Calculate regression coefficients for specific conditions in a data frame
A bunch of things to say.


You're looping 3 times on the same vector "TANK", once would have be enough.
You're growing vectors in each loop
One loop just recreate the same vector reversed (more on this later)
One loop repeat a vector as much as there's entries in TANK (which is  not what you're after I'm pretty sure).
You're creating a local data.frame just to return it.


For your questions:


  it does return the list in reverse order 30 - 1


You're making the vector by prepending the value before the actual state:

for(i in TANK)
     tank &lt;- c(i, tank)


Using tank &lt;- c(tank, i) will give you the normal order.


  it changes the treatment labels from letters ("A", "B", "C", etc...)
  to numbers ("1", "2", "3", etc...).


I highly suspect TREATMENT to be of class factor, concatenation in the vector drop the factor and keep only the indices values.

DISCLAIMER I made the assumption that your variables are:


All of length 90
Matching by the TANK name position


You code can be optimized to:

k.tank &lt;- function() {
  # allocate the vector once before the loop
  k &lt;- numeric(30) 
  treat &lt;- character(30)
  tanks=unique(TANK)

  for(i in seq_along(tanks)) { # loop on indice instead of value
    # Assign the result in the proper place directly
    treat[i] &lt;- TREATMENT[TANK == i][1]
    k[i] &lt;- coef(summary(lm(log(MASS)[TANK == tanks[i]] ~ DATE[TANK == tanks[i]])))[2, 1]
  }

  # create the data.frame, as last statement this will be the function return value
  data.frame(tanks, treat, k)
}


With sample values:

set.seed(42)
MASS=runif(90)*100
TANK=rep(1:30,3)
TREATMENT=rep(rep(LETTERS[1:5],6),3)
DATE=sort(rep(seq.Date(Sys.Date(),by=1,length.out=3), 30))


This gives:

   tanks treat           k
1      1     A -0.15155006
2      2     B  0.02382969
3      3     C  0.48811951
4      4     D -0.19125411
5      5     E  0.14033970
6      6     A -0.50391863
7      7     B -0.49942663
8      8     C  0.90820123
9      9     D  0.02682661
10    10     E -0.53769180
11    11     A -1.18268285
12    12     B -0.81647939
13    13     C -0.73156739
14    14     D  0.31479427
15    15     E -0.42545699
16    16     A -0.13376959
17    17     B -2.41040604
18    18     C  0.58095049
19    19     D  0.03985375
20    20     E -2.93855112
21    21     A -0.22053714
22    22     B  0.06480414
23    23     C -0.50659181
24    24     D -0.19135960
25    25     E  1.12094187
26    26     A  0.04589634
27    27     B -0.25630776
28    28     C -1.15457853
29    29     D -0.82633221
30    30     E -0.50380311


Now you read until there, and still assuming your variables are of same length (but if not I can't really get your code) you may try doing something more direct:

# Create a data.frame with your observations
df &lt;- data.frame(TANK,MASS,TREATMENT,DATE, stringsAsFactors=FALSE)
# Create a function for ease of use, taking a data.frame as input
tank.coef &lt;- function(int.df) {
  coef(summary(lm(log(int.df$MASS) ~ int.df$DATE)))[2,1]
}

# run the above function on global data.frame, subsetting by TANK value
k &lt;- by(df,df$TANK,tank.coef)


This will return a by object (coef by tank):

&gt; head(k)
df$TANK
          1           2           3           4           5           6 
-0.15155006  0.02382969  0.48811951 -0.19125411  0.14033970 -0.50391863 


You can get rid of the name with as.vector and recreate a data.frame with:

&gt; rdf &lt;- data.frame(tank = df$TANK[df$TANK==names(k)], treat = df$TREATMENT[df$TANK==names(k)],k = as.vector(k))
&gt; head(rdf)
  tank treat           k
1    1     A -0.15155006
2    2     B  0.02382969
3    3     C  0.48811951
4    4     D -0.19125411
5    5     E  0.14033970
6    6     A -0.50391863