < 2e-16 ***
V 0.1782144 0.0090858 19.615 < 2e-16 ***
W -0.5105135 0.0258136 -19.777 < 2e-16 ***
X -0.0583590 0.0043641 -13.373 < 2e-16 ***
Y -0.1554609 0.0042604 -36.489 < 2e-16 ***
Z 0.0064478 0.0001184 54.459 < 2e-16 ***
AA 0.3880479 0.0164929 23.528 < 2e-16 ***
AB 0.1511362 0.0050471 29.945 < 2e-16 ***
AC 0.0557880 0.0181129 3.080 0.002070 **
AD -0.6569099 0.0368771 -17.813 < 2e-16 ***
AE -0.0040679 0.0003960 -10.273 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 97109.0 on 56 degrees of freedom
Residual deviance: 5649.7 on 26 degrees of freedom
AIC: 6117.1
Number of Fisher Scoring iterations: 6
然后我认为我应该对过度分散的数据使用负二项式回归。既然你可以看到我有很多自变量，我想选择重要的变量。我决定使用逐步回归来选择自变量。首先，我创建了一个完整的模型：
full.model <- glm.nb(A~., data=d,maxit=1000)
# when not indicating maxit, or maxit=100, it shows Warning messages: 1: glm.fit: algorithm did not converge; 2: In glm.nb(A ~ ., data = d, maxit = 100) : alternation limit reached
# When indicating maxit=1000, the warning message disappear.
summary(full.model)
Call:
glm.nb(formula = A ~ ., data = d, maxit = 1000, init.theta = 2.730327193,
link = log)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.5816 -0.8893 -0.3177 0.4882 1.9073
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 11.8228596 8.3004322 1.424 0.15434
B -0.2592324 0.1732782 -1.496 0.13464
C 0.2890696 0.1928685 1.499 0.13393
D 0.3136262 0.3331182 0.941 0.34646
E 0.3764257 0.1313142 2.867 0.00415 **
F 0.3257785 0.1448082 2.250 0.02447 *
G -0.7585881 0.2343529 -3.237 0.00121 **
H -0.0714660 0.0343683 -2.079 0.03758 *
I -0.1050681 0.0357237 -2.941 0.00327 **
J 0.0810292 0.0566905 1.429 0.15291
K 0.2582978 0.1574582 1.640 0.10092
L -0.2009784 0.1543773 -1.302 0.19296
M -0.2359658 0.3216941 -0.734 0.46325
N -0.0689036 0.1910518 -0.361 0.71836
O 0.0514983 0.1383610 0.372 0.70974
P 0.1843138 0.3253483 0.567 0.57105
Q 0.0198326 0.0509651 0.389 0.69717
R 0.0892239 0.0459729 1.941 0.05228 .
S -0.0430981 0.0856391 -0.503 0.61479
T 0.2205653 0.1408009 1.567 0.11723
U 0.2450243 0.1838056 1.333 0.18251
V 0.1253683 0.0888411 1.411 0.15820
W -0.4636739 0.2348172 -1.975 0.04831 *
X -0.0623290 0.0508299 -1.226 0.22011
Y -0.0939878 0.0606831 -1.549 0.12142
Z 0.0019530 0.0015143 1.290 0.19716
AA -0.2888123 0.2449085 -1.179 0.23829
AB 0.1185890 0.0696343 1.703 0.08856 .
AC -0.3401963 0.2047698 -1.661 0.09664 .
AD -1.3409002 0.4858741 -2.760 0.00578 **
AE -0.0006299 0.0051338 -0.123 0.90234
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for Negative Binomial(2.7303) family taken to be 1)
Null deviance: 516.494 on 56 degrees of freedom
Residual deviance: 61.426 on 26 degrees of freedom
AIC: 790.8
Number of Fisher Scoring iterations: 1
Theta: 2.730
Std. Err.: 0.537
2 x log-likelihood: -726.803
不指示maxit，或者maxit=100时，显示Warning信息：1：glm.fit：算法没有收敛；2：在 glm.nb(A ~ ., data = d, maxit = 100) 中：达到交替限制。当指示 maxit=1000 时，警告信息消失。
然后我创建了第一个模型：
first.model <- glm.nb(A ~ 1, data = d)
然后我尝试了前向逐步回归：
step.model <- step(first.model, direction="forward", scope=formula(full.model))
#Error in glm.fit(X, y, wt, offset = offset, family = object$family, control = object$control) :
#NA/NaN/Inf in 'x'
#In addition: Warning message:
# step size truncated due to divergence
#What is the problem?
它给了我错误消息：glm.fit 中的错误（X，y，wt，offset = offset，family = object$family，control = object$control）：NA/NaN/Inf in 'x' 另外：警告消息：步长因发散而被截断
我还尝试了向后回归：
step.model2 <- step(full.model,direction="backward")
#the final step
Step: AIC=770.45
A ~ B + C + E + F + G + H + I + K + L + R + T + V + W + Y + AA +
AB + AD
Df Deviance AIC
<none> 62.375 770.45
- AB 1 64.859 770.93
- H 1 65.227 771.30
- V 1 65.240 771.31
- L 1 65.291 771.36
- Y 1 65.831 771.90
- B 1 66.051 772.12
- C 1 67.941 774.01
- AA 1 69.877 775.95
- K 1 70.411 776.48
- W 1 71.526 777.60
- I 1 71.863 777.94
- E 1 72.338 778.41
- G 1 73.344 779.42
- F 1 73.510 779.58
- AD 1 79.620 785.69
- R 1 80.358 786.43
- T 1 95.725 801.80
Warning messages:
1: glm.fit: algorithm did not converge
2: glm.fit: algorithm did not converge
3: glm.fit: algorithm did not converge
4: glm.fit: algorithm did not converge
我的问题是：为什么使用前向和后向逐步回归会有所不同？为什么在执行前向选择时会收到错误消息？另外，这些警告信息到底是什么意思？我应该如何处理它？
我不是统计人员，但需要对我的研究数据进行静态分析。因此，我正在努力学习如何使用真实数据进行不同的回归分析。我在网上搜索了类似的问题，但我仍然可以理解......如果我在回归分析中做错了什么，请告诉我。如果您能帮助我解决这些问题，我将不胜感激！