sql - Perform linear regression in R with data from SAP HANA database -

i trying import dataset r apply linear regression model, skeptical of code new r. dataset follows 5000+ rows of data:

power consumption cputi dbsu

as column names , followings integers values in above column:

132 25 654

the sql code call r function wrote

create column table "predictive analysis" "anagappan.power_consumption" no data;  select power_app, power_db,cputi,dbti,dbsu  "anagappan.power_consumption"; drop procedure use_lm;  create procedure use_lm( in train "anagappan.power_consumption", out result "predictive analysis")  language  rlang  begin  library(lm)  model_app <- lm( power_app ~ cputi + dbti + dbsu + kbytes_transferred, data = train )  colnames(datout) <- c("power_app", "cputi", "dbti", "dbsu", "dbsu")  predictive analysis <- as.data.frame( lm(model_app))  end; 

the result obtain says procedure created unable call linear model on data, how initiate linear model?

although i'm not familiar sap products, have stab @ r code assume between begin , end;.

library(lm) 

is incorrect, mentioned @olli. access r's linear model capabilities, have call - nothing. it's loaded default through stats package (this may not true if r called in --vanilla mode.

model_app <- lm( power_app ~ cputi + dbti + dbsu + kbytes_transferred, data = train ) 

appears ok, @ least syntax's point of view.

for

colnames(datout) <- c("power_app", "cputi", "dbti", "dbsu", "dbsu") 

i can't see define datout. if variable not created database, not exist , r should complain along lines of

error in colnames(notexist) <- "x" : object 'notexist' not found 

i assume want predict (means) based on model. line

predictive analysis <- as.data.frame( lm(model_app)) 

will not work because r's variables should not have spaces, as.data.frame not work on lm object , model_app doesn't exist (notice case). think should along lines of

# based on http://help.sap.com/hana/sap_hana_r_integration_guide_en.pdf # have specify variable result exported database result <- as.data.frame(predict(model_app)) 

you can try out.

x <- 1:10 y <- rnorm(10)  mdl <- lm(y ~ x)  as.data.frame(predict(mdl))     predict(mdl) 1    0.47866685 2    0.34418219 3    0.20969753 4    0.07521287 5   -0.05927180 6   -0.19375646 7   -0.32824112 8   -0.46272579 9   -0.59721045 10  -0.73169511