eval(predvars, data, env) 中的错误：找不到对象“下水道”

Question

I have a set of data containing species names and numbers ( spp_data ) and I am trying to test how the species are influenced by different parameters such as pH, conductivity, as well as the Sewer position (Upstream/Downstream) ( env_data1 ). When I'm trying to run the lm() I get the following error:

lm1 <- lm(specnumber ~ Sewer + pH + Conductivity, data=spp_data,env_data1)

Error in eval(predvars, data, env) : object 'Sewer' not found

Is it because the column Sewer is non-numeric? I also tried to exclude that column and run the lm() but it did not work.

species data

summary(spp_data)
    Pisidium        G_pulex         C_pseudo        A_aquatic          V_pisc       
 Min.   :0.000   Min.   : 0.00   Min.   : 0.000   Min.   :0.0000   Min.   :0.00000  
 1st Qu.:0.000   1st Qu.: 3.00   1st Qu.: 0.000   1st Qu.:0.0000   1st Qu.:0.00000  
 Median :0.000   Median : 8.00   Median : 3.000   Median :0.0000   Median :0.00000  
 Mean   :1.429   Mean   :16.86   Mean   : 4.476   Mean   :0.5714   Mean   :0.04762  
 3rd Qu.:2.000   3rd Qu.:20.00   3rd Qu.:10.000   3rd Qu.:0.0000   3rd Qu.:0.00000  
 Max.   :7.000   Max.   :68.00   Max.   :16.000   Max.   :4.0000   Max.   :1.00000  

    Taeniopt          Rhyacoph         Hydropsy        Lepidost        Glossos     
 Min.   :0.00000   Min.   :0.0000   Min.   :0.000   Min.   :0.000   Min.   : 0.00  
 1st Qu.:0.00000   1st Qu.:0.0000   1st Qu.:0.000   1st Qu.:0.000   1st Qu.: 0.00  
 Median :0.00000   Median :0.0000   Median :0.000   Median :0.000   Median : 0.00  
 Mean   :0.09524   Mean   :0.2381   Mean   :1.286   Mean   :1.238   Mean   : 1.81  
 3rd Qu.:0.00000   3rd Qu.:0.0000   3rd Qu.:3.000   3rd Qu.:2.000   3rd Qu.: 1.00  
 Max.   :2.00000   Max.   :2.0000   Max.   :5.000   Max.   :7.000   Max.   :14.00  
    Agapetus         Hydroptil          Limneph         S_person           Tipula 
 Min.   : 0.0000   Min.   :0.00000   Min.   :0.000   Min.   :0.00000   Min.   :0  
 1st Qu.: 0.0000   1st Qu.:0.00000   1st Qu.:0.000   1st Qu.:0.00000   1st Qu.:0  
 Median : 0.0000   Median :0.00000   Median :0.000   Median :0.00000   Median :0  
 Mean   : 0.5714   Mean   :0.04762   Mean   :0.381   Mean   :0.09524   Mean   :0  
 3rd Qu.: 0.0000   3rd Qu.:0.00000   3rd Qu.:1.000   3rd Qu.:0.00000   3rd Qu.:0  
 Max.   :12.0000   Max.   :1.00000   Max.   :2.000   Max.   :2.00000   Max.   :0  
    Culicida         Ceratopo     Simuliid          Chrinomi         Chrnomus     
 Min.   :0.0000   Min.   : 0   Min.   : 0.0000   Min.   : 0.000   Min.   : 0.000  
 1st Qu.:0.0000   1st Qu.: 0   1st Qu.: 0.0000   1st Qu.: 0.000   1st Qu.: 1.000  
 Median :0.0000   Median : 1   Median : 0.0000   Median : 2.000   Median : 3.000  
 Mean   :0.5714   Mean   : 7   Mean   : 0.5238   Mean   : 7.286   Mean   : 6.095  
 3rd Qu.:0.0000   3rd Qu.: 8   3rd Qu.: 0.0000   3rd Qu.: 8.000   3rd Qu.: 6.000  
 Max.   :5.0000   Max.   :31   Max.   :10.0000   Max.   :67.000   Max.   :41.000  

environmental data

summary(env_data)
    Sample             Sewer                 pH        Conductivity      
 Length:21          Length:21          Min.   :7.780   Length:21         
 Class :character   Class :character   1st Qu.:7.850   Class :character  
 Mode  :character   Mode  :character   Median :8.100   Mode  :character  
                                       Mean   :8.044                     
                                       3rd Qu.:8.270                     
                                       Max.   :8.280                     
     Depth           %rock            %mud      %sand,,         
 Min.   : 7.00   Min.   :10.00   Min.   : 0   Length:21         
 1st Qu.: 8.00   1st Qu.:10.00   1st Qu.:20   Class :character  
 Median :11.00   Median :70.00   Median :30   Mode  :character  
 Mean   :17.14   Mean   :57.14   Mean   :40                     
 3rd Qu.:28.00   3rd Qu.:80.00   3rd Qu.:90                     
 Max.   :40.00   Max.   :90.00   Max.   :90

Answer 1

Assuming that the rows of your spp_data match the rows of your environmental data ... I think if you do

lm1 <- lm(as.matrix(spp_data) ~ Sewer + pH + Conductivity, 
        data=env_data1)

you will get the results of running 44 separate linear models, one for each species. (Be careful: with 44 regressions and only 21 observations, you may need to do some multiple comparisons corrections to avoid overstating your conclusions.)

There are R packages for more sophisticated multi-species analyses such as mvabund or gllvm , but they might not apply to a data set this size ...