对比只能应用于R

时间:2015-06-20 23:21:15

标签: r

我正在尝试对我的数据进行逻辑回归,但是我收到了这个错误:

logistic <- lm(response ~., data = df_without, family='binomial')

Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : 
  contrasts can be applied only to factors with 2 or more levels

与关于此主题的其他两个问题(herehere)不同,我对所有因素都有&gt; = 2个级别,我的响应变量也有2个级别:

summary(df_without$response)
     0      1 
123534  64591 


summary(df_without[sapply(df_without, is.factor)])

enter image description here

我的数据框可用{。{3}}作为.Rdata文件。

sessionInfo()

R version 3.2.0 (2015-04-16)
Platform: x86_64-apple-darwin13.4.0 (64-bit)
Running under: OS X 10.10.1 (Yosemite)

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] parallel  splines   stats     graphics  grDevices utils         datasets  methods   base     

other attached packages:
 [1] Amelia_1.7.3        Rcpp_0.11.6         randomForest_4.6-10 e1071_1.6-4         plyr_1.8.2         
 [6] gbm_2.1.1           survival_2.38-1     glmnet_2.0-2            foreach_1.4.2       Matrix_1.2-0       
[11] caret_6.0-47        ggplot2_1.0.1       lattice_0.20-31     lubridate_1.3.3     RJDBC_0.2-5        
[16] rJava_0.9-6         DBI_0.3.1          

loaded via a namespace (and not attached):
 [1] compiler_3.2.0      nloptr_1.0.4        class_7.3-12        iterators_1.0.7     tools_3.2.0        
 [6] digest_0.6.8        lme4_1.1-7          memoise_0.2.1       nlme_3.1-120        gtable_0.1.2       
[11] mgcv_1.8-6          brglm_0.5-9         SparseM_1.6         proto_0.3-10        BradleyTerry2_1.0-6
[16] stringr_1.0.0       gtools_3.5.0        grid_3.2.0          nnet_7.3-9          foreign_0.8-63     
[21] minqa_1.2.4         reshape2_1.4.1      car_2.0-25          magrittr_1.5        scales_0.2.4       
[26] codetools_0.2-11    MASS_7.3-40         pbkrtest_0.4-2      colorspace_1.2-6    quantreg_5.11      
[31] stringi_0.4-1       munsell_0.4.2 

1 个答案:

答案 0 :(得分:3)

我知道代码只有答案会受到主持人的严厉处理和严厉处理,但这确实不言自明:

 # I did download the excessively large file
> table(df_without[ complete.cases(df_without), 'pymnt_plan'])

        n    y 
   0 1231    0