Force R include 0 as a value in regression of samples versus year

Question

Not sure if this question would be better in Cross Validated, but I think this is the same programming question as a simple statistical one.

I have a 102 x 1147 data block where there are years (between 1960 and 2016), and each entry is a scientific article. I count the number of publications published each year on specific topics (guided by the values ​​in specific columns), and I want to calculate the linear slope of the year and the annual count of the number of documents.

Here is my script, first a linear model, then a graph:

# THEME 1 (POPABU)
sub2=subset(as.data.frame(table(sysrev60[,c("YR","POPABU")])),
        POPABU==1,select=c(1,3))
sub2$YR<-as.numeric(paste(sub2$YR))

lm_eqn <- function(df){
  m <- lm(Freq ~ YR, sub2);
  eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2, 
               list(a = format(coef(m)[1], digits = 2), 
                    b = format(coef(m)[2], digits = 2),
                    r2 = format(summary(m)$r.squared, digits = 3)))
  as.character(as.expression(eq));                 
}

ggplot(sub2, aes(x=YR,y=Freq)) + 
  scale_y_continuous(limit=c(0,20),expand=c(0, 0)) +
  scale_x_continuous(breaks=c(1960,1965,1970,1975,1980,1985,1990,1995,2000,
                          2005,2010,2015),labels=c(1960,1965,1970,1975,1980,1985,
                                                   1990,1995,2000,2005,2010,2015)) +
  geom_bar(stat='identity') + 
  geom_text(x = 1960, y = 16, label = lm_eqn(df), size=5,hjust=0, parse = TRUE) +
  stat_smooth(method="lm",col="red") +
  xlab(" ") + ylab("No of papers") +
  annotate("text",x=1960,y=18,label="THEME 1",
       family="serif",size=7,hjust=0,color="darkred")

, > 0. , 0, (1960- 2016) 25 , , .. 0 0.

, , . a DPUT "sub2":

dput(sub2)
structure(list(YR = c(1960, 1961, 1962, 1963, 1964, 1965, 1966, 
1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 
1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 
1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 
2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 
2011, 2012, 2013, 2014, 2015, 2016), Freq = c(0L, 0L, 0L, 0L, 
0L, 1L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 2L, 1L, 0L, 1L, 
3L, 0L, 1L, 0L, 2L, 0L, 3L, 0L, 1L, 0L, 1L, 0L, 0L, 1L, 1L, 2L, 
0L, 2L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 2L, 0L, 1L, 
1L, 1L, 2L, 3L, 5L)), .Names = c("YR", "Freq"), row.names = 58:114, class = "data.frame")

, , -, 0, , , .

, script. ?

Answer 1

, , , , , lm() - - :

# Make sure zeros are there:
sub2$Freq
[1] 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 2 1 0 1 3 0 1 0 2 0 3 0 1 0 1 0 0 1 1 2 0 2
[39] 0 0 0 1 0 0 0 0 0 1 0 2 0 1 1 1 2 3 5
# Yep
X <- cbind(rep(1, nrow(sub2)), sub2$YR) # add a column of 1s for intercept
solve(t(X) %*% X) %*% t(X) %*% sub2$Freq # (X'X)^-1 X'Y -- OLS formula

            [,1]
[1,] -38.1778584
[2,]   0.0195748

, , , :

, , -38, 0,02. , . , , , , Freq , , , - , .