How to transpose result of colmeans in R to pass into linear regression function?-CodePudding

I have a model in R:

lm(formula = Y ~ rowmeans(df[, c(10:14)]), data=df)

I want to fit this against the means of columns 10:14 so that my output has 5 rows to predict the Y values.

I'm calculating the column means as follows:

t(as.data.frame.list(colMeans(df[, c(10:14)], na.rm=TRUE)))

This is the correct output. However, when passing this into my linear regression predict() function, I'm not receiving 5 rows as expected.

I'm trying:

fit <- lm(formula = Y ~ rowmeans(df[, c(10:14)]), data=df)
predict(fit, newdata = list(t(as.data.frame.list(colMeans(df[, c(10:14)], 
                                                     na.rm = TRUE)))))

This produces an output of 300 values...

CodePudding user response：

Do not put complicated stuff in a model formula:

df$x <- rowmeans(df[, 10:14])  ## create variable 'x'
fit <- lm(Y ~ x, data = df)

## provide variable 'x'
newdf <- data.frame(x = colMeans(df[, 10:14], na.rm = TRUE))
predict(fit, newdf)